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= The value
. The expectation value of its
= = + ah/ y.
Hence the expectation value is
=
= ah/
Since
=1 The zero point energy of a particle in a one dimensional box at a infinite height is: The occurrence of zero point energy in accordance with the Heisenberg uncertainty principle is :
Chemistry : Basic Elements Since En = the particle.
, the energy corresponding to n = 1, i.e., E1 =
is called the zero point energy of
Since this energy is finite it means that the particle inside the box is not at rest even at 0K. This being so, the position of the particle cannot be precisely known. Again since only the expectation value of the kinetic energy is known, the linear momentum of the particle is also not precisely known. Thus occurrence of zero point energy implies uncertainty in the position, Dx and also certainty in the x component of the linear momentum Dpx. This means that occurrence of zero point energy is in keeping with the Heisenberg uncertainty principle.
Analytical Chemistry 8 Analytical Chemistry On a 1000 cm wall coated tabular column of 0·25 mm bore, the helium carrier gas velocity is 40 cm/sec. The retention time tR, for decane is 1·28 min; peak width at half height is 0·88 sec. The retention time for a non-retained compound is:
The average linear velocity, u of the mobile phase is where `L' is the column length and tm is the retential time for a non retained compound.
\
sec or 0.41 min
The partition ratio, where t´R is adjusted retention time which is equal to tR tm.
\
Chemistry : Basic Elements Substances A and B were found to have retention times of 17·30 and 19·92 minutes respectively on a 25·0 cm column. The widths (at the base) for A and B were 1·10 and 1·22 minutes respectively. The average number of plates in the column and the plate height calculated resolutions. Column resolution,
Rs
where WA and WB are the widths at the base for A and B respectively. DZ is the difference in time of their arrival at the detector. DZ = (19·92 17·30) min = 2·62 min WA = 1·10 min and WB = 1·22 min
Number of plates,
Thus
and
Average Basis of Classification of Chromatographic Methods : Classification is based on the phenomenon involving the process of either partition or adsorption. Exclusion chromatography involves separation of sample components according to the molecular size. Recent techniques are gel permeation and sieving separation. Analytical Chemistry The Rf Value : Rf value is the relative rate of solute and solvent.
The substance having same Rf value can be separated by two dimensional paper chromatography. Descending technique is better because compounds with low Rf values which cannot be completely separated by ascending method can be separated by descending technique. The Rf values of individual functional groups in a chemically related compounds are very close. So the term RM is additive and is composed of the partial RM values of the individual functional groups or other groupings of atoms in the molecule. Mathematically RM = log (1/Rf 1). On a column 125 cm long, operated at 160° C, these retention times (in minutes) were obtained : air peak, 0·90, heptane, 1·22 and octane, 1·43. The base widths of the bands were 0·16 for heptane and 0·22 for octane. The relative retention and the resolution for these bands are :
The degree of separation or resolution of two adjacent bands is defined as the distance between band peaks divided by the average band width. i.e.,
Chemistry : Basic Elements Chromatography Matched : (1) Packed Column (a) Tungsten rhenium alloy (2) Capillary column (b) Volatile sulphur or phosphorus compounds (3) Thermal conductivity detector (c) Rubidium silicate bead (4) Flame Ionization detector (d) Ultraviolet radiation
(5) Thermionic Emission detector (e) Fused Silica (very high purity glass) (6) Electron capture detector (f) Electrolytical conductivity (7) Flame photometric detector (g) Stainless steel, nickel or glass (8) Photoionization detector (h) High energy electrons (Beta particles) (9) Electrolytic conductivity detector (i) Pure hydrogen/Air flame (1)(g), (2)(e), (3)(a), (4)(i), (5)(c), (6)(h), (7)(b), (8)(d), (9)(f). The problems arising from using a poorly packed HPLC column. Resettling of the packing with use that can create a void at the top of the column that would lead to broad peaks with poor symmetry. At high mobile phase velocities the equation for reduced plate height becomes approximately h/v = c. Under conditions required to obtain the maximum number of plates, the pressure drop tends to infinity. Measuring column efficiency in HPLC Column efficiency could be measured in terms of plate number, N i.e., N = 16 (V/w)2 \ Narrower the peak (low w) the higher will be the value of N and so more efficient the column will be. Analytical Chemistry Main advantages of TLC (Thin layer chromatography) are ® Simple equipment ® Short development time ® Separation of micrograms of the substances
® Highly sensitive ® Early recovery of separated components and easy visualisation of separated compounds. ® The technique may be employed for adsorption, partition or ion exchange chromatography. The cation exchanger in the H + form required to separate Ca+2 ions from 1 dm3 of 0·1 N CaC12 are : The adsorption effect of an ion is characterised by the distribution coefficient
... (1) where Cr and Cs are the equilibrium concentrations of ions in the respective phases and m is the mass of the exchanger in grams. V is the volume of aqueous phase in cm3. Therefore from equation(1)
The quantity C1/D is the static exchange capacity of the cation exchanger, hence
The mixtures of Zn (II) and Mg (II) usually separated employing anion exchanger rather than cation exchanger are: Several metal ions (e.g., Fe, Al, Zn, Co, Mn, etc.) can be absorbed from hydrochloric acid solutions on anion exchange Chemistry : Basic Elements resins owing to the formation of negatively charged chloro complexes. Each metal is absorbed over a well defined range of pH, and this property can be used as the basis of a method of separation. Zinc is absorbed from 2M acid while Magnesium is not thus by passing a mixture of Zn (II) and Mg (II) through a column of anion exchange resin a separation is effected. Zn (II) is eluted with dilute nitric acid. A cation exchange resin column is saturated with copper(II) ions recovered from rinse waters from plating operations. It is desired to recover the copper and to convert the resin
to the H+ form for reuse. Normally this might be done by washing the column with 3M H2SO4 but 6M HCI is found to be superior. 6 M HCl is found to be superior in this case because it forms tetra chlorocuprate (II) (2 ) and trichlorocuprate(II) (1 ) species with copper(II) which will not bind to the cation exchange sites. In multiple extraction of solute from aqueous to organic phase, if the volume of the organic solvent used each time remains constant and equal to the volume of aqueous phase, the weight unextracted after n such operations is :
and weight extracted in n operations is
Suppose we have a solution containing `m' gm of a substance in `V' ml of solution. This solution is repeatedly extracted using every time `v' ml of organic solvent which is immiscible with the first. Suppose `ml' be the amount of solute that remains unextracted at the end of first operation. Then the distribution coefficient may be written as Analytical Chemistry
... (1) or m1 = m (KV/KV + v) ... (2) Exactly in the similar manner, amount m2 that remains unextracted at the end of second extraction,
... (3) or, m2 = m (KV/KV + v)2 ... (4)
Therefore the weight that remains unextracted at the end of n operations will be m2 = m (KV/KV + v)n and the weight extracted in `n' operations will be given by m mn = m [1 (KV/KV + v)2] Note : The efficiency of extraction increases by increasing the number of operations using only a small quantity of the extracting liquid at each time. Synergistic Extraction The phenomenon in which two reagents when used together, extracted a metal ion with enhanced efficiency compared to their individual action is called synergism. A common form of synergistic extraction is that in which a metal ion, Mn + is extracted by a mixture of an acidic chelating reagent, HR and an uncharged basic reagent, S. The joint action of the reagents is especially pronounced in those cases where the coordination capacity of the metal ion is not fully achieved in the MRn chelate, then the extractant S gives a mixed complex, MRn Sx which is extracted with much greater efficiency than the parent chelate. An example is the synergistic influence of zinc in the extraction and AAS determination of trace cadmium in water.
Chemistry : Basic Elements The weight of Fe(III) left unextracted from 100 ml of a solution having 400 mg of Fe3+ in 6M HCl after two extractions with 25 ml of diethyl ether : D = 150, Calculated : Amount left unextracted,
where, W = amount of solute, Vw = volume of aqueous solution, D = 150, V0 = volume of organic phase.
\
In the extraction of cerium(IV) with 2-thenoyl trifluoroacetone in benzene, the distribution ratio was 999. The volume of organic phase was 20 ml and that of aqueous phase 50 ml, what was the percentage extraction : D = 999, V0 = 20 ml, Vw = 50 ml, Assume E = x is percentage of extraction
The electronic transitions will have the lowest energy : n ® p* transition will generally have the lowest energy. The effect of addition of Co2+aq to the equilibrium reaction is : Co2+(aq) + 4NCS ® [Co(NCS4)]2 (Pale Pink) (Colourless) (Blue) on the % transmittance of the solution. As Co2 + is added to the equilibrium mixture, the concentration of the complex ion in the mixture increases. The percentage of absorption by the complex increases and transmittance decreases. Analytical Chemistry The Matched : Energy changes involved Region in electromagnetic spectrum (1) Nuclear (a) Infrared. (2) Inner shell electrons (b) Radio waves. (3) Valence electrons (c) Visible.
(4) Molecular vibrations (d) Gamma rays. (5) Spin orientation (e) X-rays. (1)(d), (2)(e), (3)(c), (4)(a), (5)(b). By the use of potential energy curves for ground and excited electronic states, the sequence of steps leading to phosphorescence and fluorescence shown :
Raman and Rayleigh scatter can be identified when included in a fluorescence emission spectrum. When excitation wavelength is varied, Rayleigh Scatter will vary exactly as the excitation, whereas Raman shift is a constant energy shift. Any fluorescence peaks vary only in intensity with the wavelength maxima remaining unchanged.
Chemistry : Basic Elements After insertion or injection of the sample into the electrothermal atomizer, the temperature profile with ramp heating drown :
See the Figure Two advantages and a major limitation of each of the following methods-(a) Flame AAS, (b) electrothermal AAS, (c) FES and (d) AFS, stated : Technique Advantages Limitation 1. Flame AAS (i) Relatively inexpensive. Analyte must be in (Atomic absorption, solution. spectroscopy) (ii) Fast technique. 2. Electrothermal AAS (i) Lower detection limits. Less precision, slower. (ii) Handles solid samples directly. 3. FES (Flame emission (i) It complements AAS. It is limited to alkali spectroscopy) and alkaline earth metals. (ii) It is fast and inexpensive. 4. AFS (Atomic fluore- (i) Good for trace analysis. No instrumentation
scence spectrometry) is made commercially. (ii) It is faster than electrothermal AAS. Acetylene-nitrous oxide flame is suitable for elements such as Al, Be and rare earths: These elements form stable oxides. The temperature of Analytical Chemistry acetylene-nitrous oxide flame is about 3000° C and it has an advantage that its burning velocity is almost same as that of air-acetylene. Consequently it can be easily and safely handled in the laboratory. Aqueous solution of a drug (0·10 milli molar) shows a percentage transmission of 50 in a 1 cm cell at 250 run. The molar absorptivity, calculated : According to Beer's law
A = log
= ecl
where T is transmittance, c is concentration in moles per litre, l is path length in cm, e is molar absorptivity.
\
= e × 1 × 104 moles/lit × 1 cm
0·3010 = e × 1 × 104 × 1 e = 0·3010/1 × 104 = 3010 The ratio of number of sodium atoms in 3p excited states to the number in the ground state at 2500 K calculated. The average wavelength of the two sodium emission lines involving the 3p to 3s transition is 5892 Å : = 1/5892 Å × 108 cm = 1.698 × 104 cm1 Ej = 1·698 × 104 cm1 × 1·986 × 1016 erg cm1
= 3·372 × 10 12 ergs (since h × c = 1·986 × 1016 erg cm1) 3s and 3p levels have 2 and 6 quantum states respectively. Pj / P0 = 6/2 = 3.
Hence
or,
=1·7 × 104
Chemistry : Basic Elements The temperature at which plasma sources operate stated. The region where plasma source produces a great number of excited emitted atoms is : Between 7000 and 15000 K. In the ultraviolet region plasma source produces a great number of excited emitted atoms. The difference between emission and excitation fluorescence spectra is : In emission, excitation is carried out at a fixed wavelength and the emission intensity records as a function of wavelength. Excitation fluorescence spectra are obtained by measuring the fluorescence intensity at a fixed wavelength while the wavelength of the exciting radiation is varied. In the determination of selenium in urine by AAS it was found that Beer's law applied up to the 2 ppm range. If the 2 ppm standard was set at 100 and a blank set at zero. The selenium concentration of the sample to read at 73 would be:
Selenium concentration of urine = The advantages of molecular fluorescence methods compared to the corresponding absorption spectrophotometric methods and how do they arise are : Fluorescence is caused by the absorption of radiant energy and re-emission of some of this energy in the form of light. The advantages of molecular fluorescence methods over absorption spectrophotometric methods are
(1) The sensitivity of fluorescence methods is generally 10 to 103 times more than the sensitivity of absorption methods. (2) These methods have greater specificity and selectivity than absorptiometric techniques. This is due to the fact that in luminescence both the wavelength of excitation and emission are selected whereas in absorptiometry only the wavelength for absorption is selected. Analytical Chemistry The g-value of the methyl radical whose centre of ESR spectrum appears at 329.4 mT in a spectrometer operating at 9·233 G. Hz (h = 6·627 × 1034 Js, electron Bohr Magneton = 9·27 × 1024 JT1) is. The number of lines in the ESR spectrum of this methyl radical predicated:
or,
Hence g = 2·0048, The interaction of an unpaired electron with three equivalent protons as in CH3 radical gives four equally spaced lines according to (n + 1) rule. The Polarizability The case with which the electron distribution in a molecule or atom may be distorted by the application of an electric field. The basis of turbidimetric and nephelometric analysis is: Measurement of the intensity of the transmitted light as a function of the concentration of the dispersed phase is the basis of turbidimetric analysis and measurement of the intensity of the scattered light (at right angles to the direction of incident light) as a function of the concentration of dispersed phase constitute the basis of nephelometric analysis.
In the determination of manganese by atomic absorption method, a solution containing an aliquot of unknown strength gives a metre reading of 23. Another sample containing this solution with added 50 ppm of manganese solution gives a metre reading of 92·5. Each reading was corrected for the background. The concentration of manganese in the original solution calculated :
Chemistry : Basic Elements As absorbance is directly proportional to the concentration of manganese. . . \ 23 = mx, where `x' ppm is concentration of Mn. 92·5 = m (x + 50) 92·523 = 50 m or, m = 1·39
and x =
= 16·5 ppm.
Amperometric titration is a better method than polarographic method for quantitative analysis. Amperometric titration is a quick, accurate and convenient method similar to potentiometric and conductometric titration and at the equivalence point there is sharp change in diffusion current. The galvanometer used need not be calibrated. The specific characteristic of capillary does not influence the titration. No polarising unit is used, suitable half cell can be easily used for the purpose. The Match: List I List II Quantity Measured & Variable Controlled Name of method (1) E, i = 0 (a) Linear potential sweep voltammetry (2) E vs volume of titrant, i = 0 (b) Amperometric titrations (3) i vs E, concentration (c) Ion selective potentiometry (4) i vs E, t (d) Conductometric titrations
(5) i vs volume of titrant, E (e) Voltammetry (6) 1/R vs volume of titrant (f) Potentiometric titrations (1)(c), (2)(f), (3)(e), (4)(a), (5)(b), (6)(d). The amperometric titration of iodine with sodium thiosulphate using two indicator electrons. (a) The amperometric titration curve and correlate the several regions on this curve with the corresponding regions on the current potential curves sketched. Analytical Chemistry (b) This titration curve change if excess iodide ion were initially present would :
(a)
(b) Using equivalent conductance values, the general form of the titration curve in (a) titration of Ba(OH)2 with HCl, (b) titration of NH4Cl with NaOH sketched :
Chemistry : Basic Elements
Titration of electroactive substance A with another electroactive substance B was carried out at the potential at which both A and B are reducible at cathode. The current vs volume of titrant B curve and label the diagram accordingly sketched : In this titration the current will drop to the end point and there will be increase again to give a V-shave titration curve.
The residual and limiting current is : Residual or condenser current flowing in a polarographic cell is the sum of the Faradic current (if) and the charging current (ic). Thus Residual current (ir) = if + ic Analytical Chemistry The limiting current (il) is the sum of the diffusion current (id) and the residual current (ir). Thus il = id + ir The merits of cyclic voltammetry are : Merits are in the realm of qualitative or diagnostic experiments. A cyclic voltammogram is to the electrochemist what the frequency domain spectrum is to the spectroscopist. It gives a picture of a substance electrochemical behaviour while varying the voltage (energy) of an indicator electrode. Cyclic voltammetry is capable of rapidly generating a new oxidation state during the forward scan and then probing its fate on the reverse scan. Peak height is related to both concentration and reversibility of the reaction. Thus it yields information about reaction reversibilities and also offers a very rapid means of analysis for suitable systems. This method is particularly valuable for the investigation of stepwise reactions and in many cases direct investigation of reactive intermediates is possible. By varying the scan rate, systems that exhibit a wild range of rate constants can be studied and transient species with half lives of milliseconds are readily detected. Matched : (1) U.V. spectroscopy (a) DC Arc
(2) FT. IR (b) Interoferometer (3) AES (c) Xenon flash lamp (4) AAS (d) Thermal conductivity detector (5) Gas chromatograph (e) Hollow cathode lamp (6) Fluorescence & (f) Deuterium discharge lamp Phosphorescence spectro photometry (1)(f), (2)(b), (3)(a), (4)(e), (5)(d), (6)(c). The full form of the techniques mentioned below and the quantity measured by them given DSC, DTA, TGA, TMA, DMA, EGA.
Chemistry : Basic Elements (1) DSCDifferential Scanning Calorimetry (Heats and temperatures of transitions and reactions). (2) DTADifferential Thermal Analysis. (Temperature of transitions and reactions) (3) TGAThermo gravimetric analysis. (weight change) (4) TMAThermomechanical analysis. (Dimension and viscosity changes) (5) DMADynamic mechanical analysis. (Modulus, damping and viscoelastic behaviour) (6) EGAEvolved gas analysis (Amount of gaseous products of thermally induced reaction) The probable thermal decomposition patterns of CaC2O4·H2O deduced using the thermogram (obtained in air) given below (Atomic weights of Ca, C and O may be assumed to be 40, 12 and 16 respectively).
The probable thermal decomposition of CaC2O4·H2O according to above thermogram will be CaC2O4·H2O ® CaC2O4 + H2O CaC2O4 ® CaCO3 + CO CaCO3 ® CaO + CO2 Analytical Chemistry A purity of a compound can be checked by Differential Scannig calorimetry (DSC) this way: In cases where Van't Hoff equation is applicable and material is pure about 97 to 100% a single endotherm is obtained using a very small sample and heating at the rate of about 1·25° C/min. The curve so obtained is integrated to the vertex by parts and the temperatures are tabulated with the partial areas upto that temperature. The following equation is employed,
where Ts is the instantaneous temperature of the sample in K, To·melting point of the infinitely pure sample (solvent) in K, R = gas constant, DH = Heat of fusion of sample, F = fraction of the total sample melted at Ts. Now a plot of temperature as a function of the reciprocal of the fraction melting, 1/F to that temperature is made which should give a straight line of slope KT02/DH with an intercept of T0. AH can be determined from the DSC curve. The three types of automatic chemical analyzers with respect to the mode of operation, advantages and disadvantages compared :
Analyzer Mode of Operation Advantages Disadvantages (1) Discrete Batch, serial proce- Runs selected test on Expensive prepackaged ssing of sample. limited number of reagents and cells, samples, quick start up. mechanically complex. (2) Continuous Continuous pro- Fast, simple equipment, Difficult to vary test flow cessing of large reagent cost minimised. doing a series of number of sam- samples, tubing is ples for same critical. series of tests. (3) Centrifugal Sample mixed Fast, all samples and Loading samples is with reagent by controls measured under time consuming, can centrifugal force. the same conditions. run only a limited number of tests.
Nuclear Chemistry 9 Nuclear Chemistry Order of Radioactive Disintegration Process : Radioactive disintegration is similar to a chemical reaction of the first order, that is, a chemical reaction in which the rate of the reaction varies as the concentration of one molecular species only. Suppose such a reaction is representation by the equation A®B Let the number of atoms of A originally present be N0 and number of atoms of A, Present at any given time t, be N. Then, the rate of change of A into B is
=
…(1)
= Disintegration constant or decay constant Thus, _ldt = dN/N …(2) If, dt = 1s
l=
Chemistry : Basic Elements On Integtrating equation (2)
=
…(3)
or In N = -lt + c …(4) where c is the integration constant.
At t = 0, N = N0 \ c = In N0 Substituting in equation (4) we have In N =
…(5)
or, In (N/N0) = -lt …(6)
or,
…(7)
Relation between half-life and decay constant : Suppose, after time t1/2 half of the atoms of the radioactive substance have disintegrated , that is N = N0/2 Substituting in equation (7) we have
= or, In 0.5 = -lt1/2 or, 2.303 log 0.5 = -lt
Half-life period of a given radioactive substance is independent of the amount of substance present initially. It depends only on disintegration constant of the element. Nuclear Chemistry Half-life nuclide 220Rn is 54.5 s
l= We know, 1 mci = 3.7 × 107 disintegration per sec.
= lN
N= Mass of
= Number of Rn nuclei
mass of one Rn nucleus
= or, = An accident occurs in a laboratory in which a large amount of radioactive material with a known t1/2 of 20 days becomes embedded in floor. Tests show that the level of radiation is 32 times the permissible limit. The laboratory can be safely occupied after : Since the initial level of radiation is 32 times the permissible limit, we have to calculate the time t in which the activity drops to 1/32 of it initial value, i.e.,
= Disintegration constant,
l=
Chemistry : Basic Elements Suppose the laboratory becomes safe for use after t days. Than
=
=
=
t= Radioactive Equilibrium Radioactive equilibrium : If in a disintegration series A, B, C, D etc. are some of the intermediates consecutive atoms i.e., a stage may come when the amounts of A, B, C, D etc. become constant which is so because equal. If then
=
But -
=
=
etc., represent the number of atoms of A, B, C, D etc., at equilibrium,
Hence
=
or,
= ...
=
Nuclear Chemistry Thus the amounts present at equilibrium are inversely proportional to their disintegration constant or directly proportional to their half-lives. Radioactive equilibrium may be of two types— Secular equilibrium : It is limiting case of radioactive equilibrium, in which the half-life of the parent is many times greater than the half-life of the daughter) usually by a factor of 104 or greater). After a sufficient long time
NB = Transient equilibrium : It is similar to secular equilibrium but differs in that the half-lives differ only by a small factor. After a sufficiently long time
NB = No. of atoms of 226Ra in 1 g of
= At equilibrium =
or,
=
=
= =
Chemistry : Basic Elements Reason Having more Biochemical Importance is biochemically more important because of its shorter half-life period. Shorter halflife ensures attainment of transient equilibrium faster. barrier height for an a- particle inside the
nucleus
Barrier height =
= A thin sample of gold was irradiated in a thermal neutrons flux of 1012 neutrons cm2 sec_1 for 25.6 hrs. In the reaction the nuclide 198 Au is produced with a half-life of 64 hours. If the thermal neutron absorption cross section is 98 barns, the specific activity of the sample is:
A= Suppose 1g of the gold was irradiated,
then N = f= s= t1/2 = Substituting these values, we get
A= = Nuclear Chemistry A sample of 100 mg of a radioactive nuclide decay to 81.85 mg of the same in exactly 7 days. The decay constant for this disintegration and the half life of the nuclide. Calculated: Disintegration constant,
l= t=
l=
Half-life, t1/2 = A sample of river water was found to contain 8 × 10_18 tritium atoms,
per atoms of ordinary
hydrogen. Tritium decomposes radioactively with a half-life of 12.3 Y. The ratio of tritium to normal hydrogen atoms 49 Y after the original sample was taken if the sample is stored in a place where additional tritium atoms cannot be formed. It would be : 49 years is almost exactly four half-lives (4 × 12.3 = 49.2Y), so we can avoid the log formula. The fractions of the tritium atoms remaining after four half-lives would be (1/2)4 = 1/16. The normal hydrogen atoms are not radioactive, and therefore, their number does not change. Thus the final ratio of tritium to normal hydrogen will be
= Using the approximate equation for the radius of the nucleus, calculate the density of the nucleus of and density of metallic silver (10.5 g/cm3). Compared : The approximate equation for the radius is r = R0 A1/3
Chemistry : Basic Elements where r is the radius of nucleus of mass number A and Ro is a constant whose value is equal to 1.5 × 10_13 cm So r = 1.5 (107)1/3 × 10_13 = 7.12 × 10-13 cm The volume, mass and density of a single nucleus are calculated from the radius by the following equations
V=
m=
d= The nuclear density is 1.18 × 1014 times that of silver metal. The ratio of the nuclear radii of the s - particle and the proton, estimated : is a-particle Radius equation = R0 A1/3
=
= 41/3 = 1.587
Suppose M represents the mass of last bound neutron from
and M1 that of
Let E be the energy needed to remove the
From the law of conservation of energy
E=
E= where mn is the mass of neutron
=
= Nuclear Chemistry A nucleus splits into nuclei, which have mass numbers in the ratio of 2 : 1. Assuming R = 1.5 f, The ratio of radii of nuclei formed : Following the equation
r = R0 A1/3 R0 is given 1.5 f For the two nuclei formed r1 = R0A11/3; r2 = R0A21/3 Since the mass numbers of nuclei formed are in the ratio of 2 : 1, therefore, r1 = 1.5 f × (2 × 235/3)1/3 = 8.056 f r2 = l.5f × (l × 235/3)l/3 = 6.418 f The average mass of chlorine is 35.5. The mass numbers of two isotopes are 35 and 37. The proportion to which two isotopes are present in ordinary chlorine is : Let the proportion of CI-35 and Cl-37 be 1 : x.
Then the average mass would be But this is given to be 35.5
So
or,
= 35.5
35 + 37x = 35.5 (1 + x) 35 + 37x = 35.5 + 35.5x 1.5x = 0.5 x = 1/3
Therefore, the ratio is 1 : 1/3 or 3 : 1.
Chemistry : Basic Elements Justified Statement"We need a very powerful crane of lift microscopic nuclear mass." Suppose volume of nuclear mass of microscopic size to be lifted is 10_4 cm3. Since the mass of a nucleon is approximately 1 amu, hence the mass (m) of the nucleus of mass number 1.66 × 10_27kg. The nuclear radius is given by r = R0 A1/3 where, R0 = 1.5 × 10_15m Nuclear volume, V = 4/3 pr3 = (4/3)p(R0 A1/3)3 = (4/3)p(1.5 × 10_15)3Am3 Hence, the nuclear density,
r=
= Thus we see the nuclear density is very high and is independent of the mass number of the nucleus. Therefore, all nuclei have approximately the same density. Now volume of the nuclear mass = 10_4 cm3 = 10_10 m3 (Q 1 m3 = 106 cm3) Mass = volume × density
= 10_10m3 × 1.17 × l0I7 kg m_3 = 1.17 × l06 kg Such a high mass can only be lifted only by a powerful crane. Cross Section The probability of nuclear process is generally expressed in terms of cross-section, which is denoted by s that has the Nuclear Chemistry dimensions of an area. This originates from the simple picture that the probability for the reaction between a nucleus and impinging particle is proportional to the cross sectional target area presented by the nucleus. The cross section for a particular nuclear process is defined by the equation Ri = lnx si This equation applies when there is a well defined beam of particles incident on a target. where, R, = number of processes of the type under consideration occurring in the target per unit time, I = number of incident particles per unit time, n = number of target nuclei per cubic centimetre of target, si = cross section of the specified process, expressed is square centimetres. x = target thickness in cm. The total cross section for collision with a fast particle is never greater than twice the geometrical crosssectional area of the nucleus and therefore, fast particle cross-sections are rarely much larger than 10_24 cm2 (radii of the heaviest nuclei are about 10_12 cm). Hence a cross-section of 10_24 cm2 is considered as big as barn and 10_24 cm2 has been named the barn (b) and used in expressing cross-sections. The millibarn (mb = 10_3 b), microbarn (mb = 10_6 b) and nanobarn (nb = 10_9 b) are also commonly used. Cross Section and Reaction Rate : For a sample containing N nuclei in a flux of f particles per square centimetre per second, the rate of reaction of type i, which have a cross section a, is given by Ri = f Nfi
This is the situation when a sample embedded in a uniform flux of particles incident on it from all directions and exactly this happens in a nuclear reactor.
Chemistry : Basic Elements How long a 60 mg piece of Co wire has to be placed in a flux of 5 × 1013 thermal neutrons per square centimetre per second to make 1 mci (1 millicurie = 3.7 × 107 dis s_1) of 5.22 Y 60Co, calculated. The cross-section for the reaction 59Co (x, y) 60Co is 3.7 b. Following the equation Ri = f Nsi
R= = 1.13 × 1012 atom s_1
From equation
we calculate that 1 mci of 60Co corresponds to 8.87 × 1015 atoms.
Thus it will take 8.87 × 1015/1.13 × 1012 = 7.85 × 103 sec or, approximately 2.2 hr to produce 1 mci of 60Co. The effective neutron capture radius of a nucleus having a cross section by 1.0 barn, calculated : l.0 barn = l.0 × l0_24cm2 The area of a circle is given by A = pr2 Hence r =
=
= 5.6 × 10_13 cm.
The ratio of number of neutrons to protons decides the so called `belt of nuclear stability'. Showing this by drawing a graph and using this, explain the decay process involving :
Nuclear Stability : n/p ratio : The stability of nucleus depends on the number of neutrons and protons present. Light Nuclear Chemistry nuclei upto atomic number 20 form most stable nuclei when the number of protons p is equal to the number of neutrons n i. e., n/p = 1. For elements with higher atomic number n/p ratio increases progressively to about 1.6. In such cases the number of neutrons is greater than that of protons. With increase in the number of protons the force of repulsion between them increases and this tends to lower the stability of nucleus.
By plotting a graph between number of neutrons and protons for the nuclei of various elements it has been found that most stable nuclei (non-radioactive nuclei) lie within the shaded area which is called the zone or belt of stability because it contains the stable nuclei. Nuclei that fall above or below this belt are unstable. Nuclei that fall above the stability belt have more neutrons while those lying below have more protons. Such unstable nuclei would attain stability by undergoing change that would produce a nucleus with n/p ratio within the stability zone. Examples :
(1) Nuclei lying above the belt of stability are richer in neutrons and hence they disintegrate is such a manner that one of their neutrons is converted into a protons i.e.,
Chemistry : Basic Elements ® such nuclei emit p-particle. For example, ® Radioactive Stable or, 6p + 8n ® 7p + 7n + n/p = 8/6 = 1.33 ® n/p = 1/1= 1.00 (2) The nuclei lying below the zone of stability are deficient in neutrons and hence disintegrate in such a way that one of their protons is converted into a neutrons. The conversion of a proton into a neutron can be done by any of the following two ways (a) Emission of a positron ® (b) Electron-capture process ® ® or, 12p + 11n ® 11p + 12n + n/p = 11/12 = 0.91 ® n/p = 12/11 = 1.09
(3) Other nuclei having higher number of protons or neutrons
disintegrate by
(a),
decay. More unstable of each of the following pairs and in each case, that type of process could the unstable nucleus undergo: (a) 16C , 16N (b) 18F, 18Ne has the ratio of neutron to protons farther above the belt of stability. It would emit a (a) to set n/p ratio back to the stable range.
particle
Nuclear Chemistry (b) has a lower n/p ratio than into the range of stability.
It could emit a positron or capture a K electron to get that ratio
Each of the following nucleide classified as `probably stable', `beta emitter' or `positron emitter'
The nucleide near the belt of stability are probably stable, those above the belt are beta emitters, those below, positron emitters. Thus the stable hucleides are 208Pb and 120Sn; the beta emitters are 40Ca, 3OA1 and 94Kr, the positrons emitters are l95Hg, 8B and l50Ho. The binding energy per nucleus of 16O is 7.97 MeV and that of I7O is 7.75 MeV. The energy needed to remove a neutron from 17O, calculated : I7O
® 16O +
Energy needed to remove a neutron from 17O is given by E = Energy equivalent of (mass of i6O + mass of neutron _ mass of l7O) = (8mp + 8mn _ 7.97 × 16) + mn _ (8mp + 9mn _ 7.75 × 17)
where mp = mass of proton, mn = mass of neutron = (7.75 × 17 _ 7.97 × 16) MeV = 4.23 MeV The approximate mass of uranium that must undergo fission to produce the same energy as 105 kg of coal, calculated. Assumed that heat of combustion of coal = 8000 cal g -1 and one fission of uranium releases 200 MeV. Mass of coal = 105 kg = 108 g Heat of combustion of coal = 8000 cal g_1 Total heat produced on combustion of 108g coal = 8000 × l08 = 8 × 1011 cal = 8 × 4.184 × 1011J = 35.5 × 1011 J
Chemistry : Basic Elements Thus, we have to calculate the mass of uranium that will undergo fission to produce 33.5 × 1011J of energy. Energy released per fission = 200 MeV = 200 × l06 × l.6 × l0_19J = 3.2 × 10_11 J
Number of U-235 atoms needed to produce this amount of energy =
\ Mass of 1023 atoms of U-235 = `K-electron Capture' When an electron from the nearest orbital i.e., K-shell orbital is absorbed by the nucleus to convert a proton into a neutron without emitting any particle, the process is known as orbital electron capture. Since it is most usually a K-electron which is captured by the nucleus, the process is also known as Kelectron capture. Usually an electron from higher energy level L, M, N etc. drop back to fill the vacancy
in K-shell.
® ® ® Spallation Reactions In these reactions, a high speed projectile having high energy (400 MeV) chips a fragment off from a heavy nucleus and a large number of light particles are emitted. The daughter nucleus has atomic number 10-20 units less and mass number about 30 units less than the parent nucleus. For example, ®
Nuclear Chemistry Nuclear Reactions completed by writing equations (i)
(ii)
(iii) 3H (a, n) ... (iv) (vi) .... (D, T)O18
(v)
(viii) 30P (..., p) 33S
(vii) (i) (ii)
® ®
(iii)
®
(iv)
®
(v)
® ®
(vi) (vii)
®
(viii)
®
If an atom of , after the absorption of slow neutrons, undergoes fission to produce 139Xe and 94Sr. Other particles are also produced. The fission reaction is ® Thus, three neutrons are emitted. Nuclear reactions balanced : (i)
® X + 5a + 3b
(ii)
® X + 6a + 4b
Chemistry : Basic Elements (iii) (iv)
® X + 8a + 6b ® X + 6a + 4b
(i)
®
(ii)
®
(iii)
®
(iv)
®
Q-value of Nuclear Reactions : All types of nuclear reactions are accompanied by either absorption or release of energy. The amount of energy released or absorbed in a nuclear reaction is called Q-value of the nuclear reaction. Thus the following two cases arises (i) When the sum of the masses of the reactants (mR) is less than that of the product (mP) (mR < mP), energy is absorbed i.e., when there is an increase in mass, energy is absorbed. Q-value (Energy absorbed) = Increased in mass in amu × 931 MeV = (mP _ mR) × 931 MeV When Q is absorbed it is given negative sign and the reaction is known as endoergic. For example, ® (ii) When the sum of the masses of the reactants is greater than that of the product (mR > mP), energy is released i.e., when there is a decrease in mass; energy will be released. Q-value (Energy released) = Decrease in mass in amu × 931 MeV = (mR - mP) × 931 MeV Nuclear Chemistry Q-value which is released is generally represented by a positive sign and the nuclear reaction is known as
exoergic. For example, ® Q-value of the following nuclear reactions calculated : (i)
®
(ii)
® ®
(i)
The change in mass in atomic mass unit (amu) can be found as follows Reactant Particles Product Particles = 4.00387
= 17.00450
= 14.00753
= 1.00814
Total = 18.01140 = mR Total = 18.01264 = mP Increase in mass = (mP _ mR) amu = 18.01264 _ 10.01140 = 0.00124 amu \ Q-value = 0.00124 × 931 = 1.15 MeV Since mR < mP, energy is absorbed in this reaction and hence Q-value is negative i.e., Q-value = _1.15 MeV. (ii)
®
Here, the mass values are as under Reactant Particles Product Particles = 1.00814
= 4.00387
= 7.01822
= 4.00387
Total = 8.02636 = mR Total = 8.00774 = mP
Chemistry : Basic Elements Decrease in mass = (mR _ mP) amu = 8.02636 _ 8.00774 = 0.01862 amu \ Q-value = 0.01862 × 931 MeV = 17.3 MeV Since mR > mP, energy is released in this reaction and hence Q-value is positive i.e., Q-value = + 17.3 MeV. 14C
is believed to be made in upper atmosphere by an (n, p) process on I4N. Q for this reactionis: The reaction (n, p) is
=
Reactant Product I4N
= 14.00753 1H = 1.00814 = 1.008665 14C = 14.00324
Total = 15.016195 = mR Total = 15.01138 = mP
Decrease in mass = mR _ mP = 15.016195 _ 15.01138 = _0.004815 amu Q = 0.004815 × 931 = 4.482 MeV Q-value for the 7Li (p, n) 7Be reaction, evaluated : Reactant Product 7Li
= 7.01822 7Be = 7.019465
1H
= 1.00814 1n = 1.008665
Total = 8.02636 = mR Total = 8.02813 = mP Increase of mass = mP _ mR = 8.02813 _ 8.02636 = 0.00177 amu Q = 0.00177 × 931 = 1.64 MeV Nuclear Chemistry Q mR < mP, energy is absorbed and hence Q-value = _ 1.64 MeV. This energy is supplied as kinetic energy of the bombarding proton and is part of the acceleration requirement for the proton supplied by the particle acceleration. The Q-value for the 3He (n, p) reaction is 0.76 MeV. The nuclidic mass of 3He is : The reaction is Reactants Products
®
3He
= x 1H = 1.00814
= 1.00865 3H = 3.01605 Total = x + 1.00865 Total = 4.02419 The mass lost must be = 0.761931 = 0.0008164 Then (x+ 1.00865) _ 4.02419 = 0.0008164 x = 3.01634 Nuclear fission. Explain why is
not suitable for a chain reaction, defined :
Nuclear Fission : The splitting of heavier atom like that of U-235 into a number of fragments of much smaller mass, by suitable bombardment with sub-atomic particles with liberation of huge amount of energy (due to mass defect) is called nuclear fission. For example, + Energy The neutrons released from the fission of first uranium atom can hit other uranium atoms and which again release neutrons, each of which can further hit are uranium atom. In this way, a chain reaction is set up resulting into the liberation of a tremendous amount of energy. not suitable for chain reaction : neutrons is
nuclei do not break up unless the energy of the bombarding
Chemistry : Basic Elements above 1.2 MeV. Such neutrons are called fast neutrons. On the other hand by slow neutrons or thermal neutrons having low energy of value 0.25 MeV slow neutrons form 239U and 239Pu as follows
®
nuclei can be break up when bombarded by
® Nuclear Reactor Nuclear reactor is an arrangement in which the energy produced in a nuclear fission can be used in a controlled manner to produce steam which can run the turbine and produce electricity. Nuclear reactor consists of following parts Fuel Rods : The fuel or fissionable material used in enriched uranium-235 (in the form of U3O8). Control Rods : The energy can be controlled by controlling the fission which, in turn can be controlled by controlling the number of neutrons released during fission. To control the fission cadmium or boron rods are used which absorb the neutrons ® ® Moderator : The material used to slow down the neutrons without absorbing them is called moderator. Graphite and Heavy water are good moderator. Coolant : The material used to remove the liberated heat is called coolant. Generally heavy water is used as coolant. Breeder reactor is different from nuclear reactor : Breeder Reactor : In such type of reactors fissionable nucleides are formed by the nonfissionable nucleides. Nuclear Chemistry Plutonium-238 and uranium-233 are suitable fissionable nucleides and are produced from more abundantly available uranium-238 and thorium-232. For example, (fast) ® ®
® Nuclear fusion reaction takes at very high temperature or why nuclear fusion reactions are known as thermo nuclear reactions : The nuclear reactions in which tighter nuclei fuse together to form a heavier nuclear are called nuclear fusion reactions. Such reactions, occur at very high temperature (of the order of > 106 K) which exist only in the sun or interior of stars therefore, such reactions are also called thermonuclear reactions. Nuclear fusion can take place at high temperature only because lighter nuclei combining together do not have sufficient kinetic energy to overcome the mutual repulsion. ® ® ® Attempts to make fusion reactors are not successful : Controlled fusion can also be carried out; however, attempts are not successful because of the difficulty of producing extremely high temperature in the laboratory and more over which vessel will withstand such high temperatures. Red giant stars, which are coolar than the sun, produce energy by means of the reaction: ® From the nuclidic mass 9Be (9.0154) and 6Li (6.01702), the energy released in MeV calculated compared with the energy released in carbon cycle (30 MeV) and in solar-hydrogen cycle (26.6 MeV). (4He = 4.0015; 1H = 1.00728)
Chemistry : Basic Elements The total mass of product and reactant nuclei are Products = (6.01702 + 4.0015) = 10.0185
Reactants = (9.01504 + 1.00728) = 10.0223 Mass difference = 0.0038 Energy released = 0.0038 × 931.5 = 3.5 MeV This value is much lower than the energy released in carbon cycle or solar helium-hydrogen cycle. 13N
decays by b+ emission. The maximum kinetic energy of b+ is 1.20 MeV. The nuclidic mass of 13N
is : ® Mass difference = [M (nucleus) for 13N] _ [M (nucleus) for 13C] - M (e) = [M (13N) _ 7M (e)] _ [M (t3C) _ 6M(e)] _ M (e) = M(13N) _ M(13C) _ 2M(e) = M(13N) _ 13.00335 _ 2 (0.00055) = M(13N) _ 13.00445 This mass equal to mass expressions of the kinetic energy of the b+
= 0.00129 Then 0.00129 = M (13N) _ 13.00455 or, M(13N) = 13.00574 Nuclear Reactions Completed :
(i)
(ii) (iii) Nuclear Chemistry (i)
(ii)
(iii)
The energy released in a positron decay process calculated: ® (11C = 11.011443; 11B = 11.009305) Mass defect = Mass of reactants _ Mass of product = 11.011443 _ 11.009305 = 0.00213 \ Energy released = 0.00213 × 931 = 1.984 MeV (Positron carries no mass)
A mixture is to be assayed for penicillin. Added 10.4 mg of penicillin of specified activity 0.405 mc mg_1. For this mixture are isolated only 0.3 mg of pure crystalline penicillin and determined its specific activity to be 0.035 mc mg_1. The penicillin content of the original sample was : From the given data, m´ = 1.0 mg, S´ = 0.405 mc mg_1 S = 0.035 mc mg_1 Substituting these values in following equations
m=
=
Chemistry : Basic Elements Tracer and Tracer Technique Most of the element occurs as mixture of isotopes, the proportion of which remains constant throughout the course of physical, chemical and biological changes. Some of these isotopes are radioactive. The radioactive isotopes of a given element behave chemically in same way as non-radioactive isotopes. Hence, if a radioactive isotope is added to its non-radioactive variety the mixture will behave the same way chemically. Since radioactive isotopes can always be detected easily, they may be used as indicators or tracers in various fields. The process in which a radioactive isotope is mixed with non-radioactive variety and the activity of radioisotope is noted after physical, chemical or biological changes, is called radioactive tracing or tracer technique. The success of radioactive tracing lies in the selections of a suitable nuclide. Neutron activation Analysis It is one of the most sensitive and specific methods available for the determination of trace quantities of a wide range elements in such diverse media as terrestial, lunar and meteoritic materials, marine sediments, air borne particles, natural water, biological material, hair, drugs, coal, petroleum, blood, etc.
The essential basis of this analysis is that element to be determined is made radioactive usually by slow neutrons irradiation and then this radioactivity (radiation emitted, on, y) is a measure of the mass of the element originally present. When an element is made radioactive by placing it in a homogenous flux of energetic charged particles of neutrons, the activity produced in the element is given by A = N f s (1 = elt)
or A = where N = the number of atoms of the nuclide in the sample capable of forming radio-isotope in question Nuclear Chemistry f = the neutron flux (i.e., no. of neutrons cm_2 s_1) s = the neutron capture cross section in cm2 for the given target nuclide. t = time of irradiation t1/2 = half-life of nuclear species produced l = decay constant in s_1 of the radioactive product The cross sectional area of the nucleus of a 10B atom assuming a spherical shape calculated. Their value to its cross section for neutrons 3.99 × 10_21 cm2 compared. r = l.4A1/3 × l0_13 = l.4 × (10)1/3 × 10_13 = 1.4 × 2.15 × 10_13 = 3.0 × 10_13 cm Area = pr2 = 3.14 × (3.0 × 10_I3)2 = 2.9 × 10_25 cm2 Compare to cross section 3.99 × 10_21 cm2, in capturing neutrons 10B acts as if it has a much larger cross section than its geometric cross-section i.e., a direct hit is unnecessary.
The fission of
represented by overall reaction considered
® Calculate the total energy released in the fission reaction. The masses in atm of various nucleides are as follows : 236U
= 236.0457, 94Zr = 93.90610, 140Ce = 139.9054,
= 0.00055,
= 1.00867
Mass of the reactants = 236.0457 Mass of the product = 93.90610 139.9053 0.00055 × 6 1.00867 × 2 235.83204
Chemistry : Basic Elements Mass defect = Mass of the reactants _ Mass of the products = 236.045 _ 235.83204 Energy released = 0.21296 × 931.5MeV = 199 MeV Matched : (i)
®
(A) Hydrogen bomb
(ii)
®
(B) Positron emission
(iii)
®
(iv)
®
(D) cc-decay
®
(v) (vi)
(C) Fission reaction
®
(E) Spallation reaction (F) K-electron capture
(G) Exo-ergic reaction (H) Capture reaction (i) (H), (ii) (F), (iii) (E), (iv) (G), (v) (A), (vi) (B). The nucleus and atom can be assumed to be spherical. The radius of the nucleus of mass number A is given by 1.25 × 10_13 × A1/3 cm. The atomic radius of atom is 1 Å. If the mass number is 64, the ratio by the atomic volume that is occupied by the nucleus would be :
= where r and R the radii of the nucleus and atom respectively.
= = 1.25 × 10_13 Nuclear Chemistry If two radioactive nuclide A and B have half-life of t and 2t respectively. If we start an experiment with one mole of each of them, Their mole ratio after a time interval of 6t
would be:
Number of half-lives of A = nA =
Number of half-lives of B = nB Initial amount of A and B = 1 mole each.
NtA =
NtA =
…(i)
Similarly NtB =
Dividing (i) by (ii)
…(ii)
=
Heat energy of the sun is generated by the fusion of four nuclei into one stock of hydrogen in the sun is consumed, then energy will be obtained from :
nucleus. If the
When all the jH nuclei are exhausted by their conversion into nuclei, the temperature an pressure in the sun will decrease and the sun will shrink in its size under its own gravitational attraction. When the sun has considerably shrunk in its size, the temperature in the sun will again became so hig that nuclei will fuse together to form
isotope and a tremendous amount of energy will be liberated
In this way sun will regain its lost sun shine and energy.
Chemistry : Basic Elements An element with mass number 15 and isotropic mass 15.00486 amu has mass defect of 0.124043 amu.
If mass of a proton and a neutron is 1.008145 amu and 1.008986 amu respectively, the atomic number of the element. Mass of the electron can be neglected would be : Let the element be represented as _ a) neutrons.
, thus the nucleus of the atom of this element has a proton and (15
Now, M´ = Mass of protons + Mass of (15 _ a) neutrons = [a × 1.008145 + (15 _ a) × 1.008986] amu M = 15.00482 (given) Mass defect = M´ _ M or, 0.124043 = 1.008145a + 1.008986 × (15 _ a) _ 15.00486 On solving this equation we get a = 7
Organic Name Reactions 10 Organic Name Reactions Favorskii Rearrangement The reaction of a-haloketones (chloro, bromo or iodo) with alkoxide ions to give rearranged esters is called Favorskii rearrangement. For example, 2-chlorocyclohexanone is converted to the methyl ester of cyclopentane carboxylic acid by treatment with sodium methoxide in ether.
The Favorskii rearrangement is an example of a migration to an electron rich carbon atom. Its mechanism with 2-chlorocyclohexanone can be represented as follows—
Chemistry : Basic Elements
(ii)
Completed Reactions :
(i)
(ii)
(i)
Organic Name Reactions Explanation : Both PhCH2COCH2Cl and PhCHCICOCH3 form PhCH2CH2COOH when reacted with OH-, followed by acidification. A common intermediate is required in this Favorskii rearrangement to set the same product. Removal of an aH by OH- is followed by an SN2 displacement of C1- to give a cyclopropanone ring. Ring opening occurs to give the more stable benzylic carbanion.
Favorskii Reaction : The reaction of aldehydes with terminal acetylenes is known as Favorskii reaction.
Sodium acetylides are the most common reagents, but lithium, magnesium and other metallic acetylides have also been used. A particularly convenient reagent is lithium acetylide-ethylene diamine complex. Alternatively, the substrate may be treated with the alkyne itself in the presence of a base, so that the acetylide is generated in situ. 1,4-Diols can be prepared by treatment of aldehyde with dimetalloacetylenes.
The Stork Enamine Reaction with its mechanism : Stork Enamine Reaction : Aldehydes and ketones react with secondary amines to form compounds called enamines. The general reaction for enamine formation can be written as
Chemistry : Basic Elements Water is removed as an azeotrope or by a drying agent. Enamine formation is also catalyzed by the presence of traces of an acid. The secondary amines commonly used are pyrrolidine, piperidine and morphine.
Cyclohexanone, for example, reacts with pyrrolidine in the following way
Enamines are good nucleophiles and have both a nucleophilic nitrogen and a nucleophilic carbon.
The nucleophilicity of the carbon of enamines makes them particularly useful reagents in organic synthesis because they can be acylated, alkylated, and used in Michael addition. Organic Name Reactions When an enarnine reacts with an acyl halide or acid anyhydrides the product is the C-acylated compound. The iminium ion that forms hydrolyzes when water is added, and overall reaction provides a synthesis of b-diketones.
Although N-acylatinn may occur in this synthesis, the N-acyl product is unstable and can act as an acylating agent itself.
As a consequence, the yields of C-acylated products are generally high. Alkylation of enamines may lead to the formation of N-alkylated product, which on heating is converted to C-alkyl compound (This rearrangement is common with allylic halide, alkyl halide or a-haloacetic ester.
Chemistry : Basic Elements Michael Addition Reaction When a new carbon-carbon bond is produced by nucleophilic addition to conjugated systems, the process is called Michael addition. The generalised process involves an a, b-unsaturated compound and a compound containing an active hydrogen attached to a carbon atom (e.g., malonic ester, acetoacetic ester, nitrocompounds, aldehydes, ketones etc.) These are condensed in the presence of a base. The overall reaction and its mechanism can be represented as follows : Mechanism : Overall reaction
Mechanism : Step 1
Step 2
Organic Name Reactions Step 3
Other examples are as follows :
Chemistry : Basic Elements The products of the following Michael addition, predicted: (i) Ethyl crotonate + Malonic ester ® (ii) Ethyl acrylate + ethylacetoacetate ® (iii) Methyl vinyl ketone + malonic ester X. (iv) Benzalacetophenone + acetophenone ® Y.
(v) Acrylonitrile + allyl cyanide ®
The acrolein, H2C=CHCHO is epoxidised much more readily with a basic solution of H2O2 than with a peroxyacid : (b) What is Robinson annelation reaction ? (c) Give the steps for this transformation Organic Name Reactions
(a) Base converts H2O2 to conjugate base, HOO-, that undergoes a Michael addition with acrolein to give an a-carbanion that then displaces HO- from the HOO- group leaving epoxide.
The acid catalyzed epoxidation goes by the typical electrophilic attack by HO+ (from H2O2) on C=C. (b) Robinson annelation reaction involves a Michael addition to an a, b-unsaturated ketone immediately followed by an aldol addition. For example,
(c) The transformation can be achieved by Michael addition using methyl vinyl ketone (Robinson annelation).
The sequence that follows illustrates how a conjugate aldol addition (Michael addition) followed by a simple aldol condensation may be used to build one ring onto another. This procedure is known as the Robinson annulation (ring forming) reaction (after the English chemist Sir Robert Robinson, who won the Nobel Prize in Chemistry in 1947 for his research on naturally occurring compounds).
Chemistry : Basic Elements
The mechanism of the following reaction :
The reaction is a Michael type of addition to a base-induced dehydration product of the (4-hydroxy phenyl) methanol.
The mechanism of addition of cyciohexanone to C6H5CH = CHCOC6H5 : Conjugate addition of enolate anion to a, b-unsaturated carbonyl compounds is known as Michael addition.
Organic Name Reactions The Mannich Reaction with its Mechanism : Mannich Reaction : This reaction takes place between an amine, an aldehyde (or ketone) and a highly nucleophilic carbon atom. The reaction is the addition of a nucleophilic carbon to an imonium ion intermediate. (The reaction is usually carried out in acid solution, but may also be base catalysed. This is the condensation between aldehydes, ammonia or a primary or secondary amine and a compound containing at least one active hydrogen atom e.g., ketones, b-ketoesters, b-cyanoesters, nitroalkanes, alkynes with CºH). For example, C6H5COCH3 + CH2O + HN(CH3)2 Mechanism : Step 1
Step 2
C6H5COCH2CH2N(CH3) 2 + H2O
The products of the Mannich reaction (Mannich bases) are useful intermediate on organic synthesis e.g.
(1) PhCOCH2CH2NHR2+Cl _PhCOCH==CH2 (2) PhCOCH2CH2MR2+
PhCOCH2CH2CN
Chemistry : Basic Elements Completed reactions :
(i) C6H5COCH3
(ii) (iii) NMe2 CHNO2+HCHO+EtNH2
(iv)
(v)
PhCOCH2CH3
(vi)
(i)
(ii)
(iii)
Organic Name Reactions
(iv)
(v)
(vi)
The Sharpless Asymmetric Epoxidation with an example: Sharpless Asymmetric Epoxidation : This is a method of converting allylic alcohols to chiral epoxy alcohols with very high enantioselectivity (i.e., with preference for one enantiomer rather than formation of racemic mixture). It involves treating the allylic alcohol with tert-butyl hydroperoxide, titanium(IV) tetra isopropoxide [Ti(O—iPr)4] and a specific stereoisomer of tartaric ester. For example,
The oxygen that is transferred to the allylic alcohol to form epoxide is derived from tert-butyl hydroperoxide. The enantioselectivity of the reaction results from a titanium complex among the reagents that includes the enantiomerically pure tartrate ester as one of the ligands. The choice whether to use (+) or (-) tartrate ester for stereochemical control depends on which enantiomer of epoxide is desired. The stereochemical preferences are such that it is possible to prepare either enantiomer of a chiral epoxide in high Chemistry : Basic Elements enantiomeric excess, simply by choosing the appropriate (+) or (-) tartrate stereoisomer as the chiral ligand.
The Ene reaction with example. How an ene reaction is related to Diels-Alder reaction: Ene reaction (Hydro-allyl addition)-A reaction of an allylic compound with an olefin which resembles both a cycloaddition and a [1, 5]-sigmatropic shift of hydrogen. In this reaction an alkene having an allylic hydrogen atom reacts thermally with a dienophile (enophile) C = C, C = O, N = N, N = O, C = S
etc. with the formation of a new s bond to the terminal carbon of the allyl group. The interaction of a hydrogen atom with the HOMO of the allyl radical and the LUMO of the enophile is a symmetry allowed process.
As related to Diels-Alder reaction this also represents a 6p electron electrocyclic reaction. Like DielsAlder reaction the ene reaction is reversible. For example, the produce) 1-pentene of the ene reaction between ethene and propene, gives back ethene and propene on decomposition at 400° C.
Organic Name Reactions A reaction with structure between .malefic anhydride and propene under thermal conditions:
Propene Maleic anhydride It is an ene reaction. Barton Reaction The photolysis of nitrites which do not contain g-hydrogen atoms usually results in the elimination of nitric oxide and the formation of hydroxy and carbonyl compounds. When g-hydrogen atoms are present than the product is an oxime or the corresponding nitroso dimer. This reaction is known as Barton S reaction.
The other example of Barton reaction is :
Chemistry : Basic Elements
By Barton reaction a methyl group in the d position to an OH group can be oxidised to a CHO group. The alcohol is first converted to the nitrite ester. Photolysis of the nitrite results in conversion of the nitrite group to the OH group and nitrosation of the methyl group. Hydrolysis of the oxime tautomer gives the aldehyde. The Hofmann-Loffler-Freytag reaction : Hofmann-Loffler-Freytag Reaction : It is a photochemically initiated decomposition of N-haloamines in acidic solution. First d-halo-amines are formed, these are then converted into pyrrolidines by intramolecular nucleophilic substitution, which involves a hydrogen abstraction. The first step of the reaction is a rearrangement, with the halogen migrating from the nitrogen to the 4 or 5 position of the alkyl group and the second step, the ring closure takesplace.
A similar reaction has been carried out on N-haloamides, which give
-lactone.
Organic Name Reactions A similar reaction has been carried out on N-haloamides, which give g-lactone.
The Shapiro Reaction with its Mechanism : Shapiro Reaction : Treatment of the tosylhydrazone of an aldehyde or a ketone with a strong base leads to the formation of an olefin, the reaction being formally an elimination accompanied by a hydrogen shift. This reaction is called Shapiro reaction.
The most useful method synthetically involves treatment of the substrate with at least two equivalent of an organolithium compound (MeLi) in ether, hexane, or tetramethylenediamine. Tosylhydrazones of a, bunsaturated ketones give conjugated dienes. The mechanism of the reaction has been formulated as
Mechanism of Baeyer-Villiger Oxidation : Both aldehydes and ketones are oxidized by peroxy acids. This reaction, called the Baeyer-Villiger oxidation, is especially useful with ketones, Chemistry : Basic Elements because it converts them to carboxylic esters. For example, treating acetophenone with a peroxy acid converts it to the ester, phenyl acetate.
The oxidation may be carried out by Caro's acid (per monosulphuric acid, H2SO5) or with perbenzoic, peracetic or monoperphthalic acid. The products of this reaction show that a phenyl group has a greater tendency to migrate than a methyl group. Had this not been the case, the product would have been C6H5COOCH3 and not CH3COOC6H5. This tendency of a group to migrate Organic Name Reactions is called its migratory aptitude. Studies of the Baeyer-Villiger oxidation and other reactions have shown that the migratory aptitude of groups is H > phenyl > 3° alkyl > 2° alkyl > 1° alkyl > methyl. In all cases, this order is for groups migrating with their electron pairs, that is, as anions. Completed Reactions :
Chemistry : Basic Elements
The product is omega-caprolactone, a seven membered lactone that cannot be made by an intramolecular ring closure.
Chichibabin Reaction Pyridine and other heterocyclic nitrogen compounds an be aminated with the alkali-metal amides by chichibabin reaction. The attack is always in the 2 position unless both such positions Organic Name Reactions are filled, in which case the 4 position is attacked. Substituted alkali-metal amides have also been used.
A similar reaction of this type is
Completed Reactions :
(i)
(ii)
(iii)
(iv)
(v)
Chemistry : Basic Elements
(vi)
(vii)
(viii)
(ix)
(i)
(ii)
Reaction (i) and (ii) are Baeyer-Villiger reaction.
Organic Name Reactions
(The reaction (vii)—(ix) are examples of Mannich reaction) The product and name the reaction :
Chemistry : Basic Elements
Robinson synthesis of tropinone (example of Mannich reaction in synthesis)
Completed sequence of reaction :
(i) A reasonable meachnism for the following reaction :
(ii) The way one may link together, phenol, formaldehyde and a secondary amine. Organic Name Reactions (iii) The structure of the product Indole + CH2O + (C2H5)2NH (iv) Starting from (I and II), one can get III (4,4-dimethyl cyclohexanone)
(ii) This is Mannich reaction, the active hydrogen compound being phenol (the hydrogens in the o- and ppositions are active) and the result is amino alkylation in the ortho position which is preferred.
(iii) Indole undergoes Mannich reaction with CH2O and (C2H5)2NH to give 3-N, N´diethylamino-ethylindole.
(iv) The first reaction is the Michael addition of the enolate anions to the methyl vinyl ketone followed by intramolecular aldol condensation.
Chemistry : Basic Elements
The following canverted :
The structures of the major product in the following reactions :
This is an example of Barton reaction.
Organic Name Reactions This is an example of Shapiro reaction.
The transformations carried out :
Chemistry : Basic Elements
Reagents in Organic Synthesis 11 Reagents in Organic Synthesis Complete the Reactions :
In terms of the catalysts used, The two reactions classified. Shown the steps in the mechanism of reaction : Product in both the reactions is CH3CH2CH3. Both reactions are catalytic hydrogenations (addition of H2). (i) is heterogeneous catalytic hydrogenation and (ii) is homogeneous catalytic hydrogenation. In step-1 an H2 adds to the rhodium complex and one Ph3P ligand (L) is lost, resulting in a five coordinate rhodium complex, A (L = Ph3P). In this oxidative addition, the Rh changes oxidation state from + 1 to + 3. In step-2 the alkene reacts with A to form a p complex, B, which undergoes rearrangement (step-3) of an H to one of the C's of the double bond, the other C forming a s bond to the Rh (C).
Chemistry : Basic Elements
In the last step, a second H is transferred to the other C, and the alkane is lost with simultaneous regeneration of the metal catalyst—
The reagents for carrying out the following transformations, indicated :
Reagents in Organic Synthesis Crown Ethers Crown ethers are cyclic polyethers and their structure permits a conformation with certain sized `holes' in which cations can be trapped by co-ordination with the lone pair electrons on the oxygen atoms. These are used as phase transfer catalysts. The cyclic polymers of ethylene glycol (OCH2CH2)n are named as X-crown-Y. X refers to total number of atoms in the ring and Y to the total number of oxygens in the ring. The ability of a crown ether depends to complex a cation: Why the nucleophilic reactions
under the reaction conditions give almost same quantitative yields ? On the size of the cavity in the crown ether which can be tailored to allow for the selective binding of only certain cations, i.e., whose ionic radius is best accommodated by the polyether. Thus 18-crown_6 shows a high affinity for K+ (ionic diameter 2.66 Å) and 15-crown-5 for Na+ (ionic diameter 1.80Å).
The crown ether, 18-crown-6 acts as a phase transfer catalyst and gets the anion into the organic phase. On coordination with a metal cation the crown ethers convert the metal ion into a species with a hydrocarbon like exterior. The crown ether, Chemistry : Basic Elements 18-crown-6, e.g., coordinates effectively with potassium ions as the cavity size is correct and because the six oxygens are ideally situated to donate their electron pairs to the central ion. Crown ethers render many organic salts soluble in non-polar solvents. Therefore salts like CH3COOK, KCN can be transferred into aprotic solvents by using catalytic amounts of 18-crown-6. In the organic phase the relatively unsolvated anions of these salts bring about nucleophilic substitution reaction on an organic substrate. Phase Transfer Catalyst A compound whose addition to a two-phase organic water system helps to transfer a water soluble ionic reactant across the interface to the organic phase where a homogeneous reaction can take place is called a phase transfer catalyst. These catalysts enhance the rate of a reaction. A quaternary ammonium halide R4N+ X_ e.g., tetrabutylammonium halide is phase transfer catalyst. It can cause the transfer of the nuclepphile for example CN_ as an ion pair (Q+ CN_) into the organic phase; since the cation (Q+) of the ion pair with its four bulky alkyl groups resembles a hydrocarbon even thoughit has a positive charge. It is thus lipophilic, i.e., it prefers a non-polar environment to an aqueous one.
Prevost and Woodward Reactions Both the reactions are essentially the additions of iodine carboxylate (formed in situ) to an alkene, i.e., the reaction of an alkene with iodine and silver salt. The Prevost procedure employs iodine and silver carboxylate under dry conditions. The initially formed transiodocarboxylate (b) from a cyclic iodonium ion (a) undergoes internal displacement to a common intermediate acylium ion (c). The formation of the diester (d) with retention of configuration provides an example of neighbouring group participation. The diester on subsequent hydrolysis gives a trans-glycol. Reagents in Organic Synthesis
In the Woodward procedure, water is present which intercepts the intermediate (c) to give a cishydroxy ester (e) and a cis-glycol on subsequent hydrolysis.
Umpolung : The reversal of polarity of the carbonyl carbon atom is termed umpolung (German: for polarity reversal). Normally the carbonyl carbon atom of an aldehyde (or a ketone) is partially positive i.e., electrophilic and therefore it reacts with nucleophiles. When the aldehyde is converted to a dithiane by reaction with 1,3-propanedithiol and reacted with butyl lithium the same carbon now becomes negatively charged to react with electrophiles. This reversed polarity of the carbonyl carbon is termed umpolung which increases the versatility of the carbonyl group in synthesis. The sulphur atoms stabilize
carbanions because sulphur has the capacity to utilize 3d orbitals for bonding and to occur in valence states higher than 2.
Chemistry : Basic Elements
Dithianes : These are thioacetals which can be prepared by treating an aldehyde with 1,3-propanedithiol in the presence of trace of acid.
A 1,3-dithiane is a weak proton acid (pKa = 32) which can be deprotonated by strong bases such as nbutyllithium. The resulting carbanion is stabilized by the electron withdrawing effect of the two sulphur atoms.
The reactions completed by writing the appropriate product:
Reagents in Organic Synthesis
Reaction completed :
Peterson's Synthesis : In the Peterson's synthesis the b-hydroxysilanes are converted to alkenes in either acidic or basic solution.
Chemistry : Basic Elements
Reagents Used in Transformations
(2) (1) (CH3)2CuLi (Lithium dimethyl cuprate). (2) (CH3)3SiCl (Trimethylsilyl chloride). Lobry de Bruyn-van Ekenstein Rearrangement
(1)
(2)
(3) Reducing sugars on treatment with dicyclohexyl carbodiimide (DCC) undergo rearrangement to form a mixture Reagents in Organic Synthesis of other reducing sugars e.g., glucose forms a mixture of fructose and mannose.
Suitable Mechanism for the Reaction :
(1)
(ii) DCC is dicyclohexylcarbodiimide.
A carboxylic acid cannot be directly converted to an acid amide by the action of an amine. Hence the need of a dehydrating agent like DCC. The acid is converted to a compound with a better leaving group.
Chemistry : Basic Elements
Difference in behaviour of RMgX and R2CuLi: Since the C—to—Mg bond has more ionic character than the C—to—Cu bond, its R group is more like R : _ and is much more reactive. A set of reagents is listed under column A and their uses under column B. The reagent with its most appropriate use—
Column A Column B (1) (Ph3P)3RhCl (a) Hydroboration (2) 18-crown-6 (b) Epoxidation (3) Bu4N+ Br_ (c) Alkylation (4) Disiamyl borane (d) Hydrogenation (5) DDQ (e) Dehydration (6) DCC (f) Dehydrogenation (7) CH3CO3H (g) Cation complexing (8) LDA (h) Phase transfer catalysis (1)—(d), (2)—(g), (3)—(h), (4)—(a), (5)—(f), (6)—(e), (7)—(b), (8)—(c). Reagents in Organic Synthesis Formulas for A through D in the following reactions
LDA rather than NaOEt is used as the base in the abvoe reaction :
Structure for A, B and C
(a)
Tributylstannane (C4H9)3SnH reduces an alkyl halide to the corresponding alkane by a free radical mechanism. The initiator is an azo compound, (CH3)2C(CN)—N = N—C(CN)(CH3)2 which breaks down to N2 and a radical. A possible mechanism : The initiator gives the carbon radical (CH3)2C(CN),(Radical) which abstracts an H from the tin compound. Radical + (C4H9)3Sn—H —® (C4H9)3Sn + Radical—H
Chemistry : Basic Elements The tributyltin radical abstracts a halogen atom from the alkyl halide and the chain is propagated as follows— (C4H9)3Sn · + R—X —® (C4H9)3Sn—X + R · then R · + (C4H9)3Sn—H —®R—H + (C4H9)3Sn · Mathced :
(1) Wilkinsons catalyst (a) Cyclic polyethers
(2) DDQ (b)
(3) Prevost reagent (c) (4) Crown ethers (d) RhCl[(Ph3P)3] (5) 1, 3-dithiane (e) (i-Pr)2N _ Li + (6) disiamyl borane (f) 2,3-dichloro-5,6-dicyano-1, 4-benzoquinone (7) LDA (g) Mixture of silver salt of an acid and iodine. (1)_(d), (2)_(f), (3)_(g), (4)_(a), (5)_(b), (6)_(c), (7)_(e). Application of Phase Transfer Catalysts In water : Na +: CN_ + n-Bu4N+Cl_ _® n-Bu4N+: CN_ + Na + Cl_ In non-polar solvent : n-C8H17Cl + n-Bu4N + : C_ _® n-Bu4N + Cl_ + n-C8H17CN n-Bu4N + Cl _ goes back into H2O and the two steps repeat. Activation of the C-terminus of an AA, followed by coupling with a second AA, is accomplished with the reagent, DCC, Reagents in Organic Synthesis
C6H11—N=C=N—C 6H11. Give the structure of the product of reaction of DCC with RCOOH :
The Merrifield solid-phase process for synthesizing peptides:
The solid phase is beads of polystyrene whose mer is mainly . However occasional benzene rings have p-CH2C1 substituents projecting out to the surface. The solid phase may be indicated as. [P]—CH2C1 (A), where [P] is the polystyrene backbone. The peptide chain is started at the Cterminus by bonding the BOC protected AA to the solid phase through benzyl ester formation, followed by removal of the BOC group.
A second AA (BOC-protected so that it will not self couple) is added, along with DCC. The two steps of addition of a BOC-protected AA and regeneration of NH2 with CF3COOH are repeated as many times as required. In between each step, the [P]-growing peptide chain is washed with suitable solvents to remove excess reagents and undesirable products. Thus intermediates do not have to be isolated and purified and the yields are high. Reactant BOC-AA´s, DCC and CF3COOH are added through an automated system. Removal of the completed peptide from the polymer is accomplished with anhydrous HF, which also removes the last BOC group. The Reactions Completed :
(i)
Chemistry : Basic Elements
(ii)
(iii)
(iv)
(i)
(ii)
(iii)
(iv) Use of SeO2
Reagents in Organic Synthesis
SeO2 is used as oxidising agent, dehydrogenating agent and carbonylating agent. Selenuous acid is formed when water reacts with SeO2.
Chemistry : Basic Elements The following reactions :
Reagents in Organic Synthesis
Baker Yeast
Saccharomyces cerevisiae (Baker Yeast) are micro organisms and are used for the reduction of carbonyl group to hydroxyl group and for reduction of double bond.
Products Formed in the Reactions
Chemistry : Basic Elements In Prevost
With OsO4
In Woodward-Prevost
The Reactions Completed :
(1)
(3)
(2)
(4) E Me3Sil.
Kinetics of Reactions 12 Kinetics of Reactions Methods to Determine Rate Laws Rate law can be determined by the following methods — Differential Method : The reactants are mixed and the concentration of reactant `c' (any one) is measured at regular intervals and a graph is drawn between concentration and time. The slope of the curve is measured. If the order of the reaction is V then the value of V can be calculated from the equation
Slope (rate) =
= Kcn, where `K' is the rate constant.
Integration Method or Hit and Trial Method : Here known quantities of standard solutions of reactants are mixed in a reaction vessel and the progress of the reaction is determined by determining the amount of reactant consumed after different intervals of time. These values are then substituted in the equations of first, second, third order and so on. The order of the reaction is the order corresponding to that equation which gives a constant value of K.
Chemistry : Basic Elements Fractional Change Method : From the equations of half life tor reactions of various orders except first order reaction, time required to complete a definite fraction of the reaction is inversely proportional to a0n_1 where `n' is the order of reaction and `a0' is initial concentration.
tµ Graphical Method : This is an extension of integration method, for a reaction of nth order, the rate of reaction
is equal to K (a0 _ x)n where K, a0 and x are the rate constant, initial concentration of the
reactant and the concentration of the reactant consumed at time `t'. If the plot of straight line than the order of the reaction is `n'. The value of
vs. (a0 _ x)n is a
(rate of reaction) is obtained from the
slope of the plot between the concentration of the reactant various time intervals`t'. In a certain reaction, time, t varies arithmetically while the concentration varies geometrically time 0 t 2t 3t ... Conc. a aa a2a a3a ... Shown that reaction is of the I order. For the first order
K= Substituting values
K=
= Kinetics of Reactions In a second calculation
K=
=
= In a thud calculation
K=
=
Thus we get the same value of K, which confirms I order. Shown that t3/4 = (2n _ 1 + l) t1/2 where `n' is the order of the reaction. Using this in the reaction of pyrolysis of CH3—CHO to determine `n'. t1/2 = 420 secs. t3/4 = 1220 secs in a particular kinetic run : For a reaction of order, `n' we have
Kt =
\ Kt1/2 =
since c =
when t = t1/2
Kt3/4 =
since c =
Dividing,
when t = t3/4
=
= (2n _ 1 + 1)
\ t3/4 = (2n _ 1 + 1) t1/2
Chemistry : Basic Elements In the pyrolysis of CH3_CHO,
=
» 3 = 2n _ 1 + 1
\n=2 On the action of Br2 on fumaric acid, following data were obtained _ First Experiment: (t) min. Concentration 0 8.87 95 7.87
Mean concentration = 8.37 and
= 0.0106
Second Experiment : (t) min. Concentration 0 3.81 1.32 3.51
Mean concentration = 3.66 and
= 0.00227
The order of reaction, found out :
n=
=
= 1.87
n=2 It is a second order reaction. The rate law for the following reaction. Assuming that the rate of formation and disappearance of NOC12 are equal in the first step, derived
Kinetics of Reactions K1 NO + C12
NOCl2(Fast) ... (1)
K2 K3 NO + NOC12 ® 2NOCl (Slow) ... (2)
[NOC1] = K3 [NO] [NOC12] ... (3)
From equation (2), But from equation (1),
K1 [NO][C12] = K2 [NOC12]
or, [NOC12] =
[NO] [Cl2] ... (4)
Substituting the value of (4) in (3), we have
[NO]2 [Cl2]
[NOCl] =
Overall reaction being 2NO + C12 ® 2NOC1 The rate of the reaction is
=
[NO]2 [Cl2]
=
[NO] = _
[C12]
For the Bromide catalysed decomposition of hydrogen peroxide (at pH < 7) 2H2O2 (aq) ® 2H2O (aq) + O2 (g) a mechanism proposed involves three steps (I) a pre-equilibrium step, (II) a reaction between Br_ and [HOOH2]+ to give HOBr and H2O which is considered to be the slowest step and (III) a fast step involving the release of O2 (g) ?
Chemistry : Basic Elements It is further observed experimentally that the rate of this reaction depends on pH and Br_concentration. The complete mechanism and obtain the rate law : K1 Mechanism: H2O2 + H+
[H2OOH]+ ... (1)
K_1 (protonation equilibrium) Br_ + [H2OOH]+ ® HOBr + H2O ... (2) K2 Fast HOBr ® H+ + Br_ + ½ O2 ... (3) Adding H2O2 ® H2O + ½O2 Applying steady state approximation to the unstable intermediate [H2OOH]+ K1 [H2O2][H+]_K_1 [H2OOH]+_K2 [Br_] [H2OOH]+= 0
[H2OOH]+ =
From equation (2)
= K2[Br_] [H2OOH]+
= The step (3) involving the formation of O2 is not shown in the above expression, this step has been omitted, since it is a fast step, subsequent to the rate determining step. The expression shows that the rate of decomposition of H2O2 depends on the concentrations of the catalyst (Br_) and the hydrogen ions. A decrease in pH increases the rate. In the following reaction mechanism the expression for the rate of formation of P : Kinetics of Reactions K1 R1 + R2
I
K2 K3 I is the reactive intermediate. I_ ® P Writing the differential equation for I and equating it to zero.
= K1 [R1][R2] _ K2[I] _ K3 [I] = 0
[I] =
Now
= K3 [I] =
[R1] [R2] = K[R1] [R2]
where K is the effective rate constant.
K= The isomerization of cyclopropane follows the Lindemann mechanism and is found to be unimolecular. The rate constant at high pressure is 1.5 × 10_4 s_1 and that at low pressure is 6 × 10_6 torr_1 s_l. The pressure of cyclopropane at which the reaction changes its order, found out: K1 Lindemann mechanism : A + A ® A + A (Activation) K2 A* + A ® A + A (deactivation) K3 A* ® B (product)
Chemistry : Basic Elements For the steady state concentration of A*
= Suppose the required transition pressure is pA. If the actual pressure is much greater than pA, the reaction is of the first order, i.e.,
= K (pressure)1
If the actual pressure is << pA, the reaction is of the second order, i.e.,
= K¢ (pressure)2 It is given that at the pressure pA, the transition of order occurs, so K.pA = K¢.pA2;
=
=
or pA = 25 torr The formation of phosgene by the reaction CO + Cl2 ® COCl2 appears to follow the mechanism : K1 Cl2 ® 2Cl K2 Cl + CO ® COCl K3 COCl + Cl2 ® COCl2 + Cl K4 COCl ® CO + Cl K5 2Cl ® C12 Kinetics of Reactions
Assuming that the intermediates COCl and Cl are in a steady state, the rate law for the formation of COCl2, found: Steady state with respect to COCl K2[Cl][CO] _ K3[COCl][Cl2] _ K4[COCl] = 0 ... (1) Steady state with respect to Cl 2K1 [Cl2] + K3 [CoCl][Cl2] + K4 [COCl] _ K2 [Cl] [CO] _ K5 [Cl]2 = 0 ... (2) Substituting (1) and (2) 2 K1 [Cl2] _ K5 [Cl]2 = 0
i.e., [Cl] =
... (3)
Substituting (3) in (1)
K2
[CO] _ K3[COCl] [Cl2] _ K4 [COCl] = 0
\ [COCl] =
... (4)
Rate of formation of COCl2 is given by step 3
= K3 [COCl] [Cl2] Substituting (4) in this equation
=
=
Chemistry : Basic Elements The half life period of a substance is 50 minutes at a certain concentration. When the concentration is reduced to one half of the initial concentration, the half life period is 25 minutes. Order of the reaction, calculated : Suppose the initial concentration is `a' moles/lit. Then (t1/2)1 = 50 minutes, a1 = a (t1/2)2 = 25 minutes, a2 = a/2 Substituting the values in the expression
=
=
or, 2 = or, log 2 = (n _ 1) log 1/2 = _ (n _ 1) log 2 1 = _ (n _ 1)
or, n = 0 Hence, the reaction is of the zero order. Nitramide N2ONH2 decomposes slowly in aqueous solution according to the reaction O2NNH2 ® N2O + H2O
The experimental rate law is
=
One of the following mechanism seems most appropriate: K1 (1) O2NNH2 ® N2O + H2O Kinetics of Reactions K1 (2) O2NNH2
O2NNH_ + H+ (fast equilibrium)
K_1 K2 O2NNH_ ® N2O + OH_ (slow) K3 H+ + OH_ ® H2O (fast) K1 (3) O2NNH2 + H+
O2NNH3+ (fast equilibrium)
K_1 K2 O2NNH3+ ® N2O + H3O+ (slow) (2) Mechanism seems appropriate because for this mechanism we have
= K2 [O2 NNH_] From the fast equilibrium step, we have
K=
=
\ [O2NNH_] = With this, the rate law becomes
= The above rate law agrees with the given one and hence this mechanism seems most appropriate.
Chemistry : Basic Elements
Shown that quantum yield =
for the photolysis of acetaldehyde :
The photolysis of acetaldeyde takes place in the light of 2500-3100 Å wavelength yielding CH4, CO, C2H6 etc. CH3CHO + hv ® CH4 + C2H6 + CO + other products
Following mechanism has been proposed — K1 (1) CH3CHO + hv ® CH3 + CHO K2 (2) CH3CHO + CH3 ® CH4 + CH3CO K3 (3) CH3CO ® CO + CH3 K4 (4) CH3 + CH3 ® C2H6 Carbon monoxide is formed only in step (3) hence
= K3[CH3CO] ... (A)
IIIly
IIIly
= K2 [CH3CHO][CH3] _ K3 [CH3CO]…(B)
= K1 Iabs + K3 [CH3CO] _ K2
[CH3CO] [CH3] _ K4 [CH3]2 ... (C) Applying steady state treatment to [CH3CO] we obtain
= K2 [CH3CHO][CH3] _ K3 [CH3CO] = 0 ... (D)
Kinetics of Reactions IIIly on applying steady state treatment to [CH3] we get
= K1 Iabs _ K2 [CH3CHO] [CH3] + K3 [CH3CO] _ K4 [CH3]2 = 0 ... (E) By adding equations (D) and (E) we get K1 Iabs _ K4[CH3]2 = 0
[CH3] =
... (F)
On substituting the value of [CH3] in equation (D) we get
K2[CH3CHO]
- K3[CH3CHO] = 0
K3 [CH3CHO] = K2 [CH3CHO]
... (G)
Substituting this value of K3[CH3CO] in equation (A) we get
= K2[CH3CHO]
=
K2 [CH3CHO] (Iabs)1/2
= K [CH3CHO] (Iabs)1/2 ... (H)
where K = K2 Now if the rate of formation of CO is divided by the intensity of absorbed radiation, the value of quantum yield is obtained,
Chemistry : Basic Elements Quantum yield =
=
=
Shown the variation of rate reaction with pressure and the three explosion limits :
Main Characteristics of Arrhenius Equation (1) Larger the activation energy smaller is the value of rate constant. This follows from K = A exp (_ Ea / RT) (2) Larger the activation energy, greater is the effect of a given temperature rise on K. This follows from equation
=
as
will be larger for a large value of Ea.
(3) At lower temperature, increases in temperature causes more change in the value of K than that at higher temperatures. This also follows from above equation where T appears in the denominator. Kinetics of Reactions The above characteristics are in agreement with the experimental results. The reaction scheme for an enzyme catalysed reaction may be written as : K1 K2 E+S
ES ® E + P
K_1 (E = enzyme, S = substrate, ES = complex, P = products) Assuming that the concentration of complex reaches a steady state after an initial transient stage, shown that :
[ES] =
=
where E0 is the initial enzyme concentration. The rate of appearance of product is given by
= K2[ES] ... (1) Applying the steady-state approximation to ES, we get
= 0 = K1[E][S] _ K_1 [ES] _ K2 [ES]
[ES] =
... (2)
The proposed mechanism.is usually tested with the experimentally determined initial rate of reaction under the condition that [S]0 > [E]0. Thus in equation (2) we may replace [E] and [S] by the relations [S] = [S]0 [E] = [E]0 _ [ES] ... (3) Substituting the above relations in equation (2) we get which on rearranging gives
Chemistry : Basic Elements
[ES] =
Michaelis and Menten derived the rate equation, r0 = without the use of steady state approximation and making the assumption that K2 < K_1. For the latter assumption, the equilibrium K1 E+S
ES
K_1 may be described by the constant :
Keq =
=
. Using this equation, derive the rate equation and show that KM = 1/Keq?
We have the relation. [E]0 = [E] + [ES] Replacing [E] in terms of [ES], we get
[E]0 = Rearranging this, we get
[ES] =
=
Substituting the above relation in the rate expression r = K2 [ES], we get
r= For the initial rate [S] = [S]0 Kinetics of Reactions Comparing the above relation with the rate equation given in the question we have KM = 1/Keq The potential energy diagram of the Michaelis Menten mechanism, drawn :
The hydrolysis of p-nitrophenylacetate to p-nitrophenol is catalyzed by a-chymotrypsin enzyme. The proposed mechanismis Fast K1 K2 E+S
ES¢ ¾® ES + P1 ¾® E + P2
where ES¢, P1 and P2 are acetyl enzyme, nitrophenol and acetate ion, respectively. If K1 is much smaller than K2 a qualitative plot of potential energy vs. reaction drawn and coordinated for the above reaction : As equilibrium reaction is fast, the energy of activation for this step will have a small value in comparison to other steps. The energy of activation for the second step wil! be much larger than the third step since K1 < K2. Hence the required potential energy diagram will have an appearanceas
The effect of increase in ionic strength on the rate constant for each of the following reactions, would be :
Chemistry : Basic Elements (a) Pr (NH3)3Cl+ + NO2_ (b) PtCl42_ + OH_ (c) Pt(NH3)2 Cl2 + OH_ (a) It decreases (b) It increases (c) There is no effect
Explicitly distinguished between the rate of a chemical reaction and the rate constant for the reaction : Rate = K [A]n [B]m. The rate is a variable, proportional to the concentrations and to the rate constant. The rate constant K, is a number which depends on the temperature only and not on the concentrations. Three sets of railroad tracks are to accomodate trains which are 14 feet wide. The minimum distance between the centres of the two outermost tracks must be : 28 ft. The molecule must have clearance of o on each side. Explained why transition states cannot be isolated as independent chemical species : Transition states have maximum energies. No matter how their bonds vibrate, the resulting molecules will have lower energies and therefore be more stable. A Stoichiometric equation for the reaction whose mechanism is given below. The value of the equilibrium constant for the first step determined : A2
2A, K1 = 1010s_1 (forward)
K_1 = 1010 M_1 s_1 (reverse) A + C ® AC, K2 = 10_4 M_1 s_1 The second step is the slow step as seen by the values of the rate constants. The overall reaction, obtained by doubling the second step reaction and adding the first step, is A2 + 2C ® 2AC The equilibrium constant expression for the first step is used to calculate an expression for [A]: Kinetics of Reactions
Kl =
=
=
Rate = K2 [A] [C] =
= 1, [A] =
= K2[C][A2]1/2 (K1 = l) A mechanism for the following decomposition devised, which will fit the given rate law: COC12 ® CO + Cl2, Rate = K [COCl2] [Cl2]1/2 One possible mechanism is COCl2
COCl + Cl (fast) initiation
The key to figuring out such a mechanism is to recognize from the half power in the rate expression that a Cl atom is involved in the rate determining step and that there must be an equilibrium between the chlorine atoms and mo.lecules. The overall stoichiometry is governed by the chain steps which occur many times more often than the other steps. A and rate constant for the reaction, CH3 + H2 ® CH4 + H at 200° C calculated if the collision diameter is 3.0 × 10_10 m and it is assumed that activation energy is zero for the above reaction: Here the two species are different, so equation
A= is used.
Chemistry : Basic Elements Consider CH3 as a unit mass of 15 a.m.u. so the reduced mass is
m=
=
= 0.293 × 10_23 g mol_1 = 2.93 × 10_27 kg mol_1
A= = 6.74 × 10_16 P m3 s_1mole_1 Taking P as unity and As Ea = 0 (given) K = (6.74 × 10_13dm3mole_1s _1)(6.023 × 1023mole_1) = 4.06 × 1011 dm3 mole_1s_1 If the activation energy of a reaction is 80.9 kJ mole_1, calculated the fraction of molecules at 400° C, which have enough energy to react to form products : Those collision will yield products if the energy possessed by the molecules is equal to or greater than activation energy to react to form products. This is given as
=
= i.e., fraction of molecules which have enough energy to go for the formation of products is very small i.e., 5.257 × 10_7. Calculated DH*, DG* and DS* for the second order reaction 2NO2 (g) ® 3NO (g) + O2 (g) at 500 K. Given A = 2.0 × 109 sec_1 Ea = 111 kJ mole_1 :
Kinetics of Reactions For a bimolecular reaction Ea = DH* + 2RT \ DH* = Ea _ 2 RT DH* = 111 _ 2 × 8.314 × 500 = 102700 J mole_1 = 102.7 kJ mole_1 Now we know that
A=
\
=
\
= In
DS* = 2.303 R log
= DS* = _ 71.17 J mole_1 K_1 Since DG* = DH* _ T DS* = 102700 _ 500 × (_ 71.17) DG* = 138300 JK mole_1
From the following rate law expressions, the corresponding stoichiometric equations, written. The initial concentrations are written as a, b, c and x represents the concentration of A and have reacted :
(1)
= K (a _ x) (b _ x)
(2)
= K (a _ x) (b _ 2x)
Chemistry : Basic Elements (3)
= Kb (a _ x)
(4)
= K (a _ x) (b + x)
(5)
= K1 (a _ x) _ K2 x
The rate law expressions show that there are two reactants, say A and B. Reactants A B Initial conc. a b Conc. at time `t' (a _ x) (b _ x) (1) A + B ® Products (2) A + 2B ® Products (3) B is in excess or acting as a catalyst, Hence A + B ® Product Excess or catalyst
(4) A + B ® Product. Hence B is acting as an autocatalyst. K1 (5) A
B
K2 The gaseous decomposition of ozone 2O3 ® 3O2, obeys the rate law r = _ d[O3]/dt = K[O3]2/[O2] Shown that the following mechanism is consistent with the above rate law : K O3
O2 + O (fast equilibrium)
K1 O + O3 ® 2O2 (slow) Kinetics of Reactions From the slow rate-determining step r = _ d [O3]/dt = K1 [O] [O3] From the fast equilibrium step K = [O2][O]/[O3] or [O] = K[O3]/[O2] Substituting for [O] in the expression for r, we have r = K1K[O3]2 /[O2] = K[O3]2/[O2] where K = K1K
Thus, the given mechanism is consistent with rate law. The termolecular reaction 2NO + H2 ® 2NOH is found to be third-order obeying the rate law : r = K [NO]2 [H] Shown that it is consistent with either of the following two mechanisms : K (i) 2NO
N2O2 (fast equilibrium)
K¢ N2O2 + H2 ® 2NOH (slow) K¢ NOH2 (fast equilibrium)
(ii) NO + H2 K¢¢
NOH2 + NO ® 2NOH (slow) (i) Since the slow step is the rate determining step, hence r = K¢ [N2O2][H2] However, from the fast equilibrium step
K=
or [N2O2] = K [NO]2
Hence r = K¢K [NO]2 [H2] = K[NO]2 [H2] where K = K¢K.
Chemistry : Basic Elements (ii) Proceeding as in above, we have r = K¢¢ [NOH2] [NO]
But K¢ =
so that
[NOH2] = K¢[NO][H2] Hence, r = K¢¢ K¢ [NO]2 [H2] = K [NO]2 [H2] where K = K¢¢ K¢ Thus both the mechanisms account for the observed rate law. Out of these which one will be the correct is determined by the experiment. For a reaction, if the effective rate constant K¢ is given by, then,
K¢ = The effective activation energy and effective frequency factor, calculated. In the above expression K1, K2, K3 and K5 stand for the rate constant for the different steps of the reaction: According to Arrhenius equation K = A . e_Ea/RT or, K = A . exp (_ Ea/RT) ü ï Hence K1 = A1 exp (_ Ea(1)/RT)ï ï
K2 = A2 exp (_Ea(2)/RT) ý .. (1) ï K3 = A3 exp (_Ea(3)/RT) ï ï K5 = A5 exp (_Ea(5)/RT) þ
We know that K¢ =
...(2)
Kinetics of Reactions Combining equations (1) and (2) we get
K¢ =
K¢ = or, K¢ = A¢ . exp (_ Ea¢/RT) ... (3) So the effective energy of activation Ea¢ is given by
E a¢ = and the effective frequency factor A¢ is given by
A¢ = Value of the rate constant is predicted by the Arrhenius equation if T ® ¥? This value physically reasonable is : From the relation K = Ae_Ea/RT, we see that if T ® ¥, K ® A, so that Ea = 0. This implies that there is no activation energy for the reaction. Hence the result is not physically reasonable. The following Lindemann mechanism for the unimolecular decomposition of a molecule A in the presence of a species Y (which may be any molecule such as inert gas like Helium or even A itself), considered : K1 A + Y ® A* + Y (activation) K_1 A + Y ® A + Y (deactivation) K2 A* ® P (reaction)
Chemistry : Basic Elements Using the steady-state approximation (s.s.a.) the rate law for the formation of the product, derived : r = _ d [A]/dt = + d [P]/dt = K2[A*] ... (1) Applying s.s.a. to the transient species A* K1 [A] [Y] _ K_1 [A*][Y] _ K2 [A*] = 0 or, K1 [A] [Y] = (K_1 [Y] + K2) [A*]
[A*] = Substituting this expression in equation (1) we obtain
r= For an enzyme-substrate system obeying the simple Michaelis Menten mechanism, the rate of product formation when the substrate concentration is very large, has the limiting value 0.02 mol dm_3. At a substrate concentration of 250 mg dm_3, the rate is half this value, K1/K_1 assuming that K2 >> K_1, calculated :
r= For [S] >> Km, r = Vmax = K2 [E]0 when r is half of this maximum value, then by definition [S] = Km = (K_1 + K2)/K1 = K_1/K1] (assuming that K2 << K_1) If the activation energy of a reaction is 80.9 KJ mole_1. The fraction of molecules at 400° C which have enough energy to react to form the products, calculated : Those collisions will yield products if the energy possessed Kinetics of Reactions by the molecules is equal to or greater than activation energy. This is given as
= e_Ea/RT =
= e_14.459
= 5.257 × 10_7 i.e., the fraction of molecules which have enough energy to go for the formation of products is very small i.e., 5.257 × 10_7. Effect of Electrolytes in the Ionic Reactions : These effects are generally known as salt effects. They are of two types— Primary Salt Effect : It is defined as the effect of ionic strength on the velocity of the ionic reaction. This effect is involved in non-catalytic reactions. Secondary Salt Effect : This arises when one of the reaciants is a weak electrolyte, the extent of ionization and hence the concentration of ions are then affected by the presence of salts. The secondary salt effect is involved in catalytic reactions. Different types of methods that are used in the study of the kinetics of reactions with an indication of the range of half lives to which each method applies; mathced : Method Range of half lives, sec (1) Conventional (a) 102_ 108 (2) Flow (b) 10_15 _ 10_10 (3) Relaxation (c) 10_3 _ 102 (4) Kinetic spectroscopy (d) 10_I0 _ 1 (1)_(a), (2)_(c), (3)_(d), (4)_(b). The reaction between two complex molecules proceeds more slowly compared to the reaction between two less complex species. Explained in short using the theory of absolute reaction rates:
Chemistry : Basic Elements For Simple Atoms. If the contribution of single translational, rotational and vibrational degrees of freedom are written as fT, fR and fv, the total partition function F for a molecule may be expressedas F = fTt, fRr, fVv, where t, r, v are the numbers of the degrees of freedom contributing.
For the reaction between two atoms. FA = fT3, FB = fT3, F++ = fT3 fR2 so the rate constant of the reaction is given by
K= For complex molecules containing NA and NB atoms : FA = fT3 fR3 fV3NA _ 6 FB = fT3 fR3 fV3NB _ 6 and F++ = fT3 fR3 fV3(NA + NA) _ 7 As the activated complex has one degree of vibrational freedom less than a normal molecule with NA + NB atoms. The rate constant is
K= Therefore rate constant differs from that obtained in the atomic case by the factor (fV/fR)5 . Since fV is usually of the order of unity, whereas fR may be from 10 to 100 for a complex molecule, the ratio fV/fR is 10_1 to 10_2, so that (fV/fR)5 is 10_5 to 10_10. Thus reactions between complex molecules may therefore be expected to proceed very much slowly.
For two parallel reactions is given by Kinetics of Reactions
shown that the activation energy, E¢ for the disappearance of A
E¢ =
, where E1 and E2 are the activation energies of the two paths:
We have
=
(For the first path)
For the overall reaction K = K1 + K2
and
=
=
We write
=
Adding
(K1 + K2) = Divide by K = (K1 + K2), we have
=
i.e.,
=
Compare with
=
we have
E¢ = Oscillatory Reaction In oscillatory reactions the concentration of the intermediate will increase and decrease alternately and periodically. This is due to the autocatalytic effect of one of the intermediates.
Chemistry : Basic Elements Lotka-volterra Mechanism : K1 (I) A + X ® 2 X K2 (II) X + Y ® 2Y K3 (III) Y ® B Rate of the above steps can be given as
= _ K1 [A] [X]
= _ K2 [X] [Y]
= K3 [Y] Steps (I) and (II) are autocatalytic. The concentration of A is held constant by supplying it to the reaction. This leaves [X] and [Y] the concentrations of the intermediates, as variables. In this case we can solve the rate equations exactly for the variable concentrations of X and Y, but concentration of A is held at a
constant value since it is being supplied to the reaction. On solving the equations numerically, the plot of [X] and [Y] against time is like
According to the collision theory for bimolecular reactions of the type 2 HI (g) ® H2 (g) + I2 (g), Kinetics of Reactions The expression of collision rate with respect to temperature and reduced mass of the collision partners : If the collision is occurring between two unlike molecules A and B of masses mA and mB then the expression for the collision number
ZAB =
where `m' is the reduced mass = d ® Mean molecular diameter T ® Temperature NA and NB are the number of molecules per unit volume. Thus for the reaction between two molecules A and B the rate is expressed as
Rate =
where e_E/RT is the Arrhenius factor. Two examples of heterogeneous catalysis : (I) Decomposition of potassium chlorate in the presence of solid MnO2 MnO2 2 KCl O3 ¾¾® 2 KCl + 3O2 (II) Catalytic activity of platinum on the decomposition of hydrogen peroxide Pt 2H2O2 ¾¾® 2H2O + O2 The mechanism of enzyme catalysis drawn, using (a)random ternary complex theory, (b)ordered ternary complex mechanism and (c) ping-pong bi-bi mechanism :
Chemistry : Basic Elements
(a) (Random ternary complex mechanism) K1 [A] K2 [B] K3 (b) E
EA
EAB ——® E + Y + Z
K_1 K_2 (ordered ternary
complex mechanism) K1[A] K2 K3 [B] K4 (c) E
EA ¾® EA¢ ¾® EA¢B ¾® E + Z
K_1 Y (ping-pong bi-bi mechanism) Mechanism of Enzyme Catalysed Reaction Step 1 : Formation of Enzyme substrate complex K1 E+S
ES (fast)
K_1 (compex) Step 2 : Decomposition of the complex K2 ES ¾¾® P + E (slow) According to the slow rate determining step, the rate of the reaction is given by
r=
=
= K2 [ES] ... (1)
Using steady state approximation for ES, we have d [ES]/dt = K1[E][S] _ K_1 [ES] _ K2 [ES] = 0 ... (2) Kinetics of Reactions The equilibrium between the free and bound enzyme is given by the enzyme conservation equation,
[E0] = [E] + [ES] ... (3) where [E0] is the total enzyme concentration (which can be measured); [E] is the free enzyme concentration and [ES] is the bound (or reacted) enzyme concentration. Thus [E] = [E]0 _ [ES] ... (4) Substituting for [E] in equation (2) d[ES]/dt = K1{[E]0 _ [ES]}[S] _ K_1[ES] _ K2[ES] =0 ... (5) Collecting terms and simplifying K1[E]0 [S] = {K_1 + K2 + K1 [S]}[ES] ... (6)
... (7)
[ES] =
Substituting for [ES] in equation (1)
r=
... (8)
Dividing the numerator and denominator by K1
r=
=
... (9)
where the new constant Km called the Michaelis constant is given by Km = (K_1 + K2)/K1 ... (10) Equation (9) is known as the Michaelis-Menten equation When all the enzyme has reacted with the substrate at high concentration the reaction will be going at maximum rate.
\ [E]0 = [ES] Hence from equation (1) rmax = Vmax = K2[E]0
Chemistry : Basic Elements where Vmax is the maximum rate, using the notation of enzymology. This Michaelis-Menten equation can now be written as r = Vmax [S]/(Km + [S]) ... (11) Two case arise : (a) Km >> [S] so that [S] can be neglected in the denominator of equation (11) r = Vmax [S]/Km = K¢ [S] (First order reaction) (b) [S] >> Km, so that Km can be neglected in denominator. \ r = Vmax = constant (zero order reaction)
If Km = [S], r =
Vmax
Thus, Michaelis constant is equal to that concentration of S at which the rate of formation of product is half the maximum rate obtained at high concentration of S. At 25° C and pH 8, the turnover numbers and Michaelis constants for the fumarase reaction Fumarate + H2O
L-malate
are K3 = 0.20 × 103 s_1, K2 = 0.60 × 103 s_1 Ks = 7.0 × 10_6 mol L_1, Kp = 100 × 10_6 mol L_1. The values of K1 and K4 and the equilibrium constant [P]eq/[S]eq are :
K1 =
=
= 1.14 × 108 L mol_1 s_1
K4 =
=
= 8.0 × 106 L mol_1 s_1
Keq =
=
= 4.8
The Macromolecules 13 The Macromolecules The number average and weight average molecular weights of polymers, defined : Number-average Molecular Weight : When the total molecular weight of all the molecules of a sample is divided by the total number of molecules, the result obtain is called the number average molecular weight. Weight-average Molecular Weight : When the total molecular weight of group of molecules having particular molecular weights are multiplied with their respective molecular weight, the products are added and the sum is divided by the total weight of the all molecules, the result obtained is called weightaverage molecular weight. The expressions for the weight average and the number average molecular weights of macromolecules, derived : Number Average Molecular Weight Let N1 molecules have molecular weight M1 each. Let N2 molecules have molecular weight M2 each and so on.
Chemistry : Basic Elements Then we have Total molecular weight of all the N1 molecules = N1M1 Total molecular weight of all the N2 molecules = N2M2 and so on. \ Total molecular weight of all the molecules = N1M1 + N2M2 + N3M3 +......
= SNiMi Total number of all the molecules = N1 + N2 + N3 +......= SNi Hence, the number average molecular weight will be given by
=
= Suppose N1, N2 etc. molecules have molecular weights M1, M2 etc. respectively. Total molecular weight of N1 molecules = N1M1 Total molecular weight of N2 molecules = N2M2 and so on. The products with their respective molecular weights will be N1M1 × M1, N2M2 × M2, etc. Sum of the products = N1M1 + N2M22 + N3M32 + ...... = SNiMi2 Total weight of all the molecules = N1M1 + N2M2 + N3M3 +...... = SNiMi The Macromolecules Hence, weight-average molecular weight is given by
=
or,
=
Under the following condition M_n = M_w :
We know that
=
= =
If
, then
= The two sides can be equal only if M1 = M2 = ... i.e., all molecules should have the same molecular weight. In that case average has no significance. M_w for a system containing equal number of particles with molecular weights 10000 and 20000, calculated : In this case N1 = N2
= Substitute N1 for N2
\
=
=
Substitute for M1 and M2
=
=
= 16667
Chemistry : Basic Elements Equal number of mloecules with M1 = 100000 and M 2 = 10,000 are mixed. M_n and M_w, calculated : Let n1 = n2 = 10 (say)
Then
=
=
=
=
=
=
=
=
=
= 55000
=
= 91818
Equal masses of polymer molecules with M1 = 10000 and M2 = 100000 are mixed. M_n and M_M, calculated : Let m1 = m2 = 200000 (say)
n1 =
=
= 20
n2 =
=
=2
=
=
= 18182
The Macromolecules
=
=
= 55000
In a particular sample of a polymer, 100 molecules have molecular weight 103 each, 200 molecules have molecular weight 104 each and 200 molecules have molecular weight 105 each. The numberaverage and weight average molecular weight, calculated :
=
= 4.4 × 104 = 44000
=
=
=
= 91000
Intrinsic viscosity molecular weight relationship. The values of two constants appearing in the above relation, determined when fractions of a polymer of molecular weights 34000, 61000 and 130000 dissolved in an organic solvent gave the intrinsic viscosities as 1.02, 1.60 and 2.75 respectively at 25°C : Intrinsic viscosity molecular weight relation is [h] = kMa Substitute the values in the equation above 1.02 = k 34000a ... (1) 1.60 = k 61000a ... (2)
Chemistry : Basic Elements Divide equation (2) by (1)
=
or
=
or, 1.568 = 1.794a Taking logarithm of both sides
a log 1.794 = log 1.568
or a =
=
= 0.769
Substitute the value of a in (1) 1.02 = k 340000.769 Taking logarithm of both sides log 1.02 = log k + 0.769 log 34000 0.0086 = log k + 0.769 log (3.4 × 104) 0.0086 = log k + 3.4847 log k = 4.5239 or, k = 3.341 × 10_4 At 25°C, the density of glucose is 1.55 g cm_3, its diffusion coefficient is 6.81 × 10_6 cm2 s_1 and the coefficient of viscosity of water is 8.937 × 10_3 Poise. Assuming that the glucose molecule is spherical, estimate its molar mass : r = 1.55 gcm_3 = 1.55 × 103 kg m_3 D = 6.81 × 10_6cm2 s_1 = 6.81 × 10_10m2 s_1 h = 8.937 × 10_3 r = 8.937 × 10_4 kg m_1 s_1 From Stokes Einstein equation
r=
The Macromolecules
= = 3.59 × 10_19 J kg_1 m_1 s_2 = 3.59 × 10_10 m (... J = kg m2 s_2) Volume of glucose molecule is given by
V=
=
(p) (3.59 × 10_10m)3
= 1.94 × 10_28 m3 so that its mass is given by m = Vr = (1.94 × 10_28m3) (1.55 × 103 kg m_3) = 3.01 × 10_25 kg = 3.01 × 10_22g Hence, the molar mass of glucose is given by M = NAm = (6.022 × 1023 mol_1) (3.01 × 10_22g) = 181.3 g mol_1 A protein sample consists of an equimolar mixture of haemoglobin (Mm = 15.5 kg mol_1), ribonuclease (Mm = 13.7 kg mol_1) and myoglobin (Mm = 17.2 kg mol_1). The number average and mass average molar masses. Which is greater, calculated:
Mn = = 15.5 kg mol_1
Mw = = 15.6 kg mol_1 Thus, Mw > Mn
Chemistry : Basic Elements The molar mass of haemoglobin from the fact that in an equilibrium ultracentrifuge experiment at 20°C, C2/C1 = 9.40, r1 = 5.5 cm and r2 = 6.5 cm. The ultracentrifuge rotor is operated at 120 rps. v = 0.749 cm3 g_1 and r = 0.9982 g cm_3, calculated:
Mw =
= = 63.4 kg mol_1 Atactic and Isotactic as Applied to Polymer Structures : When a substituted vinyl monomer like styrene (C6H5CH = CH2) is polymerised the carbon atom bearing the substituent becomes asymmetric and may take three possible arrangements. In the atactic polymer a random arrangement of groups is observed as in the case of free radical polymerisation. If all the asymmetric carbon atoms have the same configuration with respect to each other the polymer is isotactic. When the configuration alternates it is syndiotatic. Isotactic and syndiotactic polymer's are usually obtained with Ziegler-Natta catalyst.
General Reaction Scheme Free radical polymerisation is catalysed by organic peroxides or other reagents which decompose to give free radicals. Following steps are involved. The Macromolecules Chain Initiation : Organic peroxides undergo homolytic fission to form tree radicals.
Chain Propagation : Free radical produced in the above step adds to an alkene molecule to form a new free radical.
This free radical can attack another alkene molecule and so on.
Chain Termination : The above chain reaction comes to halt when two free radical chains combine 2R(CH2CH2)n CH2CH2 ® R(CH2CH2)n CH2CH2 :
CH2CH2 (CH2CH2)n R Polyethylene (Polymer) Free radical polymerisation can also be promoted by a mixture of ferrous sulphate and hydrogen peroxide (FeSO4 + H2O2). These two compounds react to produce hydroxyl radicals. (OH ) which act as chain initiators. They are used as catalyst in the manufacture of Orion and Teflon. Stereo Regular Polymers Isotactic and syndiotactic polymers are collectively referred to as stereoregular polymers. They are always arranged in a head to tail (a, b) arrangement. Karl Ziegler and Giulio Natta, the co-winners of the 1963, Nobel Prize in Chemistry developed Chemistry : Basic Elements methods for synthesizing polymers with specified stereoisomerism. They found that,a mixture of TiCl4 and Al(C2H5)3 was an excellent heterogeneous catalyst called Ziegler Natta catalyst for synthesizing stereoregular polymers. The polymers have the following three stereo regular arrangements: (a) Isotactic (`tatto', a Greek word meaning `to put in order'): In this case, each carbon bearing the R group has the same configuration. (b) Syndiotactic (`Syndyo' means `every two') : In this case, the alternate carbons bearing the R group have the same configuration. (c) Atactic : Here there is a completely random arrangement of monomer units. Ziegler-Natta Catalysts Ziegler-Natta catalyst may be defined as a combination of (a) a transition metal compound of elements from groups IV to VII and (b) an organometallic compound of a metal from groups I to III of the periodic table. The transition metal compound is the catalyst and the organometallic compound is the co-catalyst. (c) Addition of amines can enhance the formation of stereoregulated polymer R2 A1X + Y TiCl3 + amine can give 90-99% of stereoregularity. If the R group is small in A1R3 the polymerisation is more stereoregulated.
Mechanism of Ziegler-Natta Polymerisation (1) Polymerisation occurs at localised active sites on the catalyst surface (2) Organometallic compound (co-catalyst) activates the site of alkylation of transition metal atom at the surface. The Macromolecules (3) Two probable mechanisms are : (a) bimetallic and (b)mono metallic mechanism, (a) Bimettalic mechanism—
The alkyl group migrates to the olefin forming a new metal alkyl i.e., a growing polymer chain, (b) Monometallic mechanism—
In both the mechanisms the polymer chain grows from the catalyst surface by successive insertion reactions of the complexed monomer, the R group originally present in the cocatalyst ending up as the terminal group of the chain.
Chemistry : Basic Elements Light Scattering Method The light scattering method is considerably more involved (theoretically and practically too) than the other methods. It can be used at extremely low concentrations of solution with sufficient accuracy. A wide range of molar mass (from a few hundreds to a few lakhs) can be determined and it is one of the most reliable methods for determining the shapes of large molecules. The kinetics of addition polymerization, given : Consider the catalytic polymerization of a monomer, M say CH2 = CHX (X = Halogen). The following stages are involved in this process. (A) The initiator molecule I yields a free radical R : I
R
The rate of the production of the free radical is given by r = K1[I] ... (1) (B) The free radical R adds to the double bond of the monomer molecule generating a new free radical: CH2 = CHX + R
RCH2 CHX
RCH2CHX + CH2 = CHX ¾¾® R— (CH2— CHX)2 and so on. The rate of the disappearance of the monomer
M is given by
= K2 [R] [M] ... (2)
These chain propagation steps are second order and very rapid. (C) Chain termination is achieved by the recombination of two free radicals i.e., R_ (CH2 _ CHX)n + (XCH _ CH2)n _ R R2 _ (CH2 _ CHX)n _ (XCH _ CH2)n _ R Since free radicals have a very low concentration, we can The Macromolecules assume that their rate of production is equal to their rate of disappearance (Steady State approximation) so that K1[I] = K3[R]2 ... (3) Hence, [R] = (K1/K3)2 [I]1/2 ... (4) Substituting for [R] in equation (2), we get
= = k [I]1/2 [M] ... (5) It is customary to define the kinetic chain length Y, as the ratio of the overall rate to the rate of the initiation step. It can also be defined as the number of repeating chain units in the polymer chain.
Y=
=
... (6)
We can define the probability p for the addition of another monomer to the growing chain radicalas
p=
=
... (7)
In Vinyl polymerisation p is of the order of 0.999. The Kinetics of condensation polymerisation, given : Consider the reaction Mm + Mn ¾¾® Mm + n where m, n are integers greater than one. The monomer M, may be removed by reaction with itself and with any other molecule present. Net rate of disappearance of the monomer is given by
= K{[M1]2+ [M1] [M2] + ....... }
Chemistry : Basic Elements =
... (1)
The net rate production of the dimer is given by
=
K[M1]2 _ K[M2]
...(2)
We can similarly show that the net rate of formation of n-mers, i.e., polymer containing n monomers, is given by
=
…(3)
Adding these equations, we get
=
... (4)
that of all the If [M1]0 is the total concentration of the monomers in the beginning and molecules at time t, the fraction f of the monomers the reaction that has occurred at time t is given by
f=
...(5)
Hence from equations (4) and (5), we get
=
...(6)
Integration yields
f= The concentration of the n-mers is given by [Mn] = [M1]0 fn _ 1 (1 _ f)2 The Macromolecules The following matched : (a) [—CH2 — CH2—]n (1) Polyvinylchloride.
(b)
(2) Nylon
(c) (d)
(4) Polyethylene.
(5) Polystyrene (a)_4, (b)_5, (c)_l, (d)_2, (e)_3. The data used and the molar mass of haemoglobin at 20° C, calculated : Density of water = 0.9982 × 103 kg m_3 Sedimentation coefficient = 4.6 × 10_13 Svedberg Diffusion coefficient = 6.9 × 10_11 m2 s_1 Velocity = 0.749 × 10_3 m3 kg_1 These physical constants are at 20°C for haemoglobin We know that
M=
=
= 64.4 Kg mol_1 = 64,400 g mol_1.
Fast Reactions 14 Fast Reactions K1 For the reaction : A B, the relaxation time t in terms K_1 of Kt and K_1, expressed : K1 B
A K_1
It is a reversible first order reaction. Let `a' be the concentration of A + B and x the concentration of B at any time.
Then
= K1 (a _ x) _ K_1 x ... (1)
The rate of change of Dx = x _ xe is accordingly,
= K1 (a _ Dx _ xe) _ K1 (Dx + xe) ... (2) At equilibrium however
= 0 = K1 (a _ xe) _ K_1 xe ... (3)
Chemistry : Basic Elements Therefore equation (2) becomes
= _ (K1 + K_1) Dx so that t = (K1 + K_1)_1 Flash Photolysis This technique for the study of a fast reaction is gas phase or liquid phase was developed by Norrish and Poster. This is an example of Pulse method which initiates a reaction by creating new reactive species—excited electronic states, radicals, ions in the system under study. The method uses a light flash of high intensity for a very short duration (10_6 s) to produce atoms or free radicals or excited species in a system. These are at a fairly high concentration and undergo further reactions which are followed spectroscopically. A spectroscopic flash of light is followed by the initial flash by some fraction of a millisecond. The absorption spectra of all the species that are formed within the system can be recorded. One cannot only get indications of what species are formed but also how these species give rise to others. Thus a very direct picture of the kinetic behaviour of a fast reaction can be obtained. The relationship between K1, K2 and t for the reaction of type found, K1 A
B+C
K2 As an example take the ionization of a weak acid in a large excess of solvent, CH3COOH + H2O
CH3COO_ + H3O+.
Let a be the concentration of CH3COOH + CH3COO_, and x the concentration of H3O+ (equal to that of CH3COO_). Including the constant water concentration in K1, we can write
= K1 (a _ x) _ K2 .x2 Fast Reactions Therefore, with
Dx = x _ xe (xe ® equilibrium concentration)
= K1 (a _ xe _ Dx) _ K2 (xe _ Dx)2 ... (1) At equilibrium
= 0 = K1 (a _ xe) _ K2 xe2 where departure from equilibrium Dx is very small, the term in (Dx)2 can be neglected compared with those in Dx, and (1) becomes
= _ (K1 + 2K2 xe) Dx so that t = (K1 + 2K2 xe)_1 ... (2)
The relaxation time for the fast reaction A B is 10 ms and the equilibrium constant is 1.0 × 10_3. The rate constant for the forward and the reverse reactions, calculated : From the equation t = l/(K1 + K_1) = 10 × 10_6 s = 10_5s ... (1) K = K1/K_1 = 1.0 × 10_3 \ K1 = 1.0 × 10_3 K_1 ... (2) Since K_1 << K1, it can be neglected in comparison with K1 in equation (1) so that
1/K1 = 10_5 s \ K1 = 105s_1 K_1 = 105 s_1/10_3 K_1 Kinetic Measurements in Flow Systems : Kinetic measurements in flow systems are generally useful when the reactants are at extremely low pressures or concentrations. Chemistry : Basic Elements Flow systems are also used for studying kinetics of fast reactions. Flow systems are of two types. In the first type, there is no stirring in the reactor and the flow through it is often called the PLUG FLOW. In the second type, there is stirring which is vigrous enough to cause complete mixing in the reactor. This is called STIRRED FLOW.
For the dissociation of water H2O H+ + OH_, the relaxation time obtained from the temperature-jump method is 40 ms, at 25° C, at the temperature : Kw = [H+] [OH_] = 1.0 × 10 _14 (mol dm_3)2 The rate constants K1 and K_2, calculated : t = 40 ms = 40 × 10_6 s = 4 × 10_5 s \ Kr = l/t = 2.5 × 104 s_1 Since Kw = [H+] [OH_] = xe2 = 1.0 × 10_14 (mol dm_3)2 \ xe = 1.0 × 10_7 mol dm_3
For water, Keq =
=
= l.8 × 10_16mol dm_3 = K1/K_2
K1 = Kr/(1 + 2xe/K)
= 2.5 × 10_5 s_1
=
\ K_1 =
=
= 1.4 × 1011 dm3 mol_1 s_1 Main Features of Relaxation Techniques Broadly speaking the technique involve (a) allowing the system to reach equilibrium, Fast Reactions (b) perturbing thfrsystem (changing the pressure or temperature) suddenly, and (c) measuring the concentrations, by optical or electrochemical methods, while the system reaches (relaxes toward) equilibrium under new conditions. The Conventional Methods Which go to equilibrium in a few seconds or less, the reason : (1) The time that it takes to mix reactauts or to bring them to a specified temperature may be significant in comparison to the half-life of the reaction. An appreciable error therefore will be made because the initial time cannot be determined accurately. (2) The time it takes to make a measurement of concentration may be significant compared to the half life. Two examples given to illustrate the importance of Picosecond and Femtosecond spectroscopy in
studying the fast reactions : Example 1—Excitation of stretching vibration of C—H bond in C2H5OH takes place at 2900 cm_1. With the help of picosecond spectroscopy it has been found that during relaxation this vibration gets transformed into two bending vibrations of the C—H bond at 1450 cm_1. Example 2—The first stage in this process of vision has been the excitation of rhodopsin. Rhodopsin partially gets deactivated forming an intermediate, prelumirhodopsin or bathorhodopsin. Picosecond spectroscopy reveals that prelumirhodopsin gets formed because of an intramolecular proton transfer—a `jump' of a proton from one position to another.
Chemistry : Basic Elements The system and rate equation with their relaxation time, matched : System Rate equation Relaxation time K1 (1) A X, = K1 (a0 _ x) _ K_1x (a) K_1 K1 (2) 2A X, = K1(a _ 2x)2 _ K_1 x (b) K_1 K1 (3) A + B Y, = K1 (a0 _ x) (b0 _ x) _ Klx (c) K_1 K1 (4) A X + Y, = K1 (a0 _ x) _ K_1 (x0 + x) (y0 + x) (d)
K_1 K1 (5) A 2X, = (e) K_1 (1)_(a), (2)_(d), (3)_(e), (4)_(b), (5)_(c) Fast Reactions When a sample of pure water in a small conductivity cell is heated suddenly with a pulse of microwave radiation, equilibrium in the water dissociation reaction does not exist at the new higher temperature until additional dissociation occurs. It is found that the relaxation time for the return to equilibrium at 25° C is 36 ms. Calculate K1 and K_1 : K1 H+ + OH_
H2O
K_1
t=
K=
=
= 1.8 × 10_6 mol L_1
Eliminating K_1, we have
t=
=
= 36 × 10_6 s
K1 = l.4 × 1011 L mol_1 s_1 K_1 = K K1 = (1.8 × 10_16 mol L_1) (l.4 × 1011 L mol_1 s_1) = 2.5 × 10_5s_1 The advantages associated with stopped flow method are: In this method the best features of static and flow methods are combined to give the best results. Other advantages are, there is no need to ensure a uniform flow, provided the mixing is satisfactory, the volumes of reactant solutions required are very small. Stopped flow methods have been used for many enzyme reactions. The method is adaptable for gas phase reactions also.
Conformational Analysis 15 Conformational Analysis The stereoisomers possible for bromochloroiodomethane are following. Their three dimensional structures and label as S and R, drawn : Only two stereoisomers, i.e., an enantiomeric pair.
Configuration (R) or (S) to the Fischer projections of lactic and shown below are assigned:
Chemistry : Basic Elements All are the same molecule R-lactic acid. In (II), group of lowest priority is on the vertical line so 1 ® 2 ® 3 is clockwise it is R. In (I and III), group 4 of lowest priority is on the horizontal line, 1 ® 2 ® 3 in each case is anticlockwise so configuration is R. The simplest alkane, optically active is :
To exhibit optical activity an alkane must contain at least one asymmetric carbon atom. Simplest optically active alkane is 3-methyl hexane.
The NMR spectra of cis and trans-1,2-dibromocyclopropane to be identical, would be :
Conformational Analysis In cis compound there, will be three NMR signals. In trans compound there will be two NMR signals. All possible conformations for the disubstituted cyclohexanes 1-Br-2-Cl, drawn :
Predominant form os (III). (I) has no elements of symmetry so the compound is optically active.
Predominant form is (VI) and is the most stable conformation of the four (II, III, M, V, VI). The equation (conformational) for the pyrolysis of cyclohexyl acetate drawn and explained: Two possible chair conformations for cyclohexyl acetate are axial-acetoxyl and equatorial-acetoxyl. Since this group is large, it will be predominantly in the equatorial form. Pyrolysis involves the formation of a cyclic structure and this is readily formed because the geometry of the molecule fits the requirement. The product is cyclohexene.
Staggered and Askew Conformation Staggered conformation is in which the bonds are equally spaced in the Newman projection. Any conformation that is not precisely staggered or eclipsed is a skew conformation.
Chemistry : Basic Elements
The rate of chair-chair interconversion in cyclohexane at room temperature is. It can be arrested this way : Approximately 1,00,000 times per second. When the solutions of the molecule are cooled to very low temperatures (~ 100°) this equilibrium is stopped. The conformation of a group can also be frozen into a desired position by putting into the ring a large bulky group e.g., t-butyl.
Optically Active Ketone Optically active ketone A undergoes racemization in basic solution. A mechanism for this process suggested and explained, whether ketone B would also racemize in basic solution :
In the optically active ketone the base abstracts the hydrogen at the chiral centre and generates a carbanion as shown. Conformational Analysis
Now when the proton gets attached to the same carbon centre, it can attach itself from the front as well as from the back thereby losing optical activity. Hence racemizes. In the case of ketone (B) there is a group CH2 in between the chiral centre and the carbonyl group. Hence such racemization in basic solution is ruled out as the hydrogen atom at the chiral centre in (B) is not acidic enough to be abstracted. The cyclohexane conformers can be isolated and identified. It is explained it with a suitable example : The two conformers of a substituted cyclohexane e.g., chlorocyclohexane interconvert very slowly at low temperatures 120° C and can be studied by 1H NMR spectroscopy. At 115° C the methine proton of chlorocyclohexane shows two distinct peaks, one at d 4·50 ppm (broad singlet) is due to the equatorial and the other (multiplets) of d 3·80 ppm are due to the axial proton. The equatorial proton is coupled with adjacent methylene protons which are all gauche (with low coupling constants) and so gives a broad signal. The axial proton shows two axial, axial couplings and two axial equatorial couplings (coupling constants are large) thus it gives well resolved multiplets. Their relative intensity provides the conformational free energy at 115° C. At 150°C the equatorial conformer crystallises out and the mother liquor can be decanted and the two conformers are thus separated. The solid when redissolved in a solvent at very low temperature gives the PMR spectrum of the equatorial conformer (d 3·80 ppm). The mother liquor gives mostly the PMR spectrum of the axial isomer (d 4·59 ppm) recorded for the CH*Cl proton.
Chemistry : Basic Elements
The following matched : (1) 1,2-disubstituted (a) meso compound. cyclohexane (2) cis-1,3-Dimethyl (b) axially chiral molecules. cyclohexane (3) Allenes, spiranes (c) Identical representation of the same compound.
(4) Homomers (d) Four stereoisomers. (5) Biphenyls (e) Two rings in perpendicular planes. (6) Atropisomers (f) Interconversion is slow. (1)—(d), (2)(a), (3)(b), (4)—(c), (5)(e), (6)—(f). The equivalent (homotopic) groups are : Atoms (including H's) or group of atoms that can be interchanged by an axis of rotation Cn (¥ > n > 1); analogous to homotopic or equivalent hydrogen atoms but generalized to cover other atoms. In cisdichloroethylene one has the following group of equivalent atoms—2 H's, 2 Cl's, 2 C's and 2 CHCI's i.e., in (Z)-dichloro ethylene, all the like atoms and groups are homotopic. Protons Encircled in Compounds The protons encircled in compounds as being stereohomotopic, enantiotopic or diastereotopic,
Conformational Analysis Compound (I)—The protons are enantiotopic because respective replacement of enantiotopic atoms by another ligand affords enantiomeric molecules. Compound (II)—The protons are homotopic. Compound (III)—The protons are diastereotopic. There is a chiral centre already present in this compound. In a prochiral assembly CX2WY, X atoms or groups are diastereotopic if either W or Y is chiral. Asymmetric synthesis and enantioselective process, discussed : An asymmetric or chiral synthesis is a reaction in which an achiral unit in an ensemble of substrate
molecules (with either enantiotopic or diastereotopic ligands or faces) is converted into chiral unit in a way so that the stereoisomers are formed in unequal amounts. Thus, the first condition of an asymmetric synthesis is the presence of a prochiral unit in the substrate molecule, it may be enantiotopic or diastereotopic ligands or faces. A process which gives an excess of one enantiomer of a pair is called an enantioselective process. Chiral chemical reagents can react with prochiral centres in achiral substrates to give partially or completely enantiomerically pure product. The free radical bromination of optically active (R)-2-chlorobutane at C-3 produces the two diastereomers of 2-bromo-3-chlorobutane in unequal amounts, explained :
This bromination at C-3 creates a second stereocentre and therefore gives rise to diastereo-mers. The two hydrogens at C-3 are diastereotopic and therefore give rise to diastereomers on substitution.
Chemistry : Basic Elements
Here diastereomers are formed in unequal proportions and this is an example of chiral synthesis. Two examples of a symmetric synthesis, given : Two examples are : (1) Reduction of 2-butanone by an enzyme NADPH.
(2) A complex of rhodium with a chiral ligand called DIOP Conformational Analysis catalyses the hydrogenation of an achiral starting material to the biologically active (-) form of dopa.
The sharpless asymmetric epoxidation is : This is a process of epoxidation of allylic alcohols with t-butyl hydroperoxide and titanium tetraisopropoxide. When enantiomerically pure tartrate esters are included, the reaction is highly enantioselective. Either the (+) or (-) tartrate ester can be used to get the either enantiomer of the product.
Cyclohexyl iodide reacts faster in Sn2 displacements than equatorial iodide, reason : For both the reactions there is no difference between the two transition states (i.e., two states have the same energies scheme). Axial iodo group raises the energy of the starting material thus lowering the activation energy.
Chemistry : Basic Elements The stereochemistry of the product, predicted :
The hydroboration oxidation reaction proceeds to give cis hydration from the less hindered side of the molecule.
The exo direction (toward the one carbon bridge) is the less hindered side of norbornane.
Conformational Analysis
Because of the axial methyl group which shields the front, attack occurs from the reverse side. The structure and configuration of the product of each of the following reactions, predicted:
These reactions are examples of the preferred trans-diaxial opening of epoxides and the analogous transdiaxial addition to double bonds.
Chemistry : Basic Elements
The following reactions give mainly one stereoisomer of the olfin, explained :
The stereochemistry of the product alkene is determined by the direction of the addition of the Grignard reagent in the first step. All the subsequent reactions require a particular conformation and are STEREOSPECIFIC. Orientation A is only one of the infinite number of conformers possible and is the Conformational Analysis most reactive because of the antiparallel arrangement of the C=O and CCl bonds allows the easy
polarization of the carbonyl group in the transition state. Thus the transition state A1 is regarded as being favoured over B because of the lesser dipole repulsion.
The products observed are those expected of attack occurs on A from the less hindered side i.e., between the two smaller groups, in this case H and Cl. Once this has occurred the later products arise by a sequence of trans displacements. As would be expected from the sequence outlined, the other alkene stereoisomer can be prepared by
Although the reaction sequence is shown for only one of the enantiomers of the racemic mixture it must hold for the other, except that the configurations of the products are reversed. This is invariably the case in the reaction of optically active compounds with symmetric reagents. Thus since the final step causes the loss of the asymmetric centres, all distinction between the path vanishes and the same product appears. Stereochemistry of Hydration The stereochemistry of hydration of fumaric acid in (1) sulphuric acid catalyzed hydration and in (2) enzyme catalyzed hydration from the following information— (i) Deuterium labelled fumaric acid traps-HO2CCD = CHCO2H
Chemistry : Basic Elements gives malic acid, HO2CCHDCH(OH)CO2H
in which the spin coupling constant of the
protons is 4·3 cps when hydration is
catalyzed by sulphuric acid. (ii) Enzymic hydration of traps-HO2CCD = CHCO2H
gives malic acid in which the spin coupling constant of the
protons is 7·1cps.
(iii) The preferred conformation of the hydration products places the carbonyl groups trans. (iv) Jtrans > Jgauche· (a) Two conclusions can be drawn from the mentioned facts— (i) traps addition in sulphuric acid catalyzed hydration; and, (ii) cis addition in enzyme catalyzed hydration. Different ways in which the configuration of one enantiomer can be affected by a chemical reaction at the chiral C are : If the product and the starting material have opposite configurations, the reaction occurred with inversion. If the product has the same configuration, the reaction went with retention of configuration. If both process occurred to the same extent, the product would be completely racemized (100% racemization) and if the two processes were not equal, partial racemization would occur. Asymmetric induction. Give an example, defined : Asymmetric induction is the use of a chiral reagent or catalyst to convert an achiral reactant to a chiral product having an excess of one enantiomer. In biochemistry the chiral catalyst is often en enzyme. For example, Conformational Analysis CH3COCOOH + Coenzyme
CH3CH(OH)COOH + oxidised coenzyme
Pyruvic acid (S) (+) lactic acid What would happen to optically active 2-iodobutane in the presence of I-, predicted :
Optically active RI should react with I with inversion at C* and eventually be racemized. The chiral C in camphor and explain why one racemic form is known, indicated : Each bridgehead C is a chiral centre. Camphor, with two different chiral C's must be expected to have 2n or four stereoisomers existing as two racemic forms but only one is known. The bridge must be cis as shown. The structural impossibility of a trans bridge eliminates a pair of enantiomers.
True or False, each of the following statement and choice, explained : (a) Racemization of an enantiomer can only occur by breaking of at least one bond to the chiral centre. (b) A reaction catalyzed by an enzyme always gives an optically active product. (c) A racemate can be distinguished from a meso or an achiral compound by an attempted resolution. (d) Conversion of an,erythro to a threo stereoisomer always occur by inversion at one chiral C. (e) A D enantiomer rotates the plane of polarized light to the right and L enantiomer to the left.
Chemistry : Basic Elements (a) True : Only by breaking a bond configuration can be changed. (b) False : The product could be meso or achiral. (c) True : Unlike a racemate meso and achiral compounds cannot be resolved because they do not consist of enantiomers. (d) True : Changing the configuration at one of the chiral C's converts one diastereomer to the other. (e) False : The term D and L do not refer to the sign of rotation. They refer to the configuration of a
stereoisomer relative to that of D-glyceraldehyde.
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