May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
Kinetic Isotope Effects :The kinetic isotope effect is a change of rate thatoccurs upon isotopic substitution and is generallyexpressed as a ratio of the rate constants,k light/k heavy. A normal isotope effect is one wherethe ratio of k light to k heavy is greater than 1. In aninverse isotope effect, the ratio is less than 1. Aprimary isotope effect is one which results from themaking or breaking of a bond to an isotopicallysubstituted atom and this must occur in the ratedetermining step. A secondary isotope effect isattributable to isotopic substitution of an atom notinvolved in bond making or breaking in the ratedetermining step. Thus when a hydrogen in asubstrate is replaced by deuterium, there is often achange in the rate. Such changes are known asdeuterium, isotope effects and are expressed by theratio k H /k D , the typical value for this ratio is 7. Theground state vibrational energy (the zero-pointvibrational energy) of a bond depends on the mass ofthe atoms and is lower when the reduced mass ishigher. Consequently, D – C, D – O, D – N bonds,etc., have lower energies in the ground state than thecorresponding H – C, H – O, H – N bonds, etc. Thus,complete dissociation of deuterium bond wouldrequire more energy than that for a correspondinghydrogen bond in the same environment. In case a H– C, H –O, or H – N bond is not broken at all in areaction or is broken in a non-rate-determining step,substitution of deuterium for hydrogen generally doesnot lead to a change in the rate, however, if the bondis broken in the rate-determining step, the rate mustbe lowered by the substitution. This helps indetermination of mechanism. In the bromination ofacetone, the rate determining step is thetautomerization of acetone which involves cleavageof a C–H bond. In case this mechanistic assignmentis correct, one should observe a substantial isotopeffect on the bromination of deuterated acetone.Indeed k H /k D was found to be around 7.CH 3 COCH 3 + Br 2 ⎯→ CH 3 COCH 2 Brrate-determining stepBromoacetoneOHCH 3 COCH 3 CH 3 C = CH 2Several mechanisms get support from kinetic isotopeeffect. Some of these are, oxidation of alcohols withchromic acid and electrophilic aromatic substitution.An example of a secondary isotope effect, where it issure that the C – H bond does not break at all in thereaction. Secondary isotope effects for k H /k D aregenerally between 0.6 and 2.0.(CZ 3 ) 2 CHBr + H 2 O → (CZ 3 ) 2 CHOH + HBrthe solvolysis of isopropyl bromide where Z = H or D, k H /k D is1.34 Secondary isotope effect.The substitution of tritium for hydrogen gives isotopeeffects which are numerically larger (k H /k T = 16).E2 elimination like S N 2 process takes place in onestep (without the formation of any intermediates). Asthe attacking base begins to abstract a proton from acarbon next to the leaving group, the C – H bondbegins to break, a new carbon-carbon double bondbegins to form and leaving group begins to depart. Inconfirmation with this mechanism, the base inducedelimination of HBr from (I) proceeds 7.11 timesfaster than the elimination of DBr from (II). ThusC–H or C – D bond is broken in the rate determiningstep. If it was not so there would not have been anyrate difference.HBase– C – CH 2 Br – CH = CH 2H1-Bromo-2-phenylethane (I)DFaster reactionBase– C – CH 2 Br – CD = CH 2DSlower reaction1-Bromo-2,2-dideuterio-2-phenylethane (II)No deuterium isotope effect is found in E1 reactionssince the rupture of C – H (or C – D) bond occursafter the rate determing step, rather than during it.Thus no rate difference can be measured between adeuterated and a non deuterated substrate.Mechanism Review : Substitution versus EliminationS N 2Primary substrateBack-side attack of Nu : withrespect to LGStrong/polarizable unhinderednucleophileBimolecular in ratedeterminingstepConcerted bond forming/bondbreakingInverse of stereochemistryFavored by polar aproticsolvent.S N 2 and E2Secondary or primary substrateStrong unhinderedbase/nucleophile leads to S N 2Strong hinderedbase/nucleophile leads to E2Low temperatrue (S N 2)/hightemperature (E2)S N 1 and E 1Tertiary substrateCarbocation intermediateWeak nucleophile/base (e.g.,solvent)Unimolecular in ratedeterminingstepRacemization if S N 1Removal of β-hydrogen if E1Protic solvent assists ionizationof LGLow temperature (S N 1)/hightemperature (E2)E2Tertiary or secondary substrateConcerted anti-coplanar TSBimolecular in ratedeterminingstepStrong hindered baseHigh temperatureXtraEdge for IIT-JEE 34 MAY 2011
UNDERSTANDINGPhysical Chemistry1. The freezing point of an aqueous solution of KCNcontaining 0.189 mol kg –1 was – 0.704 ºC. On adding0.095 mol of Hg(CN) 2 , the freezing point of thesolution became –0.530ºC. Assuming that thecomplex is formed according to the equationHg(CN) 2 + x CN – x –→ Hg (CN) x + 2Find the formula of the complex.Sol. Molality of the solution containing only KCN is(–∆Tf) (0.704 K)m = == 0.379 mol kg –1K–1f (1.86 K kg mol )This is just double of the given molality( = 0.189 mol kg –1 ) of KCN, indicating completedissociation of KCN. Molality of the solution afterthe formation of the complexm =(–∆TKff)=(0.530K)(1.86K kg mol–1)= 0.285 mol kg –1If it be assumed that the whole of Hg(CN) 2 isconverted into complex, the amounts of variousspecies in 1 kg of solvent after the formation of thecomplex will ben(K + ) = 0.189 mol,n(CN – ) = (0.189 – x) molx–n(Hg(CN) x+ 2 ) = 0.095 molTotal amount of species in 1 kg solvent becomesn total = [0.189 + (0.189 – x) + 0.095] mol= (0.473 – x) mol Equating this to 0.285 mol,we get(0.473 – x) mol = 0.285 moli.e. x = (0.473 – 0.285) = 0.188Number of CN – units combined =Thus, the formula of the complex is0.188mol0.095mol= 22–Hg (CN) 4 .2. The equilibrium constant K p of the reaction2SO 2 (g) + O 2 (g) 2SO 3 (g) is 900 atm –1 at 800K. A mixture containing SO 3 and O 2 having initialpartial pressures of 1 atm and 2 atm, respectively, isheated at constant volume to equilibrate. Calculatethe pressure of each gas at 800 K. [IIT- 1989]Sol. Since to start with SO 2 is not present, it is expectedthat some of SO 3 will decompose to give SO 2 and O 2at equilibrium. If 2x is the partial pressure of SO 3 thatis decreased at equilibrium, we would have2SO 2 (g) + O 2 (g) 2SO 3 (g)t = 0 0 2 atm 1 atmt eq 2x 2 atm + x 1 atm – 2x2(p )2SO3(1 atm − 2x)Hence, K p ==22(pSO) (p )2 O (2x) (2 atm + x)2= 900 atm –1Assuming x
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UNDERSTANDINGPhysical Chemistry1. The freezing point of an aqueous solution of KCNcontaining 0.189 mol kg –1 was – 0.704 ºC. On adding0.095 mol of Hg(CN) 2 , the freezing point of thesolution became –0.530ºC. Assuming that thecomplex is formed according to the equationHg(CN) 2 + x CN – x –→ Hg (CN) x + 2Find the formula of the complex.Sol. Molality of the solution containing only KCN is(–∆Tf) (0.704 K)m = == 0.379 mol kg –1K–1f (1.86 K kg mol )This is just double of the given molality( = 0.189 mol kg –1 ) of KCN, indicating completedissociation of KCN. Molality of the solution afterthe formation of the complexm =(–∆TKff)=(0.530K)(1.86K kg mol–1)= 0.285 mol kg –1If it be assumed that the whole of Hg(CN) 2 isconverted into complex, the amounts of variousspecies in 1 kg of solvent after the formation of thecomplex will ben(K + ) = 0.189 mol,n(CN – ) = (0.189 – x) molx–n(Hg(CN) x+ 2 ) = 0.095 molTotal amount of species in 1 kg solvent becomesn total = [0.189 + (0.189 – x) + 0.095] mol= (0.473 – x) mol Equating this to 0.285 mol,we get(0.473 – x) mol = 0.285 moli.e. x = (0.473 – 0.285) = 0.188Number of CN – units combined =Thus, the formula of the complex is0.188mol0.095mol= 22–Hg (CN) 4 .2. The equilibrium constant K p of the reaction2SO 2 (g) + O 2 (g) 2SO 3 (g) is 900 atm –1 at 800K. A mixture containing SO 3 and O 2 having initialpartial pressures of 1 atm and 2 atm, respectively, isheated at constant volume to equilibrate. Calculatethe pressure of each gas at 800 K. [IIT- 1989]Sol. Since to start with SO 2 is not present, it is expectedthat some of SO 3 will decompose to give SO 2 and O 2at equilibrium. If 2x is the partial pressure of SO 3 thatis decreased at equilibrium, we would have2SO 2 (g) + O 2 (g) 2SO 3 (g)t = 0 0 2 atm 1 atmt eq 2x 2 atm + x 1 atm – 2x2(p )2SO3(1 atm − 2x)Hence, K p ==22(pSO) (p )2 O (2x) (2 atm + x)2= 900 atm –1Assuming x
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