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1SOLVED PAPER : CSIR-UGC-NET/JRF June 2015

SOLVED PAPER : CSIR-UGC-NET/JRF June 2015CHEMICAL SCIENCES BOOKLET-[A]

Part-B

21. The biological functions of carbonic anhydrase and carboxypeptidase A, respectively, are(a) interconversion of CO2 and carbonates, and hydrolysis of peptide bond(b) gene regulation and interconversion of CO2 and carbonates(c) gene regulation and hydrolysis of peptide bond(d) interconversion of CO2 and carbonates and gene regulation

Soln. The carbonic anhydrase enzyme that catalyst the rapid interconversion of carbon-di-oxide and waterto bicarbonate and protons (and vice versa).

carbonic2 2 2 3 2anhydraseCO H O H CO in tissue high CO concentration

3 2 3 2 2HCO H H CO CO H O in lungs and nephronse of kidney

NH

HN

R

O

O

O

R'

+ H2O NH

O

R

O

+ H3NO

O

R'

Polypeptide (n-residues) Polypeptide (n-1residues) Amino acid

Where, R ' = Arg, Lys, and OrnithineCorrect option is (a)

22. The Fe–Nporphyrin bond distances in the deoxy and oxy-hemoglobin, respectively, are(a) ~ 2.1 and 2.0 Å (b) ~ 2.0 and 2.0 Å (c) ~ 2.2 and 2.3 Å (d) ~ 2.3 and 2.5 Å

Soln.

N

N N

N

protein

Fe+2 (high spin)

2.1Å

N

N N

N

Fe+2 (low spin)

2Å

Deoxy form

protein

oxy form

tense form relax form(enter the cavity)(out of cavity)

Correct answer is (a)23. The binding modes of NO in 18 electron compounds [Co(CO)3(NO)] and [Ni(5–Cp) (NO)], respec-

tively, are(a) linear and bent (b) bent and linear (c) linear and linear (d) bent and bent

Soln. 3[Co(CO) (NO)] 9 6 + 3 = 18 electron

2 SOLVED PAPER : CSIR-UGC-NET/JRF June 2015

5[Ni( – Cp) (NO)]10 5 + 3 = 18 electron

In both complexes NO donating three electron. So, NO is in linear form.Note : When NO in bent form donating 1 electron.Correct option is (c)

24. The role of copper salt as co-catalyst in Wacker process is(a) oxidation of Pd(0) by Cu(II) (b) oxidation of Pd(0) by Cu(I)(c) oxidation of Pd(II) by Cu(I) (d) oxidation of Pd(II) by Cu(II)

Soln. 24 2 4 2 3PdCl C H H O CH CHO Pd 2HCl 2Cl

Pd(II) Pd(0)

22 4Pd 2CuCl 2Cl PdCl 2CuCl

oxidation

Pd[O] Cu(II) Pd(II) Cu(I)reduction

2 2 2 212CuCl O 2HCl 2CuCl H O2

Correct option is (a)25. For typical Fischer and Schrock carbenes, consider the following statements

A. Oxidation state of metal is low in Fischer carbene and high in Schrock carbeneB. Auxilliary ligands are -acceptor in Fischer carbene and non--acceptor in Schrock carbeneC. Substituents on carbene carbon are non--donor in Fischer carbene and -donor in SchrockcarbeneD. Carbene carbon is electrophilic in Fischer carbene and nucleophilic in Schrock carbeneThe correct statements are(a) A, B and C (b) A, B and D (c) B, C and D (d) A, C and D

Soln. Correct option is (b)26. The species having the strongest gas phase proton affinity among the following,

(a) N3– (b) NF3 (c) NH3 (d) N(CH3)3Soln. Gas phase proton affinities

3N 308 kJ / mole

3NF 604 kJ / mole

3NH 872 kJ / mole

3 3N CH 974 kJ / moleProton affinity decide the energy release when a molecule/ion accept a proton. Higher the value ofgas phase proton affinities more will be basicity. Hence, 3N is most basic.Correct option is (a)

27. Consider the following statements regarding the diffusion current at dropping mercury electrodeA. It does not depend on mercury flow rateB. It depends on drop timeC. It depends on temperatureCorrect statement(s) is/are(a) A only (b) B only (c) A and B (d) B and C


Soln. According to Ilrovic equation,1/2 2/3 1/6 º

avg tid 607 n O m t CM

(1) So diffusion current depends on tm which is mass flow rate of mercury. So, option (a) is wrong.(2) Diffusion current also depends on ‘t’ which is drop time.(3) Though temperature is not directly involved in the equation but changing temperature will changeconcentration Cº as well as mt so idavg also depends on temperature.Correct option is (d)

28. Q value for the reaction 13N(n, p)13C is 3.236 MeV. The threshold energy (in MeV) for the reaction13C(p, n)13N is(a) –3.236 (b) –3.485 (c) 3.485 (d) 3.845

Soln. Threshold energy = Total mass of reactant 143.236 3.485mass of target nuclei 13

Q

Correct option is (c)29. The 119Sn NMR chemical shift (approximately in ppm) corresponding to (5–Cp)2Sn (relative to

Me4Sn) is(a) – 4 (b) + 137 (c) + 346 (d) – 2200

Soln. Correct option is (d)30. All forms of phosphorus upon melting, exist as

(a)P

PP Pn (b)

P

PP P

P

PP P

n

(c) P Pn ` (d)

P

P P P

PPP P

PPP P

PP

P P P P P PP P

P PP

Soln. All the allotropic phosphorous forms changes into white P4 discrete units. Which has structure

P P

P

P

Correct option is (a)31. For the oxidation state(s) of sulphur atoms in S2O, consider the following

A. –2 and +4 B. 0 and +2 C. +4 and 0The correct answer(s) is/(are)(a) A and B (b) A and C (c) B and C (d) C only

Soln. The structure of S2O show resonance as

SS

O

+20

0–2

SOS

+2

–2–2

+2

–2OS = 0 +2 –2 OS = –2 +4 –2

(I) (II)

Hence, the probable O.S. of ‘S’ are (0 and +2) from structure ‘I’ and (–2 and +4) from structure II.Correct option is (a)


32. The correct set of pscudohalide anions is(a) 4 4 6CN , ClO , BF , PF (b) 3 3 4 6N , NO , HSO , AsF

(c) 34 2 4 3SCN , PO , H PO , N (d) 2

3CN , N , SCN , NCN

Soln. There are many anions which show similar properties like halide ions, they are called pseudo halideand their dimers are called pseudo halogens.

Pseudo halides are 23, , , , , , CN N SCN NCN CNO NCO SeCN etc.

Correct option is (d)33. In transition metal phosphine (MPR3) complexes, the back-bonding involves donation of elec-

trons from(a) 2g 3M(t ) PR ( *) (b) 2g 3M(t ) PR ( *) (c) gM(e ) P(d) (d) 3 2gPR ( ) M(t )

Soln. The bonding in phosphine ligands, like that of carbonyls having two components. The primary com-ponent is sigma donation of the phosphine lone pair to on empty orbital on the metal. The secondcomponent is back donation from filled metal orbital to an empty orbital on the phosphine ligand.This empty phosphorous orbital has been described as being either a d-orbital or an antibonding

sigma orbital * .

•• PR3M

emptyd or p-orbital

filled-orbital

-bond

M••

PR3

filledd-orbital

empty*-orbital

(red colour)PR3

-backbond

LnM(PR3)LnM PR3

PR3

PR3

M t2g

M

So, in transition metal phosphine (M–PR3) complexes, the back bonding involves donation of elec-trons from

*2g 3M t PR

Correct option is (a)


34. The refluxing of RhCl3.3H2O with an excess of PPh3 in ethanol gives a complex A. Complex A andthe valence electron count on rhodium are, respectively,(a) [RhCl(PPh3)3], 16 (b) [RhCl(PPh3)5], 16 (c) [RhCl(PPh3)3], 18 (d) [RhCl(PPh3)5], 18

Soln. The refluxing of RhCl3.3H2O with an excess of PPh3 in boiling ethanol gives a RhCl(PPh3)3–chlorotris (triphenyl phosphine) rhodium (I) is known is known as wilkinson catalyst.

boiling ethanol3 2 3 3 3 23

A oxidized solvent

RhCl 3H O excess PPh RhCl PPh Ph P O 2HCl 2H O

PPh3 serves as the reducing agent.

RhCl

PPh3

Ph3PPPh3

It is a square planer 16-electron complex. Valence electron counting on rhodium metal centre.(i) there is no overall charge on complex(ii) there is one anionic ligand (Cl–)(iii) Rh metal atom must have +1 charge to compensate for the one negatively charged ligand.So, the oxidation state of Rh is +1.Now, we can do our electron counting.Rh (+1) d8 (8 electron)3PPh3 6 electronCl 2 electron____________________________________Total 16 electron____________________________________Correct option is (a)

35. The -hydrogen elimination will be facile in

(a) M

H

(b) M

H

(c) M

H

(d) M H

Soln. -hydrogen elimination mechanism.

M

H

M

H

empty orbital agostic interaction

#

M

HH2C CH2

M H

Since C–H, bond pair electron donate to the metal for this elimination. Therefore, as the donorability of the -electron pair increases rate of -elimination increases.Electron donor ability at bond is

C H (sp3) > C H (sp2) >C H (sp)

So, more facile -elimination occur in option (a)Correct option is (a)


36. The reaction 2 2

2 25 5Co CN H O X Co CN X H O

follows a/an

(a) Interchange dissociative (Id) mechanism (b) Dissociative (D) mechanism(c) Associative (A) mechanism (d) Interchange Associative (Ia) mechanism

Soln. Correct option is (*)

37. Correct statement on the effect of addition of aq. HCl on the equilibrium is

HOOH

O

O

O

O + CN–O

CN

..... Eq. A

..... Eq. B

(a) Equilibrium will shift towards right in case of both A and B(b) Equilibrium will shift towards left in case of both A and B(c) Equilibrium will shift towards right in A and left in case of B(d) Equilibrium will shift towards right in B and left in case of A.

Soln. In first, examples is an esterification in forward direction and an ester hydrolysis in the backwarddirection. Ester hydrolysis is catalysed by acid or base but esterfication by acid only.

The equilibrium will shift towards right in HCl.In second example is cyanohydrin formation from a ketone. In HCl solution (at pH less than 12). Theoxyanion will be protonated and the reaction driven over to the right.Correct option is (a)

38. The compound that exhibits sharp bands at 3300 and 2150 cm–1 in the IR spectrum is(a) 1-butyne (b) 2-butyne (c) butyronitrile (d) butylamine

Soln. C C H 3300 cm–1

C C 2150 cm–1

C C CH2H CH3

3300 cm–1 2150 cm–1

Correct option is (a)39. The 1H NMR spectrum of a dilute solution of a mixture of acetone and dichloromethane in CDCl3

exhibits two singlests of 1 : 1 intensity. Molar ratio of acetone to dicholromethane in the solution is(a) 3 : 1 (b) 1 : 3 (c) 1 : 1 (d) 1 : 2

Soln.C

O

CH3H3CAcetone

6 proton

+ CH2Cl2Dichloromethane

2 proton6×1=6 2×1=2 (1:1 intensity)

Both acetone and CH2Cl2 have equal intensity (1:1) for equalization. Dichorlomethene multiply by(3)

6×1 = 6 2×3 = 6Ratio = 1 : 3

Correct option is (b)


40. Intense band generally observed for a carbonyl group in the IR spectrum is due to(a) The force constant of CO bond is large(b) The force constant of CO bond is small(c) There is no change in dipole moment for CO bond stretching(d) The dipole moment change due to CO bond stretching is large.

Soln. Intensity in IR spectrum is due to change polarity and change in dipole moment and expressed interm of transition moment between two levels.

Transition moment, ' " v v vR µ d (transition moment in the form of dipole moment)

Intensity, 2' " v vI µ (intensity in the form of dipole moment)Correct option is (d)

41. The compound that gives precipitate on warming with aqueous AgNO3 is

(a)

Br

(b)

Br

(c)

Br

(d)

N

Br

Soln. Those compound gives precipitate on warming with aqueous AgNO3. Which is more stable

Br

aq. AgNO3

NO3

+ AgBr

Aromatic more stable

Correct option is (c)42. Following reaction goes through

COOAg BrBr2

(a) free radical intermediate (b) carbanion intermediate(c) carbocation intermediate (d) carbene intermediate

Soln. A free radical has been generated at bridgehead position via the Hunsdieker reaction to show that afree radical need not be planar.Correct option is (a)

43. The most stable conformation for the following compound isMeMe

Me

(a)

Me

H

MeMe

(b)

H

Me

MeMe

(c)

Me

H

Me

Me

(d)

H

Me

Me

Me


Soln. The most stable conformation for the following compound is

MeMe

Me

Me

H

MeMe

H

Me

MeMe

sterichinder

(1, 3, allylic strain occurs)(less stable)

(B)free from allylic 1, 3 strain

(most stable)(A)

(B) is less stable due to the presence of Eclipsing interaction between allylic substituent result inallylic 1, 3 strain.Correct option is (a)

44. The major product formed in the following reaction is

CHOMe

ONaBH4, CeCl3

MeOH, H2O

(a) Me

OH

OH (b)

O

Me

OH(c) Me

O

OH (d) CHO

Me

OH

Soln.CHO

Me

ONaBH4, CeCl3

MeOH, H2O CHOMe

OCeCl3+ NaBH4

CHOMe

OCeCl3

H

MeOH, H2OCHO

Me

OH

NaBH4 Na + H B

H

H

H

Correct option is (d)45. The correct relation between the following compounds is

·Cl

H H

HOMe

·H

H

Cl

Me

OH(a) enantiomers (b) diastereomers (c) homomers (identical) (d) constitutionalisomers

Soln.·

Cl

H H

HOMe

·H

H

Cl

Me

OH

SR R

S

180º

So, these two molecule are homomers (identical)Correct option is (c)


46. The correct order of heat of hydrogenation for the following compounds is

Me MeMe

Me

MeMe

MeMe

(I) (II) (III) (IV)(a) I > II > III > IV (b) I > III > II > IV (c) IV > I > III > II (d) IV > II > I > III

Soln. Correct option is (b)47. Among the following, the correct statement(s) about ribose is (are)

(A) on reduction with NaBH4 it gives optically inactive product.(B) on reaction with methanolic HCl it gives a furanoside(C) on reaction with Br2–CaCO3–water it gives optically inactive product.(D) it gives positive Tollen’s test(a) A, B and D (b) A, B and C (c) B and C (d) D only

Soln.

C

H OH

H OH

CH2OH

H OH

OH

NaBH4

CH2OH

H OH

H OH

CH2OH

H OH

Plane of symmetry is present. Hence, optically inactive

Ribose

C

H OH

H OH

CH2OH

H OH

O

H

HClMeOH

OH

HO OH

HO

O HCl

MeOH

O

HO OH

HO

OH

(Open chain)furnoside for

CHO

H OH

H OH

CH2OH

H OH

Br2–CaCO3

H2O

CHO

H OH

CH2OH

H OH + CO2

Optically active

Yes, ribose gives positive Tollen’s test.Correct option is (a)

48. Biogenetic precursors for the natural product umbelliferone among the following are

O OHOumbelliferone


(A) L-tryptophan (b) cinnamic acid (c) L-methionine (d) L-phenylalanineSoln. Umbelliferone is a phenylpso and as such is synthesized form L-phenylalanine. Which is produced

via the Shikimate path way. Phenylalanine is lysated into cinamic acid. Followed by hydrolyzed 4-hydroxylate to yield 4-coumaric acid.Correct option is (b)

49. Number of signals in the 13C{1H| NMR spectrum of (R)-4-methylpentan-2-ol are(a) 3 (b) 4 (c) 5 (d) 6

Soln. (R)-4-methylpentan-2-ol areCH3OHH

54321

5

43

21

CH3

OH CH3

6

High deshielded region (higher value)

So, total number of signal in 13C{1H| NMR spectrum is 6.Correct option is (d)


Me Me

EtO2C

H O

NaBH40ºC

MeOH/THF

(a)

Me Me

H OH

HHO

(b)

Me Me

H H

OH

HO (c)

Me Me

EtO2C

H OH

H(d)

Me Me

EtO2C

HOH

OH

Soln.

C C

EtO2C

H O

NaBH40ºC

MeOH/THF

Me Me

EtO2C

H H

H OMe

Me Me

EtO2C

H H

H

H

HH

H

Hsteric hinderence

H

OHO

Correct option is (d)



Meheat

(a)

Me

(b)

H

H Me(c)

H

H Me(d)

H

H MeSoln. The major product formed in the following reaction is

CH2 H

Ene reaction

Me

Correct option is (a)52. The major product formed in the following reaction is

O Me

Me

H2N–NH2HClEt3N, CH3CN, rt

(a)

O Me

Me

(b)

O Me

Me Me

(c)

Me

Me

HO

(d)

Me

Me

HO

Soln.

MeOO

H

H2N NH2

Me

MeO OH

NH

NH2

H

MeOOH2

NH NH2

MeO

N

H

NH2

MeON NH

H

Et3NBase

MeON

NHMe

NNH

OH NEt3

Me

NNH

HO

Me

NN

HO

Me

HO

+ H NEt3

Me

HO

Correct option is (c)


53. The magnitude of the stability constants for K+ ion complexes of the following supra-molecularhosts follows the order,

OHN

O

ONH

O

OO

O

ONH

O

OS

O

OS

O

(A) (B) (C)

(a) B > A > C (b) C > A > B (c) A > B > C (d) C > B > ASoln. The crown ether form complex with metal cation of I-st group. This dep3end upon

(i) size of cavity(ii) complexation ability.K+ is best filled in crown-6 and ‘N’ is good donar than O and S therefore the order of hosts will beA > B > CCorrect option is (c)

54. Antitubercular drug(s) among the following is (are)(A) Salbutamol (B) Ethambutanol (C) Isoniazid (D) Diazepam(a) A and B (b) B and C (c) C and D (d) D alone

Soln. (B) Ethanbutanol and (C) isoniazid are the first line drugs, they are active against mycobacteriumturbercli.Correct option is (b)

55. A particle is in a one-dimensional box with a potential V0 inside the box and infinite outside. Anenergy state corresponding to n = 0 (n : quantum number) is not allowed because(a) the total energy becomes zero(b) the average momentum becomes zero(c) the wave function becomes zero everywhere(d) the potential 0V 0

Soln. If n = 0, then function becomes not acceptable.Correct option is (c)

56. An eigenstate of energy satisfie n n nH E . In the presence of an extra constant potential V0

(a) both En and n will change (b) both En and average kinetic energy will change

(c) only En will change, but not n (d) only n will change, but not En.

Soln. n n nH E

0 n n nH E V

In presence of extra potential function will remain same but correction term will be introduced intoenergy in accordance with perturbation theory.

Correct option is (c)57. The intensity of a light beam decreases by 50% when it passes through a sample of 1.0 cm path

length. The percentage of transmission of the light passing through the same sample, but of 3.0 cmpath length, would be(a) 50.0 (b) 25.0 (c) 16.67 (d) 12.5


Soln. 1 1– log T cx ... (1)

2 2– log T cx ... (2)Substituting (2) from (1)

2 22

1 1

log 3 50log log 3 log1 log 2 3 0.3010log 1 100

T x TT x

3 0.30102 3 0.3010 0.9030

1 11010 10

T

1 1 0.125 100 12.5%7.99 8

Correct option is (d)58. The electric-dipole allowed transition among the following is

(a) 3 3S D (b) 3 3S P (c) 3 1S D (d) 3 1S FSoln. For allowed transition

0 1 SIn 3 3S P

0 1 SCorrect option is (b)

59. The product x2 xyC ( x

2C is the two-fold rotation axis around the x-axis and xy is the xy mirrorplane) is(a) xz (b) yz (c) y

2C (d) z2C

Soln. 2, , , , , , xcxyx y z x y z x y z

2 xxy xzc

Correct option is (a)60. The simplest ground-state VB wave function of a diatomic molecule like HCl is written as

1 ,1 3 , 2H Cl zs p B , where B stands for

(a) 3 , 2 1 ,1H z Clp s (b) 1 , 2 3 ,1H Cl zs p

(c) 1 , 2 3 ,1Cl Cl zs p (d) 1 , 2 3 ,1Cl H zs p Soln. HCl is a ionic compound in valency bond theory

1 3 z

HCl H Cls p

Cl

1s

3pz

H

1 1 3 2 3 1 1 2H Cl z Cl z Hs p p s



61. Heat capacity of a species is independent of temperature if it is(a) tetratomic (b) triatomic (c) diatomic (d) monatomic

Soln. For a monatomic species only translational degree of motion is occur. No vibrational degree offreeom. So, heat capacity is independent of temperarure.Correct option is (d)

62. In a chemical reaction : 5 3 2PCl g PCl g Cl g , xenon gas is added at constant volume.The equilibrium(a) will shift towards the reactant(b) will shift towards the products(c) will not change the amount of reactant and products(d) will increase both reactant and products

Soln. Addition of neutral gas xenon at constant volume will increase total pressure. But partial pressure ofgases PCl5, PCl3 and Cl2 will remain same.

For example, PCl5PCl5

nP

V as neither PCl5

n is changing nor V so PCl5P will remain same.

So, there will be no effect on equilibrium.Correct option is (c)

63. The temperature-dependence of a reaction is given by

20k AT exp E / RT

The activation energy aE of the reaction is given by

(a) 01E RT2

(b) 0E (c) 0E 2RT (d) 02E RT

Soln.0

2

ERTk AT e

0ln ln 2 ln Ek A TRT0

2

2ln 0 Ed k

dT T RT

But, 2ln aEd kdT RT 0

2 2

2 aE E

T RT RT2 0

02

2 2 a a

EE RT E RT ET RT

Correct option is (c)64. For a reaction, 2A B 3Z , if the rate of consumption of A is 4 3 12 10 mol dm s the rate of

formation of Z (in mol dm–3 s–1) will be

(a) 43 10 (b) 42 10 (c) 44 103

(d) 44 10

Soln. Rate of formation of Z =

4 4

3 rate of consumption of A2

3= 2 10 3 102



65. Dominant contribution to the escaping tendency of a charged particle with uniform concentration ina phase, depends on(a) chemical potential of that phase (b) electric potential of the phase(c) thermal energy of that phase (d) gravitational potential of that phase

Soln.

' " i iz z iµ µ z F

Electrochemical part

Chemical potential

Electrical potential part

For a electron, ' " e eµ µ F 1 iz

If is negative ' "e eµ µ• Escaping tendency of charge species moreIf the negative electrical potential is an applied to an electrode electrochemical potential is largerthen chemical potential. So, it means that tendency of an electron suffering from the electrode is inhence tendency of oxidation reaction is increase if the positive potential is applied then the tendencyof electron to escape. It means the reduction of electron is increased.Correct option is (b)

66. The intrinsic viscosity depends on the molar mass as aKM The empirical constants K and a are dependent on(a) solvent only (b) polymer only(c) polymer solvent pair (d) polymer-polymer interaction

Soln. Correct option is (c)67. The correct G for the cell reaction involving steps

2Zn s Zn aq 2e

2Cu aq 2e Cu s is

(a) 2

2

0 Zn

Cu

aG RT ln

a

(b)

20 Zn

Cu s

aG RT ln

a

(c)

2

Zn s0

Cu

aG RT ln

a

(d) 2

2

0 Zn

Cu

aG RT ln

a

Soln.

2

2

2 2

0

Zn s Zn aq 2e

Cu aq 2e Cu s

Zn Cu Zn Cu

G G RT nQ

0cell cell

RTE E log QnF

... (i)

Equation (i) is multiplied by –nF.0

cell cellnFE nFE RT log Q

20

2

Zn Cu sG G RT log

Cu Zn s

2

2

ZnQ

Cu


2

2

0 Zn

Cu

aG RT ln

a

Correct option is (d)68. The lowest energy-state of an atom with electronic configuration 1 1ns np has the term symbol

(a) 31P (b) 1

1P (c) 32P (d) 3

0P

Soln. 1 1ns npL = 1S = 1 hence, 2S + 1 = 3 3PFor J value, if orbital is less than half filled than

1 1 0 J L SHence, 3P0, correct option is (d).

69. Energy of interaction of colloidal particles as a function of distance of separation can be identified as(1) van der Waals, (2) double layer, (3) van der Waals and double layer. The correct order of interac-tions in the figure corresponding to curves (a) , (b) and (c), respectively, is

(a)(b)E

r

(a) 1, 2, 3 (b) 2, 3, 1 (c) 3, 1, 2 (d) 1, 3, 2Soln. The attractive energy due to vander waals interaction is inversely proportional to the sixth power of

the interatomic distance r i.e. 6 aG

r

E

r

The electrostatic energy of repulsion, Gel is2 2

042

khr d

elr eG

R h

E

r



70. The packing factor (PF) and number of atomic sites per unit cell (N) of an FCC crystal system are(a) PF = 0.52 and N = 3 (b) PF = 0.74 and N = 3(c) PF = 0.52 and N = 4 (d) PF = 0.74 and N = 4

Soln.1 18 6 1 3 48 2

Z

And % P.F. 33400 400 4 3.14 0.74

3 3 2 2

r rZa r


PART-C71. Differential pulse polarography (DPP) is more sensitive than D.C. Polarography (DCP). Consider

following reasons for it(A) non-faradic current is less in DPP in comparison to DCP(B) non-faradic current is more in DPP in comparison to DCP(c) polarogram of DPP is of different shape than that of DCPCorrect reason(s) is/are(a) A and C (b) B and C (c) B only (d) A only

Soln. The residual current in case of DCP is non-faradic current which should be minimum to get theaccurate result. This problem of non-faradic current is overcome by use of DPP in which non-faradiccurrent is less than DCP. That is why DPP is more sensitive than DCP.

Though the polarograph is different in both cases but it has nothing to do with sensitivity.Correct option is (d)

72. Considering the following parameters with reference to the fluorescence of a solution:(A) molar absorptivity of fluorescent molecule(B) intensity of light source used excitation(C) dissolved oxygenThe correct answer for the enhancement of fluorescence with the increase in these parameters is/are(a) A and B (b) B and C (c) A and C (d) C only

Soln. Increase of intensity of light used increases fluorescence as the fluorescence life-time increases.Molar absorptivity increases absorbance which further enhances fluorescence.Correct option is (a)

73. The geometric cross section of 125Sn (in barn) is nearly(a) 1.33 (b) 1.53 (c) 1.73 (d) 1.93

Soln. Cross-section = 2 2227

r r ... (1)

1/30r R A 1/313 131.4 10 125 7 10 cm m

Therefore, from (1),

Cross-section 26 2 26 222 49 10 154 107

cm cm

26 2 26 222 49 10 154 107

cm cm

24 21.54 1 10 ban barn cm



74. Match column A (coupling reactions) with column B (reagents)Column-A Column-B(1) Suzuki coupling (I) H2C CHCO2CH3(2) Heck coupling (II) RB(OH)2(3) Sonogashira coupling (III) PhCO(CH2)3Znl(3) Negeshi coupling (IV) CH CR

(V) SnR4The correct match is(a) 1-II, 2-I, 3-IV, 4-III(b) 1-I, 2-V, 3-III, 4-IV(c) 1-IV, 2-III, 3-II, 4-I(d) 1-II, 2-III, 3-IV, 4-V

Soln. For Suzuki coupling boron is required.For Heck coupling alkene is requiredFor Sonogashira coupling terminal alkyne is requiredFor Negeshi coupling organo zinc is required.Correct option is (a)

75. The oxoacid of phosphorus having P atoms in +4, +3, and +4 oxidation states respectively, is(a) H5P3O10 (b) H5P3O7 (c) H5P3O8 (d) H5P3O9

Soln. The average of oxidation states is 4 3 4 11

3 3

5 3 10H P O 5 3x 20 03x 15 x 5

5 3 7H P O 5 3x 14 0x 3

5 3 8H P O 5 3x 16 011x3

5 3 9H P O 5 3x 18 013x3


76. The geometries of [Br3]+ and [I5]

+, respectively, are(a) trigonal and tetrahedral (b) tetrahedral and trigonal bipyramidal(c) tetrahedral and tetrahedral (d) linear and trigonal pyramidal

Soln. 3 27 2 1Br Br Br S 4

2

Therefore, geometry = tetrahedral

[I5+]

I•••• III

I• •

I

I

II

I

••

(trigonal bipyramidal, sp3d hybridisation)



77. According to Wade’s theory the anion [B12H12]2– adopts

(a) closo-structure (b) arachno-structure (c) hypo-structure (d) nido-structureSoln. 2 2

12 12 n nB H B H Therefore, it is closo structureCorrect option is (a)

78. Considering the inert pair effect on lead, the most probable structure of PbR2[R =2. 6-C6H3(2, 6–Pr2C6H3)2] is

(a) RR

RRPb Pb (b) Pb

R

R

R

RPb

(c) R

R R

RPb Pb (d) Pb

R

R

R

RPb

Soln. Due to IPE ns2 electron-pair become inactive. In Pb

Pb =ns np

npz(n=6)Therefore, the structure must be

RR

RRPb Pb

npzns

ns npz

Correct option is (a)79. The reaction of SbCl3 with 3 equivalents of EtMgBr yields compound X. Two equivalents of SbI3

react with one equivalent of X to give Y. In the solid state, Y has a 1D-polymeric structure in whicheach Sb is in a square pyramidal environment. Compounds X and Y respectively, are

(a) 3SbEt and 2 nSb Et I (b) 2Sb Et Cl and 2 n

Sb Et Cl

(c) 3SbEt and 2 2 nSbEt Br (d) 2Sb Et Br and n

SbEt I Br

Soln. 3 3SbCl 3EtMgBr SbEt 3MgBrCl

3 3 2 2n2SbEt SbI SbEtI SbEt I

Correct option is (a)80. Match the complexes given in column I with the electronic transitions (mainly responsible for their

colours) listed in column IIColumn-I Column-II(I) Fe(II)–protoporphyrin IX (A) *

(II) [Mn(H2O)6]Cl2 (B) spin allowed d d


(III) 2 26Co H O Cl (C) spin forbidden d d

(D) M L charge transferThe correct answe is(a) I-A, II-C and III-B (b) I-D, II-B and III-C(c) I-A, II-C and III-D (d) I-A, II-B and III-C

Soln. Fe II protoporphyrin * transition

2 52 2Mn H O Cl Mn d high spin spin forbidden d d transition.

eg

t2g

hveg

t2g

S=5/2 S=3/2

S 1 . Hence, not allowed.

2 72 26

Co H O Cl Co d high spin (d–d)

eg

t2g

hveg

t2g

S=3/2

S 0 , hence allowedCorrect option is (a)

81. The following statements are given regardng the agostic interaction C H Ir observed in[Ir(Ph3P)3Cl].(A) Upfield shift of C–H proton in 1H NMR spectrum(B) Increased acid character of C–H(C) C H in IR spectrum shifts to higher wavenumberThe correct answer is/are(a) A and C (b) B and C (c) A and B (d) C only

Soln.

(Ph3P)2IrCl

PH

2

agostic interaction

Due to this agostic interaction C–H bond becomes weak and hydrogen flanked in between M and C.

As proton come in contact with metal becomes shielded and upfield shift in 1H NMR. Since v k ,bond strength of C–H bond decreases, in IR spectrum shift to lower wave number.



82. Amongst hte followng (A) 53

Mn Cp CO , (B) 5

2Os Cp , (C) 5

2Ru Cp and

(D) 5

2Fe Cp , the compounds with most shielded and deshielded Cp protons respectively, are

(a) D and A (b) D and B (c) C and A (d) C and BSoln. Here, Cp is cyclopentadienyl ligand is one of the most common and popular ligands in organometlalic

chemistry.

6 electron donor

It is an anionic ligand that normally coordinates in an n5 mode as a 6 electron donor.Now, 18-electron rule apply on the complexes

(A) 53

Mn n Cp CO Mn(II) d5 (5 electron)Cp 6 electron3CO 6 electron

____________________________________________________Total 17 electron

(B) 5

2Os n Cp

Os (II) d6 (6 electron)2Cp 12 electron________________________________________________Total 18 electron

(C) 5

2Ru n Cp

Ru (II) d6 (6 electron)2Cp 12 electron_______________________________________________Total 18 electron

(D) 5

2Fe n Cp

Fe (II) d6 (6 electron)2 Cp 12 electron______________________________________________Total 18 electron

53

Mn n Cp CO does not follow 18-electron rule. So, this complex is most deshielded com-

pound whne we going 3d to 4d and 4d to 5d series in the transition element the shielding effectdecrease because of the electron density or the distance between metal to ligand, increases. So,

5

2Fe n Cp is more shielded than other..

Thus, the compound D is most shielded and compound A is most deshielded.Correct option is (a)


83. Total number of vert ices in metal clusters 6 517 15Ru C CO , Os C CO and

5 16Ru C CO are 6, 5 and 5, respectively. The predicted structures of these complexes, respec-

tively are(a) closo, nido and nido (b) closo, nido and arachno(c) arachno, closo and nido (d) arachno, nido and closo

Soln. 6 17Ru C CO

TEC = 8×6 + 4 +17×2 = 86PEC = 86–6×12 = 14PEC 14 7

2 2

Therefore, 7 = 6 + 1 = (n + 1) closo

5 15Os C CO TEC = 8×5 + 4 + 15×2 = 74PEC = 74–12×5

PEC = 74–6 = 14 7 22

n

Therefore, nido

5 16Ru C CO

8 5 4 16 2TEC TEC = 76PEC = 76–60 = 16

16 8 5 32 2

PEC Arachno


84. Among the complexes, (A) 4 6K Cr CN , 4 6

K Fe CN , (C) 3 6K Co CN and

4 6K Mn CN , Jahn-Teller distortion is expected in(a) A, B and C (b) B, C and D (c) A and D (d) B and C

Soln. 2 44 6

K Cr CN Cr d low spin For a compound to show John Teller Distortion eg or t2g set should be electronically degenerate.

(1) Cr2+ d4 (low spin)

eg

t2g

Electronically degenerate hence show John Teller Distortion.

(2) 2 64 6

K Fe CN Fe low spin d low spin


eg

t2g

Electronically non-degenerate hence, no John Teller Distortion.

(3) 3 63 6

K Co CN Co low spin d low spin Same as above hence, no John Teller Distrotion.

(4) 24 6

K Mn CN Mn low spin

eg

t2g

electronically degenerateHence, complex will show John Teller Distortion.Correct option is (c)

85. The reductive elimination of Ar–R (coupled product) from A is facile when

(A) P

PdP

Ph Ph

PhPh

Ar

R

(a) R = CH3 (b) R = CH2Ph (c) R = CH2COPh (d) R = CH2CF3

Soln.P

PdP

Ph Ph

PhPh

Ar

R

For reductive elimination, the eliminating group should be electron releasing group and also theeliminating product should be neutral.So, only Me group is electron releasing group among all the given option.Correct option is (a)

86. The total number of metal ions and the number of coordinated imidazole units of histidine in theactive site of oxy-hemocyanin, respectively, are(a) 2Cu2+ and 6 (b) 2Fe2+ and 5 (c) 2Cu+ and 6 (d) Fe2+ and 3

Soln. Oxy-hemocyanin are proteins that transport oxygen throughout the bodies of some invertebrate ani-mals. These metalloprotein contain two copper atoms that reversibly bind a single oxygen molecule


(O2).

Cu2+ O Cu2+

N

NH

N

NH

O

NN

HNNH

HN

Oxy-hemocyaninThe total of six coordinated imidazole units of histidine present in the active site of oxy-hemocyanine.Correct option is (a)

87. Match the action of H2O2 in aqueous medium given in column A with the oxidation/reduction listedin column BA : action of H2O2 B : type of reaction

(I) Oxidation in acid (A) 3 4

6 6Fe CN Fe CN

(II) Oxidation in base (B) 4 3

6 6Fe CN Fe CN

(III) Reduction in acid (C) 24MnO Mn

(IV) Reduction in base (D) 2 4Mn Mn The correct answer is(a) I-A, II-B, III-C, IV-D (b) I-B, II-D, III-C, IV-A(c) I-C, II-D, III-B, IV-A (d) I-D, II-A, III-C, IV-B

Soln. (I) 3 4

2 2 2 26 62 Fe CN H O OH 2 Fe CN 2H O O

(alkaline medium ) Reduction

(II) 4 3

2 2 26 62 Fe CN 2H H O 2 Fe CN 2H O

(acidic medium) Oxidation

(III) 24 2 2 2 22MnO 6H 5H O 2Mn 8H O 5O

(acidic medium) ReductionCorrect option is (b)

88. The reduced form of a metal ion M in a complex is NMR active. On oxidation, the complex gives anEPR signal with ||g 2.2 and g 2.0 . Mossbauer spectroscopy cannot characteristic the metalcomplex. The M is(a) Zn (b) Sn (c) Cu (d) Fe

Soln. Correct option is (c)


89. The least probable product from A on reductive elimination is

MCH3

CH3P

P

Ph Ph

Ph Ph

(A)

(a) H3CCH3 (b) CH4 (c) H3C

CH3 (d) H3C CH3

CH3

Soln. (1) MCH3

CH3P

Preductive elimination

H3CCH3

(2) MCH3

CH3

P

P elimination

HM

CH3P

P

CH3

HR.E.

MP

P+ CH4

(3) MCH3P

P

CH3

H MP

P+

Olefin insertion

insertion intoM–H bond

MCH3P

P CH3

H3C

R.E.

CH3

H3C CH3

(4) insertion intoM–H bond

MHP

P

CH3

CH3

Olefin insertion

difficult thenM–H bond

M

CH3P

P

H3CCH3

H-elimination

MP

P

H

H+

CH3

H3C

All the product is possible but possibility (4) is least probable.Correct option is (c)

90. Water plays different roles in the following reactions.

(i) 22 22H O Ca Ca 2OH H (ii) 2 2 n

nH O Cl Cl H O

(iii) 22

2 2 66H O Mg Mg H O

(iv) 2 2 22H O 2F 4HF O

The correct role of water in each reaction is,(a) (i) oxidant, (ii) acid, (iii) base and (iv) reductant(b) (i) oxidant, (ii) base, (iii) acid and (iv) reductant(c) (i) acid, (ii) oxidant, (iii) reductant and (iv) base(d) (i) base, (ii) reductant, (iii) oxidant and (iv) base

Soln.

H2O + Ca Ca++ + 2OH + H2

+1 –2 0 –2 +1 0

ReductionOxidant


2 2 nH O Cl Cl H O

Where,Cl = Base, H2O = Acid

6H2O + Mg++

OH2

Mg

H2O

H2OOH2

OH2

OH2

++

Base(ligand)

2H2O + 2F2 4HF + O2

+1 -2 0 +1-1 0

Oxidation

same OS

Reductant


91. With respect to and bonding in Pt in the structure given below, which of the followingrepresent the correct bonding.

PtC

C

Ph

Ph

Ph3P

Ph3P1.32Å

(a) M L and *M L (b) L M and L M

(c) L M and L M (d) L M and *M L

Soln.M

(filled)

(empty)

Step-1 (-bond formed)

Step-2 (-bond formed)


92. The complex 2 2Fe phen NCS phen 1,10 phenanthroline shows spin cross-over behaviour

CFSE and µeff at 250 and 150K, respectively are(a) 00.4 , 4.90 BM and 02.4 , 0.00 BM (b) 02.4 , 2.90 BM and 00.4 ,1.77 BM

(c) 02.4 , 0.00 BM and 00.4 , 4.90 BM (d) 01.2 , 4.90 BM and 02.4 , 0.00 BM


Soln. 2 2Fe Phen NCS

2 6Fe d complex At high temperature high spin and at low temperature low spin behaviour at 250 K.

eg

t2g

CFSE = –1.6 0 0 01.2 0.4

µ n n 2 4 4 2 4.90 BM at 150 K

eg

t2g

CFSE = 0 00.4 6 2.4 µ = 0Correct option is (a)

93. Consider the following statements with respect to uranium(A) UO2

+ disproportionates more easily than UO22+

(B) U3O8 is its most stable oxide of U

(C) Coordination number of U in 2 3 2 22 2UO NO H O 4H O is six.

(D) UO22+ is linear

The correct set of statements is(a) A, B and D (b) A, C and D (c) B, C and D (d) A, B and C

Soln. 4 22 2 22UO aq 4H aq U UO aq 2H O E 0.56

2UO undergo disproportional than 22UO

• 2UO is linear 2O U O

• 2 3 22 2UO NO H O , co-ordination number = 8

• 3 8U O is stable at high temperature.


94. Et Et2 , CO2(R3P)2Ni(1, 5-cyclooctadiene)

O

Et

Et

OEt

Et


For the above conversion, which of the following statements are correct?(A) CO2 combines with Ni(PR3)2 (1, 5-cyclooctadiene)(B) Insertion of CO2 occurs(C) Insertion of Et Et takes placeThe correct answer is(a) A and B (b) B and C (c) C and A (d) A, B and C

Soln.

Et

Et

C

O

O

Et

Et

[2+2+2]

cycloadditionreaction O

Et

EtEt

OEt

The preparation of tetraethyl pyrone via [2 + 2 + 2] cycloaddition of diynes and CO2. The reactionemploys catalytic amounts of Ni(O), PR3 ligand, CO2 (1 atm).Correct option is (b)

95. Consider the following statements for 4 32 6NH Ce NO Z

(A) Coordination number of Ce is 12(B) Z is paramagnetic(C) Z is an oxidising agent(D) Reaction of Ph3PO with Z gives a complex having coordination number 10 for Ce.The correct statements are(a) A, B and C (b) B, A and D (c) B, C and D (d) A, C and D

Soln. 4 32 6NH Ce NO Z

(1) 3NO behaves as bidentate ligand. Hence, C.N = 12(2) Ce4+ has no unpaired electron hence dimagnetic not paramagnetic.(3) Ce4+ has higher oxidation state hence behave as oxidising agent.

(4) Me CO24 3 3 3Ph3PO2 6 4 2

NH Ce NO Ce NO Ph PO C.N. 10

OPPh3 = mono dentate ligand; 3NO = bidentateCorrect option is (d)

96. The major prouct formed in the following reaction sequence isHO2C

O 1. (i) SOCl2, (ii) NaN3, MeOH2. t-BuOK3. H3O+

?

(a) NH

O

HHO2C

(b) NH

O

H

HO2C


(c)

HN

O

H

HO2C

(d)

HN

O

H

HO2C

Soln. O

C

O

HO SOCl2

O

C

O

Cl NaN3 Na+ + N3–

O

C

O

N3

O

N

H

H

COtBuOK

Base O

NCO

(enolate ion)

O

HNCO

H3O+

OH–

Curtius reaction

COOH group goes to back side and CH3 group goes to above side

HN

CO

H

HN

CO

COOHH3O+

H3O+

HN

O

H

HOOCH3C

COOH

rotate


97. The major proucts A and B in the following reaction sequence areR

Br

R=MeNaNH2NH3(l)

(B)R=OHNaNH2NH3(l)

(A)

(a)

OH

NH2

A =

Me

B =

NH2

+

Me

NH2(1 : 1)

(b)

OH

A = B =

Me

NH2

NH2


(c)

OH

A =

OH

NH2

NH2

+

Me

B =

Me

NH2

NH2

+

(1 : 1) (1 : 1)

(d) A =

OH

NH2

Me

NH2

B =

Soln.

Me

Br

NaNH3

Me

NH2

Me

NH2

+

Me

NH2

50%

50%

OH

Br

NaNH2

NH3(l)

O

Br

O

NH2

O

NH2

NH3(l)

O

H

O

H

NH2

[A]

H+

working


O

AcOp-TsNH-NH2

NaBH3CN

(a)

AcO

(b) (c)

AcO

(d)

AcO


Soln. O

AcOP-TSNHNH2, AcOH

NaBH3CN HN

AcO

NH

TS N

AcO H

NH

heat N

AcO

NH

–N2

AcO

Correct option is (c)99. The major products A and B in the reactions sequence are

H2N

O+

EtO2C

O

CO2Et

aq. KOHr.t. (A) aq. KOH

reflux (B)

(a) NH

CO2H

CO2EtA =NH

CO2H

CO2HB =

(b) NH

CO2H

CO2EtA =NH

CO2H

B =

(c) NH

CO2Et

CO2HA =NH

CO2H

B =

(d) NH

CO2Et

CO2HA =NH

B =

Soln. H2N

O+

EtO2C

O

CO2Et

aq. KOHr.t.

O

N C CO2Et

CHH

CO2Et O

NH

CO2Et

CO2Et

NH

CO2EtH

OH2 CO2Et

H

NH

C

C

O

OEt

O

OEt

KOH

NH

CO2H

C

O

OEt

(A)

aq. KOHreflux

NH

COOH

(B)

H3O+



100. The major products formed in the following reaction are

OMeOH

0.5 equiv. PhC(Me)2OOH1.0 equiv. Ti(OiPr)4

1.2 equiv. (–)–DIPTCH2Cl2, –20ºC

(a) OMe

OH

O

A= B=OMe

OH

(b) OMe

OH

O

A= B=OMe

OH

(c) OMe

OH

O

A= B=OMe

OH

(d) OMe

OH

O

A= B=OMe

OH

Soln. OMeOH

0.5 equiv. PhC(Me)2OOH1.0 equiv. Ti(OiPr)4

1.2 equiv. (–)–DIPTCH2Cl2, –20ºC

OMeOH

O

Correct option is (a)101. The correct statement about hte following reaction is

N F

O

NH2 Br2

NaOH

(a) The product is 2-fluoropyridin-3-amine and reaction involves nitrene intermediate(b) The product is 2-fluoropyridin-3-amine and reaction involves radical intermediate(c) The product is 2-hydroxynicotinamide and reaction involves benzyne-like intermediate(d) The product is 2-hydroxynicotinamide and reaction involves addition-elimation mechanism


Soln.

N F

O

NH2

Hofmann rearrangement

NaOH

N F

O

NHBr2

Br Br

N F

O

NBr

H OHNaOH

N F

O

NBr

–Br

N F

O

N

N F

N

acyl nitrene intermediate

C O

isocyanate

H2O

N F

HN CO2H

–CO2

N F

NH2

2-fluoropyridin-3-amine


OAc

NH

Ph

Pd(OAc)2PPh3, Et3N

CH3CN

(a) N

Ph

(b)

NH

H

Ph

(c)

NH

H

Ph

(d)

HO

N

Ac

Ph

Soln.

OAc

NH

Ph

OAcPd(OAc)2

Pd(0)

(OAc)2Pd

NH

Ph

oxidation to Pd(II)

NH

PhH

(OAc)2Pd Pd(II)

Pd(0)N

H

H

Ph



103. The major products A and B formed in the following reactions are

Me

HO Me

MeKHTHF

18-crown-6, rt(A)

n-BuLi, 0ºC

BrPh3P(B)

(a) Me CHOMe

Me

A = MeMe

Me

B =

(b) Me CHOMe

Me

A = MeMe

Me

B =

(c) Me CHOMe

Me

A = MeMe

Me

B =

(d) Me CHOMe

Me

A = MeMe

Me

B =

Soln. Me

HO Me

MeKHTHF

18-crown-6, rt3, 3-S.T. shift

MeO Me

MeO

1

223

32

1

1

1, 2-e,e-trans(6-M.C.T.S)

OHC

Me

Me

Me

eclipsed form

Me

Me

Me

CHO

staggered formmore stable

H

ClPh3P

n-BuLi0ºC

Me

Me

Me

C

O

H

Ph3P

non-stabilized phosphour ylideform z-alkene

Me

Me

Me

OPPh3

MeMe

Me

z-alkene



104. The major products A and B formed in the following reactions areO

(i) Li, NH3(i)

(ii) allyl bromide(A)

1. PdCl2, CuClO2, DMF-H2O

2. ethanolic KOH(B)

(a)

O

A= B= O

H

(b)

O

A= B=

O

OH

Me

(c)

O

A= B=

O

Me

OH (d)

O

A=

O

B= O

H

Soln.

O

OLi

O•

Br

O•

OH3O+

O

O

O

(A)

(i) PdCl2, CuClO2, DMF, –H2O

O

O Ethanolic KOH(base)

O

O

OO

H3O+ ethanolKOH

OO

H3O+

OHO

H

O

Correct option is (d)105. An organic compounds shows following spectral data:

1IR cm :1680

13H NMR CDCl : 7.66 m,1H , 7.60 (m,1H), 7.10 m,1H , 2.50 s,3H

133C NMR CDCl : 190,144,134,132,128, 28 m/z (EI) : 126 (M+, 100%), 128 (M++2, 4.9%)


The structure of the compound is

(a) O

OAc

(b) OS

(c) O CO2Me (d) S

O

Soln.CH3

O

H

HH

190

132128

134

7.60 (m, 1H)

IR–1680 cm–1

13C NMR (28)

2.50 (s, 3H)

(7.66 (m,1H)

144

7.10 (m, 1H)

Correct option is (d)106. The correct set of reagents to effect the following transformation is

O

CO2Me

O

(a) (I) (i) NaOMe, MeI; (ii) NaCl, wet DMSO, 160ºC; (II) (i) LDA, –78ºC, TMSCl; (ii) t-BuCl,TiCl4, 50ºC(b) (I) (i) NaOMe, MeI; (ii) aq. NaOH then HCl, heat; (II) (i) Et3N, TMSCl, rt; (ii) t-BuCl, TiCl4,50ºC(c) (i) LDA, t-BuCl, (ii) LDA, MeI; (iii) aq. NaOH then HCl, heat(d) (I) (i) NaCl, wet DMSO, 160ºC; (ii) NaH, t-BuCl; (II) (i) morpholine, H+ ; (ii) MeI then H3O

+.

Soln.

O

CO2Me NaOMeMeI

O

CO2Me Me I

O

COMe

Me

O

NaClwet DMSO

160ºC

O

C

O

O –CO2

O

MeH+

O

MeLDA, –78ºC

TMSCl

O

Me

Me

O

MeTMS Cl

OSMT

SN1 reaction

t-BuCl,TiCl4, 5ºC

O

Me

SiMe3 O



107. The correct structures of the intermediates [A] and [B] in the following reactions are

N O

H

POCl3 [A] [B]Ph NH2

N NH

Ph

(a) N OP(O)Cl2H

ClA =N P

H

ClB =

O

Cl

Cl

(b) N O

P(O)Cl2

A = B =N Cl

P(O)Cl2

Cl

(c) B =N OP(O)Cl2H

ClA =

N Cl

(d) N O

P(O)Cl2

A =N P

H

ClB =

O

Cl

Cl

Soln.N O

H

+ P

O

ClCl

ClN O

H

P

O

ClClCl(A)

N O

H

P

O

ClCl

Cl

N Cl(B)

H2N Ph

NH

NH

Cl

PhN NH

Ph

Correct option is (c)108. The correct reagent combination A and the major product B in the following reaction sequence are

O

EtO2CA

O

EtO2C

O H2N-NH2B

(a) A : LiHMDS, AcCl B = NH

NEtO2C (b) A : n-BuLi, AcCl B = N

H

NEtO2C


(c) A : LiHMDS, AcOEt B =N

N

OHOH

(d) A : n-BuLi, AcOEt B =N

N

OHOH

Soln.

O

EtO2CLiHMDS

H3C C

O

Cl

OLi

EtO2C + Cl C

O

CH3 C CH3

O O

EtO

O

H2N NH2

–2H2O

N N

H

EtO2C

N NH

EtO2C(B)

1, 3-H-shift

Correct option is (a)109. The major product of the following reaction sequence is

Br

N

I

Ts

NHAc

CO2MePd(OAc)2PPh3, Et3N

OH

Pd(OAc)2PPh3, Et3N

(a)

NTs

CO2Me

AcHN

OH

(b)

NTs

CO2Me

AcHN

(c) NTs

NHAc

CO2Me

(d) NTs

NHAc

CO2Me

OH

Soln.

Br

N

I

Ts

NHAc

CO2MePd(OAc)2PPh3, Et3N

Br

N

Ts

CO2Me

NHAc

(Heck reaction)NTs

NHAc

CO2Me

OHPd(OAc)2PPh3.Et3

OH

(Heck reaction)



110. The major product formed in the following reaction isO

O

HO

COOEtPhMe2Si

TiCl

AlCp

Cp

Me

Mepyridine, toluene, –40ºC

(a)

O

O

PhMe2Si

EtOOC

H(b)

O

O

PhMe2Si

EtOOC

H

(c)

O

O

PhMe2Si

EtOOC

H(d)

O

O

PhMe2Si

EtOOC

H

Soln.

O

O

HO

COOEtPhMe2Si

O

H2C

HO

PhMe2Si

COOEt

(4+2) cyclooxidationTebbe reagentOlifination

O

O

PhMe2Si

EtOOC

H

This above two steps are taken from “Total synthesis of Azadiractin”.Correct option is (c)

111. The major products A and B in the following synthetic sequence are

Me

O

(i) PhMgBr, CuI

(ii) H3O+ (A)NaOEtBr2

(B)

(a) Me

O

Ph

A =CH2Br

O

Ph

B =


(b) Me

O

Ph

A =CH2Br

O

Ph

B =

(c) Me

O

Ph

A =Me

O

Ph

B =

Br

(d) Me

O

Ph

A =Me

O

Ph

B =

Br

Soln. Me

O

• •O

Me

PhMgBr

Micheal addition

H3O+

Ph

O Me

H

(A)Ph

C

O

(A)

CH3

NaOEtEtO H

OEt

Isomericeto gain the stability

C

O

CH2H

BaseNaOEt

PhC

O

CH2

Br2

PhC

O

Br Br CH2Br

O

PhPh

C

O

CH2Br

(B)


O

H

hv, acetone

(a)

O

Me

(b) O

Me(c)

O

Me(d)

O


Soln.

OCH3

H

hv, acetone

OCH3

H

••

OCH3

••

hv, acetone hv, acetone

OCH3

••6

43

21

O

CH3

6 5 4

12

35

Correct option is (a)113. The hydrocarbon among the following having conformationally locked chair-boat-chair form is

(a) H H

H H

(b) H H

H H

(c) H H

H H

(d) H H

H H

Soln.H H

H HH

H

H4

3

5 6H

1

2

Correct option is (d)114. The major product formed in the following reaction sequence is

HO

O

NH2

(i) (Boc)2O, pyridine

(ii) TBSCl, Imidazole

(iii) LiAlH(Ot-Bu)3EtOH, –78ºC

(a) TBSO

OH

NHBoc

(b) BocO

OH

NHTBS

(c) TBSO

OH

NHBoc

(d) BocO

OH

NHTBS

Soln. HO

O

NH2O

OOO

OHO

O

NH C

O

O

(Boc)2

HO

O

NHBoc

TBSCl

(Bn)3Si Cl


TBSO

O

NHBocH

*LLiAlH(Ot-B)3

OH

OSTB

BocHNNHBoc

H OH

BTSOS

attack from above side

Correct option is (c)115. The major product in the following reaction sequence is

ON2

hv, Me

MeOTIPS

vycor filterClCH2CH2Cl, 80ºC

(a)

HO

MeMe

OTIPS

H

H

(b)

O

MeMe

OTIPS

(c)

HO

MeMe

OTIPS

(d)

HO

MeMe

OTIPS

Soln.

ON2

hv

–N2

OC

O

ketene

OTIPS[2+2]

O

Me Me

OTIPS

[1, 3] C–Csigmatropic

hv

hvO OTIPS

MeMe

H

54

321

[1, 5] H-shift80ºC

Cl CH2 CH2 Cl(solvent)

O OTIPS

MeMe

HHO OTIPS

MeMe

tautomerizationor aromatization



116. Structures of A and B in the following synthetic sequence are

AcO NH

O

CHO(i) Ph3P CHCO2Me

(A)(ii) heat(i) LiAlH4

(ii) H3O+ (B)

(a) A =N

O

CO2Me

OAc

B =N

O

(b) A =N

O

CO2Me

H

B =N CH2OH

H

(c) A = AcO NH

O

HOH2C

H

B = AcO NH

O

HOH2C

H

(d) A =N

O

CO2Me

H

B =N CH2OH

H

O

Soln. Ph3P CHCO2Me Ph3P CHCO2MeStabilized phosphorous ylide to form (E alkene)

AcO NH

O

CO

H+ CH

PPh3

CO2Me

witting reactionAcO N

H

O

CH

H

CO2Me

AcO NH

MeO2C

NC

O

MeO2C

–AcO CO2MeN

O

HCO2Me

N

H

LiAlH4



117. In the following reaction, the ratio of A : B : C is (*indicates labelled carbon)

* NBSAlBNCCl4heat

Br

(A)

*+

Br

(B)

*+

Br

(C)*

(a) 1 : 1 : 1 (b) 1 : 2 : 1 (c) 2 : 1 : 1 (d) 3 : 2 : 1

Soln.* NBS

AlBNCCl4heat

*+

*

Br

(A)

*+

Br

(C)*

+

Br

(A)

*

•

•

*

•

*•

21

All are same amount

Br*

+

(B)

43

Br

So, A : B : C = 2 : 1 : 1.


118. Structure of the major product in the following synthetic sequence isCO2Me

N2

(i) CuI

(ii) SeO2

(a) HO

CO2Me

MeH

H

(b) CO2Me

MeH

H

OH

(c) CO2Me

MeH

H

OH

(d) CO2Me

MeH

H

HO


Soln.

CO2Me

N2

(i) CuI

(ii) SeO2

N2 CO2Me CuI

HCCO2Me CO2Me

MeH

H CO2Me

MeH

H

HO

SeO2[2, 3] sigmatropic

rearrangement


119. Major product formed in the following synthetic sequence on the monoterpene pulegone is

O

(i) Br2(ii) NaOEt, EtOH(iii) KOH, EtOH

(a) CO2H (b)

COOH

(c) (d) OHO

OEt

Soln.

O

Br2

O

BrNaOEt, EtOH

O

Br

O

Br

OH

Br

O OH

COOHKOH, EtOH



120. Optically pure isomers A and B were heated with NaN3 in DMF. The correct statement from thefollowing is

NMe2

Br(A)

NMe2

Br(B)

NMe2

N3(C)

NMe2

N3(D)

(a) A gives optially pure D and B gives optically pure C(b) A gives racemic mixture of C and B gives optically pure C(c) A gives optically pure C and B gives racemic C(d) A gives optically pure D and B gives racemic D

Soln. A gives racemic mixture of C and B gives optically pure C.

NMe2

Br

Br

NMe NMe2

Me2N

N3N3

NMe2

N3

ORN3

NMe2

racemicmixture

(A)

NMe2

Br(B)

NMe2Br

N3–

NMe2

N3

NMe2Br

(Optically pure)


121. A molecular orbital of a diatomic molecule changes sign when it is rotated by 180º around themolecular axis. This orbital is

(a) (b) (c) (d)

Soln. –

+180º

–

+

-BMO

180º+

–

–

+ +

–

–

+

-ABMOIf the sign is changed it is called ‘ ’ otherwise ‘ ’.Correct option is (b)


122. IR active normal modes of methane belong to the irreducible representation:

3 2 42 2 2

1

22 2 2 2 2

1

2

8 3 6 61 1 1 1 11 1 1 1 12 1 2 0 0 2 ,3 0 1 1 1 , ,3 0 1 1 1 , , , , ,

d d

x y z

T E C C SA x y zAE z x y x yT R R RT x y z xy yz zx

(a) 1E A (b) 2E A (c) T1 (d) T2

Soln.3 2 48 3 6 6

15 0 1 1 3 dE C C S

IRR

1 2

2 1

1, 1, 30, 1

nA nE nTnA nTIRR = A1 + E + T1 + 3T2Translation = T2Rotational = T1Therefore, normal modes of vibration = A1 + E + 2T2.Now, IR active modes = T2Correct option is (d)

123. The symmetric rotor among the following is

(a) CH4 (b) CH3Cl (c) CH2Cl2 (d) CCl4Soln. Symmetric rotor is CH3Cl

H

C

HH

ClIA

IB IC=

Symmetric tops, , 0 B C A AI I I where IThe moment of inertia about the C–Cl bond axis is now not negligible because it involves the rota-tion of three compartively massive hydrogen atoms off this axis such a molecule spinning about thisaxis can be imagined as a top and called symmetric top.Correct option is (b)

124. The nuclear g-factors of 1H and 14N are 5.6 and 0.40 respectively. If the magnetic field in an NMRspectrometer is set such that the proton resonates at 700 MHz, the 14N nucleus would resonate at

(a) 1750 MHz (b) 700 MHz (c) 125 MHz (d) 50 MHz

Soln. 0 B 2

e gm

0 gB g


1 1

2 2

gg

1 0.4

700 5.6

MHz

10.4700 505.6

MHz MHz

Correct option is (d)125. The spectroscopic technique, by which the ground state dissociation energies of diatomic molecules

can be estimated, is(a) microwave spectroscopy (b) infrared spectroscopy(c) UV-visible absorption spectroscopy (d) X-ray spectroscopy

Soln. The ground state energies of the atomic molecule can be estimated by infrared spectroscopy.

Ener

gy

Illustration of dissociation

EexD0

D0 Dg

continum

" 1continuum limit 0

exD E cm ( 0 0' and "D D are dissociation energies)

Correct option is (b)126. The term symbol for the first excited state of Be with the electronic configuration 1s2 2s1 3s1 is

(a) 3S1 (b) 3S0 (c) 1S0 (d) 2S1/2

Soln. 2 1 1Be 1s 2s 3s

2s 3s

1 1S 12 2

Multiplicity 2S 1 2 1 1 3

L = 0 + 0 = 0 S – term

J L S — L S 0 1 — 0 1 i.e.1

Hence, terms is 3S1.Correct option is (a)


127. Which of the following statement is INCORRECT?(a) A Slater determinant is an antisymmetrized wavefunction(b) Electronic wavefunction should be represented by Slater determinants(c) A Slater determinant always corresponds to a particular spin state(d) A Slater determinant obeys the Pauli exclusion principle

Soln. In quantum mechanics, a slater determinant is an expression that describe the wave function of amultifermionic system that satisfies antisymmetry requirements and consequantly the Pauli prin-ciple by changing sign upon exchange of two electrons. Slater determinants is a means of ensuringthe antisymmetry of a wave function through the use of matrices. A Slater determinant always corre-sponds to a particular spin state is not true.Correct option is (c)

128. Compare the difference of energies of the first excited and ground states of a particle confined in (i)

a 1-d box 1 , (ii) a 2-d square box 2 and (iii) a 3-d cubic box 3 . Assume the length of each

of the boxes is the same. The correct relation between the energy differences 1 2, and 3 for thethree states is(a) 1 2 3 (b) 1 2 3 (c) 3 2 1 (d) 3 1 2

Soln. Energy in 1-D box 2 2

28n hm

2 – first excited state energy n = 22

2

48

hm

1 – ground state energy 2

218

hnm

Difference, 2 1

2 2 2

2 2 2

4 38 8 8

h h hm m m

Energy in 2-D box, 2 2 2

28x yn n h

m

2 – first excited state any (2, 1) (1, 2) 2

2

58

hm

1 – ground state energy (1, 1) 2

2

28

hm

Difference, 2 1

2 2 2

2 2 2

5 2 38 8 8

h h hm m m

3D-box energy 2 2 2 2

28x y zn n n h

m


2 – first excited state energy (2, 1, 1) (1, 2, 1) (1, 1, 2) 2 2

2 2

4 1 1 68 8

h hm m

1 – ground state energy (1, 1, 1) 2 2

2 2

1 1 1 38 8

h hm m

Difference, 2 2 2

2 1 2 2 2

6 3 38 8 8

h h hm m m


129. The correct statement about both the average value of position x and momentum p of a 1-d harmonic oscillator wavefunction is

(a) 0x and 0p (b) 0x but 0p

(c) 0x and 0p (d) 0x but 0p

Soln. One dimensional simple harmonic oscillator

1/42 /2

0

x

x e

1/4 1/42 2* /2 /2

x xx x dx e x e dx

1/22 21

x n xx e dx x e dx

In the above equation, n = 1 if n is oddn is oddTherefore, the vlaue = 0

1/4 1/42 2* /2 /2x x

x xp p dx e i e dxx

1/2

2 2/2 /2x xi e e dxx

1/2

2 2/2 /222

x xxi e e dx

1/2

2 20x n xi x e dx x e dx

if n = odd

1/2

0 0i

Hence, 0xp



130. The value of the commutator , , xx x p is

(a) i x (b) i (c) i (d) 0

Soln. , , xx x p ... (1)

, x x xx p xp p x x i i x

x x

i x i x xi i x i i

x x x x

, xx p iFrom equation (1), we get

, 0 x i xi i x xi xi Correct option is (d)

131. The equilibrium constants for the reactions 4 2 2 2CH g 2H O g CO g 4H g and

2 2 2CO g H O g CO g H g are K1 and K2, respectively. The equilibrium constant for

the reaction 4 2 2CH g H O g CO g 3H g is

(a) 1 2K K (b) 1 2K K (c) 1 2K / K (d) 2 1K K

Soln. 4 2 2 2 1

2 2 2 2

CH g 2H O g CO g 4H g k 1

Co g H O g CO g H g k 2

Equation (1) substracting from equation (2)

4 2 2 2

2 2 2

4 2 2 2

CH g 2H O g CO g 4H g

CO g H O g CO g H g

CH g H O g CO g 3H g k1/ k

and the equalibrium constant 1

2

kk


132. Consider the progress of a system along the path shown in the figure S B C for one mole of anideal gas is then given by

A(T , V )1 1

C(T , V )3 1 B(T , V )2 2

Adiabaticprocess

P

V


(a) 1

3

TR lnT (b)

3

1

TR lnT (c)

2

1

VR lnV (d)

1

2

VR lnV

Soln. Correct option is (*)133. A thermodynamic equation that relates the chemical potential to the composition of a mixture is

known as(a) Gibb’s-Helmholtz equation (b) Gibbs-Duhem equation(c) Joule-Thomson equation (d) Debye-Huckel equation

Soln. For a multicomponent open system, we have

i ii

G n µ

Differentiating the above equation, we get

i i i ii i

dG µ dn dµ n

But, by the fundamental equation,

i ii

dG SdT Vdp µ dn

Subtracting the two equations, we get

i ii

n dµ SdT Vdp

Where,µi = Chemical potentialni = number of molesI = components

Correct option is (b)134. According to transition state theory, the temperature-dependence of pre-exponential factor (A) for a

reaction between a linear and a non-linear molecule, that forms products through a non-linear tran-sition state, is given by(a) T (b) T2 (c) T–2 (d) T–1.5

Soln.

A–B

Linear

+D

C ENon-linear

D

C E

A Bnon-linear

transition state

#

3 11#

3 1 3 3

t r v non linearA B A B

A B CDE t r v t r vlinear non linear

q q qN k T N k TqA Ah q q h q q q q q q

7

3 vA B

t r linear

qN k Th q q


70

3/2

A BTN k T

h T T 3/2

TAT T 3/2 A T

1.5 A TCorrect option is (d)

135. For a given ionic strength, (I) rate of reaction is given by

1/2

0

klog 4 0.51 Ik

. Which of the following reactions follows the above equation?

(a) 22 8S O I (b) 2

3 5Co NH Br OH

(c) 3 2 5CH COOC H OH (d) 2 2H Br H O

Soln.0

log 2

A Bk AZ Z Ik

Given, 2 4 0.51 A BAZ Z I I

2 A BZ Z


136. For a reaction on a surface

H2 + S S S

H

S

H

S

H

S

Hslow S

H

S + HAt low pressure of H2, the rate is proportional to

(a) 2H (b) 21/ H (c) 1/22H (d) 1/2

21/ H

Soln.2

1

kP

1/2

1

kP

1/2

1/21

kPkP

At low P, 1/21 kP

½1/22 kP H

Correct option is (c)137. The temperature-dependence of an electrochemical cell potential is

(a) G

nFT

(b) H

nF

(c) S

nF

(d) S

nFT

Soln. cellG nFEGibb’s-Helmholtz equation,


P

GG H T

T ... (1)

G nFETherefore, we get

PP

G FnFT T ... (2)

From equation (1) and (2), we get

P

EnFE H nFTT ... (3)

Rearranging we have, P

EH nF E TT

But, G H T S ... (4)

H GST

From equation (3) and equation(4), we get P

ET S nFTT

P

ES nFT

P

E ST nF

Temperature dependence = SnF

Correct option is (c)138. The single-particle partition function (f) for a certain system has the form BTf AVe . The average

energy per particle will then be (k is the Boltzmann constant)(a) BkT (b) 2BkT (c) kT / B (d) 2kT / B

Soln. BTf AVe ... (1)

Since, average energy 2

V

nfNkTT

Therefore, average energy per particle 2

V

nfkTT

... (2)

From equation (1), we get, 0 0

BT

V V V V

nf nA nV ne BT T T T

From equation (2), average energy per particle = 2 2kT B BkTCorrect option is (b)


139. The indistinguishability correction in the Boltzmann formulation is incorporated i the following way: (N = total number of particles, f = single-particl partition function)(a) replace by f/N! (b) replace fN by fN/N!(c) replace f by f/ln(N!) (d) replace by fN by fN/ln(N!)

Soln. Since Boltzmann formulation deals with whole system containing N number of particles

Therefore, fN should be replaced by !

NfN

. {Since on incorporating condition of indistinguishability

total number of arrangements changes by 1

!N }

Correct option is (b)140. In a photochemical reaction, radicals are formed according to the equation

2

4 10 2 5

k2 5 2 5 2 6 2 4

C H h 2C H

C H C H C H C H

If I is the intensity of light absorbed, the rate of the overall reaction is proportional to

(a) I (b) 1/2I (c) 4 10I C H (d) 1/21/24 10I C H

Soln. Rate of reaction, 22 2 5r k C H ... (1)

SSA on C2H5, Rate of formation of C2H5 = Rate of dissociation of C2H5

2 21 1 2 5 2 2 52 2 2 ak I k C H k C H

21 1 2 2 5 ak I k k C H

2 12 5

1 2

ak IC Hk k

Therefore, from equation (1),

2 1

1 2

ak k Irk k ar I


141. Conductometric titration of a strong acid with a strong alkali (MOH) shows linear fall of conduc-tance up to neutralization point because of(a) formation of water(b) increase in alkali concentration(c) faster moving H+ being replaced by slower moving M+.(d) neutralization of acid.

Soln. When a strong alkali e.g. sodium hydroxide is added to a solution of a strong acid, e.g. hydrochloricacid, the reaction.

2H Cl M OH M Cl H O

• During the reaction highly conducting H are replaced by M which has much lower conduc-tance , so addition of alkali to acid solution is accompanied by a decrease of conductance.


• When neutralisation is complete then the further addition of alkali results in an increase ofconductance, since the sodium ion and hydroxyl ion are no longer used up in the chemical reac-tion.

• At the neutral point the conductance of the system will have a minimum value.Correct option is (c)

142. Find the probability of the link in polymers where average values of links are (A) 10, (B) 50 and (C)100(a) (A) 0.99, (B) 0.98, (C) 0.90 (b) (A) 0.98, (B) 0.90, (C) 0.99(c) (A) 0.90, (B) 0.98, (C) 0.99 (d) (A) 0.90, (B) 0.99, (C) 0.98

Soln.1 11

1

avav

k pp k

11 0.9

10 p

11 1 0.02 0.9850

p

11 1 0.01 0.99100

p


143. The stability of lyophobic colloid is the consequence of(a) van der waals attraction among the solute-solvent adducts(b) Brownian motion of the colloidal particles(c) insolubility of colloidal particles in solvent(d) electrostatic repulsion among double-layered colloidal particles

Soln. Lyophobic colloids are thermodynamically unstable but they are kinetically stable does or does notoccur depend on the balance of attractive and repulsive force. For stabilisation to occur, the repul-sive forces must dominate.Correct option is (d)

144. In a conductometric experiment for estimation of acid dissociation constant of acetic acid, the fol-lowing values were obtained in four sets of measurements.

5 5 51.71 10 ,1.77 10 ,1.79 10 and 51.73 10The standard deviation of the data would be in the range of(a) 5 50.010 10 0.019 10 (b) 5 50.020 10 0.029 10 (c) 5 50.030 10 0.039 10 (d) 5 50.040 10 0.049 10


Soln. Four sets of measurements are 5 5 51.71 10 ,1.77 10 ,1.79 10 and 51.73 10

Therefore, average value (Mean value) 5 51.71 1.77 1.79 1.73

10 1.75 104

Standard deviation,

2 210 10

2 2 210 10

1.71 1.75 10 1.77 1.75 10

1.79 1.75 10 1.73 1.75 104

ix µ

N

10 50.0010 10 0.0316 10

which is in range of 5 50.030 10 0.039 10 Correct option is (c)

145. Silver crystallizes in face-centered cubic structure. The 2nd order diffraction angle of a beam of X-ray

1Å of (111) plane of the crystal is 30º. Therefore, the unit cell length of the crystal would be(a) a = 3.151Å (b) a = 3.273 Å (c) a = 3.034Å (d) a = 3.464 Å

Soln. 2 sin d n

2 2 22 sin

a nk

2 sin 30º 2 13

a Å

3 3 2 2 1.732 3.464sin 30º 1

a


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