csir-ugc-net/jrf chemical sciences paper june 2014

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CSIR-UGC-NET/JRF CHEMICAL SCIENCES PAPERJUNE 2014

PART-A

1. The following diagram shows 2 perpendicularly inter-grown prismatic crystals (twins) of identical shape andsize. What is the volume of the object shown (units are arbitrary)?

10

22

(a) 60 (b) 65 (c) 72 (d) 80Soln. Volume of both cuboid = (10×2×2) + (10×2×2) – (2×2×2) = 40 + 40 –8 = 72 unit3.

Correct answer is (c)

2. Suppose in a box there are 20 red, 30 black, 40 blue and 50 white balls. What is the minimum number of ballsto be drawn, without replacement, so that you are certain about getting 4 red, 5 black, 6 blue and 7 white balls?(a) 140 (b) 97 (c) 104 (d) 124

Soln. So, minimum number of balls = 50 + 40 + 30 + 4 = 124Correct answer is (d)

3. In the growing years of a child, the height increases as the square root of the age while the weight increases indirect proportion to the age. The ratio of the weight to the square of the height in this phase of growth.(a) is constant(b) reduces with age(c) increases with age(d) is constant only if the weight and height at birth are both zero

Soln. H AW A

H k. A

1W k .A

Now, 22 2H k. A k .A k '.A (where k ' is other constant)

and 2

21 1

H k 'A k ' kw k .A k

(other constant)

Hence, correct answer is (d)4. Students in group A obtained the following marks : 40, 80, 70, 50, 60, 90, 30. Students in group B obtained

40, 80, 35, 70, 85, 45, 50, 75, 60 marks. Define

dispersion (D) = (maximum marks – minimum marks dispersionmean

RD

Then,

[Booklet-C]



2

(a) RD of group A = RD of group B (b) RD of group A > RD of group B(c) RD of group A < RD of group B (d) D of group A < D of group B

Soln. Group A : 40, 80, 70, 50, 60, 90, 30Group B : 40, 80, 35, 70, 85, 45, 50, 75, 60Dispersion of Group A = 90–30 = 60 Group B = 85–35 = 50

RD of Group A 60 60 140 80 70 50 60 90 30 607

RD of Group B 50 50 140 80 35 70 85 45 50 75 60 609

RD of Group A > RD of Gropu BCorrect anwer is (b)

5. In 450 g of pure coffee powder 50g of chicory is added. A person buys 100g of this mixture and adds 5 g ofchicory to that. What would be the rounded-off percentage of chicory in this final mixture?(a) 10 (b) 5 (c) 14 (d) 15

Soln. After mixing chicory new mixture = 450 + 50 = 500 gramSince, in 500 gram chicory is 50 gram

Therefore, in 1g chicory is =50

500

Therefore, in 100 gram chicory is 50 100 10 gram500

After adding 5g chicoryNew mixture = 100 + 5 = 105 gramand quantity of chicory = 10 + 5 = 15 gram

Hence, 15 100% 100 14.28 14.28%

105 7


6. The time gap between the two instants, one before and one after 12.00 noon, when the angle between the hourhand and the minute hand is 66º, is(a) 12 min (b) 16 min (c) 18 min (d) 24 min

Soln. Solve by hit and trial

Apply between hands of watch = 1Hour 30 min2

in a gap of 24 min, there is angle between hands of watch is 66º.Correct answer is (d)

7. Suppose

2x y x y

2x y x y

1x y x y x y x y +, – and × have their usual meanings. What is the value of 197 315 197 315 197*315 ?

(a) 118 (b) 512 (c) 2 (d) 4



3

Soln. 2 2 1197 315 197 315 . 197 315

22512 118 630 394 4197 315 197 315

Correct answer is (d)

8. If A×B = 24, B×C = 32, C×D = 48, then A×D(a) cannot be found (b) is a perfect square(c) is a perfect cube (d) is odd

Soln. A×B = 24 ... (1)B × C = 32 ... (2)C × D = 48 ... (3)(1) × (2) × (3)A×B×B×C×C×D = 24×32×48A×B2×C2×D = 24×32×48

Divide it by (B×C)2 –

2

2

A B C D 24 32 48 3632 32B C

Which is perfect square. Hence, correct answer is (b)

9. If all horses are donkeys, some donkeys are monkeys, and some monkeys are men, then which statement mustbe true?(a) All donkeys are men (b) Some horses are men(c) Some horses are men (d) All horses are also monkeys.

Soln. Horse

Donkey

Men

D

H M

According to both figure,

only (ii) is true

10. A rectangular area of sides 9 and 6 units is to be covered by square tiles of sides 1, 2 and 5 units. The minimumnumber of tiles needed for this is(a) 3 (b) 11 (c) 12 (d) 15

Soln. Area of rectangle = 9×6 = 54 unit2.2 2

2

2

2

5

5

11

1 5 5 6 2 2 5 1 1 12 Correct answer is (c)

11. Suppose n is a positive integer. Then 2 2 1n n n

(a) may not be divisible by 2(b) is always divisible by 2 but may not be divisible by 3(c) is always divisible by 3 but may not be divisible by 6.(d) is always divisible by 6.



4

Soln. These type of question can be solved by hit and trial method.Put n = 2, 3, 4, ............ in (n2 + n) (2n + 1)For n = 2 (22 + 2) (2×3 +1) = 6×5 = 30 n = 3 (32 + 3) (2×3 + 1) = 12×7 = 84 n = 4 (42 + 4) (2×4 + 1) = 20×9 = 180and we know that n ( n + 1) (2n + 1) is always divisible by 6Correct answer is (d)

12. There is a train of length 500 m, in which a man is standing at the rear end. At the instant the rear end crossesa stationary observer on a platform, the man starts walking from the rear to the front and the front to the rear ofthe train at a constant speed of 3 km/hr. The speed of the train is 80 km/hr. The distance of the man from theobserver at the end of 30 minutes is(a) 41.5 km (b) 40.5 km (c) 40.0 km (d) 41.0 km

Soln. In 30 minute train will cover = 180 40 k.m.2

Now, speed of man = 3 km/h

In 12 hour man will cover

13 1.5 km 1500 m2

Length of train = 500 m.It means man will go ends to start, start to end and again ends to start. So, distance between observer and manwill be 40 + 500 m = 40.5 km.Correct answer is (b)

13. Three identical flat equivalent-triangular plates of side 5 cm each are placed together such that they form atrapezium. The length of the longer of the two parallel sides of this trapezium is

(a) 354

cm (b) 5 2 cm (c) 10 cm (d) 10 3 cm

Soln.

A D

CB E

5

5

55 5

55Hence, length of BE = 5 + 5 = 10 cmCorrect answer is (c)

14. An archer climbs to the top of a 10 m high building and aims at a bird atop a tree 17m away. The line of sightfrom the archer to the bird makes an angle of 45º to the horizontal. What is the height of the tree?(a) 17 m (b) 27 m (c) 37 m (d) 47 m

Soln.

A

B D

E

C

10m

17m

17

45º

Let AB is a building and CD is tree. Bird is located at point C and archer in point A. Draw a line parallel to BDi.e. AE.

AE BD 17 Now, in AEC ,



5

CE tan 45ºAE

CE 117

CE = 17Hence, length of tree = 17 + 10 = 27 m.Correct answer is (b)

15. Consider a right-angled triangle ABC where AB = AC = 3. A rectangle APOQ is dranw inside it, as shown,such that the height of the rectangle is twice its width. The rectangle is moved horizontally by a distance 0.2 asshown schematically in the diagram (not to scale)

C

Q

AP B

TSQ

What is the value of the ratio Area of ABC ?Area of OST

(a) 625 (b) 400 (c) 225 (d) 125Soln. In this question, firstly we proof that ABC and OST are similar. Therefore, in ABC and OST

ABC OST 90º

ACB OST Alternate interior Hence, ABC OST

ABC OST AC AB BCST OS OT

Now, OS = 0.2AC AB 3 3ST OS ST 0.2

ST 0.2

Now,

1 1AB AC 3 3ar ABC 92 2 2251 1ar OST 0.04OS ST 0.2 0.22 2


16. 80 gsm paper is cut into sheets of 200 mm×300mm size and assembled in packets of 500 sheets. What willbe the weight of a packet? (gsm = g/m2)(a) 1.2 kg (b) 2.4kg (c) 3.6 kg (d) 4.8 kg

Soln. Weight of 1m2 sheet = 80 gram

Now, area of all 500 sheets = 200×300×500 mm2 2200 300 500 30 m1000 1000

Hence, total weight = 80×30 = 2400 gram 2400 2.4 kg1000

Correct answer is (b)



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17. Find the missing letterA B C DF I L OK P U ZP W D ?

(a) P (b) K (c) J (d) L

Soln.A

F

1

6

K11

16P

+5

+5

+5 B

I

2

9

P16

23W

+7 C

L

3

12

U21

D

+9 D

O

4

15

Z26

K/37

+11

+7

+7

+9

+9

+11

30+11

Put the place value of alphabates – in first column difference is 5, in 2nd –7, in 3rd –9, in 4th –11.18. A merchant buys equal numbers of shirts and trousers and pays Rs. 38000. If the cost of 3 shirts is Rs. 800

and that of a trouser is Rs. 1000, then how many shirts were bought?(a) 60 (b) 30 (c) 15 (d) 10

Soln. Let he purchase y shirts and y trousers

then cost of 1 shirt = 800 rs

3and cost of 1 trousers = 1000 rs

Hence, 800 y 1000 y 38000

3

800y 3000y 3800 3 3800y 38000 3 y = 30Correct answer is (b)

19. Consider the set of numbers {171, 172, ....., 17300}. How many of these numbers end with the digit 3?(a) 60 (b) 75 (c) 100 (d) 150

Soln. Unit of 171 = 7; 172 = 9; 173 = 3, 174 = 1; 175 = 7; 176 = 9; 177 = 3; 178 = 1Hence, 173, 177, 1711, .......... will ends with 3 as unit.Therefore, 4 300 75

282020×

×

So, 75 is our answer.Correct answer is (b)

20. Find the missing number in the triangle

90 13 ?

7

3 5 6 4 2 6

1 8

(a) 16 (b) 96 (c) 50 (d) 80

Soln.

7 5 3 7 5 3 90

6 1 4 6 1 4 13

8 2 6 8 2 6 80




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PART-B

21. The correct order of basicity for the following anions is

O

NO2

O O

NO2

NO2(I) (II) (III)(a) II > III > I (b) I > II > III (c) II > I > III (d) III > II > I

Soln.

OH

NO2

O

NO2

OH

O

NO2

NO2

OH

O

NO2

NO2

NO2

In case of ortho and para resonance as well as inductive effect both operate. Thus, electron withdrawl willbe more in the case of ortho and para substituted nitro phenol.Thus, conjugate base of ortho and para will be less basic.Meta substituted phenoxide (II) will be most basic.Now, in between I and III, since H-bonding (intramolecular) occurs in the case of I, it is less acidic as com-pared to para, thus it will be more basic as compared to III. II > I > III.Correct answer is (a).

22. The major product formed in the reaction of 2, 5-hexanedione with P2O5 is

(a) O

(b)

O

(c) O

O

O

(d)

O

O

Soln.CO

O

P2O5

O

Mechanism:P2O5 is a dehydrating agent. Reacts with 1, 4-dicarbonyl compound to form substituted furan.Therefore, Paal known synthesis.Correct answer is (a)



8

23. The absolute configuration of the two stereogenic (chiral) centres in the following molecule is

O

(a) 5R, 6R (b) 5R, 6S (c) 5S 6R (d) 5S, 6S

Soln.

O

(5S, 6R)

6

23

4

5

IIIV

I

1

IIIIII

III I


24. The correct statement about the following molecule is

Br

Br(a) Molecular is chiral and possesses a chiral plane(b) Molecule is chiral and possesses a chiral axis.(c) Molecule is achiral as it possesses a plane of symmetry.(d) Molecule is achiral as it possesses a centre of symmetry.

Soln. In Biphenyl system chirality is due to presence of plane that is why it comes under planar chirality.Since, the given molecule possesses a chiral plane. Thus it is optically active.Correct answer is (a)

25. Consider the following statements about cis- and trans-decalins(A) cis-isomer is more stable than trans-isomer(B) trans-isomer is more stable than cis-isomer(C) trans-isomer undergoes ring-flip(D) cis-isomer undergoes ring-flipThe correct statements among the above are(a) B and D (b) A and C (c) A and D (d) B and C

Soln. • Trans-Decalin is more stable about 2.7 kcal/mole as compared with cis-decalin.• Trans-Decalin has a unique and rigid conformation ring fliping is not possible as it would other wise

afford a highly strained system with one ring attached to other by two axial bond.• Cis-Decalin exists as an equilibrium between two enantiomeric all chair conformation. So, flip into one

other.

H

HTrans-Decalin

FlipH

HCis-Decalin

HH

FlipX

Correct answer is (a)



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26. In bis(dimethylglyoximato)nickel(II), the number of Ni–N, Ni–O and intramolecular hydrogen bond(s), re-spectively are(a) 4, 0 and 2 (b) 2, 2 and 2 (c) 2, 2 and 0 (d) 4, 0 and 1

Soln. Ni++ is tested by dmg.

2Red

Ni 2dmg Ni dmg Tschugaeff test

The structure of complex is

C N

C N

Me

Me

O

OHdmg

O

NNi

N

OH

N

OH

O

N

C

C C

CMe

Me

Me

Me

II

Hydrogen bond

Therefore,Ni – N bonds 4

Ni – O ZeroHydrogen bonds twoCorrect answer is (a)

27. Among the following species, (A) Ni(II) as dimethylglyoximate, (B) Al(III) as oxinate, (C) Ag(I) as chloride,those that precipitate with the urea hydrolysis method are(a) A, B and C (b) A and B (c) A and C (d) B and C

Soln. Correct answer is (b)

28. If an enzyme fixes N2 in plants by evolving H2, the number of electrons and protons associated with that,respectively are(a) 6 and 6 (b) 8 and 8 (c) 6 and 8 (d) 8 and 6

Soln. 2 3 2N 8H 8e 16 ATP 2NH H 16ADP 16Pi Correct answer is (b)

29. The particles postulated to always accompany the positron emission among(A) neutrino, (B) anti-neutrino, (C) electron,are(a) A, B and C (b) A and B (c) A and C (d) B and C

Soln. Particles that are always to emit during position emission

1 1 01 0 1 antineutrinoP n e v

Correct answer is (c)30. Toxicity of cadmium and mercury in the body is being reversed by proteins, mainly using the amino acid

residue,(a) Glycine (b) Leucine (c) Lysine (d) Cysteine

Soln. Toxicity of cadmium and mercury in the body is being reversed by cysteine amino acid.Correct answer is (d)

31. NiBr2 reacts with (Et)(Ph2)P at –78ºC in CS2 to give red compound ‘A’, which upon standing at room tem-perature turns green to give compound, ‘B’ of the same formula. The measured magnetic moments of ‘A’ and‘B’ are 0.0 and 3.2 BM, respectively. The geometries of ‘A’ and ‘B’ are(a) square planar and tetrahedral (b) tetrahedral and square planar(c) square planar and octahedral (d) tetrahedral and octahedral



10

Soln. 2

2

Et. Ph .P2 78º/CS

Re d GreenNiBr A B

µ 0 BM µ 3.2 BM

n = unpaired n 0 µ n n 2

n = near about two. electrons

Ni++ = 3d84s04p0 =3d 4s 4p

Hence, (A) will be square planar.

For (A), Ni++ =

dsp2 hybridisation

For (B), Ni++ =

sp3 hybridisation3d 4s 4p

Hence, (B) will be tetrahedral.

Therefore, NiBr22 EtPh2P

NiBr

Br

PEtPh2

Ph2EtP

Ni

Br

BrPEtPh2

Ph2EtPSquare planar Tetrahedral

(A) (B)(A) and (B) are polytopal isomers.Correct answer is (a)

32. The correct non-linear and iso-structural pair is(a) SCl2 and I3

– (b) SCl2 and I3+ (c) SCl2 and ClF2

– (d) I3+ and ClF2

–

Soln. (I) SCl2 b.p. = 2 l. p. = 2 Stearci number (S) = 4(II) 3I

b.p. =2 l.p. = 3 S = 5

(III) 3I b.p. = 2 l.p. = 2 S = 4

(IV) 2ClF b.p. = 2 l.p. = 3 S = 5Shapes of species are

SCl

Cl

••

••

SCl2

Angular

I

I

I

•• ••••••

I3

I

I

II

••

••

I3

Cl

F

F

••••••

ClF2

Linear Angular Linear

Correct answer is (b)33. Ozone present in upper atmosphere protects people on the earth

(a) due to its diamagnetic nature(b) due to its blue colour(c) due to absorption of radiation of wavelength at 255nm(d) by destroying chlorofluoro carbons

Soln. Ozone is a diamagnetic gas which is of dark blue coloured due to absorption of red light. 557 and 602 nm



11

Ozone depliction discovered by J.C. Farman over Halley Bay in Antarctica.Ozone also show strong absorption in 255 UV which is good for earth and living beings as this ‘UV-b’ ismost dangerous

255 UV b Correct answer is (c)

34. If L is a neutral monodentate ligand, the species, 22 34 6 4AgL , AgL and AgL , respectively are

(a) paramagnetic, paramagnetic and dimagnetic(b) paramagnetic, diamagnetic and paramagnetic(c) diamagnetic, paramagnetic and diamagnetic(d) paramagnetic, diamagnetic and diamagnetic

Soln. In 24AgL and 26AgL , AgAg2+ has d9 configuration, hence have unpaired electron. Hence paramagnetic.

34AgL 3Ag has d8 configuration and dsp2 hybridization (square planar so pairing of electron). Hence,diamagnetic in nature.Correct answer is (a)

35. Chromite ore on fusion with sodium carbonate gives(a) 2 4 2 3Na CrO and Fe O (b) 2 2 7 2 3Na Cr O and Fe O

(c) 2 3 33Cr CO and Fe OH (d) 2 4 2 3 3Na CrO and Fe COSoln. Chromium is extracted from chromite ore:

(I) fusion2 4 2 3 2 2 4 2 3 21100ºC4FeCr O 8Na CO 7O 8Na CrO 2Fe O 8CO

(II) 2 4 2 4 2 4 2 2 7 22Na CrO H SO Na SO Na Cr O H O

(III) 2 2 7 2 7 2 3Na Cr O 2C Cr O Na CO CO

(IV) 2 3 2 3Cr O 2Al Al O 2Cr Correct answer is (a)

36. The ligand(s) that is (are) fluxional in 5 15 5 5 5 2C H C H Fe CO in the temperature range 221–

298K, is (are)(a) 5

5 5C H (b) 15 5C H (c) 5

5 5C H and CO (d) 15 5C H and CO

Soln. 221 298K5 1 5 55 5 5 5 5 5 5 52C H C H Fe CO C H C H Fe

Correct answer (b)

37. 36CoL is red in colour whereas 36CoL ' is green. L and L ' respectively corresponds to,(a) NH3 and H2O (b) NH3 and 1, 10-phenanthroline(c) NH3 and 1, 10-phenanthroline (d) H2O and NH3

Soln. 36CoL red colour absorbs green radiations.

36CoL ' green colour absorbs red radiations.

BVR

O

GY

Energy of green radiations > enegy of red radiations.Therefore, L will be stronger ligand and than L ' . Thus, L and L ' are NH3 and H2O respectively..Correct answer is (a)



12

38. The oxidation state of Ni and the number of metal-metal bonds in 22 6Ni CO

that are consistent with the

18 electron rule are(a) Ni(–II), 1 bond (b) Ni(IV), 2 bonds (c) Ni(–I), 1 bond (d) Ni(IV), 3 bonds

Soln. 2

2 6Ni CO

2x 0 2

Ni N

CO

CO

CO

OC

OC

OC

–2

2x2

M–M bond =1

x 1 Correct answer is (c)

39. Structures of SbPh5 and PPh5 respectively are(a) trigonal bipyramidal, square pyramidal (b) square pyramidal, trigonal bipyramidal(c) trigonal bipyramidal, trigonal bipyramidal (d) square pyramidal, square pyramidal

Soln. SbPh5 is square pyramidal while PPh5 is trigonal bipyramidal.

Ph

Ph Ph

Ph

Sb

Ph Ph

Ph

PhP

Ph

PhThis is due to size of central atom. ‘Sb’ is bigger than ‘P’, therefore favour SP while small sized ‘P’ favour TBP.Correct answer is (b)

40. The typical electronic configurations of the transition metal centre for oxidative addition(a) d0 nad d8 (b) d6 and d8 (c) d8 and d10 (d) d5 and d10

Soln. Most commonly the metal in the complexes in their low oxidation state with d8 or d10 configuration undergooxidative addition.

Ir Ph3PPPh3

OC

Cl+H2O

Ir Ph3PPPh3

OC

HH

Cl (+3)

(+1)

(16 electron)

0 PhBr3 34 22PPh318 electron

Pd PPh Pd PPh PhBr

Correct answer is (c)41. Gelatin added during the polarographic measurement carried out using dropping mercury electrode

(a) reduces streaming motion of Hg drop(b) decreases viscosity of the solution(c) eliminates migrating current(d) prevents oxidation of Hg

Soln. Correct answer is (a)42. The pKa values of the following salt of aspartic acid are indicated below. The predominant species that would

exist at pH = 5 is

(pKa = 9.9) H3N COOH

COOH

(pKa = 2.0)

(pKa = 3.9)



13

(a) H3N COO

COOH

(b) H3N COO

COO

(c) H2N COO

COO

(d) H3N COOH

COO

Soln. H3N COOH

COOH

(pKa = 9.9) (pKa = 2.0)

(pKa = 3.9)

It is aspartic acid (amino acid) which contains ionizable side chain.

When pH = 5; since both COOH group have pKa = 2.0, pKa = 3.9. It will be deprotonated form.

Thus the predominant species that would exist at pH = 5 is

H3N COO

COOCorrect answer is (b)

43. The major product formed in the following photochemical reaction is

t-Bu

t-Bu

t-Buhv

(a)

t-Bu

Ht-Bu

t-Bu

(b)

t-Bu

t-But-Bu

(c)

t-Bu

Ht-Bu

t-Bu

(d)

t-Bu

t-But-Bu

Soln. Correct answer is (a).

44. The pair of solvents in which PCl5 does NOT ionize, is(a) CH3CN, CH3NO2 (b) CH3CN, CCl4(c) C6H6, CCl4 (d) CH3CN, C6H6

Soln. PCl5 is a polar solvent which will dissolve only in polar solvent.Therefore, since benzene and carbon tetrachloride both are non-pllar solvent. Thus, PCl5 will not dissolve in itCorrect answer is (c)

45. The major product formed in the following reaction isMe

OTs

H

H

Ph

Me

AcOH

(a)

Me

OAc

H

H

Ph

Me

(b)

Me

OAc

H

Me

Ph

H

(c) Me

Ph H

Me

(d) Me

Ph Me

H



14

Soln. Ph

Me H

H

OTs

Me

?

In the presence of acid, protonation takes place, thus reaction will proceed via SN1 mechanism.Thus, retenation of configuration will be observed and hence the product formed will be

Ph

Me H

H

OTs

Me

AcOHPh

Me H

H

OHTs

Me

Ph

Me H

H Me

AcO+Ph

Me H

H

OAc

Me


46. The correct order for the rates of electrophilic aromatic substitution of the following compound is

N

MeN

Cl

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

Soln. Since N is a electronegative element, thus it decreases the electron density in the ring and thus the ring is lessprone to be attacked by the electrophile in electrophilic aromatic substitution reaction.Hence, pyridine (I) will be least reactive in which the electronegative element N is present in the same ring.

Since in structure (I) N is outside the ring and thus it increases the electrophilicity of the ring by donatingelectron pair through its lone pair.Thus it will be most prone to electrophilic attack.Correct order will be I > III > IICorrect option is (d)

47. The commutator of the kinetic energy operator, xT̂ and the momentum operator, xp̂ for the one-dimensionalcase is

(a) i (b) di

dx (c) 0 (d) i x

Soln. ˆ ˆ, x xT p

2 2

2 ,2

im xx

2 2 2 2

2 22 2i i

m x x mx x

2 2 2

22 2i im x m x x xx

2 3 2 3

3 32 2i im mx x

3 3 3 3

3 3 0 zero2 2i i

m mx x




15

48. The major product formed in the reaction of trans-1-bromo-3-methylcyclobutane with sodium iodide in DMFis

(a) Me

(b) Me

(c)

Me

I

(d)

Me

I

Soln.

Me

Br

NaIDMF

cis

Me

Br

NaIDMF

cis

? ?

Strong base and polar solvent (aprotic) favour SN2 mechanism.• Since, SN2 mechanism proceed via inversion of configuration.• Thus, the product formed will have confirmation opposite to that of reactant i.e.

Me

Br

NaIDMF

Me

ICorrect answer is (c)

49. When Si is doped with a Group V element,(a) donor levels are created close to the valence band(b) donor levels are created close to the conduction band(c) acceptor levels are created close to the valence bond(d) acceptor levels are created close to the conduction band

Soln. When Si is doped with group V elements like P, then at each P centre tehre is extra electron which constitute adonar band that lies close to empty conduction band.VB = Valence bond, CBDB = Donar bond, DB

CB = Conduction bond VBCorrect answer is (b)

50. The symmetry point group of propyne is(a) C3 (b) C3V (c) D3 (d) D3d

Soln.

C

HH

H

C

C

H Plane and axes

C3V

H




16

51. For a first order reaction A products , the plot of

t

0

Aln

A

vs time, where 0A and tA refer to

cocentration at time t = 0 and t respectively, is(a) a straight line with a positive slope passing through origin(b) a straight line with a negative slope passing through origin.(c) an exponential curve asymptotic to the time axis

(d) a curve asymptotic to the

t

0

Aln

A

axis.

Soln.

0 0f

1 t

AcA ln k lnc A

t

0

Aln kt

A

y mx c

Therefore, straight line with negative slope and intercept = 0


52. In radical chain polymerization, the quantity given by the rate of monomer depletion, divided by the rate ofpropagating radical formation is called(a) kinetic chain length (b) propagation efficiency(c) propagation rate constant (d) polymerization time

Soln. Kinetic chain length describes the number of chain propagation steps in between the chain initiation step andthe chain termination step.Mathematically it is defined as

Kinetic chain length = rate of product formation

rate of initial stepCorrect answer is (a)

53. Number of rotational symmetry axes for triclinic crystal system is(a) 4 (b) 3 (c) 1 (d) 0

Soln. The triclinic crystal with parameters a b c and 90º is least symmetric.Therefore, no axis of symmetryCorrect answer is (d)

54. Generally, hydrophobic colloids are flocculated efficiently by ions of opposite type and high charge number.This is consistent with the(a) peptization principle (b) krafft theory(c) Hardy-Schulze rule (d) Langmuir adsorption mechanism

Soln. Correct answer is (c)

55. Examine the following first order consecutive reactions. The rate constant (in s–1 units) for each step is givenabove the arrow mark

(A) 5 810 10P Q R (B)

5 310 10P Q R

(C) 7 710 10P Q R (D)

2 610 10P Q R

Steady-state approximation can be applied t.o(a) Aonly (b) C only (c) B and C only (d) A and D only

Soln. Steady state approximation for conservative reactions1 2k kP Q R

is applied when first step is slow and second step is fast.Correct answer is (d)



17

56. The figure below represents the path followed by a gas during expansion from A B . The work done is (Latm)

A5

4321

1 2 3 4 5

P(atm)

V(L)

B

(a) 0 (b) 9 (c) 5 (d) 4Soln. 2 1W P V W P V V

A – B WorkA I I B A IWork, P = 5 atm

V = 1 to 1W = 5(1–1) = 5(0) = 0

I B ,P = 1

2 1 5 1 4V V V

1 4 4W P V Correct answer is (d)

57. An aqueous solution of an optically pure compound of concentration 100 mg in 1 mL of water and measuredin a quartz tube of 5 cm length was found to be –3º. The specific rotation is(a) –30º (b) –60º (c) –6º (d) +6º

Soln.03 300 60º

C 0.01 5 5

Correct answer is (b)58. Two phases and of a species are in equilibrium. The correct relations observed among the variables, T,,

p and µ are(a) T T , p p , µ µ (b) T T , p p , µ µ

(c) T T , p p , µ µ (d) T T , p p , µ µ

Soln. If two phases , of a species are in equilibrium then temperature, pressure and chemical potential for bothphases are in same type.

T T ; P P ; µ µ Correct answer is (c)

59. The number of configurations in the most probable state, according to Boltzmann formula, is(a) BS/ke (b) BS/ke (c) BE/k Te (d) BG/k Te

Soln. BS k nW

arrangementBS k n

BS k nW / BS k nW



18

/ BS kW eCorrect answer is (a)

60. The correct match of the 1H NMR chemical shifts of the following species/compounds is

+

(I) (II) (III)(a) I : 5.4; II : 7.2; III : 9.2 (b) I : 9.2; II : 7.2; III : 5.4(c) I : 9.2; II : 5.4; III : 7.2 (d) I : 7.2; II : 9.2; III : 5.4

Soln. 1H NMR chemical shift of these species can be explained on the basis of ring current leads to deshieldingeffect.

When these species are placed in magnetic field, the -electrons in the aromatic ring system areinduced to circulate around the ring. Thus, hydrogen corresponds to these ring are said to be deshielded by thediamagnetic anirotropy of the ring.

Since, in cylopentadiene ring 4 e are there and also it is not aromatic, thus, iits proton are leastdeshielded compared to rest two ring. Thus value is lowest among three.

Since, benzene and tropylium cation both are aromatic and contains 6 e , but since in the case oftropylium cation positive charge is there. Thus, its proton are more deshielded as compared to benzene. Thus, value will be in order..III > II > ICorrect answer is (a)

61. The major prorudcts formed in the following areO

OEt(i) Ph3C–Na+

(ii) H3O+

(a)

O

OH+ Ph3C–CH2CH3 (b)

CPh3

CPh3+

HOEtOH

(c)

O O

OEt+ EtOH (d)

O

OH + H2C CH2

Soln.

O

OEt(i) Ph3C–Na+

(ii) H3O+C

O

OEt

C

O

OEt

C

O

CO OEt

OEt

–EtO–

H3O+

C

O

OEt

CO

C

O

C

O

OEt+ EtOH




19

62. In a Diels-Alder reaction, the most reactiv diene amongst the following is(a) (4E)-1, 4-hexadiene (b) (4Z)-1, 4-hexadiene(c) (2E, 4E)-2, 4-hexadiene (d) (2Z, 4Z)-2, 4-hexadiene

Soln.

12

34

65

(4E)-1, 4-hexadiene

12

3

4

5

6

(4Z)-1, 4-hexadiene

12

34

56

(2E, 4E)-2, 4-hexadiene

12

3

4 5

6

(2Z, 4Z)-2, 4 hexadiene

Now, for better overlap in Diels-Alder reaction, the diene must be in cis-oid fastion. Thus, favourable configu-ration is (2E-4E)-2, 4 hexadiene.Correct answer is (c)

63. Consider the statements about the following structures X and Y

Ph CN Ph •NHX Y

(A) X andy are resonance structures (B) X and Y are tautomers(C) Y is more basic than X (D) X is more basic than YThe correct statement(s) among the above is/are(a) A and C (b) C (c) B and D (d) B and C

Soln.Ph CN Ph •

NHX Y

Ph C Ph CNX YN

H

Ph •

Y NH

Thus, X and Y are tautomers.Also, it is clear from the structure of X and Y that X is more basic than Y.Correct answer is (d).

64. Pericyclic reaction involved in one of the steps of the following reaction sequence is

S

Ph

O

(i) heat

(ii) (EtO)3P

OH

(a) [1, 3] sigmatropic shift (b) [3, 3] sigmatropic shift(c) [1, 5] sigmatropic shift (d) [2, 3] sigmatropic shift

Soln.S

Ph

O

(i) heat

(ii) (EtO)3P

OH2

3

211

[2, 3] sigmatropic shift

O

S

Ph

(ii) (EtO)3P




20

65. Atorvastatin (structure given below) is a

HOOC

OH

OH

N

F

Ph

O

PhHN

Me

Me

(a) cholesterol lowering drug (b) blood sugar lowering drug(c) anti-plasmodial drug (d) anti-HIV drug

Soln. Atorvastatin is a cholesterol owering drug.Correct answer is (a)

66. The maximum bond order obtained from the molecular orbitals of a transition metal dimer, formed as linearcombinations of d-orbitals alone, is(a) 3 (b) 4 (c) 5 (d) 6

Soln. A maximum five ‘d’ orbitals can form bonding with each otherM M bonds

bonds bonds

122

Firstly reported in 2007 in chromium complexes.

Cr ArAr Cr

Ar = Substituted aromatic rings.Correct answer is (c)

67. The term symbol that is NOT alllowed for the np2 configuration is(a) 1D (b) 3P (c) 1S (d) 3D

Soln. For np2 configurations (microstates)L M S

(I) l = +1 0 –1 1D 2 1 0

(II) 2P 1 3 1

(III) 1S 0 1 0

(IV) 3D 1 3 1The configuration (IV) violets the Pauli’s Principle, Hence term symbol 3D is not possible.Correct answer is (d)

68. If the ionization energy of H atom is x, the ionization energy of Li2+, is(a) 2x (b) 3x (c) 9x (d) 27x

Soln. The formula for ionisation energy is2

213.6 z eVI.E. | z atomic number, n shell number

atomn

This is for one electron ‘H’ atom like systemHence, for Li++ = 1s1 (one electron system)

2

213.6 3I.E. 13.6 9

1

. For ‘H’ atom, I.E. = 2

213.6 1I.E. 13.6 x

1

Therefore, I.E. of Li++ = 9xCorrect answer is (c)



21

69. If temprature is doubled and the mass of the gaseous molecule is halved, the rms speed of the molecular willchange by a factor of

(a) 1 (b) 2 (c) 12 (d)

14

Soln.3

msRTrM

msTrM

T = 2T, m = m/2

2/ 2

Trms

M4

Trms

M

2 TrmsM

(Two time increase)


70. In the graph below, the correct option, according to Kohlrausch law, is

BA

C

D

A

C(a) A is a weak electrolyte and B is a strong electrolyte(b) A is a strong electrolyte and B is a weak electrolyte(c) C is a strong electrolyte and D is a weak electrolyte(d) C is weak electrolyte and D is a strong electrolyte

Soln. According to Kohlrausch law molar conductance of strong electrolyte decreases linearly with square root ofconcentration while that of weak electrolyte decreases exponentially. Hence, (c) is a strong electrolyte whileD is a weak electrolyte.Correct answer is (c)



22

PART-C

71. Reaction of [Ru(NH3)5(isonicotinamide)]3+ with [Cr(H2O)6]2+ occurs by inner sphere mechanism and rate of

th reaction is determined by dissociation of the successor complex. It is due to the(a) Inert ruthenium birdged to inert chromium centre(b) Inert ruthenium bridged to labile chromium centre(c) Labile ruthenium bridged to inert chromium centre(d) Labile ruthenium bridged to labile chromium centre

Soln.CNO

NH2(H2N)5Ru

3+

+ [Cr(H2O)6]2+

InertLabile

–H2O

CNO

NH2

(H2N)5RuCr(H2O)S +5

Electron exchange

CN O

NH2

(H2N)5Ru Cr(H2O)S

+5+3+2

Successor complex

In the successor complex inert rutherm bridged to inert chromium.


72. Consider the second order rate constants for the following outer-sphere electron transfer reactions:

3 22 26 6Fe H O / Fe H O

4.0 M–1 sec–1

3 23 3Fe phen / Fe phen

3.0×107 M–1

(phen = 1, 10-phenanthroline)The enhanced rate constant for the second reaction is due to the fact that(a) The ‘phen’ is a -acceptor ligand that allows mixing of electron donor and acceptor orbitals that enhancesthe rate of electron transfer(b) The ‘phen’ is a -donor ligand that enhances the rate of electron transfer(c) The ‘phen’ forms charge transfer complex with iron and facilitates the eletron transfer(d) The ‘phen’ forms kinetically labile complex with iron and facilitates the electron transfer.

Soln. (Phen) is an -accepter ligands hence there is mixing of donor and acceptor orbital having similar symmetrythis leads to fast transfer of electron leading to enhance rate of reaction.

33

5 02g g

*5 *0

Fe phen

t e

2

3

6 02g g

*6 0g

Fe phen

t e

e

Electron transfer takes place from *6 2 *5 3Fe Fe




23

73. The compound [Re2(Me2PPh)4Cl4] (M) having a configuration of 2 4 2 *2 can be oxidized to M+ and M2+.The formal metal-metal order in M, M+ and M2+ respectively, are(a) 3.0, 3.5 and 4.0 (b) 3.5, 4.0 and 3.0 (c) 4.0, 3.5 and 3.0 (d) 3.0, 4.0 and 3.5

Soln. [Re2(Me2PPh)4Cl4] – (M)

M =

*

8 2 6B.O. 32 2

M+ =

*

8 1 7B.O. 3.52 2

M+2 =

*

8B.O. 42


74. In low chloride ion concentration, the anticancer drug cis-platin hydrolyses to give a diaqueo complex and thisbinds to DNA via adjacent guanine

N

NHN

NH NH2

O

(guanine)The coordinating atom of guanine to Pt(II) is(a) N1 (b) N3 (c) N7 (d) N9

Soln. Correct answer is (c)75. The 19F NMR spectrum of ClF3 shows

(a) doublet and triplet for a T-shaped structure(b) singlet for a trigonal planar structure(c) singlet for a trigonal pyramidal structure(d) doublet and singlet for a T-shaped structure



24

Soln. Cl

F

F

F c2 ans planea

b

b

FCl

F

F

••

••19 F 2F doublet19 F 1F tripletT shaped structureCorrect answer is (a)

76. The low temperature (–98ºC) 19F NMR spectrum of SF4 shows doublet of triplets. It is consistent with thepoint group symmetry.(a) C3V (b) C4V (c) Td (d) C2v

Soln.

SF F

F F

SF4 Sea saw shape

c2 and plane

C2V


77. Amongst organolithium (A), Grignard (B) and organoaluminium (C) compounds, those react with SiCl4 to givecompound containing Si-C bond are(a) A and B (b) B and C (c) A and C (d) A, B and C

Soln.Si

Cl

ClClCl

+-

+ RMgX

RLiR3Al

In all case there is electrostatic interaction between Si and R+

Si

Cl

Cl ClClXMgR

R SiCl3


78. In its electronic spectrum, 32 6V H O

exihibits two absorption bands, one at 17, 800 (v1) and the second

at 25, 700 (v2) cm–1. The correct assignment of these bands, respectively, is(a) 3 3 3 3

1 1g 2g 2 1g 1gv T F T F , v T F T P

(b) 3 3 3 31 1g 1g 2 1g 2gv T F T P , v T F T P

(c) 3 3 3 31 2g 1g 2 2g 2gv A T F , v A T F

(d) 3 3 3 31 2g 2g 2 2g 1gv A T F , v A T F



25

Soln. 3 3 2 2 02 2g6V H O V 3d t eg

eg

t2g

hveg

t2g

hv

one electron transfer3T1g3T1g+ 3T2g

eg

t2g

3A2g

two electron transfer

Two transitions.

3 31 1g 2gV T F T F ; 3 3

2 1g 1gV T F T P

3P

3F

3T1g

3T2g

3T1g (P)

3A2g

Correct answer is (a)79. Reactions of elemental as with hot and conc. HNO3 and H2SO4, respectively, give

(a) As4O6 and As2(SO4)3 (b) As(NO3)5 and As2(SO4)3(c) As4O6 and H3AsO4 (d) H3AsO4 and As4O6

Soln.

AsHNO3

Hot. and conc. H3AsO4 Arsenic acid (V)

H2SO4[H3AsO3]

H2SO4As4O6

Arseneous acid (III)The reason is that HNO3 is better oxidising agent than H2SO4 also acts as dehydrating agent.Correct answer is (d)

80. The total valence electron count and the structure type adopted by the complex [Fe5(CO)15C)] respectively,are(a) 74 and nido (b) 60 and closo (c) 84 and arachno (d) 62 and nido

Soln. [Fe5(CO)15C)]Total valency electron = 8×5 + 15×2 + 4 = 74PEC = TEC – n × 12PEC = 74–5×12 = 74–60 = 14

14 72 42

PEC

7 = n + 2 where, n = number of metal in electron. = 5 + 2 = (n + 2) NidoCorrect answer is (a)



26

81. 1H NMR spectrum of 55 5 2 4 2C H Rh) C H at –20ºC shows a typical AA ' XX ' pattern in the olefinic

region. On increasing the temperature to ~70ºC, the separate lines collapse into a single line which is due to(a) free rotation of the ethylene ligand about the metal-olefin bond(b) interamolecular exchange between the ethylene ligands(c) intermolecular exchange between the ethylene ligands(d) change in hapticity of the cyclopentadienyl ligand

Soln. Correct answer is (a)

82. The nuclides among the following, capable of undergoing fission by thermal neutrons, are(A) 233U (B) 235U (C) 239Pu (D) 232Th(a) A, B and D (b) A, C and D (c) B, C and D (d) A, B and C

Soln. Correct answer is (d)

83. The use of dynamic inert atmosphere in thermogravimetric analysis (TGA)(a) decreases decomposition temperature (b) decrease weight loss(c) reducds rate of decomposition (d) increases weight loss

Soln. In thermogravimetric analysis (TGA) use of dynamic innert atmosphere is decrease decomposition with re-spect to temperature.Correct answer is (a)

84. The correct statements for hollow cathode lamp (HCL) from the following are(A) HCL is suitable for atomic absorption spectroscopy (AAS)(B) lines emitted from HCL are very narrow(C) the hardening of lamp makes it unsuitbale for AAS(D) transition elements used in lamps have short life(a) A, B and C (b) B, C and D (c) C, D and A (d) D, A and B


85. Identify the correct statement about

22 6Ni H O

and 2

2 6Cu H O

(a) All Ni-O and Cu-O bond lengths of individual species are equal(b) Ni-O(equatorial) and Cu-O(equatorial)(c) All Ni-O bond lengths are equal whereas Cu-O (equatorial) bonds are shorter than Cu-O(axial) bonds(d) All Cu-O bond lengths are equal whereas Ni-O(equatorial) bonds are shorter than Ni-O(axial) bonds.

Soln.

H2O

H2O OH2

OH2

Ni

OH2

OH2

H2O

H2O OH2

OH2

Cu

OH2

OH2

equatorial bond short

Ni2+

t2g

eg

Cu2+

t2g

eg

Symmetrical electronicenvironment so all bond length same

There is large electron density at z-axisSo, on z-axis increases




27

86. Reaction of nitrosyl tetrafluoroborate to Vaska’s complex gives complex A with M N O 124º . Thecomplex A and its N-O stretching frequency are, respectively(a) [ IrCl(CO)(NO)(PPh3)2]BF4, 1620 cm–1

(b) [IrCl(CO)(NO)2(PPh3)](BF4)2, 1730 cm–1

(c) [IrCl(CO)(NO)2(PPh3)](BF4)2, 1520 cm–1

(d) [IrCl(CO)(NO)(PPh3)2], 1820 cm–1

Soln.Ir

Cl

PPh3

CO

Ph3P

NOBF4

–Ir

Cl

PPh3

CO

Ph3P

NO124º

BF4

16 electron 18 electron

NO = 1650 cm–1 Bent

NO = 1520 cm–1 Linear


87. The correct order of decreasing electronegativity of the following atoms is,(a) As > Al > Ca > S (b) S > As > Al > Ca (c) Al > Ca > S > As (d) S > Ca > As > Al

Soln. The electronegativities of elements areCa Al As S

1.0 1.5 2.0 2.5


88. A 1 : 2 mixture of Me2NCH2CH2CH2PPh2 and KSCN with K2[PdCl4] gives a square planar complex A.Identify the correct pairs of donor atoms trans to each other in complex A from the following combinations.(a) P, N (b) N, S (c) P, S (d) N, N

Soln. s and p both form -bonding with complex and -bonding capacity of sulphur is greater than phosphorus dueto smaller size of d-orbital of sulphur. Hence, in presence of sulphur trans to phosphorus donor atom phospho-rus-metal bond will be weak hence they do not lie trans to each other in the complex. As nitrogen does notinvolvent in -bonding with complex hence when nitrogen atom is trans to phosphorus, phosphorus becomeable to form efficient -bond with metal hence become stable thats why P and N are trans to each other

SCNPd

N

P

MeMe

NCSPh

PhCorrect answer is (a)

89. For a low energy nuclear reaction, 2224 Mg d, Na, the correct statements from the following are(A) Kinetic energy of d particle is not fully available for exciting 24Mg.(B) Total number of protons and neutrons is conserved(C) Q value of nuclear reaction is much higher in magnitude relative to heat of chemical reaction(D) Threshold energy is Q value.(a) A, B and C (b) A, B and D (c) B, C and D (d) A, C and D




28

90. At pH 7, the zinc(II) ion in carbonic anhydrase reacts with CO2 to give

(a) Zn O

H

O

O

(b) Zn C

OH O

O

(c) Zn

OH

O

C O(d)

O

Zn

H

O

C O

Soln.

Zn

OH H

His94

His96

His119

H+

Zn

OH

His94

His96

His119

CO2

Zn

OH

His94

His96

His119

OC

O

Zn

OH

His94

His96

His119

O

O

Zn

OH

His94

His96

His119

O

OH2O

HCO3–


91. Molybdoenzymes can both oxidize as well as reduce the substrates, because(a) Mo(VI) is more stable than Mo(IV)(b) Mo(IV) can transfer oxygen atom to the substrate and Mo(VI) can abstract oxygen atom from the sub-

strate(c) Conversion of Mo(VI) to Mo(IV) is not favoured(d) Mo(VI) can transfer oxygen atom to the substrate and Mo(IV) can abstract oxygen atom from the sub-

strate.Soln. Correct answer is (d)

92. A comparison of the valence electron configuration of the elements, Sm and Eu suggests that(a) Sm is a better one electron reductant than Eu(b) Sm is a better one electron oxidant than Eu(c) Facile oxidation state is +2 for both the elements(d) Both of these display similar redox behaviour.

Soln. Sm 4f6 5s0 6s2

Eu 4f7 5s0 6s2

Sm can accept one electron and become half field. So, it is better oxidant.Correct answer is (b)

93. The cooperative binding of O2 in hemoglobin is due to(a) a decrease in size of iron followed by changes in the protein conformation(b) an increase in size of iron followed by changes in the protein conformation(c) a decrease in size of iron that is NOT accompanied by the protein conformational changes(d) an increase in size of iron that is NOT accompanied by the protein conformational changes

Soln. The movement of iron atom and imidazole side chain of histidine F8 toward the porphyrin plane results in



29

breaking of some of the salt bridges. The breaking of these salt bridges reduces the strain in hemoglobinmolecule. Therefore, the oxyform of hemoglobin is called relaxed state (i.e., R state). The T form ofdeoxyhemoglobin discourages the addition of first dioxygen molecule.

The bonding of one dioxygen molecule to a subunit of hemoglobin reduces the steric hindrance in theother subunits (due to breaking of salt bridges) and therefore encourages the third as well as fourth subunits.This is called cooperative mechanism.Correct answer is (a)

94. Amongst the following which is not isolobal pairs(a) Mn(CO)5, CH3 (b) Fe(CO)4, O (c) Co(CO)3, R2Si (d) Mn(CO)5, RS

Soln. Total electronCo(CO)3 15 electron not isolobalR2Si 6 electronMn(CO)5 17 electron isolobalCH3 7 electronFe(CO)4 16 electron isolobalO 6 electronMn(CO)5 17 electron isolobalRS 7 electronCorrect answer is (c)

95. The correct order of the size of S, S2–, S2+ and S4+ species is,(a) 2 4 2S S S S (b) 2 4 2S S S S (c) 2 2 4S S S S (d) 4 2 2S S S S

Soln. As positive charge increases the size decreases while with increase in negative charge increase the size. This isdue to increase in Zeff in former case while decrease in Zeff in later case.Hence, order of size is 2 2 4S S S S Correct answer is (c)

96. The major product formed in the following reaction isBr H

n-Bu3SnH, AlBNbenzene reflux

H

(a) H

H

H

(b) H

H

H

(c)

Br

H

SnBu3

H

(d)

H

H

Soln.

Br

CC

H

n-Bu3SnH, AlBNbenzene reflux

HHomolytic cleavage

CC

H

H

•C

C

H

H

•

nBu3SnHAIBN

C

H

H

•

HH

H

Product

Correct answer is (a).



30

97. The correct combination of reagents to effect the following conversion isO

H

H

COOH

H

H

(a) (i) Ph3P+CH2OMeCl–, BuLi, (ii) H3O

+, Jones’ reagent(b) (i) H2N

–NHTs; (ii) BuLi (2 equiv); (iii) DMF(c) (i) H2N

–NHTs; (ii) BuLi (2 equiv); (iii) CO2(d) (i) ClCH2CO2Et, LDA; (ii) BF3.OEt2; (iii) DMSO, (COCl)2, Et3N, –78ºC to rt.

Soln.

O

H2N NHTs

NH

HN Ts

Bu

N N H

Bu

Li

CO O

(i)

(ii) H+

CO2H

Correct answer is (c)98. The major product formed in the following reaction is

anhyd. CF3COOHCH2Cl2

Bu4N+Br–

(a)

BrH

(b)

H

Br(c)

H

COOCF3

(d) O

Soln. anhyd. CF3COOHCH2Cl2

Bu4N+Br–

H+

BrH

Br


99. Consider the following reaction,O

+ Ph–N3CF3COOH N

OPh

The appropriate intermediate involved in this reaction is

(a)

OHN

Ph (b) N

HO

H

Ph

(c)

NOPh

(d)

HO NN

Ph

N



31

Soln.

O

+ Ph N N N

O N

Ph

N N

CF3COOH

HO N

Ph

N N

Intermediate

O N

Ph

N N

–N2N

OPh

Product

O

N Ph

Correct answer is (d)100. The correct 13C NMR chemical shift values of carbons labeled a-e in the following ester are

Me O

O

Mee d

b

a

(a) a : 19; b : 143; c : 167; d : 125; e : 52 (b) a : 52; b : 143; c : 167; d : 125; e : 19(c) a : 52; b : 167; c : 143; d : 125; e : 19 (d) a : 52; b : 167 ; c : 125; d : 143 ; e : 19

Soln. Me O

OMee d

b

a

c

Me O

OMee d

b

a

c

143

19 52

167

125

Correct answer is (d)101. The products A and B in the following reaction sequence are

OH

OO

ClMeO

Et3N(A)

NH2

(B)

(a) O

O

A : OMe

O

NH

O

B :

(b) O

O

A : Cl

O

O

O

B : NH

O



32

(c) O

O

A : OMe

O

MeOB : NH

O

(d) O

O

A : Cl

O

B : O

O

O

Soln.C

OH

O C

O

ClMeO

Et3N(A)

NH2

(B)

Since less basic group are better leaving group. Thus, Cl– will be a better leaving group as compared tomethoxy group.Thus, formation of A takes place according to following mechanism.

COH

O C

O

ClMeO

Et3NC

O

OC O

O

C

O

OMe

NH

O

NH2

(B)

(A)


102. The biosynthesis of isopentenyl pyrophosphate from acetyl CoA involves:A. Four molecules of acetyl CoAB. Three molecules of ATPC. Two molecules of NADPHD. Two molecules of lipoic acidThe correct options among these are(a) A, B and D (b) A and B (c) B and C (d) A, C and D

Soln. The biosynthesis of isopentenyl pyrophosphate from acetyl CoA consumes three molecule of ATP and twomolecule of NADPH.Correct answer is (c)

103. Amongst the following, the major products formed in the following photochemical reactions areO hv

HO

OO

OA B C D



33

(a) A and C (b) B and C (c) A and D (d) A and B

Soln.

O hv OH•

HO

OH

•

•

OH

CO

tautomerisation


104. The products A and B in the following reaction sequence are

N

O

O

OMe

O

BnO

H2N NH2 (A) heatB

(a) O

MeO

BnO

H2N

A: B:NH

OBnO

(b) A: B:NO

O

HO

MeOO

NO

O

O

O

(c) A: B:NO

O

BnO

HOO

NO

O

O

O



34

(d) O

MeO

BnO

H2N

A: B:

HN

OMeO

Soln.

NO

O

C

BnO

O

OMe

NH2NH2 NH

CO

C

BnO

O

OMe

NHNH2

O

OMe

O

BnO

NH

H

NH2

O

BnO–H

NH

O

BnO


105. Anthranilic acid, on treament with iso-amyl nitrite furnishes a product which displays a strong peak at 76(m/e) in its mass spectrum. The strucrture of the product is

(a) (b)

NO

COOH

(c)

OH

COOH(d)

COOH

Soln.

C

NH2

OH

O

iso-amyl nitrite

ON

O

C

N

O

O

N

–CO2

–N2

Benzene diazonium2-carboxylate

Benzene

Now two molecule of benzene combines to form biphenyl (dimerises)

Biphenyl

This benzyne has m/z value 76.




35

106. The organoborane X, when reacted with Et2Zn followed by p-iodotoluene in the presence of catalytic amountof Pd(PPh3)4 furnishes a tri-substituted alkene. The intermediate and the product of the reaction, respectively,are

H

)3B

X

(a)

H Znand

H

(b)

H

Zn

andH

(c) H

Zn

andH

(d) H Zn

andH

Soln.H

)3BEt2Zn

H Zn

I

CH3

Pd(PPh3)4 Hattack fromsame side


107. Using Boltzmann distribution, the probability of an oscillator occupying the first three levels (n = 0, 1 and 2) isfound to be p0 = 0.633, p1 = 0.233 and p2 = 0.086.The probability of finding an oscillator in energy levels in n 3 is(a) 0.032 (b) 0.048 (c) 0.952 (d) 1.000

Soln. Total probability = PT = 1Probability of occupying first three levels = P0 + P1 + P2

PT = P0 + P1 + P2 + probability of finding an oscillator in energy levels in n 31 = 0.952 + n 3P 1 0.952 0.048 Correct answer is (b)

108. The major products A and B in the following reaction sequence are

NO2

(i) PhNCO Et3N

(ii)(A)

H2, Rancy Ni

MeOH, H2O, AcOH

(B)

(a) A: ON

B:O OH



36

(b) A: ON

B:O OH

(c) A: ON

B:O O

(b) A: ON

B:O

OH


109. The correct combination of reagents required to effect the following conversion is

COOCH3

COOCH3 O

OH(a) (i) Na, xylene, Me3SiCl, heat; (ii) H3O

+

(b) (i) Na, xylene, heat; (ii) H2O2, NaOH(c) (i) NaOEt, EtOH; (ii) Na, xylene, heat(d) (i) TiCl3, Zn–Cu, Me3SiCl, heat; (ii) H3O

+

Soln.C

C O

OH

O

OCH3

O

OCH3

This reaction is acyloin condensation.The acyloin condensation of diesters favours intramolecular cyclisation over intermolecular polymerisation.The mechanism of acyloin condensation consist of 4 steps, which is as follows.(i) Oxidative ionization of two sodium atoms on the double bond of two ester molecules

COCH3

OCH3C

O

O

C OCH3

OCH3C

ONa

ONa

•

•

Na

Na

•

•

(ii) Intramolecular free radical coupling, followed by alkoxy elimination in both side, producing a 1, 2 diketone.

C OCH3

OCH3C

ONa

ONa

•

• C OCH3

OCH3C

O

O

O

O

(iii) Oxidative ionization of two sodium atoms on both diketone double bonds. The sodium enodiokete isformed.



37

ONa

ONa

O

O

Na•

Na•

••

ONa

ONa

(IV)

ONa

ONa

H2O–2NaOH

OH

OH

OH

O

tautomerisation

Note : Here along with Na, xylene TMsCl has been used to improve the yield of the product. This is becauseby removing the alkoxide ion formed, thereby presenting the base catalyzed side reaction.Correct answer is (a)

110. An organic compound gives following spectral data:IR : 2210, 1724 cm–1, 1H NMR : 1.4 (t, J = 7.1 Hz, 3H), 4.4 (q, J = 7.1 Hz, 2H); 13C NMR : 16, 62,118, 119, 125, 127, 168.The compound is

(a) NC

O

O (b)

O

ONH2

(c) NC

O

O

(d) O

CN

O

Soln. IR rate indicates towards a C N group ( ~ = 2210 cm–1) and a carbonyl group CO

(~ = 1724)

3 21.4 , 7.1 , 3 CH – CH

1.4 ,3

t J Hz H a

t H

2 2(Further deshielded due to the presence of

electron with drawing group)

4.4 , 7.1 , 2 CH – CH – O – EWGq J Hz H b

Thus the required structure will be

CN

O

O

= 16

= 62

= 168 = 127

= 119

= 125

= 118

Correct answer is (c).



38

111. The major product formed in teh following reaction is

N

Ph

COOCH3H3COOC

+ PhPh

(a) N

Ph

Ph Ph

H3COOC COOCH3(b)

N

Ph

Ph Ph

H3COOC COOCH3

(c) N

Ph

H3COOC Ph

H3COOC Ph(d)

N

Ph

H3COOC Ph

H3COOC Ph

Soln.

N

Ph

COOCH3H3COOC

+ C C PhPh

N

Ph Ph

Ph

H3COOC COCH3

O

ProductCorrect answer is (b)

112. The correct combination of reagents for effecting the following sequence of reactions is

AO

O

O

OB

(a) A = O3/O2; B = K+ –OOC-N=N-COO–K+, AcOH(b) A = O2, Rose Bengal, hv; B = K+ –OOC-N=N-COO–K+, AcOH(c) A = O2, Rose Bengal, hv; B = H2, Pd/C(d) A = O2, Rose Bengal; ; B = H2, Pd/C

Soln.A

O

O B

O

O

Rose Bengal is a dye used for catalytic purpose.It is [4+2] cycloaddition reaction, which is favourable in photochemical condition.

O O

[4+2] cycloadditionO2 Rose Bengal, hv

O

OKOOC N N COOK

AcOHO

O

Note : H2/Pd will lead to the formation of dihydroxy compound

O

O H2/Pd

OH

OHThus, it will not be the suitable reagent.Correct answer is (b)



39

113. The correct combination of reagents required to effect the following conversion is

O

N

O

N

I

(a) I2, HNO3 (b) s-BuLi, –78ºC followed by KI(c) NaOEt followed by ICH2CH2I (d) s-BuLi, –78ºC followed by ICH2CH2I

Soln.

O

N

H BuLi

O

N

Li

Ortho lithiation

I CH2 CH2 I

O

N

I+ H2C CH2 + LiI


114. Consider a particle confined in a cubic box. The degeneracy of the level, that has an energy twice that of thelowest level, is(a) 3 (b) 1 (c) 2 (d) 4

Soln.

0 l

(2,2,2)(g=1)

(g=3)

(g=3)

(g=3)

(g=1)

(3,1,1),(1,1,3),(1,3,1)

(2,2,1),(2,1,2),(1,2,2)

(1,1,2),(1,2,1),(2,1,1)

(1,1,1)

12h8m

2

2l

11h8m

2

2l

9h8m

2

2l

6h8m

2

2l 3h8m

2

2l

Cubic box energy, x y z

22 2 2

n n n x y z2h n n n

8m

2

26h

8m(for n = 2) energy is double from

2

23h

8m(ground state)

So, degeneracy is 3Correct answer is (a)

115. Only two products are obtained in the following reaction sequence. The structures of the products from the listI-IV are

O

(i) NaNH2

(ii) BrCH2CH2Br



40

O O

OO

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

Soln.O

(i) NaNH2(ii) BrCH2CH2Br

O

NaNH2

O(a)

O

O O(b)

From (a),

O

Br CH2CH2Br

OCH2

CH2

Br NaNH2

O

Br

O(P)

From (b),

O

Br CH2CH2

O

Br NaNH2

O

Br

O(Q)

Br

Thus, P and Q are the required product.Correct answer is (a)

116. The major product A formed in the following reaction is

OCOOMeMeOOC

HeatA

(a)

COOMeMeOOC

H

H

O(b)

O

COOMeMeOOC



41

(c)

COOMeMeOOC

O

HH

(d)

COOMe

COOMe

OH

Soln.O

C C COOMeMeOOC

Heat

MeOOCMeOOC

O

H

HCorrect answer is (a)

117. The products A and B in the following reaction sequnce are

SeO2

dioxane reflux(A)

aq. NaCN, MnO2

i-PrOH, Me2NH(B)

(a) A :CHO

B :NO

(b) A :CHO

B :

OO

(c) A : CHO B :O

O

(d) A : CHO B :O

N

Soln. SeO2

dioxane reflux(A)

SeO2 is most commonly used for allylic oxidation. First formation of allylic alcohol takes place, which can beoxidized easily to , unsaturated carbonyl compound if desired.

The oxidation are belived to involve an ene reaction, between the alkene and the hyrated form of thedioxide, followed by a [2, 3]-sigmatropic rearrangement of the resulting allyl seleninic acid and finalhydrolysisof Se(II) ester to the allylic alcohol. Further, oxidation of the alcohol gives the , unsaturatedcarbonyl compound. A notable application of this reaction is in the oxidation of 1, 1 dimethyl alkene to thecorresponding E allylic alcohols or aldehydes by selective attack on the E-methyl group.

H

enereaction

SeHO OH

OHSe O

OH

OH

–H2O

SeHO O

OH

[2. 3]-sigmatropicrearrangement

OSeHOOHCHO

furtheroxidation



42

CHO

Aq. NaCNMnO2 CH

O–Na+

CNCyanohydrin

MnO2

C O

CN

i-PrOH

Me2NH

C O

N

Me

Me

The aldehyde formed reacts with cyanide to give cyanohydrin. Oxidation of the cyanohydrin with manganesedioxide gives the acyl nitrile. Which then reacts with the alcohol solvent to give the ester.Correct answer is (d)

118. The spatial part of the wave function of the atom in its ground state is 1s(1) 1s(2). The spin part would be(a) 1 2 (b) 1 2

(c) 1 1 2 1 22

(d) 1 1 2 1 22

Soln. According to the pauli exclusion principle the wavefunction of fermions (here electrons) must be antisymmet-ric i.e., 1, 2 2,1 .

For a triples state, spin wavefunctions can be 1 2 or 1 2 or [ 1 2 2 1 ] , which

are symmetric in nature. As total space spin . So, for total to be antisymmetrical w.r.t. electron exchange,

space the spatial part of wavefunction is symmetrical so the spin part will be antisymmetric

1 1 2 1 22


119. The number of phases, components and degrees of freedom, when Ar is added to an equilibrium mixture ofNO, O2 and NO2 in gas phase are, respectively,(a) 1, 3, 5 (b) 1, 4, 5 (c) 1, 3, 4 (d) 1, 4, 4

Soln. 2 2g g g

NO O NO add Ar g

(1) Phase only (gas phase) all are gasesP = 1

(2) ComponentC = N–EC = 4–1 = 3

(3) F = C–P+2F = 3–1+2 = 4

Correct answer is (c)120. The major product formed in the following reaction is

OH

TsClpyridine



43

(a) TsO

(b)

OTs

(c) OTs

(d) OTs

Soln.

OH

TsClpyridine

OTs

•

OTs OTs

Correct answer is (b)121. A particle in a one dimensional harmonic oscillator in x-direction is perturbed by a potential x is a number .

The first-order correction to the energy of the ground state(a) is zero (b) is negative(c) is positive (d) may be negative or positive but NOT zero.

Soln.2 2

1/4 1/4/2 /2

x xe x e dx

21/2

xx e dx

21/2

xx e dx

n is odd. So, total integration is zero.2

0

n xx e dx if n is odd

20

n xx e dx if n is even.Correct answer is (a)

122. The points A and B in the following sequene of reactions are

CHOOH

OH

OH

OH

HOMeOH/H+

reflux, 24h(A)

PhCH(OMe)2 (1 equiv)

H+B

(a) A= B=OHOHO

OH

OMeOH

OOHO OMe

OH

O

Ph

(b) A= B=OHO

HO

OH

OMeHO

OOHO

OMeHO

OPh

(c) A= B=OHO

HO

OH

OMe

OH

OOO OMe

HO

HOPh

H



44

(d) A= B=OHOHO

OH

OMe

OH

OHOO OMe

O

HO

PhH

Soln.CHOHO

OH

OH

OH

OH

CHOH OH

HO HH OHH OH

CH2OH

H OH

HO HH OHH

CH2OH

O HOHO

OH

HO OH

Thus, this nothing but D glucose. (cyclic form of D glucose).

Now, methylation with MeOH in the presence of acid leads to the formation of mono-methylated D glucose.

OHO

HO

OH

HO OH

MeOH/H+

reflux, 24h

OHO

HO

OH

HO OMePhCH(OMe)2(1 eq)

H+OO

HO

OMe

OMePh

Ph CH OMe

OMe

OO

OMe

HOHO

OPh

Product


123. The mass spectrum of the product A, formed in the following reaction, exhibits M, M+2, M + 4 peaks in theratio of about 1 : 2 : 1. The reagent HX and the product A are

Br

OH

HX A

(a) Br

F

HX = HF and A = (b) Br

Cl

HX = HCl and A =

(c) Br

Br

HX = HBr and A = (d) BrHX = HBr and A =Br



45

Soln. Natural abundance of Cl35 is 100% while C37 is 32.5. While natural abundance of 79Br is 100 while 81Br is98.0.Thus, relative intensities of isotope peaks for various combination of bromine and chlorine will be as follows.

M M + 2 M + 4Br 100 97.7 –Cl 100 32.6 –Br2 100 199.0 95.4Since in given problem M, M+2, M+4 peak is in ratio 1 : 2 : 1.Thus, two Br atom is present in compound A

Br

OH

HBr

Br

OH2

Br

BrBenzyl cation

Br

Br

(A)Br–H2O


124. Match the following natural products in column A with their structural features in column B Column A Column B(I) Colchicine (A) Tetrahydrooxepine(II) Strychnine (B) Phenanthrene(III) Quinine (C) Tropolone(IV) Ephedrine (D) Phenylethylamine

(E) Quinoline(F) Benzofuran

Identify the correct match from the following(a) I-C, II-A, III-E, IV-D (b) I-F, II-A, III-B, IV-E(c) I-A, II-D, III-F, IV-D (d) I-C, II-A, III-E, IV-F

Soln. Correct answer is (a)125. A particle in a one-dimensional box (potential zero between to a and infinite outside) has the ground state

energy 2

0 20.125hE

ma . The expectation value of the above Hamiltonian with x x x a yields an

energy E1. Using a linear combination of two even functions x x a and 22 x x a , we obtain variationalmainimum to the ground state energy as E2. Which of the following relations holds for E0, E1 and E2?(a) 0 1 2 E E E (b) 0 2 1 E E E (c) 1 0 2 E E E (d) 2 0 1 E E E

Soln. When the variational term are increased the energy come closer to ground state energy. So, the energy of two

even function x x a and 22 x x a correspond to the energy E2 come closer to E0. Because the systemare very complicated and not solve easily.Correct answer is (b)

126. The dissociation constant of a weak acid HX at a given temperature is 2.5×10–5. The pH of 0.01 M NaX atthis temperature is(a) 7.3 (b) 7.7 (c) 8.3 (d) 8.7

Soln.

52 a HXNaX H O HX NaOH k 2.5 10

C 0 0C 1 C C

2

bCK

1 1

2bK C



46

b

a

K KC K C

146

5a a

K CK 0.01 10OH C C 2 10K C K 2.5 10

OH 610 10p log OH log 2 10 5.698

pH 14 5.698 8.301 8.3 Correct answer is (c)

127. The ground state energy of hydrogen atom is –13.598 eV. The expectation values of kinetic energy, T and

potential energy, V , in units of eV, are

(a) 13.598, 27.196 T V (b) 27.196, 13.598 T V

(c) 6.799, 6.799 T V (d) 6.799, 20.397 T VSoln. For hydrogen atom.

Virial Theorem.2 T V

2 kinetic energy potential energy Ground.state energy = –13.598

potential energy 2 kinetic energy

potential energy 2 13.598

potential energy 27.196

kinetic energy 13.598Correct answer is (a)

128. If 0.8 0.4 A B is a normalizdd molecular orbital of a diaotmic molecule AB, constructed from A and

B which are also normalized, the overlap between A and B is(a) 0.11 (b) 0.31 (c) 0.51 (d) 0.71

Soln.

2 21 2 1 2

2 2

c c 2c c s 1

0.8 0.4 2 0.4 0.8 s 10.64 0.16 0.64s 1

0.64s 1 0.80 0.64s 0.20

0.20s0.64

5s 0.3116


129. At a given temperature consider

2 3 2 1Fe O s 3CO g 2Fe s 3CO g ; K 0.05

122 2 22CO g 2CO g O g ; K 2 10



47

The equilibrium constant for the reaction

2 3 22Fe O s 4Fe s 3O is

(a) 131 10 (b) 382 10 (c) 154 10 (d) 242 10

Soln.

2 3 2 1

122 2 2

Fe O s 3CO g 2Fe s 3CO g K 0.05 1

2CO g 2CO g O g K 2 10 2

Equation (1) × (2)Equation (2) × (3)Adding both

22 3 2 1

22 2 2

2 3 2

2Fe O 6CO 4Fe s 6CO g K

6CO g 6CO 3O g K

2Fe O 4Fe s 3O

2 31 2K K K

32 12 36 36 36 380.05 2 10 0.05 0.05 8 10 0.4 0.05 10 0.020 10 2 10


130. In a bomb calorimeter, the combustion of 0.5 g of compound A (molar mass = 50 g mol–1) increased thetemperature by 4K. If the heat capacity of the calorimeter along with that of the material is 2.5 kJ K–1, themolar internal energy of combustion, in kJ, is(a) 1000 (b) –1000 (c) 20 (d) –20

Soln. Molar internal energy of combution = Mzm

50 4 2.5 50 10 50 10 10 5000 1000 kJ

0.5 0.5 0.5 5

Energy of combution is negative – 1000Correct answer is (b)

131. The translational, rotational and vibrational partition functions for a molecule are

10 1 13translation rotation vibration Bf 10 m , f f 1, k T / h 10 at room temperature, 23

AN 6 10Using the approximate data given above, the frequency factor (A) for a reaction of the type:atom + diatomic molecule non-linear transtion state product, according to the conventional transitionstate theory is

(a) 32 10 (b) 76 10 (c) 122 10 (d) 136 10

Soln. A is given by #

A

A BC

N kT QA A BC non-linear transition stateh Q Q

3 3 2rot trans vibA

3 3 2 2trans trans rot vib

q q qN kT .h q q q q

rot vib 23 13 7A3 30

trans

q q 10 1N kT 6 10 10 6 10h 10q




48

132. The interplanar spacing of (110) planes in a cubic unit cell with lattice parameter a = 4.242Å is(a) 5Å (b) 6Å (c) 7.35Å (d) 2.45Å

Soln. hk 2 2 2

adh k

1104.242d 3Å1 1 0

Correct answer is (d)133. A compound AxBy has a cubic structure with A atoms occupying all corners of the cube as well as all the face

centre positions. The B atoms occupy four tetrahedral voids. The values of x and y respectively, are(a) 4, 4 (b) 4, 8 (c) 8, 4 (d) 4, 2

Soln. A= corners + face centres 1 18 6 1 3 48 2

And B = 4 (Full 1 contribution in tetrahadral void)Correct answer is (a)

134. The number of lines in the ESR spectrum of CD3 is (the spin of D is 1)(a) 1 (b) 3 (c) 4 (d) 7

Soln. CD3 (ID = 1)(2nI + 1) = (2×3 × 1 + 1) = 7Correct answer is (d)

135. The C = O bond length is 120 pm in CO2. The moment of inertia of CO2 would be close to (masses of C andO are 1.9×10–27 kg and 2.5×10–27 kg, respectively)(a) 45 21.8 10 kgm (b) 45 23.6 10 kgm

(c) 45 25.4 10 kgm (d) 45 27.2 10 kgmSoln. The moment of inertia of CO2.

O C O

mA mAR

mB

2AI 2m R

227I 2 2.5 10 kg 120 pm 27 12 12 2I 2 2.5 10 kg 120 120 10 10 mtr

51 2I 5 14400 10 kg mtr 51 2I 72000 10 kg mtr

47 2I 7.2 10 kg mtr Correct answer is (d)

136. The fluorescence lifetime of a molecule in a solution is 5×10–9 s. The sum of all of the non-radiative rateconstants nrk for the decay of excited state is 8 11.2 10 s . The fluorescence quantum yield of the mol-ecule is(a) 0.1 (b) 0.2 (c) 0.4 (d) 0.6

Soln.f

ffk

... (i)



49

ff IC ISC

1k k k

... (ii)

Given, 8 1IC ISCk k 1.2 10 sec and 9

f 5 10 sec

Therefore, (ii)

98 1

f

15 10 seck 1.2 10 sec

8 1f 9

1k 1.2 10 sec5 10 sec

1 9 1 8 1f

10k 10 10 sec 1.2 10 sec5

8 1 8 1 8 12 10 sec 1.2 10 sec 0.8 10 sec

Therefore, (i) 9 8 1f f fk 5 10 sec 0.8 10 sec 0.4


137. Solutions of three electrolytes have the same ionic strength and different dielectric constants as 4, 25 and 81.The corresponding relative magnitude of Debye-Huckel screening, lengths of the three solutions are(a) 4, 25 and 81 (b) 2, 5 and 9 (c) 1/2, 1/5 and 1/9 (d) 1, 1 and 1

Soln. Debye-Huckel screening lengths

2

2i 0

0 r B

e NA c zk T

0 r B2 2

A i i

k T1e N c z

01

1 4 2 1 25 5 1 81 9


138. Simple Huckel molecular orbital theory(a) considers electron-electron repulsion explicitly(b) distinguishes cis-butadiene and trans-butadiene(c) disinguishes cis-butadiene and cyclobutadiene(d) has different coulomb integrals for non-equivalent carbons.

Soln. In simple Huckel molecular orbital theory only distinguishes between cis-butadiene and cyclobutadiene on thebases of energy correction. In cyclobutadiene delocalization energy is zero and cis butadiene is 0.472Correct answer is (c)

139. For the non-dissociative Langmuir type adsorption of a gas on a solid surface at a particular temperature, thefraction of surface coverage is 0.6 at 30 bar. The Langmuir isotherm constant (in bar–1 units) at this temperatureis(a) 0.05 (b) 0.20 (c) 2.0 (d) 5.0



50

Soln.p

pp

k or k

1 k 1

1 1 0.6k 0.0 5p 1 30 bar 1 0.6


140. For a set of 10 observed data points, the mean is 8 and the variance is 0.04. The ‘standard deviation’ and the‘coefficient of variation’ of the data are, respectively(a) 0.005, 0.1% (b) 0.02, 0.2% (c) 0.20, 2.5% (d) 0.32, 1.0%

Soln. n = 10, variance = 0.04, µ = 8

Standard deviation, variance 0.04 0.20

Coefficient of variation = 0.2 0.025

µ 8

In percentage (coefficient of variation) = 0.025×100% = 2.5%Correct answer is (c)

141. In the Lineweaver-Burk plot of (initial rate)–1 vs. (initial substrate concentration)–1 for an enzyme catalyzedreaction following Michaelis-Menten mechanism, the y-intercept is 5000 M–1 s. If the initial enzyme concentra-tion is 91 10 M, the turnover number is(a) 32.5 10 (b) 41.0 10 (c) 42.5 10 (d) 52.0 10

Soln. Lineweaver-Burk plot 1 1vsr s

Slope max

kmr

, max 2 0r k E

and Intercept = max

1r TON (Turn Over Number) = k2

Given, Intercept = 5000 M–1 s1

max

1 500 M sr

max 11r

5000 M s

2 101k E

5000 M s

91

1TON 1 10 M5000 M s

65 5

3 91 10 10TON 10 2 10

5 55 10 10




51

142. The E E direct product in D3 point group contains the irreducible representations

3 3 2

1

2

2

D E 2C 3CA 1 1 1A 1 1 1E 2 1 0

(a) 1 2A A E (b) 12A E (c) 22A E (d) 1 22A 2ASoln. E×E = 2×2 (–1) (–1) 0×0

= 4 1 0

1A

1n 4 2 0 16

2A

1n 4 2 0 16

E1n 8 2 10 16

Therefore, E × E = A1 + A2 + ECorrect answer is (a)

143. The result of the product 2 2C x C y is

(a) E (b) xy (c) 2C z (d) i

Soln. The product of 2 2C x C y is C2(z)

HO

H1 2

y

x

z

c2(x) HO

H2c2(y)

HO

H12

HO

H12

HO

H1 2

c2(z)

2 2 2c x c y c zCorrect answer is (c)

144. Given;

02

3 0

0

A. Fe OH s 2e Fe s 2OH aq ; E 0.877V

B. Al aq 3e Al s ; E 1.66V

C. AgBr aq e Ag s Br aq ; E 0.071V

The overall reaction for the cells in the direction of spontaneous change would be(a) Cell with A and B : Fe reduced

Cell with A and C : Fe reduced(b) Cell with A and B : Fe reduced

Cell with A and C : Fe oxidized(c) Cell with A and B : Fe oxidized

Cell with A and C : Fe oxidized(d) Cell with A and B : Fe oxidized

Cell with A and C : Fe reducedSoln. (I) Cell A and B



52

02

03

0 0 0right left

Fe OH s 2e Fe s 2OH E 0.877V

E 1.66VAl 3e Al

E E E

–0.87 1.66 0.79V

G nFE E ve

G ve spontaneous Fe reduced.(II) A and C

02

0

Fe OH s 2e Fe s 2OH aq E 0.877

AgBr e Ag s Br aq E 0.071

0 0 0right leftE E E

0.071 0.877 0.948

0 0G nFE E ve

G ve

(Reaction spontaneous)

In this cell Fe carried negative reduction potential goes to left hand side of the half cell and in this half celloxidation occur.Fe oxidisedNote : In electrochemical series (Al+3, Al) stay below to (Fe+2, Fe) and (AgBr(s), Br– : Ag) stay above. So,(Al+3, Al) reduced and (AgBr, Br–.Ag) oxidised by (Fe+2, Fe).Correct answer is (b)



53

145. The reagent A used and the major product B formed in the following reaction sequence are

N

O

Me

AN

Me

BrCN (B)

(a) 4A : LiAlH B: Br NCN

(b) 4A : LiAlH B: NC NBr

(c) 4A : NaBH B: Br NCN

(d) 2A : H Pd C N

CN

Me

Soln. N

O

Me

AN

Me

BrCN (B)

Catalytic hydrogenated works in the case of alkene and alkyne. Thus, H2|Pd will not work here.Therefore, NaBH4 reduces aldehyde and ketones to corresponding alcohol.Therefore, reduction of amide to amine can be beautifully done by LiAlH4, but it does not follow the samepattern as observed for ester and acids.

N

O

Me

LiN

O

Me

Li

N

OAlH3Li

Me

H AlH3H

N

H

Me

HN

Me

N

M

CNBr

Br N

CNN

Me

Br CN


csir-ugc-net/jrf chemical sciences paper june 2014

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