CSIR-UGC-NET/JRF CHEMICAL SCIENCES PAPER

(1)

CSIR-UGC-NET/JRF CHEMICAL SCIENCES PAPER JUNE 2014

PART-A

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

1

0

2 2

(a) 60 (b) 65 (c) 72 (d) 80

Soln. 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 balls to 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 = 124

Correct 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 in direct 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_ _A WA H k. A   1 Wk .A Now, 2





2 2

H  k. A k .Ak '.A (where k ' is other constant)

and 2 2 1 1 H k ' A k ' k w  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





dispersion mean

RD 

Then,

(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, 30 Group B : 40, 80, 35, 70, 85, 45, 50, 75, 60 Dispersion of Group A = 90–30 = 60 Group B = 85–35 = 50 RD of Group A _{40 80 70 50 60 90 30}60 60 1 60 7          RD of Group B _{40 80 35 70 85 45 50 75 60}50 50 1 60 9            RD of Group A > RD of Gropu B Correct 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 of chicory 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 gram Since, in 500 gram chicory is 50 gram

Therefore, in 1g chicory is = 50 500

Therefore, in 100 gram chicory is 50 100 10 gram 500 

After adding 5g chicory

New mixture = 100 + 5 = 105 gram and quantity of chicory = 10 + 5 = 15 gram Hence, % 15 100 100 14.28 14.28%

105 7

    

Correct answer is (c)

6. The time gap between the two instants, one before and one after 12.00 noon, when the angle between the hour hand 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 = Hour 30 min 1 2

  

in a gap of 24 min, there is angle between hands of watch is 66º.

Correct answer is (d) 7. Suppose





2 x y  xy





2 x y  xy





1 x y  x y  x y  x y

+, – and × have their usual meanings. What is the value of



 





197 315  197 315







197 *315



?

(3)

Soln.





197 315



2 



197 315



2



. 197 315







1







₂ 2



512 118 _{630 394} 4 197 315 197 315   _     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 = 24×32×48 A×B2_×C2_{×D = 24×32×48} Divide it by (B×C)2_–









2 2 A B C D 24 32 48 36 32 32 B 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 must be 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 minimum number 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 1 1













1 5 5 6 2 2 5 1 1 12 Correct answer is (c)

11. Suppose n is a positive integer. Then



n2n





2n1



(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 = 180}

and we know that n ( n + 1) (2n + 1) is always divisible by 6

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 crosses a stationary observer on a platform, the man starts walking from the rear to the front and the front to the rear of the train at a constant speed of 3 km/hr. The speed of the train is 80 km/hr. The distance of the man from the observer 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 = 80 1 40 k.m. 2

 

Now, speed of man = 3 km/h In 1

2 hour man will cover 1

3 1.5 km 1500 m 2

   

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 man will 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 a trapezium. The length of the longer of the two parallel sides of this trapezium is

(a) 5 3 4 cm (b) 5 2 cm (c) 10 cm (d) 10 3cm Soln. A D C B E 5 5 5 5 5 5 5 Hence, length of BE = 5 + 5 = 10 cm Correct 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 sight from 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 BD i.e. AE.

AE BD 17

  

(5)

CE tan 45º AE CE 1 17   CE = 17

Hence, length of tree = 17 + 10 = 27 m.

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 as shown schematically in the diagram (not to scale)

C Q A P B T S Q

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

 

(a) 625 (b) 400 (c) 225 (d) 125

Soln. 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 BC ST OS OT    Now, OS = 0.2 AC AB 3 3 ST  OS ST 0.2 ST 0.2   Now,









1 1 AB AC 3 3 ar ABC ₂ ₂ 9 225 1 1 ar OST _{OS ST} _{0.2 0.2} 0.04 2 2           _ _ _ _ Correct answer is (c)

16. 80 gsm paper is cut into sheets of 200 mm×300mm size and assembled in packets of 500 sheets. What will be 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 200 300 500 2 30 m 1000 1000

 

 

 Hence, total weight = 80×30 = 2400 gram 2400 2.4 kg

1000

 

(6)

17. Find the missing letter A B C D F I L O K P U Z P W D ? (a) P (b) K (c) J (d) L Soln. A F 1 6 K 11 16 P +5 +5 +5 B I 2 9 P 16 23 W +7 C L 3 12 U 21 D +9 D O 4 15 Z 26 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 3}rd _{–9, in 4}th _–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 = 800rs

3

and cost of 1 trousers = 1000 rs Hence, 800 y 1000 y 38000 3     800y 3000y 3800 3 3800y38000 3 y = 30 Correct answer is (b)

19. Consider the set of numbers {171_{, 17}2_{, ..., 17}300_{}. How many of these numbers end with the digit 3?}

(a) 60 (b) 75 (c) 100 (d) 150

Soln. Unit of 171_{= 7; 17}2_{= 9; 17}3_{= 3, 17}4_{= 1; 17}5_{= 7; 17}6_{= 9; 17}7_{= 3; 17}8_{= 1}

Hence, 173_{, 17}7_{, 17}11_{, ... will ends with 3 as unit.}

Therefore, _{4 300 75} 28 20 20 × ×

So, 75 is our answer.

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                   Correct answer is (d)

(7)

PART-B

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

NO2

O O

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 will be 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 comcom-pared 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. C O O P₂O₅ O Mechanism:

P₂O₅ is a dehydrating agent. Reacts with 1, 4-dicarbonyl compound to form substituted furan. Therefore, Paal known synthesis.

(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 2 3 4 5 II IV I 1 III III III I Correct answer is (c)

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-flip

The 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 H Trans-Decalin Flip H H Cis-Decalin H H Flip X

(9)

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.}





 





2 Red Ni2dmg _{ }Ni dmg _ Tschugaeff test The structure of complex is

C N C N Me Me O OH dmg O N Ni N O H N O H O N C C C C Me Me Me Me II Hydrogen bond Therefore, Ni – N bonds 4 Ni – OZero Hydrogen bonds two

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. N₂8H8e16 ATP2NH₃H₂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 0

1 0 1 antineutrino

P  n  e v

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.

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 .P 2 _{78º/ CS} Re d Green NiBr 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 hybridisation

3d 4s 4p

Hence, (B) will be tetrahedral.

Therefore, NiBr2 2 EtPh₂P Ni Br Br PEtPh2 Ph2EtP Ni Br Br PEtPh₂ Ph2EtP

Square planar _Tetrahedral

(A) (B)

(A) and (B) are polytopal isomers.

32. The correct non-linear and iso-structural pair is (a) SCl₂ and I₃– _{(b) SCl} 2 and I3 + _{(c) SCl} 2 and ClF2 – _{(d) I} 3 +_{and ClF} 2 – Soln. (I) SCl₂ b.p. = 2 l. p. = 2 Stearci number (S) = 4

(II) I₃ b.p. =2 l.p. = 3 S = 5 (III) I₃ b.p. = 2 l.p. = 2 S = 4 (IV) ClF₂ b.p. = 2 l.p. = 3 S = 5 Shapes of species are

S Cl Cl

••

SCl₂ Angular I I I

••

•

I₃ I I I I

••

•

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

(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’ is most dangerous

255 UV b

   

34. If L is a neutral monodentate ligand, the species,

_

AgL₄

_

2, AgL

_

₆

_

2 and AgL

_

₄

_

3, respectively are (a) paramagnetic, paramagnetic and dimagnetic

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

Soln. In



AgL₄



2 and



AgL₆



2, AgAg2+_{has d}9_{configuration, hence have unpaired electron. Hence paramagnetic.}



AgL4



3

 ₃

Ag 

 has d8_{configuration and dsp}2_{hybridization (square planar so pairing of electron). Hence,}

diamagnetic in nature.

35. Chromite ore on fusion with sodium carbonate gives

(a) Na CrO and Fe O₂ ₄ ₂ ₃ (b) Na Cr O and Fe O₂ ₂ ₇ ₂ ₃ (c) Cr CO2



3



₃ and Fe OH





₃ (d) Na CrO and Fe2 4 2



CO_{3 3}



Soln. Chromium is extracted from chromite ore:

(I) 4FeCr O2 48Na CO2 37O2 _1100ºCfusion8Na CrO2 42Fe O2 38CO2 (II) 2Na CrO₂ ₄H SO₂ ₄ Na SO₂ ₄Na Cr O₂ ₂ ₇H O₂

(III) Na Cr O2 2 72CCr O2 7Na CO2 3CO (IV) Cr O₂ ₃2AlAl O₂ ₃2Cr

36. The ligand(s) that is (are) fluxional in _



 5 C H5 5



 1 C H5 5



Fe CO





₂_ in the temperature range 221– 298K, is (are)

(a)  5 C H₅ ₅ (b)  1 C H₅ ₅ (c)  5 C H and CO₅ ₅ (d)  1 C H and CO₅ ₅

Soln.



5 C H5 5



1 C H5 5



Fe CO





₂ 221 298K



5 C H5 5



5 C H5 5



Fe 

 _{ } _{ } _{} _{ } _{ } 

   

Correct answer (b)

37.



CoL₆



3 is red in colour whereas



CoL '₆



3 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.



CoL₆



3 red colour absorbs green radiations.



CoL '6



3



 green colour absorbs red radiations.

B V R O G Y 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..

(12)

38. The oxidation state of Ni and the number of metal-metal bonds in ₂

_

_

2 6 Ni 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 6 Ni CO 2x 0 2         Ni N CO CO CO OC OC OC –2 2 x 2   _{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 Ph P Ph Ph

This is due to size of central atom. ‘Sb’ is bigger than ‘P’, therefore favour SP while small sized ‘P’ favour TBP.

40. The typical electronic configurations of the transition metal centre for oxidative addition (a) d0_{nad d}8 _{(b) d}6_{and d}8 _{(c) d}8_{and d}10 _{(d) d}5_{and d}10

Soln. Most commonly the metal in the complexes in their low oxidation state with d8_{or d}10_{configuration undergo}

oxidative addition. Ir Ph₃P PPh₃ OC Cl +H₂O Ir Ph₃P PPh₃ OC H H Cl ₍₊₃₎ (+1) (16 electron)  







 



0 PhBr 3 ₄ _2PPh3 3 ₂ 18 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 pK_a values of the following salt of aspartic acid are indicated below. The predominant species that would exist at pH = 5 is

(pK_a = 9.9) H₃N COOH

COOH

(pK_a = 2.0)

(13)

(a) H3N COO COOH (b) H₃N COO COO (c) H₂N COO COO (d) H₃N COOH COO Soln. H3N COOH COOH (pK_a = 9.9) (pK_a = 2.0) (pK_a = 3.9)

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

When pH = 5; since both COOH group have pKa = 2.0, pK_a = 3.9. It will be deprotonated form. Thus the predominant species that would exist at pH = 5 is

H₃N COO

COO

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

t-Bu t-Bu t-Bu hv (a) t-Bu H t-Bu t-Bu (b) t-Bu t-Bu t-Bu (c) t-Bu H t-Bu t-Bu (d) t-Bu t-Bu t-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 it

45. The major product formed in the following reaction is Me 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 AcOH Ph 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 Me

N

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 less prone 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 donating electron pair through its lone pair.

Thus it will be most prone to electrophilic attack. Correct order will be I > III > II

Correct option is (d)

47. The commutator of the kinetic energy operator, T and the momentum operator, ˆ_x ˆp for the one-dimensional_x case is (a) i (b) i d dx  (c) 0 (d) i x Soln. _Tˆ_x, pˆ_x_ 2 2 2, 2      _  _       i m x x  2 2 2 2 2 2 2m x i x i x 2m x            _  _      _  _     2 2 2 2 2 2 i i m x x m x x x                  2 3 2 3 3 3 2 2 i i m x m x            





3 3 3 3 3 3 0 zero 2 2 i i m x m x            Correct answer is (c)

(15)

48. The major product formed in the reaction of trans-1-bromo-3-methylcyclobutane with sodium iodide in DMF is (a) Me (b) Me (c) Me I (d) Me I Soln. Me Br NaI DMF cis Me Br NaI DMF 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 NaI DMF Me I Correct 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 a donar band that lies close to empty conduction band.

VB = Valence bond, _CB

DB = Donar bond, _DB

CB = Conduction bond VB

50. The symmetry point group of propyne is

(a) C₃ (b) C_3V (c) D₃ (d) D_3d Soln. C H H H C C

H Plane and axes C_3V

H

(16)

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

 

t 0 A ln A        

vs time, where

 

A ₀ and

 

A _t 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 A ln A         axis. Soln.

 

0 0 f 1 _t A c A ln k ln c A      _ _         _ _

 

t 0 A ln 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 of propagating 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 and the chain termination step.

Mathematically it is defined as Kinetic chain length =

rate of product formation rate of initial step Correct 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 symmetry

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 given}

above the arrow mark

(A) P105 Q108 R (B) P105 Q103 R (C) P107 Q107 R (D) P102 Q106 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 reactions

1 2

k k

PQR

is applied when first step is slow and second step is fast.

(17)

56. _{The figure below represents the path followed by a gas during expansion from A}__B. The work done is (L atm) A 5 4 3 2 1 1 2 3 4 5 P(atm) V(L) B (a) 0 (b) 9 (c) 5 (d) 4 Soln. W  P V W P V



₂V₁



A – B Work A  I I B AI Work, P = 5 atm V = 1 to 1 W = 5(1–1) = 5(0) = 0 I B, P = 1 2 1 5 1 4 V V V      

 

1 4 4 W  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 measured in 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. 0 3 300 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 both phases are in same type.

T_ T_; P_ P_; µ_ µ_

59. The number of configurations in the most probable state, according to Boltzmann formula, is (a) _eS/kB (b) _eS/kB (c) _eE/k TB (d) _eG /k TB Soln. S k_BnW



arrangement



B S k n B S k nW / _B S k nW

(18)

/ _B S k W e

60. The correct match of the 1_{H 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. 1_{H NMR chemical shift of these species can be explained on the basis of ring current leads to deshielding}

effect.

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

Since, in cylopentadiene ring ₄__e_{are there and also it is not aromatic, thus, iits proton are least} deshielded 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 of tropylium cation positive charge is there. Thus, its proton are more deshielded as compared to benzene. Thus,

 value will be in order.. III > II > I

61. The major prorudcts formed in the following are O OEt (i) Ph3C–Na+ (ii) H₃O+ (a) O OH + Ph3C–CH2CH3 (b) CPh₃ CPh₃ + HO EtOH (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 C O OEt OEt –EtO– H₃O+ C O OEt C O C O C O OEt + EtOH Correct answer is (c)

(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. 1 2 3 4 6 5 (4E)-1, 4-hexadiene 1 2 3 4 5 6 (4Z)-1, 4-hexadiene 1 2 3 4 5 6

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

1 2 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.

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

Ph CN Ph •

NH

X 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 Y The 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 • _NH X Y Ph C Ph C N X N Y 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 OH 2 3 2 1 1  [2, 3] sigmatropic shift O S Ph (ii) (EtO)3P Correct answer is (d)

(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.

66. The maximum bond order obtained from the molecular orbitals of a transition metal dimer, formed as linear combinations of d-orbitals alone, is

(a) 3 (b) 4 (c) 5 (d) 6

Soln. A maximum five ‘d’ orbitals can form bonding with each other

M M  bonds  bonds  bonds 1 2 2 Firstly reported in 2007 in chromium complexes.

Cr Ar Ar Cr

Ar = Substituted aromatic rings.

67. The term symbol that is NOT alllowed for the np2_{configuration is}

(a) 1_D _(b)3_P _(c)1_S _(d)3_D Soln. For np2_{configurations (microstates)}

L M S

(I) _l_{= +1 0 –1} 1_D ₂ ₁ ₀

(II) 2_P ₁ ₃ ₁

(III) 1_S ₀ ₁ ₀

(IV) 3_D ₁ ₃ ₁

The configuration (IV) violets the Pauli’s Principle, Hence term symbol 3_{D 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 is 2

2

13.6 z eV

I.E. | z atomic number, n shell number atom

n 

  

This is for one electron ‘H’ atom like system Hence, for Li++_{= 1s}1_{(one electron system)}

2 2 13.6 3 I.E. 13.6 9 1 

   _{. For ‘H’ atom, I.E. =}

2 2 13.6 1 I.E. 13.6 x 1     Therefore, I.E. of Li++_{= 9x} Correct answer is (c)

(21)

69. If temprature is doubled and the mass of the gaseous molecule is halved, the rms speed of the molecular will change by a factor of (a) 1 (b) 2 (c) 1 2 (d) 1 4 Soln. ms 3 RT r M  ms T r M  _{T = 2T, m = m/2} 2 / 2  T rms M 4 rms T M 2 rms T

M (Two time increase) Correct answer is (b)

70. In the graph below, the correct option, according to Kohlrausch law, is B A 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 of concentration while that of weak electrolyte decreases exponentially. Hence, (c) is a strong electrolyte while D is a weak electrolyte.

(22)

PART-C

71. Reaction of [Ru(NH₃)₅(isonicotinamide)]3+_{with [Cr(H} 2O)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. C N O NH₂ (H₂N)₅Ru 3+ + [Cr(H2O)6]2+ Inert Labile –H2O C N O NH₂ (H₂N)₅Ru Cr(H2O)S +5 Electron exchange C N 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:



2



₆ 3



2



₆ 2 Fe H O / Fe H O          4.0 M–1 sec–1





₃ 3





₃ 2 Fe 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 enhances the 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 symmetry this leads to fast transfer of electron leading to enhance rate of reaction.





₃ 3 5 0 2g g *5 *0 Fe phen t e        





₃ 2 6 0 2g g *6 0 g Fe phen t e e       

Electron transfer takes place from *6



Fe2



 *5



Fe3



(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 M}2+_{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 6 B.O. 3 2 2     M+ = *    8 1 7 B.O. 3.5 2 2     M+2 = *    8 B.O. 4 2  

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

N NH N N H 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 19_{F NMR spectrum of ClF}

3 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 plane a b b F Cl F F

••

19 F 2F doublet 19 F 1F triplet T shaped structure

76. The low temperature (–98ºC) 19_{F NMR spectrum of SF}

4 shows doublet of triplets. It is consistent with the

point group symmetry.

(a) C_3V (b) C_4V (c) T_d (d) C_2v

Soln.

S

F F

SF₄ Sea saw shape

c₂and plane C_2V

77. Amongst organolithium (A), Grignard (B) and organoaluminium (C) compounds, those react with SiCl₄ to give compound 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 Cl Cl Cl + - + RMgX RLi R₃Al

In all case there is electrostatic interaction between Si and R + Si Cl Cl _Cl Cl XMgR R SiCl3 Correct answer is (d)

78. In its electronic spectrum,

_

₂

_

3 6 V 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) v₁3T_1g

_{ }

F 3T_2g

_{ }

F , v₂ 3 T_1g

_{ }

F 3T_1g

_{ }

P (b) v₁3T_1g

 

F 3T_1g

 

P , v₂ 3T_1g

 

F 3T_2g

 

P (c) v₁3A_2g 3T_1g

 

F , v₂ 3A_2g 3T_2g

 

F (d) v₁ 3A_2g 3T_2g

_{ }

F , v₂ 3A_2g 3T_1g

_{ }

F

(25)

Soln.

_

₂

_

3 3 2 2_2g 0 6 V H O  V 3d t eg       e_g t_2g hv e_g t_2g hv

one electron transfer 3_T 1g 3_T 1g+ 3T2g e_g t_2g 3_A 2g two electron transfer

Two transitions.

 

3 3 1 1g 2g V  T F  T F ; V2 3T1g

 

F 3T1g

 

P 3 P 3 F 3 T1g 3 T2g 3 T1g(P) 3 A2g

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.

As HNO3

Hot. and conc. H3AsO4 Arsenic acid (V) H₂SO₄

[H₃AsO₃]

H₂SO₄

As₄O₆

Arseneous acid (III)

The reason is that HNO₃ is better oxidising agent than H₂SO₄ also acts as dehydrating agent.

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 = 74 PEC = TEC – n × 12 PEC = 74–5×12 = 74–60 = 14 14 7 2 42  PEC

7 = n + 2 where, n = number of metal in electron. = 5 + 2 = (n + 2) Nido

(26)

81. 1_{H NMR spectrum of} 5





5 5 2 _{4 2} C 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

82. The nuclides among the following, capable of undergoing fission by thermal neutrons, are (A) 233_U _(B)235_U _(C)239_Pu _(D)232_Th

(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.

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



2



₆ 2 Ni H O      and





2 2 ₆ Cu 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 electronic environment so all bond length same

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

(27)

86. Reaction of nitrosyl tetrafluoroborate to Vaska’s complex gives complex A with M N O 124º  . The complex 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 NO BF₄– Ir Cl PPh3 CO Ph₃P N O 124º BF₄ 16 electron 18 electron NO = 1650 cm–1 _Bent NO = 1520 cm–1 _Linear Correct answer is (a)

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 are

Ca 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 due to 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 not involvent in -bonding with complex hence when nitrogen atom is trans to phosphorus, phosphorus become able to form efficient -bond with metal hence become stable thats why P and N are trans to each other

SCN Pd N P Me Me NCS Ph Ph

89. For a low energy nuclear reaction, 24Mg d,

_



_

22Na, the correct statements from the following are (A) Kinetic energy of d particle is not fully available for exciting 24_Mg.

(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 O H H His94 His₉₆ His119 H+ Zn O H His94 His₉₆ His119 CO₂ Zn O H His₉₄ His₉₆ His119 O C O Zn O H His₉₄ His96 His₁₁₉ O O Zn O H His₉₄ His96 His₁₁₉ O O H₂O 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.

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

(29)

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

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

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 electron

Co(CO)₃ 15 electron not isolobal R₂Si 6 electron

Mn(CO)₅ 17 electron isolobal CH₃ 7 electron

Fe(CO)₄ 16 electron isolobal O 6 electron

Mn(CO)₅ 17 electron isolobal RS 7 electron

95. The correct order of the size of S, S2–_{, S}2+_{and S}4+_{species is,}

(a) S S 2 S4 S2 (b) S2 S4 S2 S (c) 2 2 4

S   S S  S  (d) S4 S2  S S2

Soln. As positive charge increases the size decreases while with increase in negative charge increase the size. This is due to increase in Z_eff in former case while decrease in Z_eff in later case.

Hence, order of size is S2 SS2 S4

96. The major product formed in the following reaction is

Br _H n-Bu3SnH, AlBN benzene reflux H (a) H H H (b) H H H (c) Br H SnBu3 H (d) H H Soln. Br C C H n-Bu3SnH, AlBN benzene reflux H Homolytic cleavage C C H H • C C H H • nBu3SnH AIBN C H H • H H H Product

(30)

97. The correct combination of reagents to effect the following conversion is O H H COOH H H (a) (i) Ph₃P+_CH 2OMeCl –_{, BuLi, (ii) H} 3O +_{, Jones’ reagent}

(b) (i) H₂N–_{NHTs; (ii) BuLi (2 equiv); (iii) DMF}

(c) (i) H₂N–_{NHTs; (ii) BuLi (2 equiv); (iii) CO} 2

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

Soln. O H2N NHTs N H H N Ts Bu N N H Bu Li C O O   (i) (ii) H+ CO2H Correct answer is (c)

anhyd. CF₃COOH CH₂Cl₂ Bu₄N+Br– (a) Br H (b) H Br (c) H COOCF3 (d) O

Soln. anhyd. CF3COOH

CH₂Cl₂ Bu4N+Br– H+ Br H Br Correct answer is (b)

99. Consider the following reaction,

O

+ Ph–N₃ CF3COOH

N O

Ph

The appropriate intermediate involved in this reaction is

(a) O H N Ph (b) N HO H Ph (c) N O Ph (d) HO _N N 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 –N₂ N O Ph Product O N Ph Correct answer is (d)

100. The correct 13_{C NMR chemical}

 

_ _{shift values of carbons labeled a-e in the following ester are}

Me O O Me e 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 O Me e d b a c Me O O Me e 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 O O Cl MeO Et₃N (A) NH₂ (B) (a) O O A : OMe O N H O B : (b) O O A : Cl O O O B : NH O

(32)

(c) O O A : OMe O MeO B : N H O (d) O O A : Cl O B : O O O Soln. C OH O C O Cl MeO Et₃N (A) NH₂ (B)

Since less basic group are better leaving group. Thus, Cl–_{will be a better leaving group as compared to}

methoxy group.

Thus, formation of A takes place according to following mechanism.

C OH O C O Cl MeO Et₃N C O O C O O C O OMe N H O NH₂ (B) (A)

102. The biosynthesis of isopentenyl pyrophosphate from acetyl CoA involves: A. Four molecules of acetyl CoA

B. Three molecules of ATP C. Two molecules of NADPH D. Two molecules of lipoic acid The 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 two molecule of NADPH.

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

O _hv HO O O O A 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 C O tautomerisation Correct answer is (d)

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

N O O OMe O BnO H₂N NH₂ (A) heat B (a) O MeO BnO H2N A: _B: NH O BnO (b) A: N B: O O HO MeO O N O O O O (c) A: N B: O O BnO HO O O _N O O O

(34)

(d) O MeO BnO H2N A: B: HN OMe O Soln. N O O C BnO O OMe NH₂NH₂ NH C O 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 O N O C N O O N –CO2 –N₂ Benzene diazonium 2-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 amount of Pd(PPh₃)₄ furnishes a tri-substituted alkene. The intermediate and the product of the reaction, respectively, are H )₃B X (a) H Zn and H (b) H Zn and H (c) H Zn and H (d) H Zn and H Soln. H )₃B Et₂Zn H Zn I CH3 Pd(PPh3)4 _H attack from same side Correct answer is (d)

107. Using Boltzmann distribution, the probability of an oscillator occupying the first three levels (n = 0, 1 and 2) is found 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 = P_T = 1

Probability of occupying first three levels = P₀ + P₁ + P₂

P_T = P₀ + P₁ + P₂ + probability of finding an oscillator in energy levels in n3 1 = 0.952 + P_{n 3}_  1 0.9520.048

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: O N B: O OH

(36)

(b) _A: O N B: O OH (c) A: O N B: O O (b) A: O N 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

C OCH3 OCH₃ C O O C OCH3 OCH₃ C ONa ONa • • Na Na • •

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

C OCH3 OCH₃ C ONa ONa • • C OCH₃ OCH₃ C O O O O

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

(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 because by removing the alkoxide ion formed, thereby presenting the base catalyzed side reaction.

110. An organic compound gives following spectral data: IR : 2210, 1724 cm–1_,1_{H NMR :}  1.4 (t, J = 7.1 Hz, 3H), 4.4 (q, J = 7.1 Hz, 2H); 13_{C NMR :}  16, 62, 118, 119, 125, 127, 168. The compound is (a) NC O O (b) O O NH2 (c) _NC O O (d) O CN O

Soln. IR rate indicates towards a _C _N group ( ~ = 2210 cm–1_{) and a carbonyl group}

C O (~ = 1724)





 





3 2 1.4 , 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 q J, 7.1Hz, 2H b CH – CH – O – EWG

   

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 COOCH₃ H₃COOC + Ph Ph  (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 COOCH3 H3COOC + Ph C C Ph  N Ph Ph Ph H₃COOC _C OCH3 O Product Correct answer is (b)

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

A O O O O B

(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] cycloaddition O₂ Rose Bengal, hv O O KOOC N N COOK AcOH O O

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

O

O H2/Pd

OH

Thus, it will not be the suitable reagent.

(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} Correct answer is (d)

114. Consider a particle confined in a cubic box. The degeneracy of the level, that has an energy twice that of the lowest 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) 12h 8m 2 2 l 11h 8m 2 2 l 9h 8m 2 2 l 6h 8m 2 2 l 3h 8m 2 2 l

Cubic box energy, _x _y _z





2 2 2 2 n n n ₂ x y z h n n n 8m      2 2 6h

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

2 3h

8m (ground state) So, degeneracy is 3

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

O

(i) NaNH2

(40)

O O

O

(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 O C H₂ 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.

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

O COOMe MeOOC Heat A (a) COOMe MeOOC H H O (b) O COOMe MeOOC

(41)

(c) COOMe MeOOC O H H (d) COOMe COOMe OH Soln. O C C COOMe MeOOC Heat MeOOC MeOOC O H H

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 : N O (b) _{A :} CHO B : O O (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 be oxidized easily to  , unsaturated carbonyl compound if desired.

The oxidation are belived to involve an ene reaction, between the alkene and the hyrated form of the dioxide, followed by a [2, 3]-sigmatropic rearrangement of the resulting allyl seleninic acid and final

hydrolysisof Se(II) ester to the allylic alcohol. Further, oxidation of the alcohol gives the  , unsaturated carbonyl compound. A notable application of this reaction is in the oxidation of 1, 1 dimethyl alkene to the corresponding E allylic alcohols or aldehydes by selective attack on the E-methyl group.

H ene reaction Se HO OH OH Se O OH OH –H₂O Se HO O OH [2. 3]-sigmatropic rearrangement O Se HO OH CHO further oxidation

(42)

CHO Aq. NaCN MnO2 _CH O–Na+ CN Cyanohydrin 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 manganese dioxide gives the acyl nitrile. Which then reacts with the alcohol solvent to give the ester.

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 2 2      (d)

   

1 1 2 1 2 2     

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 2

2      

119. The number of phases, components and degrees of freedom, when Ar is added to an equilibrium mixture of NO, 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  2

 

g _g _g

NO O _NO _add Ar g _ (1) Phase only (gas phase) all are gases

P = 1 (2) Component C = N–E C = 4–1 = 3 (3) F = C–P+2 F = 3–1+2 = 4 Correct answer is (c)

OH

TsCl pyridine