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CSIR-NET-JRF Dec-2012 Chemical SciencesPaper-Code(A) with solutions

Part-A

1. Which of the following numbers is the largets?4 3 4 2 3 23 4 2 4 2 32 , 2 , 3 , 3 , 4 ,4

(a) 432 (b)

243 (c) 234 (d) 324

Soln. 43 812 2 ; 34 642 2 ; 42 163 3 ; 24 163 3 , 32 8 164 4 2 ; 23 9 184 4 2

Correct answer is (a)

2. The cube ABCDEFGH in the figure has each edge equal to a. The area of the triangle with vertices at A, C andF is:

A

D C

G

FE

H

B

(a) 23 a4

(b) 23 a2

(c) 23 a (d) 22 3 a

Soln.A C

F

a 2

a 2 a 2

Area of equilated triangle ACF = 23 side4

= 23 a 24

= 23 a2

Correct answer is (b)

3. What is the number of distinct arrangements of the letters of the word UGCCSIR so that U and I cannotcome together?(a) 2520 (b) 720 (c) 1520 (d) 1800

Soln. Total number of arrangements = 7! 25202!

Total number of arrangment when UZ comes together = 6! 2 7202!

Desire arrangement = 2520 –720 = 1800Correct answer is (d)

4. Suppose the sum of the seven positive numbers is 21. What is the minimum possible value of the average of thesquares of these numbers?(a) 63 (b) 21 (c) 9 (d) 7

Soln. Lets number are a, b, c, d,e, f, g

Given: a + b + c + d + e + f + g = 21

We know that Arithmetic Progression Geometrical Progression.

2

Arithmetic Progression = Geometrical Progression (When all the numbers are equal)

Two or more positive numbers which sum is constant, then the square of numbers minimum when the numbersare equal.

a + b + c + d + e + f + g = 21 a + b + c + d + e + f + g 3

7

For the given requirement a = b = c = d = e = f = g = 3 2 2 2 2 2 2 2a b c d e f g 9

7

Correct answer is (c)

5. Let 13 13 13 13 13 13 13 13 13 13 13 131 2 3 ...... 100 1 3 5 ...... 99 2 4 6 ..... 100A , B , C

100 50 50

Which of the following is true?(a) B < C < A (b) A < B < C (c) B < A < C (d) C < A < B

Soln. 13 13 13 1350B 1 3 5 ...... 99 ... (i)13 13 13 1350C 2 4 6 ...... 100 ... (ii)

Each term of (ii) is greater to term of (i)Then 50C > 50 B. So, C > B

13 13 13 13100A 1 2 3 ........10 100A 50B 50C; 2A B C Thus, 2A 2B and 2A 2C

A B A C


6. A circle of radius 5 units in the XY plane has its centre in the first quadrant, touches the x-axis and has a chordof length 6 units on the y-axis. The coordinates of its centre are(a) 4, 6 (b) 3, 5 (c) 5, 4 (d) 4, 5

Soln. Let co-ordinate of the centre is O(h, 5)

Q

P

O

x

I quadrant

y

553M3 5

Applying Pythagoras theorem is in right angle triangle QOM2 2 2 2 2 2OQ OM MQ ; 5 h 3 ; 2h 25 9; h 4

Cente is lie in first quadrant then negative value is not possible.Correct anwer is (d)

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7. A wire of length 6cm is used to make a tetrahedron of each edge 1m, using only one strand of wire for eachedge. The minimum number of times the wire has to be cut is

(a) 2 (b) 3 (c) 1 (d) 0

Soln.

A

B C

D

It required atlest a single cut. First wire ABCDB = 5 cm, second wire AD = 1 cmCorrect answer is (c)

8. If the sum of the next two terms of the series below is x, what is the value of log2x?2, –4, 8, –16, 32, –64, 128, ……..(a) 128 (b) 10 (c) 256 (d) 8

Soln. It is a geometrical series first term a = 2, Geometrical ratio = –2T8 = ar7 = 2(–2)7 = –256log of negative number not possible.Correct answer is (none).

9.30º

A conical vessel with semi-vertical angle 30º and height 10.5 cm has a thin lid. A sphere kept inside ittouches the lid. The radius of the sphere in cm is(a) 3.5 (b) 5 (c) 6.5 (d) 7

Soln. r

QO

90

P

30

Let radius of sphere is ‘r’ then OP = 10.5 – r

In triangle POQ ; OQsin30OP

; r0.5

10.5 r

; r = 3.5


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10. Amar, Akbar and Anthony are three friends, one of whom is a doctor, another is an engineer and the third is aprofessor. Amar is not an engineer. Akbar is the shortest. the tallest person is a doctor. The engineer’s height isthe geometric mean of the heights of the other two. Then which of the following is true?(a) Amar is a doctor and he is the tallest (b) Akbar is a professor and he is the tallest(c) Anthony is an engineer and he is shortest (d) Anthony is a doctor and he is the tallest

Soln. Correct answer is (a)

11. If 100 cats catch mice in 100 minutes, then how long will it take for 7 cats to catch 7 mice?(a) 100/7 minutes (b) 100 minuts (c) 49/100 minutes (d) 7 minutes

Soln. Correct answer is (b)

12. What does this diagram demonstrate?

(a) n n 1

1 2 3 .... n2

(b) 2 2 2 2 n n 1 2n 1

1 2 3 .... n6

(c) 21 3 ...... 2n 1 n (d) 22 2 2.n n 1 2n 12 4 .... 2n

3

Soln. The each half diagram demonstrate gradually increasing of unit square by one unit.Correct answer is (a)

13. Suppose therea are socks of N different colors in box. If you take out one sock at a time, what is the maximumnumber of socks that you have to take out before a matching pair is found? Assume that N is na even number.(a) N (b) N + 1 (c) N–1 (d) N/2

Soln. Socks in pair thus atleast ‘n+1’ number of socks would be taken out.Correct answer is (b)

14. At what time after 4 O’ clock, the hour and the minute hands will lie opposite to each other?(a) 4 50' 30" (b) 4 52' 51" (c) 4 53' 23" (d) 4 54' 33"

Soln. Correct answer is (d)

15. Which of the following curves just touches the ‘x’ axis?

(a) 2y x x 1 (b) 2y x 2x 2 (c) 2y x 10x 25 (d) 2y x 7x 12

Soln. Curve touch the ‘x’ axis where y = 0 and for that condition atleast a real ‘x’ should be exists.Equation, 2y x 10x 25 = 0, (x–5)2 = 0, x = 5.Correct answer is (c)

16.

C D

BA

O

If AB is parallel to CD and AO = 2OD, then the area of triangle OAB is bigger than the area of triangle OCDby a factor of(a) 1 (b) 3 (c) 4 (d) 8

Soln. Let OD = x; CD || AB. Thus, COD AOB & DCO ABO OCD & OBA are symmetric triangle.

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Thus,2Area of triangle OAB 2 4

Area of triangle OCD 1

Correct answer is (c).

17. *

A semi-circular arch of radius R has a vertical pole put on the ground together with one of its legs. An ant on thetop of the arch finds the angular height of the tip of the pole to be 45º. The height of the pole is

(a) 2 R (b) 3 R (c) 4 R (d) 5 R

Soln. Correct answer is (c)

18. Suppose we make N identical smaller spheres from a big sphere. The total surface area of the smaller spheresis X times the total surface area of the big sphere, where X is(a) N (b) 1 (c) N1/3 (d) N3.

Soln. Let big sphere is ‘R’ and small sphere radius is ‘r’.

Thus, 3 34 4R N r3 3 ;

33 3 r 1R Nr ;

R N

2 24 R X N 4 r ; 2 2R X Nr ; 2 13 3

2rX N N N NR


19. What is the next number in the sequence 24, 30, 33., 39, 51, .......?(a) 57 (b) 69 (c) 54 (d) 81

Soln.

24 3033 39 51 ...................

2+4=(+6)3+0=(+3) 3+3 = (+6) 3+2 = (+12) 5+1=(+6)

Thus the next number must be 51 + 6 = 57Correct answer is (a)

20. Four lines are drawn on a plane with no two parallel and no three concurrent. Lines are drawn joining the pointsof intersection of the previous four lines. The number of new lines obtained this way is:(a) 3 (b) 5 (c) 12 (d) 2

Soln.

(3)

(2)

(1)

(4)P Q R

UT S

New lines, PU, QS, RTT


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Part-B

21. For an odd nucleon in ‘g’ nuclear orbital and parallel to I, spin and parity are(a) 9/2 and (+) (b) 7/2 and (+) (c) 9/2 and (–) (d) 7/2 and (–)

Soln. In g nuclear orbital total subshell is nine.+½

One odd nucleon finding in g-subshellTotal spin = 9/2

4Parity 1 1

Correct answer is (a)22. For the deposition of Pb by electroplating, the best suited compound among the following is

(a) PbCl2 (b) PbSO4 (c) Pb(Et)4 (d) Pb(BF4)2.

Soln. PbCl2 and PbSO4 are ionic compounds and insoluble in cold water. Therefore, cannot be used for deposi-tion of Pb and Pb(Et4) is toxic.Correct answer is (d)

23. Appropriate reasons for the deviation form the Beer’s law among the following are(A) Monochromaticity of light (C) Very high concentration of analyte(B) Association of analyte (D) Dissociation of analyte.(a) A, B and D (b) B, C and D (c) A, C and D (d) A, B and C

Soln. • Beer’s law is subjected to certain real and apparent deviation.

• Real deviations are most usually encountered in relatively concentrated solutions of the absorbing com-pound (> 0.01 M). These deviations are due to interactions between the absorbing species and to atterationsof the refractive index of the medium.• Most common are the apparent deviations. These deviations are due to:Chemical reasons arising when the absorbing compound dissociates, associates or reacts with a solvent toproduce a product having a different absorption spectrum.• Strict adherence to Beer’s law is observed only with truly monochromatic radiation.Correct answer is (b)

24. Which one of the following shows the highest solubility in hot concentrated aqueous NaOH?(a) La(OH)3 (b) Nd(OH)3 (c) Sm(OH)3 (d) Lu(OH)3.

Soln. Since size of Lu3+ is smallest, therefore Lu(OH)3 complex most easily formed with NaOH and dissolves.

3

3 6Lu OH 3OH Lu OH

Correct answer is (d)

25. In the vibrational spectrum of CO2, the number of fundamental vibrational modes common in both infraredand Raman are(a) Three (b) Two (c) One (d) Zero

Soln. CO2 has centre of symmetry so on per exclusion principle all the IR active vibration will be Raman inactive.Hence, fundamental vibrational modes common in both IR/ and Raman will be zero.Correct answer is (d)

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26. The light pink color of 2

2 6Co H O

and the deep blue color of 24CoCl are due to

(a) MLCT transition in the first and d-d transition in the second(b) LMCT transition in both(c) d-d transitions in both(d) d-d transition in the first and MLCT transition in the second.

Soln. [Co(H2O)6]2+ t2g5 eg2

eg

t2g

electron transition fromt2g to eg. Therefore, d-d transition.

[CoCl4]2– e4 t23

t2

e

electron transfer from e to t2therefore, d-d transition.


27. In 2

2 2 6Mo S

cluster the number of birdging 22S and coordination number of Mo respectively, are

(a) 2 and 8 (b) 2 and 6 (c) 1 and 8 (d) 1 and 6

Soln. 2

2 2 6Mo S

The structure of 2

2 2 6Mo S

cluster is:

Mo

S

S

SS

S

Mo

S S S

S

S

S

S

So, the bridging S22– is two and co-ordination number of Mo is 8.


28. 1H NMR spectrum of HD would show(a) a singlet (b) a doublet(c) a triplet with intensity ratio 1:2:1 (d) a triplet with intensity ratio 1:1:1

Soln. Duterium D I = 1

Multiplicity = 2nI + 1 = 2×1×1+1=3Where, n = number of duterium atom

I = spin of duterium atom.Intensity ratio will be non-pascal.Hence, 1H NMR of HD will appear as triplet with intensity ratio 1:1:1.Correct answer is (d)

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29. The number of possible isomers of 3 22Ru PPh acac acac acetylacetonate is:(a) 2 (b) 3 (c) 4 (d) 5

Soln.Ru

PPh3

PPh3

acac acac

Trans

Ru

PPh3

PPh3

acac

acac

Ru

PPh3

PPh3

acac

acaccis-d

Mirrorcis-l


30. The total number of Cu–O bonds present in the crystalline copper(II) acetate monohydrate is:(a) 10 (b) 6 (c) 8 (d) 4

Soln.

Cu

Cu

OH2

OH2

O

O

O

O

H3CC

H3CC

O

O

O

O

CCH3

CCH3

10 Cu–O bonds

Correct answer is (a).

31. The electronegativity differences is the highest for the pair(a) Li, Cl (b) K, F (c) Na, Cl (d) Li, F

Soln. Among these elements K is least electronegative and F is most electronegative. Therefore electronegativitydifference is highest for the pair K, F.


32. Which ones among 23 3 3 3CO , SO , XeO and NO and NO3

– have planar structure?

(a) 23 3 3CO , SO and XeO (b) 3 3 3SO , XeO and NO

(c) 23 3 3CO , XeO and NO (d) 2

3 3 3CO , SO and NO

Soln.CO3

2–

O C

O

O

sp2-hybridization(Trigonal planar)

SO3

S

O

O O


Xe

OO

O

XeO3

sp3-hybridization(Trigonal pyramidal)

NO3–

N

O

O


O

23 3 3CO , SO and NO have planar structure.


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33. The substitution of 5 Cp group with nitric oxide is the easiest for

(a) 52Cp Fe (b) 5

2Cp CoCl (c) 52Cp Ni (d) 5

2Cp Co

Soln.

NiNO

Ni

N

O

Stable complex

20 e– 18e–


34. The molecule

(CO)5M C

OCH3

Phobeys 18 e rule. The two ‘M’ satisfying the condition are(a) Cr, Re (b) Mo, V (c) V, Re (d) Cr, V

Soln. (OC)5M COCH3

Ph

2 × 5 + M + 2 = 18, M = 18–12 = 6So, M = Cr, and Re+.Correct answer is (a)

35. The correct ser of the biologically essential elements is,(a) Fe, Mo, Cu, Zn (b) Fe, Cu, Co, Ru (c) Cu, Mn, Zn, Ag (d) Fe, Ru, Zn, Mg

Soln. (i) Fe, Mo, Cu, Zn

MetalloenzymesHemocyanin

Hb & Mbcytochromes

Nitrogen fixation


36. The number of lines exhibited by a high resolution EPR spectrum of the species,[Cu(ethylenediamine)2]2+ is [Nuclear spin (I) of Cu = 3/2 and that of N = 1](a) 12 (b) 15 (c) 20 (d) 36

Soln. Cu

N

NCH2

CH2N

N

H2C

H2C

2+ N

Cu

N

Cu

n number of equivalent nitrogen atomsn number of Cu atomI spin of one nitrogenI spin of copper

10

s1m2

, NmI 1 ; Cu3mI2

Multiplicity N N N Cu32n I 1 2n I 1 2 4 1 1 2 1 1 9 4 362


37. Degradation of penicillin G

N

SHN

OPh

OOH

O

H H

gives penicillamine that can utilize nitrogen, oxygen or sulfur atoms as donors to bind with lead(II), mercury(II) or copper(II). The structure of penicillamine is

(a) H2N

SH

O OH

(b)

HN

OPh

O OH

SHH

(c)

O

HN SHHO

(d)

H2N

HO

O HN

S

OHO

H H

Soln. N

SHN

OPh

OOH

O

H H

H2N

SH

OH

O

donor atom

donor atomdonor atom


38. The molecular that has an S6 symmetry element is(a) B2H6 (b) CH4 (c) PH5 (d) SF6

Soln. B2H6 has D2h point group and S2 axis

CH4 has Td point group and S4 axisPH5 has D3h point group and S3 axisSF6 has Oh point group andS6 axisSF6 has S6 symmetry element.Correct answer is (d).

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39. The electric dipole allowed transition in a d2 atomic system is(a) 3 1F D (b) 3 1F P (c) 3 3F D (d) 3 3F P

Soln. Electric dipole allowed transition in d2

For electronic transition selection rule, spin multiplicity S 0 ; 1, J 0, 1 3 3F D allowed transition because S 0, 1 Correct answer is (c)

40. When a hydrogen atom is placed in an electric field along the y-axis, the orbital that mixes most with the groundstate 1s obrital is(a) 2s (b) 2px (c) 2py (d) 2px

Soln. When a hydrogen atom is placed in an electric field along the y-axis. The magnetic moment i.e. electron densitywill be oriented along y-axis and therefore the orbital that mixes most with the ground state 1s orbital is 2py.


41. For water, 1vapH 41 kJ mol . The molar entropy of vaporization at 1 atm pressure is approximately

(a) 410 J K–1 mol–1 (b) 110 J K–1 mol–1 (c) 41 J K–1 mol–1 (d) 11 JK–1 mol–1.

Soln. For water vapH 41 kJ / mol

Molar entropy of vaporization, vap

b

Hs

T

For water Tb = 373 K (at 1 atm)3

1 1 1 141 10 J / mols 109.9 Jk mol 110 Jk mol373

Correct answer is (b).

42. If A and B are non-commuting hermitian operators, all eigenvalues of the operator given by the commutator [A,B] are(a) complex (b) real (c) imaginary (d) zero

Soln. The eigenvalues of an anti-Hermitian operator are either purely imaginary or equal to zero. If A and B are non-commuting hermitian operator. So, all eigenvalues are purely imaginary or zero. [Theorem: 2.4 J.T.Lee, Page101].

43. The value of commutator 2xx, p is given by

(a) 2i (b) 2i (c) 2i x (d) x2i p

Soln. The value of n n 1xx.p ni px

2x x x x x x x x x xx, p ] x, p p x, p p p x, p i p i p 2i p


44. The correlation coefficient between two arbitrary variables x and y is zero, if

(a) xy yx (b) 22x x (c) 22y y (d) xy x y

Soln. If two arbitrary variables x and y then measure the intensity of correlation coefficient

2 22 2

n xy x yr

n x x n y y

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If xy x y xy x y r 0They may be related in a non-linear fashion.Correct answer is (d)

45. A carnot takes up 90 J of heat from the source kept at 300K. The correct statement among the following is(a) It transfers 60 J of heat to the sink at 200K(b) It transfers 50 J of heat to the sink at 200K(c) It transfers 60 J of heat to the sink at 250 K(d) It transfers 50 J of heat to the sink at 250 K

Soln. c hh h

h c

Q TQ 90J T 300K 1 1Q T

c ch h

h c h c

Q QT T1 1 1Q T Q T

60 200 6 2 2 290 300 9 3 3 3


46. The relative population in two states with energies E1 and E2 satisfying Boltzmann distribution is given by

1 2 1 2 Bn /n 3/2 exp E E /k T . The relative degeneracy g2/g1 is:(a) 2 (b) 2/3 (c) 3/2 (d) 3

Soln. The relative population in two states with energy E1 and E2 satisfy Boltzmann distribution law, 1 2E E

1 1 kBT

2 2

n g en g

where, g1 and g2 are degeneracy 1

2

g 3g 2

E2

E1The value of 2

1

g 2g 3


47. The Daniel cell is

(a) 2 2I IIPt s | Zn s | Zn aq || Cu aq | Cu s | Pt s

(b) 2I IIPt s | Zn s | Zn aq || Ag aq | Ag s | Pt s

(c) 2 2I IIPt s | Fe s | Fe aq || Cu aq | Cu s | Pt s

(d) 2I 2 2 4 IIPt s | H s | H SO aq || Cu aq | Cu s | Pt s

Soln. Daniel cell consists of two half cells. The half cell on the left contains a zinc metal electrode dipped in ZnSO4solution. The half cell on the right consits of copper metal electrode in a solution of CuSO4. The half cell joinedby a salt bridge.

Oxidation half reaction, 2Zn s Zn aq 2e

Reduction half cell, 2Cu aq 2e Cu s

__________________________________________

Net 2 2Zn s Cu aq Zn Cu s Correct answer is (a).

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48. If the concept of half-life is generalized to quarter-life of a first order chemical reaction, it will be equal to(a) n 2/k (b) n 4/k (c) 4/k (d) 1/4k

Soln. Quarter life is the time in which the concentration decreases to 14 th of its initial value.

Rate law expression for first order reaction is

akt na x

... (1)

Therefore, Quarter life, 14

3ax & t t4

So, (1) becomes,

14

a akt n n n 43a a/4a4

; 14

n4tk

Therefore, correct answer is (b)

49. Kohlrausch’s law is applicable to a dilute solution of(a) Potassium chloride in hexane (b) Acetic acid in water(c) Hydrochloric acid in water (d) Benzoic acid in benzene

Soln. Due to Kohlrausch’ law at infinite dilution, when dissociation is complete, each ion makes a definite contribu-tion towards molar conductance of the electrolyte irrespective of the nature of the other ion with which it isassociated and that the molar conductance at infinite dilution for any electrolyte is given by the sum of thecontribution of the two ions. Only hydrochloric acid in water completely dissociation and contribution withequal ion.Correct answer is (c).

50. A dilute silver nitrate solution is added to a slight excess iodide solution. A solution of AgI is formed whosesurface adsorbs.

(a) I (b) 3NO (c) Na (d) Ag

Soln. A dilute solution of siliver nitrate is add a slight excess of a dilute solution of sodium iodide, a negatively chargedsolution of silver iodide is formed. This is due to the adsorption of iodide ions.

AgI

Na+

I–

I–

I–

I–

I–

I–I– I–

I–

I–

I–

I–

Na+

Na+

Na+

Na+Na+

Na+

Na+

This is the electrical property of colloid based on the concept of electrical double layer

3 3AgNO NaI AgI NaNO The ions preferentially adsorbed on the surface of a particle of a colloidal system are called potential determin-ing ions. The negatively charged surface of AgI particle attracts the positive ions (Na+) and repels the negativeions (NO3

–). The positive Na+ ions tend to form a compact layer in the vicinity of the potential determining I–

ion layer. This is called stern layer. The ion present in the stern layer are called counter ions.


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51. The absorption spectrum of O2 shows a vibrational structure that becomes continuum at 56875 cm–1. At thecontinuum, it dissociates into one ground state atom (Og) and one excited state atom (Oe). The energy differ-ence between Oe and Og is 15125 cm–1. The dissociation energy (in cm–1) of ground state of O2 is:

(a) 5687515125 (b)

1512556875 (c) 72000 (d) 41750

Soln.

Ener

gy

Illustration of dissociation

EexD0

D0 Dg

continum

Dissociation energy, gD 56875 15125 41750


52. The angle between the two planes represented by the Miller indices (1 1 0) and (1 1 1) in a simple cubic latticeis:(a) 30º (b) 45º (c) 60º (d) 90º

Soln. The angle between two planes having Miller Indices 1 1 1 2 2 2h k & h k is given by

1 2 1 2 1 22 2 2 2 2 21 1 1 2 2 2

h h k kcosh k h k

2 2 2 2 2 2

1 1 1 1 0 1 2 2cos2 3 31 1 0 1 1 1

cos 0.816 ; 1cos 0.816

35.27º

Actually all options are incorrect.

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53. The correct representation of the variation of molar conductivity (y-axis) with surfactant concentration (x-axis)is [CMC = critical micelle concentration].

(a)

CMC

x

y

(b)

CMC

x

y

(c)

CMC

x

y

(d)

CMC

x

y

Soln. The molar conductance of anionic surfactant of the type Na+R– in water is plotted against the square root ofthenormality of the solution. The curve obtained, instead of being the smoothly decreasing curve characteristic ofionic electrolytes of this type, has a shart break in it, at low concentrations. This sharp break in the curveaccompanied by reduction in the conductance of the solution, indicating a sharp increase in the mass per unitcharge of the material in solution, is interpreted as evidence for the formation of micelles at that point from theunassociated molecules of surfactant with part of the charge of the micelle neutralized by associated counterions. The concentration at which this phenomenon occurs is called the critical micelle concentration (CMC)

The effect of concentration of electrolyte is given by

ilog CMC a log c b

For homologous ionic surfactant

log CMC A BN [Refer: PSP, page 1114, 1115, 1116]Correct answer is (b)

54. The major product formed in the following reaction is

N2Ph

Me

O

Ag2O

MeOHhv

(a) OMePh

Me O

(b) OMePh

Me O

(c) O

Me

Ph

(d) OMePh

Me

O

16

Soln. Ph N2

Me

O

Ag2O

MeOH or hv–N2

PhCH

Me

O rearrangement C OPh

Me

MeOH

Wolf

C OPh

Me

HOMe

protontransferC OHPh

Me

OMe

Ph OH

Me OMe

Ph

Me

O

OMe


55. If the pKa value for p-methoxybenzoic acid is 4.46 and that of benzoic acid is 4.19, the para for methoxy groupis:(a) 8.65 (b) 4.32 (c) 0.27 (d) – 0.27

Soln.P–OMe

H

klogk

P–OMe Hlog k log k = H P–OMe a P OMeHlog k log k pk pk = 4.19 - 4.46 = – 0.27Correct answer is (d)

56. The biosynthetic precursor of cadinene is:

H

H

cadinene

(a) shikimic acid (b) mevalonic acid (c) arachidonic acid (d) prephenic acid.

Soln. Correct answer is (b)

57. The correct order of acidity of the compounds A – C is:

O

O

O

O

O

OHO

HO

A B C

(a) A > B > C (b) B > C > A (c) C > A > B (d) B > A > C

17

Soln.

O

O

–H+

O

O

O

O

–H+

O

O

Here Phenyl ring increasesThe E.W. tendency

OHO

HO O

squaric acid is acidiclike H2SO4

–H+

OO

O O

Sqaric acid ion is aromatic in nature.

Acidity E.W. tendency..So, the correct order of acidity is C > A > B.Correct answer is (c).

58. The mechanism involved in the following conversion is:

OHN

OCOOEt

Ph piperidine+ H2N

Ph

COOEt

+ CO2

(a) E2-elimination (b) E1-elimination (c) syn-elimination (d) E1 cb-elimination.

Soln.

O

HN

O

Ph

COOEt

peperidineO

HN

O

Ph

COOEt

Carbanion(stable)

E1CB

+ H2NPh

COOEt+ CO2


18

59. The correct statement(s)-A-D are given for the following reaction. The correct one(s) is (are)

MeO2S2

F

+NH

ODMSOK2CO3

N OMeO2S

(a) aromatic ipso substitution reaction (b) aromatic nucleophilic substitution(c) aromatic electrophilic substitution (d) aromatic free radical substitution.

Soln. MeO2S

F

+

N

O

H

DMSOK2CO3

NMeO2S O

It is an aromatic ipso substitution reaction and aromatic nucleophilic substitution.Correct answer is (b)

60. The following photochemical transformation proceeds through

Ph

O Ph

Mehv

N

OH

Ph

Ph

(a) Norrish type I reaction (b) Norrish type II reaction(c) Barton reaction (d) Paterno-Buchi reaction

Soln.

O

PhN

Me

Phhv

O

PhN

Ph

CH3•

•OH

PhN

Ph

CH2

••

1, 4-biradical N

PhOH

Ph

Norrish-type-II reaction


61. A tripeptide gives the following products on Edman degradation.

HN

N

Ph

O

Ph

S + H2N

HN

OH

O

O

The tripeptide is(a) Phe-Ala-Gly (b) Phe-Gly-Ala (c) Ala-Gly-Phe (d) Gly-Ala-Phe

19

Soln. HN

N

Ph

S O

Ph

+HN

H2N

O

OH

O

H2N

OH

O

Ph

Phe

OHH2N

O

Ala

+H2N

O

OH

Gly

Since, Edman degradation occur from N–T–AA. So, the sequence is Phe–Ala–Gly.Correct answer is (a).

62. In the 1H NMR spectrum recorded at 293 K, an organic compound (C3H7NO), exhibited signals at 7.8(1H, s), 2.8 (3H, s) and 2.6 (3H, s). The compound is

(a)

O

H NMe2(b)

H

N NMe2Me (c)

O

NH

MeMe (d)

NH

Me OMe

Soln. At lower temperature DMF remains in form-2 prefentially

C

O

NHCH3

CH3

C N

O

H

CH3

CH3

Form-1 at higher tempeature

Form-2 at 293 K

In form-2 methyl groups are chemically non-equivalent and they appears as singlets at 2.8 and 2.6 ppm.Singlet at 7.8 ppm appears due to olefinic proton of form-2.Correct answer is (a)

63. In the IR spectrum of p-nitrophenyl acetate, the carbonyl absorption band appears at(a) 1670 cm–1 (b) 1700 cm–1 (c) 1730 cm–1 (d) 1760 cm–1.

Soln. C = O structure occurs at cm–1.1750 – 1735 for aliphatic esters.1740 – 1715 if C = O conjugated with aromatic.1765 – 1762 if oxygen atom is conjugated with alkene or aromatic.Example: Phenyl acetate 1765 cm–1

p-nitrophenyl acetate 1761 cm–1.

O C

O

CH3O2N


20

64. The absolute configuration at the two chiral centres of (–)–camphore is:

O1

4(a) 1R, 4R (b) 1R, 4S (c) 1S, 4R (d) 1S, 4S

Soln. CIP Priorities

H3C CH3

CH3

H

O1

4

•

C

CCC

CC

CC

O

O•

C

CCC

CC

CC

O

O

CCC

1 43

1

1

2 2

4th in backR-configuration

4th in frontS-configuration



NH

1. Me2NCHOPOCl3

2. H3O+

(a) NH

CHO

(b) NH

OHC

(c) NH

CONMe2

(d) NH

Me2NOC

Soln.

N

H

Position 3 is more susceptible for electrophilic attack in the case of indole.Since, it is Vilsmeyer reaction.So, the major product will be

N

H

N

H

O

H

O

H NCH3

CH3

POCl3H3O+


21

66. The first person to separate a racemic mixture into individual enantiomers is(a) J, H van’t Hoff (b) Pasteur (c) H.E. Fischer (d) F. Wohler

Soln. L.Pasteur has isolated individual isomers first time. Pasteur separated the left and right crystal sphapes fromeach other to form two piles of crystals.

Mirror image crystals


67. Consider the following statements for [18]-annulene(A) It is aromatic(B) The inner protons resonate at 9.28 in its 1H NMR spectrum(C) There are six protons in the shielded zone.(a) A, B, C (b) A and B only (c) B and C only (d) A and C only

Soln. [18] annulene is aromatic compound and shows ring current

• The ring current produces strong induced magnetic field.• Outer 12 protons appears at 8.9 ppm because they remains in deshielding zone.Inner six protons appears at – 1.8 ppm because they remains in shielding zone (up-field)Hence, statement A and C are correct.

HH H

HH

H

H H

H

H

H

H

HH

H

H

H

H

outer hydrogen H~8.9 ppm

inner hydrogen H~ -1.8 ppm

Correct answer is (d).

68. In the compound give below, the relation between HA, HB; and between Br1, Br2 is:

HA HB

Br1 Br2

(a) HA, HB are enantiotropic; and Br1 Br2 are diastereotopic(b) HA, HB are diastereotopic; and Br1, Br2 are enantiotropic(c) HA, HB are diastereotopic; and Br1, Br2 are homotopic(d) HA, HB are enantiotropic; and Br1, Br2 are homotopic.

22

Soln.CH3

Br

CH3

Br

HB

HA

21

Molecule has a plane of symmetry bisecting HA and HB reflecting Br1/Br2 and CH3/CH3.• Br1 and Br2 are reflected by plane so they are enantiotopic to each other.• HA and HB are bisecting by plane. Hence, they are diesterotopic to each other.Correct answer is (b).

69. The most appropriate reagent to effect the following chemoselective conversion is

NH

OAcBoc

OTBDMS

NH

OHBoc

OTBDMS(a) HCl, EtOH, reflux (b) Bu4NF(c) K2CO3, MeOH (d) CF3COOH, EtOH, rt.

Soln.NH

COB

OTBDMS

O

O

CH3 NHCOB

OTBDMS

OH

Ester group

The hydrolysis of ester is faster in alkaline medium. So, the correct answer will be K2CO3, MeOH.Correct answer is (c).

70. Among the following, an example of a “Green Synthesis” is(a) Synthesis of malachite green(b) Friedel-Craft’s acylation of anisole with Ac2O/anhydrous AlCl3.(c) Jones’ oxidation of benzyl alcohol to benzoic acid.(d) Diels-Alder reaction of furan and maleic acid in water.

Soln. (i) Synthesis of Malachite green

CHO + NMe22H2SO4 CH

NMe2

NMe2

Malachite (Toxic)

(ii) Friedel-Craft’s acylation of anisole with Ac2O/AlCl3.O CH3

+ H3C C

O

O C

O

CH3AlCl3

O CH3

COCH3

23

In this reaction using reagent as well as product is toxic.(iii) Jones oxidation of benzyl alcohol to benzoic acid.H2C OH

Acetone/H2SO4/CrO3

COOH

In this reaction using reagent is hazardous for body.(iv) Diels-Aldes reaction of Furane and Maleic acid in water.

O +OH

OH

O

O

H2OO

COOH

COOHendo

Diels-Alder reaction is the case of Green synthesis.Correct answer is (d)

Part-C

71. The recoil energy of a Mossabauer nuclide of mass 139 amu is 2.5 MeV. The energy emitted by the nucleusin keV is:(a) 12.5 (b) 15.0 (c) 20.5 (d) 25.0

Soln. Recoil energy 2

2R2Mc

2

6

63

536E2.5 10

139 931.5

2.5 139 931.5 10E 24.57 10 eV536

Recoil energy

2536 ER

M eV


72. Complexes of general formula, fac-[Mo(CO)3(phosphite)3] have the C—O stretching bands as given below.Phosphines: PF3(A); PCl3(B); P(Cl)Ph2(C); PMe3(D)v(CO), cm–1: 2090(i); 2040(ii); 1977(iii); 1945(iv)The correct comibination of the phsphine and the streching frequency is,(a) (A–i), (B–ii), (C–iii), (D–iv) (b) (A–ii), (B–i), (C–iv), (D–iii)(c) (A–iv), (B–iii), (C–ii), (D–i) (d) (A–iii), (B–iv), (C–i), (D–ii)

Soln. As the -accepting abilities of phosphine increases vC–O stretching frequency of the complex increases. So,the order of -accepting abilities among the given phosphine is:

PF3 > PCl3 > PClPh2 > PMe3.i.e. A > B > C > D.So, the correct combination for vC–O streching is:A 2090 C 1977B 2040 D 1945Correct answer is (a).

73. On subjecting 9.5 ml solution of Pb2+ of X M to polorographic measurements, Id was found to be 1 A. When0.5 mL of 0.04 M Pb2+ was added before the measurement, the Id was found to be 1.25 A.(a) 0.0035 (b) 0.0400 (c) 0.0067 (d) 0.0080

Soln. Correct answer is (c).

24

74. Match each item from the List-I (compound in solvent) with that from the List-II (its behaviour) and select thecorrect combination using the codes given below.

List-I List-IIA. CH3COOH in pyridine (i) strong acid.B. CH3COOH in H2SO4 (ii) weak acidC. HClO4 in H2SO4 (iii) strong baseD. SbF5 in HF (iv) weak base(a) (A–i), (B–ii), (C–iii), (D–iv) (b) (A–ii), (B–i), (C–iii), (D–iv)(c) (A–iii), (B–iv), (C–ii), (D–i) (d) (A–iv), (B–ii), (C–iii), (D–i)

Soln. 3 5 5 3 5 5CH COOH C H N : CH COO C H NH solvocation

3 2 4 3 2 4CH COOH H SO CH COOH HSO solvent anion

4 2 4 4 3 4HClO H SO ClO H SO solventcation weak acid

5 6 2SbF 2HF SbF H F strong acid


75. Structure of a carborane with formula, C2B4H8 is formally derived from(a) Closo-borane (b) Nido-borane (c) Arachno-borane (d) Conjuncto-borane

Soln. C2B4H8(BH)2B4H8; B2H2B4H8

Or, 44H6 10 6 6B H B H nido borane


76. Boric acid is a weak acid in aqueous solution. But its acidity increases significantly in the presence of ethyleneglycol, because(a) ethylene glycol releases additional H+

(b) B(OH)4– is consumed in forming a compound with ethylene glycol.

(c) ethylene glycol neutralizes H+ released by boric acid.(d) Boric acid dissociates better in the mixed-solvent.

Soln. 3 4reversible reaction

B OH OH B OH

In presence of ethylene glycol, B(OH)4– is consumed as shown below and boric acid behaves as strong acid.

BOH

OH

HO

HO

+

H2C OH

H2C OH

–

BOH

OH

O

O

H2C

H2C

–H2O

CH2

CH2

+

BO

O

O

O

H2C

H2C

–H2O

CH2

CH2

HO

HO


25

77. Coordination number of “C” in Be2C3 whose structure is correlated with that of CaF2, is:(a) 2 (b) 4 (c) 6 (d) 8

Soln. In 2 3Be C ; Be+ = In Td voids

Be+=C3

– – =

3C = In FCC arrangement (Lattice point at corner + at each face centre)Therefore, coordination number of Be+ = 4

And coordination number of 3C 8 Correct answer is (d)

78. For the molecule below,

MoCO(H3C)2HN

CO

consider the following statements about its room temperature spectral data.(A) 1H NMR has singlets at 5.48 and 3.18 ppm(B) 1H NMR has multiplet at 5.48 and singlet at 3.18 ppm(C) IR has CO stretching bands at 1950 and 1860 cm–1

(D) IR has only one CO stretching band at 1900 cm–1.The correct pair of statement is,(a) A and C (b) B and C (c) A and D (d) B and D

Soln.Mo

COCONH

H3C

H3C

H

H

H

HH

all 5H proton gives on signal in1H NMR at 5.48

Since, the molecule is C2–symmetric. So, two CO’s will give two absorption band in IR spectrumCorrect answer is (a)

79. In the cluster 3 9Co CH CO obeying 18e rule, the number of metal-metal bonds and the bridgind ligandsrespectively, are(a) 3 and 1 CH (b) 0 and 3 CO (c) 3 and 1 CO (d) 6 and 1 CH

Soln. The structure of cluster [Co3(CH)(CO)9]

26

Co

OC

OC

OC Co

CO

CO

COC

CoOC CO

CO

H

So, the number of M–M bond = 3 and bridging ligand = 1 CHCorrect answer is (a)

80. Consider the ions Eu(III), Gd(III), Sm(II) and Lu(III). The observed and calculated magnetic moment valuesare closest for the pair(a) Gd(III), Lu(III) (b) Eu(III), Lu(III) (c) Sm(III), Gd(III) (d) Sm(III), Eu(III)

Soln. The electronic configuration of Gd3+ and Lu5+ are 4f7 and 4f14, for this ions total angular momentum is zero i.e.there is no orbital contribution. Therefore, calculated and observed values of magnetic moment are closest forthis pair.


81. Silicates with continuous 3D frame work are(a) Neso-silicates (b) Soro-silicates (c) Phyllo-silicates (d) Tecto-silicates

Soln. Correct answer is (d)

82. The correct spinel structure of Co3O4 is:

(a) 2 34t o

Co 2Co O (b) 2 3 34t o

Co 2Co Co O

(c) 2 3 34t o

Co Co Co O (d) 3 24t o

2Co Co O

Soln. Co3O4 is a normal spinel. In normal spinel the Co2+ ions occupy tetrahedral voids and Co3+ ions occupyoctahedral voids. Therefore, spinel structure of Co3O4 is (Co2+)t(2Co3+)oO4.Correct answer is (a).

83. In the solid state, the 35CuCl ion has two types of bonds. These are

(a) Three long and two short (b) Two long and three short(c) One long and four short (d) Four long and one short

Soln. In trigonal bipyramidal complexes, the two ligands lie on z-axis and the three in xy plane somewhere in between

the axes. In xy plane, there are four electrons and on z-axis there is only one electron in 2dz orbital (E.C.2 2 2 2 1xz yz xy 2 2 2x y z

d d d d d

)


84. In metalloenzymes, the metal centres are covalently linked through the side chains of the amino acid residues.The correct set of amino acids which are involved in the primary coordinates spheres of metalloenzymes is(a) Ala, Leu, His (b) Glu, His, Cys (c) Leu, Glu, Cys (d) Ala, His, Glu

Soln. So, it creat chiral environment across the catalyst.


27

85. Consider the catalyst in column-I and reactin in column-IIColumn-I Column-II

A. [(R)-BINAP]Ru2– (i) hydroformylationB. [Rh(CO)2I2]

– (ii) asymmetric hydrogenation.C. Pd(PPh3)4 (iii) asymmetric hydrogen transfer

D. Ru

N NH2Ts

Cl(iv) heck coupling.

The best match of a catalyst of column-I with the reaction nuclear column-II is(a) (A–ii), (B–i), (C–iv), (D–iii) (b) (A–i), (B–ii), (C–iii), (D–iv)(c) (A–iii), (B–i), (C–iv), (D–ii) (d) (A–iv), (B–iii), (C–ii), (D–i)

Soln. [Rh(CO)2I2]– Hydroformylation.

[Pd(PPh3)4] Heck coupling (It is a catalyst for Heck coupling)

RuN NH

Ts Asymmetry hydrogen transfer.


86. A solution of 2.0 g of brass was analysed for Cu electrogravimetrically using Pt-gauze as electrode. The weightof Pt-gauze changed from 14.5g to 16.0 g. The percentage weight of Cu in brass is(a) 50 (b) 55 (c) 60 (d) 75

Soln. The weight of Pt-gauze as electrode increases by 16–14.5 = 1.5 g because 1.5 g of Cu is deposited on Pt-gauze. Therefore, weight of Cu in 2.0 gram brass is 1.5g.

Therefore, percentage weight of Cu in brass = 1.5 100 752.0


87. The platinum complex of NH3 and Cl– ligands is an anti-tumour agent. The correct isomeric formula of thecomplex and its precursor are(a) cis-Pt(NH3)2Cl2 and PtCl4

2– (b) trans-Pt(NH3)2Cl2 and PtCl42–

(c) cis-Pt(NH3)2Cl2 and Pt(NH3)42+ (d) trans-Pt(NH3)2Cl2 and Pt(NH3)4

2–

Soln. Trans effect of Cl– > NH3 Antitumer agent is cis-[Pt(NH3)2Cl]

The precursor of this complex is [PtCl4]2–

PtC l

C l

C l

C l

2–

+ N H 3P t

C l

C l

N H 3

Cl

–+ N H 3

PtC l

C l

N H 3

N H 3


28

88. Successive addition of NaCl, H3PO4, KSCN and NaF to a solutin of Fe(NO3)3.9H2O gives yellow, colourless,red and again colorless solutions due to the respective formation of:

(a) 2 2 2

2 2 4 2 25 5 5 5Fe H O Cl , Fe H O PO , Fe H O SCN , Fe H O F

(b) 2 2

2 2 4 2 24 5 5 5Fe H O Cl OH , Fe H O PO , Fe H O SCN , Fe H O F

(c) 2 3 2 2

2 2 2 25 6 5 5Fe H O Cl , Fe H O , Fe H O SCN , Fe H O F

(d) 2 2

2 2 4 2 25 5 5 4Fe H O Cl , Fe H O PO , Fe H O SCN , Fe H O SCN F

Soln. When Fe(NO3)3.9H2O is dissolved in water, the complex ion [Fe(H2O)6]

3+ is formed.

[Fe(H2O)6]3+

+Cl–

+PO43–

SCN–

F–

[Fe(H2O)5Cl]2+

[Fe(H2O)5(PO4)]

[Fe(H2O)5(SCN)]2+

[Fe(H2O)3F]2+


89. Which one of the following will NOT undergo oxidative addition by methyl iodide?

(a) 22Rh CO I

(b) 3 2Ir PPh CO Cl

(c) 22CpRh CO (d) 5

2Cp Ti Me Cl Soln. [Rh(CO)2I2]

– 16e– and a better cordidate for O.A.[Ir(PPh3)2(CO)Cl] 16e– species (Vaska’s complex) O.A. is possible.[Cp2Ti(Me)(Cl)] d0- system O.A. is not possible.[CpRh(CO)2] 18e– species as such O.A. is not possible but after the dissociation of

ligand CO’ O.A. is possible.Correct answer is (d)

90. In hydrofomylation reaction using 3 3Rh PPh CO H as the catalyst, addition of excess PPh3 would

(a) increase the rate of reaction (b) decrease the rate of reaction.(c) not influence of the rate of reaction (d) stop the reaction.

Soln. Rh

CO

H

Ph3P

PPh3

PPh3

18e– species in excess PPh3 it becomes 20e– species but the active catalyst is 16e– species. So, in presence ofexcess PPh3. The rate of hydroformylation will be decreases.Correct answer is (b).

29

91. Find out the number of lines in the 31P NMR signal for

P

NH H

N

HF

F

H H

15

15

(a) 3 (b) 6 (c) 18 (d) 90

Soln. P

NH H

HF

FN

H H

1 1

1 1

2

15

15

1 15 191 1 1H I ; N I ; F I2 2 2

Four H1 are equivalent. Two nitrogen are equivalent.Two fluorine are equivalent. One 2H is separate.

Multiply =

1 15

1 1 1 12 4 1 2 2 1 2 2 1 2 1 12 2 2 2

4H 2N 2F 1H

= 5 3 3 2 90


92. The rate of exchange of OH2 present in the coordination sphere by 18OH2 of, (i) [Cu(OH)2)6]2+, (ii) [Mn(OH2)6]

2+,(iii) Fe(OH2)6]

2+, (iv) [Ni(OH2)6]2+, follows an order

(a) (i) > (ii) > (iii) > (iv) (b) (i) > (iv) > (iii) > (ii)(c) (ii) > (iii) > (iv) > (i) (d) (iii) > (i) > (iv) > (ii)

Soln. The rate of water exchange in [Cu(H2O)6]2+ is fastest due to John Teller distortion. For other three complexes

of 3d-series dipositive metal cation. The rate of water exchange decreases with increase in effective nuclearcharge and decrease in size.Correct answer is (a)

93. Based on the behaviour of the metalloenzymes, consider the following statements(A) In the enzymes, the zinc activates O2 to form peroxide species.(B) In the enzymes, the zinc activates H2O and provides a zinc boud hydroxide.(C) In the oxidases, the iron activates O2 to break the bonding between the two oxygens(D) Zinc ion acts as a nucleophile and attacks at the peptide carbonylThe set of correct statements is,(a) A and B (b) B and C (c) C and D (d) A and D

Soln. Because zinc is a Lewis acid and hence,

O

H

H

O

H

H

Zn–H+

Zn(OH)Zn2+

So, correct option is (b)However, oxidase are enzyme that catalyse the reduction of O2 H2O or H2O2.

2 2O g 4e 8H inside 2H O 4H outside Correct answer is (b)

30

94. Fe2+-porphyrins fail to exhibit reversible oxygen transport and cannot differentiate CO from O2. However, thehemoglobin is free from both these pit falls. Among the following(A) Fe2+ - porphyrins undergo -oxodimer formation and the same is prevented in case of the hemoglobin.(B) Fe–CO bond strength is much low in case of hemoglobin when compared to the Fe2+ - porphyrins.(C) While Fe–CO is linear, Fe–O2 is bent and is recognized by hemoglobin(D) The interlinked four monomeric units in the hemoglobin are responsible to overcome the pitfalls.The correct set of statements is(a) A and B (b) A and C (c) C and D (d) B and D

Soln. Free heme (i.e. without globular protein chain) it foms-oxodimer) i.e.

OFeFe(III) (III) -oxo-dimer (hematin).

Which is unable to bind O2.Correct answer is (b)

95. Reactions A and B are, termed as respectively.

(A) SnCl2 + Co2(CO)8 Sn

Cl

Cl

Co(OC)4(CO)4Co

(B) Me2SnCl2 + Sn

Me

Me

Re(OC)5(CO)5Re2NaRe(CO)5 + 2NaCl

(a) Insertion, Metathesis (b) Metathesis, insertion(c) Oxidative, addition, metathesis (d) Oxidative addition, insertion

Soln. Since SnCl2 will behaves as a carbene and it insert into the M–M bond. So, it is a kind of insertion reaction.

(OC)4Co Co(CO)4SnCl2

insertion Co(CO)4 Co(CO)4Sn

Cl

Cl

Re(CO)5 Re(CO)5Sn

CH3

CH3

Sn

H3C

H3C

Cl

Cl

Re(CO)5

Re(CO)5

Metathesis


31

96. A metal crystallizes in fcc structure with a unit cell side of 500 pm. If the density of the crystal is 1.33 g/cc, themolar mass of the metal is close to(a) 23 (b) 24 (c) 25 (d) 26

Soln. We have, 3

03

0

N aZM ; MZN a

33 23 1 101.33 g cm 6.023 10 mole 500 10 cmM

4

= 25


97. The activation energy for the bimolecular reaction A BC AB C is E0 in the gas phase. If the reaction iscarried out in a confined volume of 3, the activation energy is expected to(a) remain unchanged (b) increase with decreasing .(c) decrease with decreasing . (d) oscillate with decreasing .

Soln. Smaller the volume the greater will be possibility of collisions and therefore the activation energy will decrease.Correct answer is (c).

98. In a many-electron atom, the total orbital angular momentum (L) and spin (S) are good quantum numbersinstead of the individual orbital (l1, l2) and spin (s1, s2) angular momenta in the presence of(a) inter-electron repulsion (b) spin-orbit interaction(c) hyperfine coupling (d) external magnetic field.

Soln. In many electron atom, there is strong electron-electron i.e. inter-electron repulsion. Due to inter-electronrepulsion, the orbital angular momentum (l1, l2) as well as spin angular momenta (s1, s2) are coupled togetherto give total orbital angular momentum (L) and spin (S).Correct answer is (a)

99. The packing fraction of a simple cubic lattice is close to(a) 0.94 (b) 0.76 (c) 0.52 (d) 0.45

Soln. Packing fraction = volume occupied by particlestotal volume of the unit cell =

= No. of particles per unit cell Volume of 1 particle

Total volume of the unit cell

... (1)

For simple cubic lattice,Number of particles per unit cell = 1

Volume of one particle = 34 r3

where, r = radius of one particle and a = unit cell edge length.Then the total volume of the unit cell = a3.

Packing fraction =

33

3 3

41 r 4 r3 3.14 0.523a 2r

for SCC, a 2r


32

100. The number of IR active vibrational modes of pyridine is:

'2v 2 2 v v

1

2 z

1 y

2 x

C E CA 1 1 1 1 zA 1 1 1 1 RB 1 1 1 1 x, R

B 1 1 1 1 y, R

(a) 12 (b) 20 (c) 24 (d) 33

Soln. C5H5N

C

CN

C

CCH

H

H

H

H

z

xy

C2V point group• Reducible represented by 3N coordinates as basis sets.

2 V v 'E CR.R. 33 3 3 11

A11n 33 3 3 11 114

; A21n 33 3 3 11 44

; B11n 33 3 3 11 74

B21n 33 3 3 11 114

1 2 1 2R.R. 11A 4A 7B 11B Rot. = A2 + B1 + B2.Tran. = A1 + B1 + B2Vib. = 10A1 + 3A2 + 3B1 + 9B2.Total = 24 vibrations.Correct answer is (c).

33

101. One of the excited states of Ti has the electric configuration 2 1 1Ar 4s 3d 4p . The number of microstates withzero total spin (S) for this configuration is(a) 9 (b) 15 (c) 27 (d) 60

Soln.+2 +1 0 –1 –2 +1 0 –1 +2 +1 0 –1 –2 +1 0 –1

Total arrangement = 15


102. For the reaction 2A 2A in a closed container, the relation between the degree of dissociation and

the equilibrium constant Kp at a fixed temperature is given by

(a) p pK / K 4p (b) 1

2p pK / K 4p

(c) p pK 4p / K (d) 1

2p pK 4p / K

Soln. 2A 2A

1 01 2

Total moles at equilibrium = 1

2 2 2

2

22 2 2AA A

P 2A A A

2X PP X 4 P1K P

1P X P X 11

34

22 2

P P p24 PK K K 4 P1

2P PK K 4P

12

P

P

KK 4P


103. The fugacity of a gas depends on pressure and the compressibility factor Z pV/RT through the relation

V is the molar volume

For most gases at temperature T and up to moderate pressure, this equation shows that(a) f p, if T 0 (b) f p, if T (c) f p, if T 0 (d) f p, if T 0


104. The internal pressure TU/ V of a real gas is related to the compressibility factor Z pV/RT by

V is the molar volume

(a) T TU/ V RT Z/ V (b) TU/ V RT / V Z

(c) 2T VU/ V RT / V Z/ V (d) 2

T VU/ V V / RT Z/ T

Soln.ZRTP

V ... (1)

T V

dU dPT PdV dT

I II

V V

dP d ZRT R d Z TdT dT V V dT

V V

dP R dZT ZdT V dT

T V V

dU dP R dZT P T T Z PdV dT V dT

2

V ||P

RT dZ ZRT PV dT V

2

T V

dU RT dZdV V dT


105. Suppose, the ground stationary state of a harmonic oscillator with force constant ‘k’ is given by2

0 exp Ax Then, A should depend on k as

(a) 1

2A k (b) A k (c) 12A k (d)

13A k

Soln. Ground state wave function for SHO1 2

4 x2

0 e

, Given function: 2

0 exp Ax

So, A 2A2

mk

35

1 12 2mk m k2A A2

12A k


106. Combining two real wave functions 1 2and , the following functions are constructed: 1 2A ,

1 2 1 2 1 2B i , C i , D i . The correct statement will then be(a) A and B represent the same state (b) A and C represent the same state.(c) A and D represents the same state (d) B and D represent the same state.

Soln. Combining two real wave function 1 2and constructed functions 1 2 and 1 2i represent thesame state because if these two function multiplied by any constant, no effect on the state and on the energybecause multiply only in the eigenvalue, no effect on the energy and no affect on the wavefunction. If we useany eigenfunction which given eigenvalue. If multiply by any constant not change in the eigenfunction and nochange in the energy of the state. Energy is same, so state is same only change in eigenvalue.


107. Crystal A diffracts from (1 1 1) and (2 0 0) planes but not from (1 1 0) plane, while the crystal B diffracts from(1 1 0) and (2 0 0 ) planes but not from the (1 1 1) plane. From the above, we may conclude that(a) A has fcc lattice while B has bcc lattice (b) A has bcc lattice while B has fcc lattice(c) A and B both have fcc lattice (d) A and B both have bcc lattice.

Soln. We know that, selection rule for allowed reflection for BCC and FCC lattice is

BCC h + k + l should be evenFCC h, k, l all either even or odd.Therefore, A is FCC and B is BCCCorrect answer is (a)

108. The decomposition of NH3 on Mo surface follows Langmuir-Hinshelwood mechanism. The decompositionwas carried out at low pressures. The initial pressure of NH3 was 10–2 torr. The pressure of NH3 was reducedto 10–4 torr in 10 minutes. The rate constant of decomposition of NH3 is:(a) 4 19.9 10 torr min (b) 0.4606 min–1

(c) 3 19.9 10 torr min (d) 10.693 min

Soln. At low pressure Langmuir Hinshelwood mechanism follows first order kinetics.

210

10 10 4t

P2.303 2.303 10k log log 0.4606 mint P 10 10


109. A polymer sample has the following composition.

Number of molecules Molecular weight10 100050 200040 4000

The polydispersity index (P.D.I) of the polymer is

(a) 85000

27 (b) 8581 (c)

850729 (d)

729850

36

Soln. We know that,1 1 2 2 3 3i i

ni 1 2 3

N M N M N MN MMN N N N

10 1000 50 2000 40 400010 50 40

= 2700

2 2 221 1 2 2 3 3i i

mi i 1 1 2 2 3 3

N M N M N MN MMN M N M N M N M

2 2 2 710 1000 50 2000 40 4000 85 1010 1000 50 2000 40 4000 270000

m

n

M 850P.D.I.M 729


110. The equilibrium constant for an electrochemical reaction,3 2 2 42Fe Sn 2Fe Sn

is 0 3 2 0 4 2E Fe /Fe 0.75 V, E Sn / Sn 0.15V, 2.303RT / F 0.06V (a) 1010 (b) 1020 (c) 1030 (d) 1040.

Soln. 3 2 2 42Fe Sn 2Fe Sn

3 2

2 4

3 2 2 4

Fe e Fe 2 ... 1

Sn Sn 2e ... 2

2Fe 2e Sn 2Fe Sn 2e

G RT nK nFe RT nK nFelog K

2.303RT

0 0right leftE E E ; E 0.75 0.15; E 0.60

20

log K 2 0.60 / 0.062.303RT F 1log K 20 0.06;

F 2.303RT 0.06K 10


111. A bacterial colony grows most commonly by cell division. The change in the population due to cell division in

an actively growing colony is gdN N dt . The population of bacterial colony at time ‘t’ is 0N N t 0

(a) 0 gN t (b) 0 gN exp t (c) 0 gN exp t (d) 20 gN t

Soln. Radioactive process follows first order kinetics and rate law expression is,

t 0N N exp t (for concentration with increasing time)

And t 0N N exp t (for concentration increasing with increasing time)Here, Since bacterial colony is growing i.e. concentration is increasing with increasing time.Correct answer is (c)

37

112. The Arrhenius parameters for the thermal decomposition of NOCl, 22NOCl g 2NO g Cl g , are13 1 1 1

aA 10 M s , E 105 kJ mol and 1RT 2.5 kJ mol . The enthalpy (in kJ mol–1) of the activatedcomplex will be(a) 110 (b) 105 (c) 102.5 (d) 100

Soln. We have, #a gH E n 1 RT ... (1)

Here, #gn 1 Bimolecular reaction

1 1H 105 kJ mole 1 1 2.5 kJ mole

1 1105 5 kJ mole 100 kJ mole


113. The rotational partition function of H2 is:

(a) hcBJ J 1

J 0, 1, 2.....2J 1 e

(b) hcBJ J 1

J 1,3, 5,......2J 1 e

(c) hcBJ J 1

J 0,2,4......2J 1 e

(d) hcBJ J 1 hcBJ J 1

J 0,2,4...... J 1,3,5.......

1 2J 1 e 3 2J 1 e4

Soln. The rotational partition function for H2 assuming that the high temperature limit is valid given by

r1 1qhcB 2 hcB

with B = 60.589 cm–1.

33

r 34 10

1 kT 1.38 10 1000q 5.742 hcB 2hcB 2 6.626 10 3.00 10 60.589

Evaluation of the rotational partition by direct summation is performed as follows

hcBJ J 1 BhcBJ J 1r

J 0, 2, 4, 6...... J 1, 3, 5.......

1q 1 2J 1 e 3 2J 1 e 5.914


114. The potential in Debye-Huckel theory is proportional to

(a) 1/ r (b) exp r (c) exp r / r (d) r

Soln. From debye huckel limiting law, 22n echarge density

kT

According to Poission equation of electrostatics 2

2

r r

4 4 2nekT

38

22

r

8 nekT

2 2

2

1 rr r r

The equation may be written as 2 2k , where 2

2 8 n ekkT

The solution of this equation is equal to kr

r

ze er

= kr

r

ze er

, the term is total potential in Debye Huckel

it means electrical potential.

Expand kre

r

term

1kr r r r r

I IIPotential due to Potential on the ionthe given ion itself due to its ionic atmosphere

ze ze ze k ze ze1 krr r r


115. The vibrational frequency and anharmonicity constant of an alkali halide are 300 cm–1 and 0.0025 respectively.The positions (in cm–1) of its fundamental mode and first overtone are respectively.(a) 300, 600 (b) 298.5, 595.5 (c) 301.5, 604.5 (d) 290, 580

Soln. 1e e300 cm ; X 0.0025

Fundamental mode = e e1 2X

First overtone = e e2 1 3X

Fundamental 300 1 2 0.0025 = 300 (1–0.0025) = 300×0.995 = 298.5

First overtone = 2×300 (1–3×0.0025) = 600(1–0.0075) = 595.5

Correct answer is (b).116. The adsorption of a gas is described by the Langmuir isotherm with the equilibrium constant 1K 0.9 kPa at

25ºC. The pressure (in kPa) at which the fractional surface coverage is 0.95, is(a) 1/11.1 (b) 21.1 (c) 11.1 (d) 42.2

Soln. Langmuir isotherm is given as kP ; kP kP; kP 1 ; kP1 kP 1

1 1 0.95Pk 1 0.9 1 0.95

= 21.1


117. The energy of a harmonic oscillator in its ground state is 12

. According to the virial theorem, the average

kinetic (T) and potential (V) energies of the above are

(a) 1 1T ; V4 4

(b) 1 3T ; V8 8

(c) 1T ; V2

(d) 3 1T ; V8 8

39

Soln. The energy of a harmonic oscillator in its ground state is 12

According to Virial Theorem for SHOEvery contribution is equal for kinetic energy operator and potential energy operator.

E K V K VE K K E

K2

2K ; K

2 4

Energy 2

only possible if k 4

and V 4

energy contribution is equal for k and V.V.


118. The energy of a hydrogen atom in a state is HH

hcR R Rydberg constant25

. The degeneracy of the state

will be(a) 5 (b) 10 (c) 25 (d) 50

Soln. Energy of a hydrogen in a state is 2

c H2 2

h R z13.6n n

... (1)

Degeneracy gn = n2.

If spin include = 2n2 Given: c hh R25

Compare with equation (1), 2n 25 Degeneracy = 25


119. The trial wave function of a system is expanded as t 1 1 2 2c c . The matrix elements of the Hamiltonian

are 1 1H 0; 1 2 2 1H 2.0 H and 2 2H 3.0 . The approximate ground-stateenergy of the system from the linear variational principle is(a) –1.0 (b) –2.0 (c) +4.0 (d) +5.0

Soln.11 11 12 12

11 22 12 1221 21 22 22

H ES H ESS 1; S 1; S S 0

H ES H ES

11 1 1H H 0 ; 12 21 1

2 1

H H H 2.0

H

; 22 2 2H H 3.0

20 E 2 03E E 4 0

2 0 3 E

2 2E 3E 4 0; E 4E E 4 0; E E 4 1 E 4 0

E 1 E 4 0; E 4 1

Ground state is lower energy = –1.Correct answer is (a).

40

120. One molecular orbital of a polar molecule AB has the form A A B Bc c , where A Band are normalizedatomic oribitals centred on A and B, respectively. The electron in this orbital is found on atom B with a prob-ability of 90%. Neglecting the overalp between A Band , a possible set of cA and cB is:

(a) A Bc 0.95, c 0.32 (b) A Bc 0.10, c 0.90

(c) A Bc 0.95, c 0.32 (d) A Bc 0.32, c 0.95

Soln. A A B BPolar molecule AB A B has the form C C 2 2A BC C 1 for normalization.

Probability on atom B find the electron = 90%2 2

B BC 90% 0.9 C 0.9 2 2

B AC 0.9 C 0.1

B AC 0.9 0.95 C 0.1 0.32

A BC 0.32; C 0.95 Correct answer is (d)

121. 4-Hydroxybenzoic acid exhibited signals at 171, 162, 133, 122 and 116 ppm in its broadband decoupoled13C NMR spectrum. The correct assignment of the signals is

(a) 171 C 4 ,162 COOH ,133 C 3 & 5 ,122 C 1 and116 C 2 & 6

(b) 171 COOH ,162 C 4 ,133 C 2 & 6 ,122 C 1 and 116 C 3 & 5

(c) 171 C 4 ,162 COOH ,133 C 2 & 6 ,122 C 1 and 116 C 3 & 5

(d) 171 COOH ,162 C 4 ,133 C 3 & 5 ,122 C 1 and 116 C 2 & 6

Soln.

OH

+

C O

OH

OH

COHO

m-directing

– –

–O/P directing

162116

133122

171

OH

CHO

O

–

–– – –4 3

1

2

• Positions-3 is most shielded because of ortho-to OH and meta to COOH. So, it will appear at 116 ppm.• Position-1 is shielded because it is para-to OH. It will upper 122 ppm.If we compare position 2 and 4 position 4 is deshielded due to directly attached OH and it will appear at 162ppm.• Position is normal and apparent at 133 ppm.Correct answer is (b)

41

122. An organic compoound (C9H10O3) exhibited the following spectral data:IR: 34000, 1680 cm–1;1H NMR: 7.8(1H, d, J = 8 Hz), 7.0 (1 H, d, J = 8Hz), 6.5 (1 H, s), 5.8 (1 H, s, D2O exchangeable), 3.9(3H,s), 2.3 (3 H, s).The compound is

(a)

HOO

OMeMe

(b)

O

OMeMe

HO (c)

O

MeOMe

HO

(d) O

MeOMe

HO

Soln. All of the compounds given 1-4 will give same number of signals and their multiplicity in their NMR spectrum.

The key parameter to assign the correct structure will be chemical shift of aromatic protons.• The most shielded singlet 1H signal in aromatic region is at 6.5 ppm is possible.• In structure 4 only. Because

Ortho to OH Ortho to OMe Meta to COMe

CO

MeHO

H OMesinglet 6.5 ppm


123. The []D of a 90% optically pure 2-arylpropanoic acid solution is +135º. On treatment with a base at RT forone hour, []D changed to +120º. The optical purity is reduced to 40% after 3 hours. If so, the optical purityof the solution after 1 hour, and its []D after 3 hours, respectively, would be(a) 80% and 60º (b) 70% and 40º (c) 80% and 90º (d) 70% and 60º

Soln. • Optically purity also known as enantiomeric excess a compound of 90% B.E. shows 135º specific rotation.

So, its 100% optically pure isomer will show 135 100 150º specific rot.90

• At one hour the specific rotation reduced to 120º. So, E.E, or optical purity = 120 100 80%150

• At three hours optical purity is 40%. So, specific rotation = 150 40 60º100

The over all relation of optical purity, composition and specific rotation can be summarized in the table.Optical purity % dext. 10 levo. sp. rot.100% 100% 0% 150º90% 95% 5% 135º

80% 90% 10% 120º

40% 70% 30% 60º


42

124. In the following pericyclic reaction, the structue of the allene formed and its configuration are

O

MePh

PhNEt2Ac2O

(optically pure)

(a) C

HPh

Me

AcO

R (b) C

HPh

Me

AcO

S

(c) C

HMe

Ph

AcO

R (d) C

HMe

Ph

AcO

S

Soln. O

MePh

PhNEt2Ac2O

?

O

MePh

3, 3 sigmatropic

321

12

O

H C C

Me

Ph

OH

C C C

Me

Ph

PhNEt2Ac2O

OAc

C C C

Me

Ph

3, 3 sigmatropic belong to 4n+2 system and 4n+2 system under thermal condition retension in configurationtakes place. So, the stereochemistry of –me and –Ph group mentain.Correct answer is (a).

43

125. In the following sequence of pericyclic reactions X and Y are

COOH

PhX

Y

H

H

Ph

COOH

hv

(a)

X Y

Ph

COOH

hv/DIS (b)

X Y

Ph

COOH

hv/CON

(c)

X Y

Ph

COOH

/DIS (d)

X Y

Ph

COOH

/CON

Soln.COOH

Ph

Con

H

Ph

COOH

H

dis

H

H

Ph

COOH

4n 4n + 2



OH1. KH/THF,

2. H3O+

(a) (b) O

(c) O (d) O

44

Soln.

OH

KH, THFO

O

1

3

3 2

2 1

3, 3, sigmatropic

O

H3O+

OHketo-enol

O


127. The following conversion involves

NO2

NPh

Cl

Et3N

H COOEt

NO2

HN

Ph

COOEt

(a) a 1, 3-dipolar species as reactive intermediate, and a cycloaddition.(b) a carbenium ion as reactive intermediate, and a cycloaddition.(c) a 1, 3-dipolar species as reactive intermediate, and an aza Witting reaction.(d) a carbanion as reactive intermediate, and an aza Cope rearrangement.

Soln. NPh

Cl NO2Et3N

C NPh

Cl NO2

C NPh

Cl NO2

1, 3 dipolar

45

C NPh

Cl NO2

+

COOEt

NPh

Cl

COOEt

NO2

NPh

Cl

NO2Base

NPh

COOEt

NO2

NPh

COOEt

NO2H

H3O+


128. The following transformation involves

N

CHO

+HO Me

MeHN H+

N

N

O

Me

(a) an iminium ion, [3, 3]-sigmatropic shift and Mannich reaction.(b) a nitrenium ion, [3, 3]-sigmatropic shift and Michael reaction.(c) an iminium ion, [1, 3]-sigmatropic shift and Mannich reaction.(d) a nitrenium ion, [1, 3]-sigmatropic shift and Michael reaction.

Soln.

N

C

O

H

+ MeHN

OH Me

N

CH

OH

NMe

MeOH

N

HC

NMe

MeOH

3

32

2

3, 3 sigmatropic

N

CHN

Me

OH

Me

In manich reaction this tyupe intermediate form iminium ion

N

CH NMe

Me OH

–H+

N

CH NMe

O


46

129. With respect to the following biogenetic conversion of chorismic acid (A) to 4-hydroxyphenylpyruvic acid(C), the correct statement is

OH

COOH

O COOH

(A)

XB

Y

HO

COOH

O

(C)

(a) X is Claisen rearrangement; Y is oxidative decarboxylation.(b) X is Fries rearrangement; Y is oxidative decarboxylation.(c) X is Fries rearrangement; Y is dehydration.(d) X is Claisen rearrangement; Y is dehydration.

Soln.

COOH

OH

O

COOH

32

1

3

21

Claisen

rearrangement

HOOC

OH

O

COOHoxidation

HOOC

O

O

COOH

decarboxidation

O

O

COOH

OH

O

COOH

keto form

So, this reaction is Claisen rearrangement and oxidative decarboxilation.Correct answer is (a)

130. Match the following(i) -amyrin (A) alkaloid; secondary alcohol(ii) squalene (B) alkaloid, phenol(iii) morphine (C) triterpene, secondary alcohol(iv) ephedrine (D) acyclic triterpene, polyene(a) (i)-(C), (ii)-(D), (iii)-(B), (iv)-(A) (b) (i)-(B), (ii)-(A), (iii)-(C), (iv)-(D)(c) (i)-(C), (ii)-(B), (iii)-(D), (iv)-(A) (d) (i)-(A), (ii)-(D), (iii)-(B), (iv)-(C)

Soln. ephedrine

CH3

H NHCH3

Ph

H OH

alkaloid secondary alcohol

47

O

HO

HO

N CH3

Phenolic part

Morphine

alkaloid, phenol

squalene

acylic triterpene polyene

-amyrin triterpene and secondary alcohol.

HO-amyrin


131. In the following reaction, the structure of B, and the mode of addition are

Me

Ph CHO

+ B Ph

OH O

(a)

OLi

Re-Si facial(b)

OLi

Re-Re facial

(c)

OLi

Re-Si facial(d)

OLi

Si-Si facial

Soln.Ph CHO

Me

+ B

Ph

Me

OOH

48

The above case belong to Aldol condensation. In aldol condensation two type enolate form as a intermediate(i) Z enolate (ii) E enolateThe formation of a particular enolate depends upon the carbonyl compound. For example.

O

Base

O O

E enolate

O O O

Z enolate

Z enolate give sign product. E enolate give anti product.

In the product the stereochemistry of –OH and CH3 is syn. So, Z enolate participate in reaction.

O

H3C

Re face

1

3 2NaBH4 (Si face attack) H3C

H OH

NaBH4 (Re face attack)

H3C

H OH

Ph

Me

O

H2

1

3

Si Face

Nu Ph

Me

H

Si Face

Nu OH

Re face attack

Ph

O2

1

3

Si Face

LDA Ph

OPh

OLi

49

Ph

Me

O

H+

O–Li+Ph

Me

OH

Me

O

The enolate is Z. So, syn product will be form and Re face of carbonyl compound and Si face of enolate.Correct answer is (c)

132. In the following reaction A and B are

O

N i-PrMgClA

O

NOMe

MeB

(a) O

NMgCl

O

N O

(A) (B)(b) O

N

O

N

(A) (B)

MgClO

(c) O

N

O

N

(A) (B)

N

Me

O (d) O

N

O

N

(A) (B)

O

Soln. O

N i-PrMgCl

O

N

MgCl

Grignard reagentThe Grignard reagent give monosubstitution with Wein-Reb amide.

O

NOMe

MeWeinReb amide

O

NOMe

Me

+O

N

MgCl O

N

C

O


50

133. Match the following biochemical transformations with coenzymes involved(i) -ketoglutarate to glutamic acid (A) tetrahydrofolate(ii) uridine to thymidine (B) NADH(iii) pyruvic acid to acetyl coenzyme A (C) thiamine pyrophosphate.

(D) pyridozamine(a) (i)-(D), (ii)-(A), (iii)-(C) (b) (i)-(A), (ii)-(B), (iii)-(D)(c) (i)-(B), (ii)-(A), (iii)-(C) (d) (i)-(D), (ii)-(B), (iii)-(C)

134. The structure of major product B formed in the following reaction sequence is

CHO

H OH

H OH

CH2OH

H OH

Br2

H2O(A)

H2O2

Fe2(SO4)3(B)

(a)

CHO

H OH

CH2OH

H OH(b)

O O

HO OH

HO

(c)

COOH

H OH

H OH

COOH

H OH(d)

COOH

H OH

COOH

H OH

Soln.

H OH

H OH

CH2OH

H OH

H O

Br2/H2OMild oxidising agent

H OH

H OH

CH2OH

H OH

HO O

CaCO3H OH

H OH

CH2OH

H OH

O O 2Ca2+

H2O2/Fe3+ H OH

CH2OH

H OH

O H

(Ruff degradations)So, correct answer is (a).

135. Given the energy of each gauche butane interaction is 0.9 kcal/mol, G value of the following reaction isMe

Me

Me

MeH2, Pd, 250ºC

(e, e conformer)

(a) 0.9 kcal/mol (b) 1.8 kcal/mol (c) 2.7 kcal/mol (d) 3.6 kcal/mol

Soln.

CH3CH3

No 1, 3 interaction

HH CH3

CH3

cistwo H/Me 1, 3 interaction

one 1, 3 interaction energy is 0.9 kcal/mol. So, Total energy 2×0.9 = 1.8 kcal/mol


51

136. In the following reaction, the reagent A and the major product B are

MeO

A

MeO

COOEt H2, 10% Pd/C

MeOHB

(A) (B)

(a) N2CHCOOEt, Cu(acac)2

MeO

COOEt

(b) N2CHCOOEt, Cu(acac)2 MeO

COOEt

(c) NaH, S

OMe

MeCOOEt

Br MeO

COOEt

(d) NaH, S

OMe

MeCOOEt

Br MeO

COOEt

Soln. 2Cu acac2N CH – COOEt : CH – COOEt

(A)

MeO

:CHCOOEt

MeO

COOEt

H2, 10% Pd/C

MeO

COOEt

(B)

The formation of product B takes place via hydrogenolysis by H2, 10%, Pd/C. Hydrogenolysis of cyclopro-pane takes place via H2, 10% Pd/C and least hindered bond will be break.


52

137. The major product formed in the following reaction sequence is

O

Me1. LiAlH4, Et2O, –20ºC2. Ac2O, Py

3. Me2CuLi, Et2O

(a) Me

Me

(b)

Me

Me

(c)

Me

Me

(d)

Me

Me

Soln.

O

Me

Chair form OMe

LiAlH4, Et2O, –20ºC

OHMe

H

Ac2O, PyOH3C C

O

CH3

Flliping

O

Me

C

O

CH3

Me2CuLi

MeMe

LiAlH4 is a smaller reagent. So, in the above existingstructure axial attack takes place

1,3, a,e 1, 3 trans

H3C

H3C

Equatorial attack


138. 12.0 g of acetophenone on reaction with 76.2 g of iodine in the presence of aq. NaOH gave solid A in 75%yield. Approximate amount of A obtained in the reaction and its structure are(a) 80 g, Cl4 (b) 40 g, Cl4 (c) 60 g, CHI3 (d) 30 g, CHI3.

Soln. Iodoform reaction, 2NaOH I NaOI HI

Ph C

O

CH3 + 3NaOI Ph C

O

CI3 + 3NaOH

Ph C

O

I3 + NaOH Ph C

O

ONa + CHI3

53

1 mole of acetophenone = 1 mole of iodoform mol. wt. = 120 mol. wt. = 394

120 g acetophenone will give 394g iodoform.

12 g acetophenone will give 394 12 39.4 g120

If yield is 75% then 12 g acetophenone will give 39.4 75 29.5 g iodoform

100


139. Consider the following reaction mechanism

The steps A, B and C, respectively, are(a) Oxidative addition; transmetallation; reductive elimination.(b) Oxidative addition; carbopalladation; -hydride elimination.(c) Carbopalladation; transmetallation; reductive elimination.(d) Metal halogen exchange; transmetallation; metal extrusion.

Soln. Step (A) is O.A. because the oxidation state of the metal is increased by two units.

L2Pd(O) PhI L2Pd

Ph

I

+2Step(B) is transmetallation because the ligand is transferring from one metal to another metal.Step(C) is reductive elimination because the oxidation state of the metal is increased by two units.

L2Pd

Ph

+2

REPh + L2Pd(0)


140. The major product formed in the following reaction sequence is

OH1. CHCl3, NaOH

2. H2O2, NaOH

(a) OH

COOH

(b) OH

OH

(c) HOOC OH (d) HOOC

O

54

Soln.

OH

CHCl3, NaOH

OH

CHO

Reimer-Tiemman reaction

H2O2 + NaOH O O OH

OH

C

O

H

+ O O H

OH

OOH

O H

OH

OC

O

H

OH

OC

O

H

OH

OH

OH

O

+ O C

O

HH

OH

OH

+ HCOO


141. The major product B formed in the following reaction sequence is

Ph D1. O

BO

D

2. H3O+(A) (B)

I Ph

Pd(OAc)2

PPh3, Et3N

(a) Ph

D

D Ph

(b) Ph

D D

Ph

D

55

(c) Ph

D

Ph

D

(d) Ph

D

Ph

D

Soln.Ph C C D

O

O

B D

Syn Addition

Ph

D

D

B

O

O

H3O+

Ph

D

D

B(OH)2

I Ph

Pd(OAc)2PPh3, Et3N

Ph

D

DPh

Suzuki Coupling(A)

The structure A also can written as Ph

D

D Ph

.


142. The major product B formed in the following reaction sequence is

O

O

1. CH2N2, Et2O

2. MeMgCl, Et2O3. H3O+

B

(a)

O

Me

(b)

O

Me

OMe (c)

O

O

(d)

O

Me

Me

Soln.

O

O

(i) CH2N2Et2O

(ii) Me MgCl, Et2(iii) H3O+

O

Me

56

Mechanism:Step-I: This ‘H’ in between two carbonyl hence more.

O

O

HH2C N2 CH2N2

Et2O

O

O

This 'H' in between two carbonyl hence more acid.

+ H3C N2

O

O+ H3C N2

O

OMe

+ N2

Step-II:

O

OMe

MeMgCl, Et2O

OMCl

OMe

MeH+/H2O

OH

OMe

Me

Note: Grignard reagent give 1, 2 addition not 1, 4.

Step-III:

OH

O

Me

H+/H2O

CH3

OH2

O

Me

CH3 O

Me

Me

H2O

O

Me

Me

O

Product


57

143. The osazone A could be obtained from

HO H

H OH

CH2OH

H OH

NNHPh

NNHPh

(A)(a) glucose and mannose (b) mannose and galactose(c) gulcose and fructose (d) galactose and fructose

Soln.H OH

HO H

H OH

H OH

CH2OH

H O

D-glucose

H OH

HO H

H OH

H OH

CH2OH

NNHPh

O

HO H

H OH

H OH

CH2OH

NNHPh

NH2NHPh NH2NHPh NH2NHPhHO H

H OH

CH2OH

H OH

Ozazone of D-glucose

NNHPhNNHPh

O

HO H

H OH

H OH

CH2OH

NNHPh

NNHPh

HO H

H OH

H OH

CH2OH

NNHPh

(A)

O

HO H

HO H

H OH

NNHPh

NH2NHPh

OHH

CH2OH

O

HO H

HO H

H OH

NNHPh

OHH

CH2OH

NH2NHPh NH2NHPh

So, the osazone (A) could be obtained from D-glucose and D-mannose.


144. The major product formed in the following reaction is:

N

1. Li, liq. NH3, t-BuOH

2. H3O+

3. KOH, MeOH

(a) N

OH (b)

Me

O

(c)

O

(d) NH2

O

58

Soln.N

(i) Li. liq. NH3, t-BuOH

(ii) H3O+

(iii) KOH, MeOH

O

Me

Mechanism:

Step-I: N

(i) Li. liq. NH3, t-BuOH

N

H

Step-II: N

H+, H2O

H

N

H

H

N

H

HH3O+

H

NH2O

H+, H2O

viasame mechanismO O

Step-III:

OO OOH

KOH, MeOH

OO O O

H+

HO O

H

KOHMeOH

O

O


145. In the following enantioselective reaction, the major product formed is

COOEt 1. n-BuLi,

NOMe

HN SiMe3

2. H3O+

3. H2, Raney Ni

(a) COOEt

NH2

(b) COOEt

NH2

(c) COOEt

NH2

(d) COOEt

NH2


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Documents

b c c b

b c d e f g

b c d e f g

c andf

positive numbers

following numbers

cmcorrect answer

ccorrect answer