KVPY-SX_2018 (CHEMISTRY)

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KVPY-SX_2018 (CHEMISTRY)

PART-I

1. The amount (in mol) of bromoform (CHBr3) produced when 1.0 mol of acetone

reacts completely with 1.0 mol of bromine in the presence of aqueous NaOH is

(A) 1

3 (B)

2

3 (C) 1 (D) 2

2. The following compound can readily be prepared by Williamson ether synthesis by

reaction between

(A) and (B) and

(C) and (D) and

3. X and Y

are:

(A) enantiomers (B) diastereomers

(C) constitutional isomers (D) conformers

4. The higher stabilities of tert-butyl cation over isopropyl cation, and trans-2-butene

over propene, respectively, are due to orbital interactions involving (A)  →  and → * (B) → vacant p and  → *

(C) →  * and → (D) → vacant p and → *

H Cl CH3 H Cl H3C Cl Cl H H H3C CH3 X Y OH I OH Cl I OH Cl OH

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KVPY-SX_2018 (CHEMISTRY)

5. Benzaldehyde can be converted to benzyl alcohol in concentrated aqueous NaOH

solution using

(A) acetone (B) acetaldehyde (C) formic acid (D) formaldehyde

6. The major product of the following reaction

is:

(A) (B)

(C) (D)

7. Among the following species, the H–X–H angle (X = B, N or P) follows the order

(A) PH3< NH3<NH4+< BF3 (B) NH3< PH3<NH₄+< BF3

(C) BF3< PH3<NH₄+< NH3 (D) BF3<NH₄+< NH3< PH3

8. The ionic radii of Na+_{, F}–_{, O}2–_{, N}3–_{follow the order}

(A) O2–_{> F}–_{> Na}+_{> N}3– _{(B) N}3–_{> Na}+_{> F}–_{> O}2– (C) N3–_{> O}2–_{> F}–_{> Na}+ _{(D) Na}+_{> F}–_{> O}2–_{> N}3–

9. The oxoacid of phosphorus having the strongest reducing property is:

(A) H3PO3 (B) H3PO2 (C) H3PO4 (D) H4P2O7

10. Among C, S and P, the element(s) that produce(s) SO2 on reaction with hot conc.

H2SO4 is/are:

(A) only S (B) only C and S (C) only S and P (D) C, S and P

CO2H OH O OH OH OH CO2H O CO2H NaBH4

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KVPY-SX_2018 (CHEMISTRY)

11. The complex that can exhibit linkage isomerism is

(A) [Co(NH3)5(H2O)]Cl3 (B) [Co(NH3)5(NO2)]Cl2

(C) [Co(NH3)5(NO3)](NO3)2 (D) [Co(NH3)5Cl]SO4

12. The tendency of X in BX3 (X = F, Cl, OMe, NMe) to form a  bond with boron follows

the order

(A) BCl3< BF3<B(OMe)3< B(NMe2)3 (B) BF3< BCl3< B(OMe)3< B(NMe2)3

(C) BCl3<B(NMe2)3< B(OMe)3< BF3 (D) BCl3< BF3< B (NMe2)3< B(OMe)3

13. Consider the following statements about Langmuir isotherm:

(i) The free gas and adsorbed gas are in dynamic equilibrium (ii) All adsorption sites are equivalent

(iii) The initially adsorbed layer can act as a substrate for further adsorption

(iv) The ability of a molecule to get adsorbed at a given site is independent of the occupation of neighbouring sites

The correct statements are

(A) (i), (ii), (iii) and (iv) (B) only (i), (ii) and (iv)

(C) only (i), (iii), and (iv) (D) only (i), (ii) and (iii)

14. Among the following, the plot that correctly represents the conductometric titration

of 0.05 M H2SO4 with 0.1M NH4OH is: (A) (B) (C) (D) Volume of NH4OH Cond uc ta nce (s) Volume of NH4OH Cond uc ta nce (s) Volume of NH4OH Cond uc ta nce (s) Volume of NH4OH Cond uc ta nce (s)

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KVPY-SX_2018 (CHEMISTRY)

15. The correct representation of wavelength-intensity relationship of an ideal

blackbody radiation at two different temperatures T1 and T2 is:

(A) (B)

(C) (D)

16. The pressure (P)-volume (V) isotherm of a van der Waals gas, at the temperature at

which it undergoes gas to liquid transition, is correctly represented by

(A) (B) (C) (D)

17. A buffer solution can be prepared by mixing equal volumes of

(A) 0.2 M NH4OH and 0.1 M HCl (B) 0.2 M NH4OH and 0.2 M HCl

(C) 0.2 M NaOH and 0.1 M CH3COOH (D) 0.1 M NH4OH and 0.2 M HCl

V P V P V P V P Wavelength Int en sit y T1 T2 T2> T1 Wavelength Int en sit y T1 T2 T2> T1 Wavelength Int en sit y T2 T1 T2> T1 Wavelength Int en sit y T2 T1 T2> T1

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18. The plot of total vapour pressure as a function of mole fraction of the components of

an ideal solution formed by mixing liquids X and Y is:

(A) (B)

(C) (D)

19. On complete hydrogenation, natural rubber produces

(A) polyethylene (B) ethylene-propylene copolymer

(C) polyvinyl chloride (D) polypropylene

20. The average energy of each hydrogen bond in A-T pair is x kcal mol–1_{and that in G-C}

pair is y kcal mol–1_{. Assuming that no other interaction exists between the}

nucleotides, the approximate energy required in kcal mol–1_{to split the following}

double stranded DNA into two single strands is

[Each dashed line may represent more than one hydrogen bond between the base pairs] (A) 10x + 9y (B) 5x + 3y (C) 15x + 6y (D) 5x + 4.5 y A–T–A–T–G–C–A–G T–A–T–A–C–G–T–C Mole fraction of X 0 1 To ta l v ap ou r pr essur e Mole fraction of X 0 1 To ta l v ap ou r pr essur e Mole fraction of X 0 1 To ta l v ap ou r pr essur e Mole fraction of X 0 1 To ta l v ap ou r pr essur e

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PART-II

21. For the electrochemical cell shown below

Pt|H2 (P = 1 atm)|H+(aq., x M)||Cu2+ (aq., 1.0 M)|Cu(s)

The potential is 0.49 V at 298 K. The pH of the solution is closest to

[Given: Standard reduction potential, E° for Cu2+_{/Cu is 0.34 V}

Gas constant, R is 8.31 J K–1_mol–1

Faraday constant, F is 9.65 × 104_{J V}–1_mol–1_]

(A) 1.2 (B) 8.3 (C) 2.5 (D) 3.2

22. Consider the following reversible first-order reaction of X at an initial concentration

[X]0. The values of the rate constants are kf = 2s–1 and kb = 1s–1

A plot of concentration of X and Y as function of time is

(A) (B) (C) (D) t Concen trati on [X]0 [Y]eq [X]eq t Concen trati on [X]0 [Y]eq [X]eq t Concen trati on [X]0 [Y]eq [X]eq t Concen trati on [X]0 [Y]eq [X]eq

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23. Nitroglycerine (MW = 227.1) detonates according to the following equation:

2C3H5(NO3)3() ⎯→ 3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g)

The standard molar enthalpies of formation of o

f

H

 for all the compounds are given

below: o f H  [C3H5(NO3)3] = –364 kJ/mol o f H  [CO2(g)] = –393.5 kJ/mol o f H  [H2O(g)] = –241.8 kJ/mol o f H  [N2(g)] = 0 kJ/mol o f H  [O2(g)] = 0 kJ/mol

The enthalpy change when 10 g of nitroglycerine is determined is

(A) –100.5 kJ (B) –62.5 kJ (C) –80.3 kJ (D) –74.9 kJ

24. The heating of (NH4)2Cr2O7 produces another chromium compound along with N2

gas. The change of the oxidation state of Cr in the reaction is

(A) +6 to +2 (B) +7 to +4 (C) +8 to +4 (D) +6 to +3

25. The complex having the highest spin-only magnetic moment is

(A) [Fe(CN)6]3– (B) [Fe(H2O)6]2+ (C) [MnF6]4– (D) [NiCl4]2–

26. Among Ce(4f1_5d1_6s2_{), Nd(4f}4_6s2_{), Eu(4f}7_6s2_{) and Dy(4f}10_6s2_{), the elements having}

highest and lowest 3rd_{ionization energies, respectively, are}

(A) Nd and Ce (B) Eu and Ce (C) Eu and Dy (D) Dy and Nd

27. The major product of the following reaction sequence

is: (A) (B) (C) (D) Ph OH Me Ph Ph Me Ph Me (i) B2H6 (ii) H2O2/NaOH Ph (iii) conc. H2SO4

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28. Among the following reactions, a mixture of diastereomers is produced from

(A) _⎯⎯→HBr _(B) _⎯⎯⎯H /Pt₂ _→ (C) HBr ROOR , h ⎯⎯⎯→ (D) 2 6 2 2 B H H O /NaOH ⎯⎯⎯→

29. Reaction of phenol with NaOH followed by heating with CO2 under high pressure,

and subsequent acidification gives compound X as the major product, which can be purified by steam distillation. When reacted with acetic anhydride in the presence of

a trace amount of conc. H2SO4, compound X produces Y as the major product.

Compound Y is

(A) (B) (C) (D)

30. A tetrapeptide is made of naturally occurring alanine, serine, glycine and valine. If the

C-terminal amino acid is alanine and the N-terminal amino acid is chiral, the number of possible sequences of the tetrapeptide is

(A) 12 (B) 8 (C) 6 (D) 4 OH O O O O CO2H O OH O O O O CO2H O Me H Me H Me H Me H

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KVPY-SX_2018 (CHEMISTRY)

ANSWER KEY

1. (A) 2. (B) 3. (D) 4. (D) 5. (D)

6. (A) 7. (A) 8. (A) 9. (B) 10. (D)

11. (B) 12. (A) 13. (B) 14. (B) 15. (A) 16. (B) 17. (A) 18. (B) 19. (B) 20. (A) 21. (C) 22. (B) 23. (B) 24. (D) 25. (C) 26. (B) 27. (C) 28. (A) 29. (A) 30. (D)

SOLUTIONS

PART-I

1. (A)

Reaction of acetone with Br2 in the presence of NaOH follow as

CH3 _C CH3+3Br2 O ⎯⎯⎯→NaOH CH3 C O O Na + CHBr3+ 3HBr

Here 3 moles of Br2react with 1 mole acetone, produce 1 mole of CHBr3

 1 moles of Br₂ 1mole of CHBr₃

3 =

Hence, the moles of CHBr3 = 1moles

3

Therefore, the correct option is (A).

2. (B)

Williamson ether synthesis→

R2–X

R1–O–H+Na or k→ R – O–Na

R1–O–R2

Order of reactivity of halide→

R–I > R–Br > R–Cl > R–F (1° > 2° > 3°)

Hence, in the following reaction, we can readily prepare the above ether compound.

O | H + _I _SN2 Re action ⎯⎯⎯⎯→ O

Because, stability of carbocation (C+₎__{Rate of SN}2_

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3. (D) (1) H C H CH3 CH3 C

Convert to fischer projection

⎯⎯⎯⎯⎯⎯⎯⎯→ C C Me Me H H (2) View side CH3 ⎯⎯⎯→Rotate 180° C C Me Me H H CH3 C C H H C C H H CH3 CH3

Convert into Fischer projection

(1) & (2) are identical

→Hence, they are conformers

Therefore, the correct option is (D).

4. (D)

Hint:- See the key point and definition of hyperconjugation

(1) _CH₃_CH CH3 (6H)  CH3 C CH3 (9H)  CH3

Hyperconjugation structure = (Number of H + 1)

As the Hydrogen increases number of Hyperconjugation ( bond vacant p orbital

overlapping) structure increases and stability of carbocation also increases. Hence, tert-butyl carbocation is more stable compared to isopropyl carbocation.

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(2) CH3 CH CH2

(3-H) Or CH3 CH ₍₆__H)CH CH3

Hyperconjugation structure = (Number of H + 1)

As the  Hydrogen increase in alkene number of Hyperconjugation ( bond *

Interaction) structure increase and stability of alkene also increase. Hence, 2 butene is more stable compare to propene.

5. (D)

It is an example of cannizzaro reaction

 the reaction mechanism follows as

Reactivity order towards Nucleophilic addition→ HCHO >PhCHO

H O–CH – Ph₂ O C H O +  OH H C H O H _{PH C} H O + Ph C O O CH2 OH H C O O + H – H

6. (A)

Solution → NaBH4ideal reducing agent for reduction of carbonyl group to

corresponding alcohol without affecting other groups. Reaction follow as 4 NaBH ⎯⎯⎯→ O OH COOH COOH

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7. (A)

Bond angle

1. PH3→no hybridisation (Drago’s rule) 93°

2. NH3→ sp3hybridisation (with one lone pair) 107°

Due to lone pair-bond pair repulsion, bond angle will reduce from 109°28’ to

107°.

3. NH₄→ sp3hybridisation 109°28’

4. BF3→ sp2hybridisation 120°

8. (A)

Ionic sizewith addition of electron and size  with loss of e.

 Nitride ion has 10 electrons in contrast with only 7 protons due to the above given

reason. It has highest inter electronic repulsions. Hence its size or radius is maximum.

Note:For isoelectronic species.

Ionic size negative charge on anion.

Hence, correct order follows as  –3 –2

N O F Na+ Therefore, the correct option is (A).

9. (B) P O H3PO3 O O H H H Reducing Hydrogen 1 Hydrogen (P–H bounded) → (A) H3PO2 → P O O–H _{(2, Hydrogen)} H H (C) P HO OH (0, Hydrogen) OH H3PO4→ O (D) P HO _OH (0, Hydrogen) H4P2 O7 O P OH OH O O →

Reducing Hydrogen  P–H (Bounded Hydrogen)

→ Hence, in H3PO2 have (2) reducing H due to this it has strongest reducing

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10. (D)

Reaction of C, S & P with H2SO4 follows as

C + H2SO4⎯→ CO2 + H2O + SO2

S + H2SO4⎯→ SO2 + H2O

P + H2SO4⎯→ SO2 + H3PO4 + H2O

 Hence, here all produced SO2 gas

So, correct options is (D).

11. (B)

Linkage Isomerism→

Linkage isomerism is the existence of coordination compounds that have the same composition differing with the connectivity of metal to ligand.

 Here, in case (B) NO2 is a ambidentate ligand So it can bind with metal as N(NO2)

& O(ONO).

Hence, compound [Co(NH3)5(NO2)]Cl2 show linkage as follow:

[Co (NH3)5 (NO2)]Cl2 [Co (NH3)5 (ONO)]Cl2

Co–N Linkage Co–O

Linkage Isomers

Linkage

Therefore, the correct option is (B).

12. (A)

In All, highest back bonding in the case of B(NMe2)3 due to less E.N & small size and N

have good e density for back bonding due to +I effect of attached group.

 E.N back bonding

edensity back bonding

 Size back bonding

Hence, the overall order of back bonding follow as

B (NMe2)3> B (OMe)3> BF3> BCl3

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13. (B)

According to Langmuir isotherm→

H of adsorption of each binding site is same because surface ishomogenous.

There is a dynamic equilibrium between free gas and adsorbed gas.

The ability of a molecule to get adsorbed at a given site is independent of the

occupation of neighbouring site. Therefore, the correct option is (B).

14. (B)

→ Reaction of H2SO4 with NH4OH follow as.

2 4 4 4 2 4 2

Salt

H SO +NH OH⎯⎯→(NH ) SO +H O

Initial → H2SO4 is a strong acid so dissociation 100%

Hence due to small size of (H+_{) conductivity is maximum.}

 Upon addition of NH4OH in H2SO4, salt [(NH4)2SO4] will form and small H ion

replace by larger Ion (NH4+) so conductivity .

After salt formation, if we add Excess of NH4OH, there is no change in conductivity

because NH4OH is a week base. Hence dissociation is very less

15. (A )

From wein’s displacement law

mT =constant

1 i.e,

T



   [At higher temperature will be small]

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16. (B) P  TC→ critical + temp  vap V→ vapour

Vapour → liquid (volume )

At critical temperature ( vapour, in equilibrium)

17. (A)

(A) NH4OH + HCl ⎯→NH4Cl + H2O

Initial 0.2 mmol 0.1 mmol

After reaction 0.1 0 0.1

NH4OH & NH4Cl → from basic buffer solution.

(B) NH4OH + HCl ⎯→ NH4Cl + HCl Initial 0.2 0.2 0 After reaction 0 0 0.2 → No buffer formed (C) NH4OH + CH3COOH ⎯→ CH3COONa + H2O Initial 0.2 0.1 After reaction 0.1 0 0.10

There is no buffer formed

(D) NH4OH + HCl⎯→NH4Cl + H2O

Initial 0.1 0.2 0

After reaction 0 0.1 0.1

→ No buffer form

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18. (B)

 We know that P  x(mole fraction of solute)

For ideal solution, according to Raoult’s law

→   →  A T A B B P PartialPof A P = P +P P PartialPof B 0 0 A A A A A A 0 B P P X P = P X initialP of solvent P    _  Similar 0 B B B P = P X A B A B 0 0 T A A B B B A X + X = 1 if X = 1 X = 0 P =P X +P X X = X = 0 orP x   _   _   _ 

19. (B)

Natural rubber formed by the polymerization of isoprene and on hydrogenation provide copolymer of ethylene & propylene.

Isoprene → CH–HC CH2 Natural rubber Polymerisation CH3 H2C H2/Ni Polymerisation (ethylene) (Propylene) CH2 = CH2 + CH2 = CH –CH3

Copolymer → Polymer which is synthesized from more than one species of

monomers.

Ex. [A–B–A–B–A–B–A–B ---]n

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Hint → See the hydrogen bounded structure of A–T (Adenine –Thymine) & –

G–C(Guanine– Cytosine)

Solution

→ No O H bond between A–T =2

Average energy of each H bond in A–T pair = X Kcal mol–1

Total no of A–T pair in structure = 5

Hence, total energy = 2x ×5 = 10 x

→ No of H bond between G–C =3

Average energy of each H-bond in G–C pair = Y Kcal mol–1

Total no of G–C pair in structure = 3

Hence total energy = 3y×3 = 9y

→Now, total energy required to split all double stranded DNA in to two single

strands. =10 x + 9y

Part-II

21. (C)  Reduction potential of 2 o H /H E + = 0 V & 2 o Cu /Cu E + = 0.34V

Hence, anode & cathode reaction follow as

o cell E = o C E – o A E = 0.34 – 0 = 0.34 V  We know that Ecell = Eo_cell – 0.0591 n logQ {Q = 2 2 [H ] [Cu ] + + , [H+] = x, [Cu2+] = 1M}  0.49 = 0.34 – 0.06 2 log[H+]2  0.15 = +2 × 0.06 2 [–log[H ]] pH +  pH 0.15 0.06 = {RT f = 0.0591, n = 2}  pH = 2.5

Therefore, the correct option is (C).

22. (B)

Anode reaction → H2(g)⎯→ 2H+(aq) + 2e–

Cathode reaction → Cu2+(aq) + 2e–_⎯→_Cu_(s)

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Keq = f b K K = x –₀   = 2 {Keq = _bf K K = 2 1 = 2}   = 2x0– 2  3 = 2x0   = 2 3x0 At equilibrium [x] = xo 3 [y] = 2 3xo

Hence, here concentration ‘y’ equal to concentration ‘x’

So, graph (B) best represents this. Therefore, the correct option is (B).

23. (B)

Reaction

2C3H5(NO3)3() ⎯→ 3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g)

 We know that o f ( H ) = o f P ( H ) – (Hf)Reactant = 3 × o ₂ f N ( H ) + o ₂ f O 1_{( H )} 2  + 2 o f CO 6( H ) + o ₂ f H O 5( H ) – o _{3 5} _{3 3} f C H (NO ) 2( H ) = 0 + 0 + 6 × (–393.5) + 5 × (–241.8) – 2 × (–364) o f ( H ) = –2842 kJ

 For one mole nitroglycerine o

f ( H ) = 2812 2 − = –1421 kJ mole–1 Wt of nitroglycerine = 10 g Mw of nitroglycerine = 227 Moles of nitroglycerine = 10 227moles Hence, o f ( H ) for 10 227moles = 1421 × 10 227.1 = 62.5 kJ

Kf→ Rate constant for forward reaction

Kb→ Rate constant for backward reaction

x Kf y

Kb

x0 0

At t = 0

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Heating reaction of (NH4)2Cr2O7 follow as

Hence, change in oxidation state of Cr → +6 to +3

25. (C)

(1)[Fe(CN)6]3–

Fe3+_→_d5

(d2_sp3→_{Hybridisation)}

CN_→_{Strong field ligand, hence, pairing occur.}

Unpaired electron = 1

MM = n(n 2)+ = 1.732 BM

(2) [Fe(H2O)6]2+

Fe2+_→_d6

(sp3_d2→_{Hybridisation)}

H2O→Weak field ligand, hence, pairing does not occur.

Unpaired electron = 4 MM = 4.90 BM

(3) [MnF6]4–

Mn2+_→_d5

F–_→_{Weak field ligand}

Hence, pairing not occur Unpaired electron = 5 MM = 5.92 BM

(4) [NiCl4]2–

Cl–→_{Weak field ligand}

Hence, paring not occar Ni2+_→_d8

MM = 2.83 BM

Therefore, the correct option is (C).

3d 4s 4p 4d

3d5 _4s _4p

(NH4)2Cr2O7⎯→ N2 + Cr2O3 + 4H2O

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26. (B)

Ionisation energy→Minimum amount of energy required to remove the most loosely

bonded electron of an isolated neutral gaseous atoms or molecules.

1. Ce

4f1_5d1_6s2

Ce2+

4f2

• Least stable

• Hence, need least amount of energy to remove electron

2. Nd 4f4_6s2 Nd2+ 4f4 3. Eu 4f7_6s2 Eu2+ 4f7

• Half field →most stable, hence need more energy to

remove electron

4. Dy

4f10_6s2

Dy2+

4f10

27. (C)

Therefore, the correct option is (C). Me (B2H6) Ph H B H H CH3 Ph BH2 H2O2/NaOH CH3 Ph OH Ph 1/2H conc. H2SO4 [H+_] H  + H2O Shift Ph  H Most stable –H Ph

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28. (A)

(A)

Diastereomers→ Stereoisomers that are not mirror images, different compounds

with different physical properties. (B) ⎯⎯⎯H /Pt2 → (C) HBr ROOR , h ⎯⎯⎯→ (D) 2 6 2 2 B H H O /NaOH ⎯⎯⎯→

29. (A)

Therefore, the correct option is (A). O–H NaOH acid-base reaction ONa (i) CO2 (ii) H+ OH COOH (X) or (CH3CO)2O CH3–C–O–C–CH3 O O O–C–CH3 COOH (Y) O Aspirin + CH3COOH Me H OH Me H Me H Br H H Me H Me H Me H Me H HBr Me H Br + Me H Br Diastereomers

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30. (D) Serine (S) → Valine (V) → Alanine (A) → Chiral Compounds H NH2–C–COOH CH3 * H NH2–C–COOH CH * Me Me H NH2–C–COOH CH2–OH * Glycine → Possible sequence: Ser – Val – Gly – Ala Ser – Gly – Val – Ala Val – Ser – Gly – Ala Val – Gly – Ser – Ala Total 4.

Therefore, the correct option is (D). H

NH2–C–COOH → Achiral compound