JEE Advanced 2018 · Paper 2 · Chemistry Q16

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ChemistryList match+3 / -1Question 34 of 54

The desired product X can be prepared by reacting the major product of the reactions in LIST-I with one or more appropriate reagents in LIST-II.

(given, order of migratory aptitude: aryl > alkyl > hydrogen)

The correct option is

X: 2,2-diphenylpropanoic acid, Ph₂C(Me)-CO₂H

LIST-I

P
$\mathrm{Ph_2C(OH)\text{-}C(OH)Me_2}$ + $\mathrm{H_2SO_4}$
Q
$\mathrm{Ph_2C(NH_2)\text{-}CH(OH)Me}$ + $\mathrm{HNO_2}$
R
$\mathrm{PhC(OH)(Me)\text{-}C(OH)(Me)Ph}$ + $\mathrm{H_2SO_4}$
S
$\mathrm{Ph_2C(Br)\text{-}CH(OH)Me}$ + $\mathrm{AgNO_3}$

LIST-II

1
$\mathrm{I_2}$ , NaOH
2
$[\mathrm{Ag(NH_3)_2}]\mathrm{OH}$
3
Fehling solution
4
HCHO, NaOH
5
NaOBr

P → 1; Q → 2, 3; R → 1, 4; S → 2, 4

P → 1, 5; Q → 3, 4; R → 4, 5; S → 3

P → 1, 5; Q → 3, 4; R → 5; S → 2, 4

P → 1, 5; Q → 2, 3; R → 1, 5; S → 2, 3

Answer and worked solution

Official answer: D

Target X is $\ce{Ph2C(CH3)COOH}$ . It can come from the methyl ketone $\ce{Ph2C(CH3)COCH3}$ by the haloform reaction ( $\ce{I2/NaOH}$ or NaOBr), or from the aldehyde $\ce{Ph2C(CH3)CHO}$ by Tollens' or Fehling's oxidation.

P. $\ce{Ph2C(OH)-C(OH)Me2}$ with $\ce{H2SO4}$ (pinacol): the more stable cation $\ce{Ph2C+}$ forms, and a methyl migrates from the other carbon. This gives $\ce{Ph2C(Me)COMe}$ , so haloform: 1, 5.
Q. $\ce{Ph2C(NH2)-CH(OH)Me}$ with $\ce{HNO2}$ (semipinacol): $\ce{Ph2C+}$ forms, and from $\ce{CH(OH)Me}$ the methyl (alkyl > H) migrates. This gives $\ce{Ph2C(Me)CHO}$ , so Tollens or Fehling: 2, 3.
R. $\ce{PhC(OH)Me-C(OH)MePh}$ with $\ce{H2SO4}$ : a benzylic cation forms, and phenyl (aryl > alkyl) migrates. This gives $\ce{Ph2C(Me)COMe}$ , so haloform: 1, 5.
S. $\ce{Ph2C(Br)-CH(OH)Me}$ with $\ce{AgNO3}$ : $\ce{Ag+}$ removes $\ce{Br-}$ to give $\ce{Ph2C+}$ , and methyl migrates as in Q. This gives $\ce{Ph2C(Me)CHO}$ : 2, 3.

(D)

Topics:Named reactions Alcohols Oxidation and reduction

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JEE Advanced 2018 · Paper 2, Chemistry question 16