JEE Advanced 2018 · Paper 2 · Chemistry Q17 · Dronacharya Exams

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ChemistryList match+3 / -1Question 35 of 54

LIST-I contains reactions and LIST-II contains major products.

Match each reaction in LIST-I with one or more products in LIST-II and choose the correct option.

Skeletal structures. List-I: P tert-butoxide sodium + tert-butyl bromide; Q tert-butyl methyl ether + HBr; R tert-butyl bromide + NaOMe; S sodium tert-butoxide + MeBr. List-II: 1 tert-butanol; 2 tert-butyl bromide; 3 tert-butyl methyl ether; 4 isobutylene; 5 di-tert-butyl ether.

LIST-I

P
$\mathrm{(CH_3)_3CONa} + \mathrm{(CH_3)_3CBr} \rightarrow$
Q
$\mathrm{(CH_3)_3COMe} + \mathrm{HBr} \rightarrow$
R
$\mathrm{(CH_3)_3CBr} + \mathrm{NaOMe} \rightarrow$
S
$\mathrm{(CH_3)_3CONa} + \mathrm{MeBr} \rightarrow$

LIST-II

1
$\mathrm{(CH_3)_3COH}$
2
$\mathrm{(CH_3)_3CBr}$
3
$\mathrm{(CH_3)_3COMe}$
4
$\mathrm{(CH_3)_2C{=}CH_2}$
5
$\mathrm{(CH_3)_3C\text{-}O\text{-}C(CH_3)_3}$

A

P → 1,5; Q → 2; R → 3; S → 4

B

P → 1,4; Q → 2; R → 4; S → 3

C

P → 1,4; Q → 1,2; R → 3,4; S → 4

D

P → 4,5; Q → 4; R → 4; S → 3,4

Answer and worked solution

Official answer: B

P. t-BuO⁻ with t-BuBr: substitution at a tertiary carbon is impossible, so the bulky base eliminates (E2). This gives isobutylene, and t-BuO⁻ becomes t-BuOH: 1, 4.
Q. t-BuOMe with HBr: protonation, then loss of $\ce{MeOH}$ gives the stable t-Bu⁺ (SN1), which takes $\ce{Br-}$ : t-BuBr, 2.
R. t-BuBr with $\ce{MeO-}$ , a strong base and a tertiary halide: E2 gives isobutylene, 4.
S. t-BuO⁻ with MeBr: SN2 at the methyl (Williamson), giving t-BuOMe, 3.

(B)

Topics:Elimination reactions SN1 and SN2 Ethers

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