2-Methyl propyl bromide reacts with $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$and gives 'A' whereas on reaction with $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ it gives ' $B$ '. The mechanism followed in these reactions and the products 'A' and 'B' respectively are:

(A) $\mathrm{S}_{\mathrm{N}} 2 . \mathrm{A}=$ iso-butyl ethyl ether; $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=$ tertbutyl ethyl ether

(B) $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~A}=$ tert-butyl ethyl ether; $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=2$ butyl ethyl ether

(C) $\mathrm{S}_{\mathrm{N}} 1, \mathrm{~A}=$ tert-butyl ethyl ether; $\mathrm{S}_{\mathrm{N}} 2, \mathrm{~B}=$ isobutyl ethyl ether

(D) $\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}=2$-butyl ethyl ether; $\mathrm{S}_{\mathrm{N}} 2, \mathrm{~B}=$ isobutyl ethyl ether

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Haloalkanes and Haloarenes

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2-Methyl propyl bromide reacts with

C_{2} H_{5} O^{-}

and gives 'A' whereas on reaction with

C_{2} H_{5} OH

it gives '

B

'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are:

(A) $S_{N} 2. A =$ iso-butyl ethyl ether; $S_{N} 1, B =$ tertbutyl ethyl ether✓ Correct
(B) $S_{N} 1, A =$ tert-butyl ethyl ether; $S_{N} 1, B = 2$ butyl ethyl ether
(C) $S_{N} 1, A =$ tert-butyl ethyl ether; $S_{N} 2, B =$ isobutyl ethyl ether
(D) $S_{N} 2, A = 2$ -butyl ethyl ether; $S_{N} 2, B =$ isobutyl ethyl ether

Answer: (A)

Solution

With C₂H₅O⁻ (strong nucleophile), 2-methylpropyl bromide undergoes SN2 mechanism to give iso-butyl ethyl ether (A). With C₂H₅OH (weak nucleophile), the reaction proceeds via SN1 mechanism: the primary bromide first rearranges to form a more stable tertiary carbocation, which then reacts with ethanol to give tert-butyl ethyl ether (B).

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Haloalkanes and Haloarenes

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