The major product $[\mathrm{C}]$ in the following reaction sequence will be: \[ \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHO} \xrightarrow[\text { (ii) } \mathrm{SOCl}_{2}]{\text { (i) } \mathrm{NaBH}_{4}}[\mathrm{~A}] \xrightarrow[\substack{\mathrm{Anhy}_{3} \\ \mathrm{AlCl}_{3}}]{\ominus}[\mathrm{B}] \xrightarrow{\mathrm{DBr}}[\mathrm{C}] \]

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Hydrocarbons

The major product

[C]

in the following reaction sequence will be:

CH_{2} = CH - CHO (i) NaBH_{4} (ii) SOCl_{2} [A] ⊖ Anhy_{3} AlCl_{3} [B] DBr [C]

Answer: (A)

Acrolein,

C H_{2} = CH - CHO

, is first reduced by

NaB H_{4}

to allyl alcohol, and

SOC l_{2}

then converts the

- OH

- Cl

, giving allyl chloride

[A]

. Friedel-Crafts alkylation of benzene with this chloride and anhydrous

AlC l_{3}

attaches the three-carbon group to the ring to give

[B]

. Finally

HBr

adds across the double bond following Markovnikov's rule, placing the bromine on the more substituted carbon, so this bromide is the major product

[C]

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option