The major product formed in the following reaction is: $\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CHCH}\left(\mathrm{CH}_{3}\right)_{2} \xrightarrow{\mathrm{HBr}}$

(B) $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{C}(\mathrm{Br})\left(\mathrm{CH}_{3}\right)_{2}$

(A) $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}(\mathrm{Br})\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$

(C) $\mathrm{Br}\left(\mathrm{CH}_{2}\right)_{3}\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$

(D) $\mathrm{CH}_{3}\mathrm{CH}(\mathrm{Br})\mathrm{CH}_{2}\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Hydrocarbons

The major product formed in the following reaction is:

CH_{3} CH = CHCH (CH_{3})_{2} HBr

(A) $CH_{3} CH_{2} CH (Br) CH (CH_{3})_{2}$
(B) $CH_{3} CH_{2} CH_{2} C (Br) (CH_{3})_{2}$ ✓ Correct
(C) $Br (CH_{2})_{3} CH (CH_{3})_{2}$
(D) $CH_{3} CH (Br) CH_{2} CH (CH_{3})_{2}$

Answer: (B)

Solution

Addition of

HBr

across the double bond follows Markovnikov's rule by way of the more stable carbocation. Protonation first gives a secondary cation, which, sitting next to the isopropyl branch, relieves itself by a hydride shift to a more stable tertiary cation. Bromide then captures this tertiary centre, so the major product carries bromine on the fully substituted carbon,

CH X_{3} CH X_{2} CH X_{2} C (Br) (CH X_{3}) X_{2}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Hydrocarbons

Solution

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