The major product of the following reaction is: \[ \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH} \xrightarrow[\text { (2)DI }]{\text { (i) DCI(1equiv.) }} \]

(B) $\mathrm{CH}_{3} \mathrm{C}(\mathrm{I})(\mathrm{Cl}) \mathrm{CHD}_{2}$

(A) $\mathrm{CH}_{3} \mathrm{CD}(\mathrm{I}) \mathrm{CHD}(\mathrm{Cl})$

(C) $\mathrm{CH}_{3} \mathrm{CD}_{2} \mathrm{CH}(\mathrm{Cl})(\mathrm{I})$

(D) $\mathrm{CH}_{3} \mathrm{CD}(\mathrm{Cl}) \mathrm{CHD}(\mathrm{I})$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Hydrocarbons

The major product of the following reaction is:

CH_{3} C \equiv CH (i) DCI(1equiv.) (2)DI

(A) $CH_{3} CD (I) CHD (Cl)$
(B) $CH_{3} C (I) (Cl) CHD_{2}$ ✓ Correct
(C) $CH_{3} CD_{2} CH (Cl) (I)$
(D) $CH_{3} CD (Cl) CHD (I)$

Answer: (B)

Solution

The terminal alkyne

CH X_{3} C \equiv CH

first meets one equivalent of DCl. Addition is Markovnikov, set by the more stable vinyl cation: the positive charge sits on the internal carbon next to the methyl group, so

Cl

lands there and

D

on the terminal carbon, giving

CH X_{3} C (Cl) = CHD

. The second equivalent, DI, now adds across this double bond, again Markovnikov:

D

goes to the

= CHD

carbon to build a

CHD X_{2}

group while

I

joins the carbon already carrying the methyl and chlorine. The product is

CH X_{3} C (I) (Cl) CHD X_{2}

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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