Hydrocarbons 2 Question 4
4. The major product of the following reaction is
$$ \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH} \xrightarrow[\text { (ii) } \mathrm{DI}]{\stackrel{\text { (i) } \mathrm{DCl}}{\mathrm{DI}}} \xrightarrow{(1 \text { equiv. })} $$
(2019 Main, 9 April I)
(a) $\mathrm{CH}_{3} \mathrm{CD}(\mathrm{Cl}) \mathrm{CHD}(\mathrm{I})$
(c) $\mathrm{CH}_{3} \mathrm{CD}(\mathrm{I}) \mathrm{CHD}(\mathrm{Cl})$
(b) $\mathrm{CH}{3} \mathrm{CD}{2} \mathrm{CH}(\mathrm{Cl})(\mathrm{I})$
(d) $\mathrm{CH}{3} \mathrm{C}(\mathrm{I})(\mathrm{Cl}) \mathrm{CHD}{2}$
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Answer:
Correct Answer: 4. (c)
Solution:
- The major product obtained in the given reaction is $\mathrm{CH}{3} \mathrm{C}$ (I) (Cl) $\mathrm{CHD}{2}$.
$\mathrm{CH}{3} \mathrm{C} \equiv \mathrm{CH} \xrightarrow[\text { 2. DI }]{\text { 1. DCl (1 equiv.) }} \mathrm{CH}{3} \mathrm{C}(\mathrm{I})(\mathrm{Cl}) \mathrm{CHD}_{2}$
Addition in unsymmetrical alkynes takes place according to Markovnikov’s rule.
Reaction proceed as follows :
