JEE PYQ: Hydrocarbons Question 29
Question 29 - 2019 (09 Apr Shift 1)
The major product of the following reaction is:
$\text{CH}_3\text{C}\equiv\text{CH} \xrightarrow{(i)\ \text{DCl (1 equiv.)}}$ $\xrightarrow{(ii)\ \text{DI}}$
(1) $\text{CH}_3\text{CD}(\text{I})\text{CHD}(\text{Cl})$
(2) $\text{CH}_3\text{CD}(\text{Cl})\text{CHD}(\text{I})$
(3) $\text{CH}_3\text{CD}_2\text{CH}(\text{Cl})(\text{I})$
(4) $\text{CH}_3\text{C}(\text{I})(\text{Cl})\text{CHD}_2$
Show Answer
Answer: (4)
Solution
$\text{CH}_3-\text{C}\equiv\text{C}-\text{H} \xrightarrow{\text{DCl (1 eq)}} \text{CH}_3-\text{C}(=\text{CH})(-\text{Cl})(-\text{D})$ (Markovnikov). Then $\xrightarrow{\text{DI}}$ $\text{CH}_3-\text{C}(\text{Cl})(\text{I})-\text{CHD}_2$ (Markovnikov). Both additions follow Markovnikov’s rule.
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