JEE PYQ: Hydrocarbons Question 21
Question 21 - 2020 (04 Sep Shift 2)
The major product [R] in the following sequence of reactions is:
$\text{HC}\equiv\text{CH} \xrightarrow{(i)\ \text{LiNH}_2/\text{ether}}$ $\xrightarrow{(ii)\ \text{H}_3\text{C}-\text{CH}(-\text{Br})-\text{CH}(\text{CH}_3)_2}$ [P] $\xrightarrow{(i)\ \text{HgSO}_4/\text{H}_2\text{SO}_4}$ $\xrightarrow{(ii)\ \text{NaBH}_4}$ [Q] $\xrightarrow{\text{Conc.}\ \text{H}_2\text{SO}_4/\Delta}$ [R]
(1) $\text{H}_2\text{C}=\text{C}(-\text{CH}_2-\text{CH}_3)-\text{CH}(\text{CH}_3)_2$
(2) $\text{H}_3\text{C}-\text{C}(=\text{CH}-\text{CH}_3)-\text{CH}(\text{CH}_3)_2$
(3) $\text{H}_3\text{C}-\text{C}(=\text{C}(\text{CH}_3)_2)-\text{CH}_2\text{CH}_3$ (trisubstituted alkene)
(4) $\text{H}_3\text{C}-\text{CH}(-\text{CH}=\text{CH}_2)-\text{CH}(\text{CH}_3)_2$
Show Answer
Answer: (3)
Solution
$\text{HC}\equiv\text{CH} \xrightarrow{\text{LiNH}_2/\text{ether}} \text{HC}\equiv\text{C}^-\text{Li}^+$. Alkylation with $\text{H}_3\text{C}-\text{CH}(-\text{Br})-\text{CH}(\text{CH}_3)_2$ gives [P]. $\text{HgSO}_4/\text{H}_2\text{SO}_4$ hydrates the triple bond (Markovnikov) to give a ketone. $\text{NaBH}_4$ reduces ketone to alcohol [Q]. Conc. $\text{H}_2\text{SO}_4/\Delta$ dehydrates [Q]. Secondary carbocation rearranges to tertiary, giving the trisubstituted alkene [R].
BETA
