Question

Predict the correct intermediate $\left(A\right)$ and product $\left(B\right)$ from the options for the given reaction.
$C{H}_3-C\equiv CH\underset{HgS{O}_4}{\overset{H_{2}S{O}_4,H_{2}O}{\to }}\left(A\right)\to \left(B\right)$

• A. $A:C{H}_3-\underset{OH}{\underset{|}{C}}=C{H}_2;B:C{H}_3-\underset{S{O}_4}{\underset{|}{C}}=C{H}_2$
• B. $A:C{H}_3-\underset{O}{\underset{||}{C}}-C{H}_3;B:C{H}_3-C\equiv CH$
• C. $A:C{H}_3-\underset{OH}{\underset{|}{C}}=C{H}_2;B:C{H}_3-\underset{O}{\underset{||}{C}}-C{H}_3$
• D. $A:C{H}_3-\underset{S{O}_4}{\underset{|}{C}}=C{H}_2;B:C{H}_3-\underset{O}{\underset{||}{C}}-C{H}_3$

Answer 1

The correct option is C $A:C{H}_3-\underset{OH}{\underset{|}{C}}=C{H}_2;B:C{H}_3-\underset{O}{\underset{||}{C}}-C{H}_3$

For unsymmetrical alkene, addition of $H_{2}O$ takes place according to Markovnikov rule.
$C{H}_3-C\equiv CH\underset{HgS{O}_4}{\overset{H_{2}S{O}_4,H_{2}O}{\to }}\underset{\left(A\right)}{C{H}_3-\underset{OH}{\underset{|}{C}}=C{H}_2}$

$A\stackrel{\text{tautomerisation}}{\to }\underset{\left(B\right)}{C{H}_3-\underset{O}{\underset{||}{C}}-C{H}_3}$