Question

When $2 -$heptyne was treated with aqueous sulphuric acid containing mercury$(II)$sulphate, two products, each having the molecular formula ${C_7}{H_{14}}O$, were obtained in approximately equal amount. What are these two compounds$?$
$A)$ $3 -$heptanone
$B)$heptan$3,4$dione
$C)$ $2 -$heptanone
$D)$heptan$3,4$diol

Answer 1

This is an oxymercuration reaction. Oxymercuration is not limited to an alkene reacting with water. Using an alkyne instead of an alkene yields an enol, which tautomerizes into a ketone. Using alcohol instead of water yields an ether. In both cases, Markovnikov's rule is observed.

Complete answer:
Oxymercuration involves mercury acting as a reagent attacking the alkene double bond to form a mercurinium ion bridge. A water molecule will then attack the most substituted carbon to open the mercurium ion bridge, followed by proton transfer to solvent water molecule.

When $2 -$heptyne was treated with aqueous sulphuric acid containing mercury$(II)$ sulphate it gives $3 - heptanone$ and $2 - heptanone$.
Oxymercuration is very regioselective and is a Markovnikov reaction ruling out extreme cases, the water nucleophile will always preferentially attack the more substituted carbon, depositing the resultant hydroxy group there.
$HgS{O_4}$ adds first, in an anti-Markovnikov fashion, forcing a $H$ from ${H_3}{O^ + }$ to add on the terminal carbon and thus water to add to the other side. ${H_2}S{O_4}$ makes the formation of ${H_3}{O^ + }$ more favorable.
So the correct answer is $A)$ $3 -$heptanone and $C)$ $2 -$heptanone.

Note:
During the oxymercuration reaction the mercury adduct product is almost always treated with sodium borohydride in aqueous base in a reaction called demercuration. In demercuration, the acetyl mercury group is replaced with a hydrogen in a stereochemically insensitive reaction known as reductive elimination.