Question

A carbonyl compound P ( from above given i-iii), which gives positive iodoform test, undergoes reaction woth $MeMgBr$ followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.

$P\underset{ii.H^{+},H_{2}O}{\overset{i.MeMgBr}{\to }}\underset{\Delta }{\overset{H_{2}S{O}_4}{\to }}Q\underset{ii.Zn-H_{2}O}{\overset{i.O_3}{\to }}R\underset{\Delta }{\overset{O{H}^{-}}{\to }}S$

The structure of Q and R are respectively:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Since compound gives iodoform reaction. Hence it should have $-\stackrel{O}{\stackrel{||}{C}}-C{H}_3$ group.
The compound P must be from (ii) and (iii).



Here R can give intramolecular aldol reaction.

Following reaction sequences using compound (iii), we get

Here R formed does not have $\alpha -$hydrogens and does not undergo aldol reaction.
Therefore P must be (ii).