Step 1: Formation of A.
Acetophenone reacts with \( PCl_5 \) to give a geminal dichloride:
\[
Ph-\overset{O}{\underset{||}{C}}-CH_3 \xrightarrow{PCl_5} Ph-CCl_2-CH_3 \quad (A)
\]
Step 2: Formation of B.
Reaction with 3 equivalents of \( NaNH_2/NH_3 \) leads to elimination of HCl and formation of an alkyne:
\[
Ph-CCl_2-CH_3 \xrightarrow{3eq. \, NaNH_2/NH_3} Ph-C \equiv C-CH_3 \quad (B)
\]
Step 3: Formation of C.
Acidification does not change the structure of the alkyne. Thus, C is also \( Ph-C \equiv C-CH_3 \).
Step 4: Formation of D (Hydroboration-oxidation).
Hydroboration-oxidation of a terminal alkyne with \( B_2H_6 \) followed by \( H_2O_2/OH^- \) proceeds with anti-Markovnikov regioselectivity, placing the hydroxyl group on the less substituted carbon after tautomerization.
\[
Ph-C \equiv C-CH_3 \xrightarrow[1. \, B_2H_6]{2. \, H_2O_2/OH^-} Ph-\overset{OH}{C}=CH-CH_3 \xrightarrow{Tautomerization} Ph-CO-CH_2-CH_3 \quad (D)
\]
The final product D is 1-phenylpropan-2-one.
Step 5: Determine the number of \( sp^2 \) hybridized carbon atoms in D.
The structure of 1-phenylpropan-2-one is:
\[
\underbrace{\overset{sp^2}{C}_6H_5}_{\text{6 } sp^2 \text{ carbons}} - \overset{sp^3}{CH_2} - \overset{sp^2}{C} = O - \overset{sp^3}{CH_3}
\]
The phenyl ring has 6 \( sp^2 \) hybridized carbon atoms. The carbonyl carbon is also \( sp^2 \) hybridized. The \( CH_2 \) and \( CH_3 \) carbons are \( sp^3 \) hybridized.
Total number of \( sp^2 \) hybridized carbon atoms in product D = 6 (from the phenyl ring) + 1 (carbonyl carbon) = 7.