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(A) Step $1$: $Ph-CO-CH_3$ reacts with $PCl_5$ to form $Ph-CCl_2-CH_3$ $(A)$.Step $2$: $A$ reacts with $3 	ext{ eq. } NaNH_2/NH_3$ to undergo dehydro

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Step $1$: $Ph-CO-CH_3$ reacts with $PCl_5$ to form $Ph-CCl_2-CH_3$ $(A)$.Step $2$: $A$ reacts with $3 	ext{ eq. } NaNH_2/NH_3$ to undergo dehydrohalogenation to form the acetylide ion $Ph-C \equiv C^- Na^+$ $(B)$.Step $3$: Acidification of $B$ gives phenylacetylene $Ph-C \equiv CH$ $(C)$.Step $4$: Hydroboration-oxidation of $C$ $(1. B_2H_6, 2. H_2O_2/OH^-)$ gives an enol $Ph-CH=CH-OH$,which tautomerises to phenylacetaldehyde $Ph-CH_2-CHO$ $(D)$.Product $D$ is $Ph-CH_2-CHO$.The phenyl ring contains $6$ $sp^2$ carbon atoms,and the carbonyl carbon in the aldehyde group is also $sp^2$ hybridised.Total $sp^2$ carbon atoms $= 6 + 1 = 7$.

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