CH_3-CO-CH_2-CH_2-CH_2-CH_2-CHO xrightarrow{K | vedclass

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Question

(B) The given compound is $CH_3-CO-CH_2-CH_2-CH_2-CH_2-CHO$. Intramolecular aldol condensation occurs between the carbonyl carbon of the aldehyde and

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The given compound is $CH_3-CO-CH_2-CH_2-CH_2-CH_2-CHO$. Intramolecular aldol condensation occurs between the carbonyl carbon of the aldehyde and the $\alpha$-hydrogens of the ketone or vice versa. $1$. The aldehyde group $(-CHO)$ has an $\alpha$-carbon at position $2$ (relative to $CHO$),but it has no $\alpha$-hydrogens. $2$. The ketone group $(-CO-)$ has $\alpha$-hydrogens at the $CH_3$ group and the $CH_2$ group. $3$. Attack of the enolate formed from the $CH_2$ group (adjacent to the ketone) on the aldehyde carbonyl leads to a $6$-membered ring. $4$. Attack of the enolate formed from the $CH_3$ group on the aldehyde carbonyl leads to a $8$-membered ring. $5$. Attack of the enolate formed from the $CH_2$ group on the ketone carbonyl is not possible as it would form a highly strained $3$-membered ring. Considering the stability of the rings formed,the $6$-membered ring product is the major one. However,in the context of intramolecular aldol condensation for this specific molecule,there are $2$ possible cyclic products formed via different enolate attacks on the aldehyde carbonyl. Thus,the correct answer is $2$.

$CH_3-CO-CH_2-CH_2-CH_2-CH_2-CHO \xrightarrow{KOH} ?$ The number of intramolecular aldol condensation products is:

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