The product (B) in the above reaction is: | vedclass

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(A) $1$. The starting material is $CH_3(^{14}CH)-CH_2-CHO$. Upon treatment with base and heat,it undergoes dehydrohalogenation to form crotonaldehyde,

Vedclass · Accepted Answer

$CH_3-CH=CH-CH=CH-CHO$

Vedclass · Answer

(A) $1$. The starting material is $CH_3(^{14}CH)-CH_2-CHO$. Upon treatment with base and heat,it undergoes dehydrohalogenation to form crotonaldehyde,$CH_3-CH=CH-CHO$,where the $^{14}C$ is at the $\alpha$-position relative to the aldehyde group,i.e.,$CH_3-CH=^{14}CH-CHO$ is not correct,it is $CH_3-CH=CH-CHO$ with $^{14}C$ at the methyl group. $2$. The reaction of $(A)$ $(CH_3-CH=CH-CHO)$ with $CH_3CHO$ in the presence of dilute $HO^-$ (aldol condensation) leads to the formation of a conjugated system. $3$. The enolate formed from $(A)$ attacks the carbonyl carbon of acetaldehyde. $4$. The final product $(B)$ is $CH_3-CH=CH-CH=CH-CHO$ where the $^{14}C$ label is maintained at the terminal methyl group.

The product $(B)$ in the above reaction is:

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