Acetaldehyde reacts with semicarbazide,the pr | vedclass

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Question

(A) Acetaldehyde $(CH_3CHO)$ reacts with semicarbazide $(NH_2NHCONH_2)$ to form acetaldehyde semicarbazone $(CH_3CH=NNHCONH_2)$.This is a nucleophilic

Vedclass · Accepted Answer

$CH_3CH=NNHCONH_2$

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(A) Acetaldehyde $(CH_3CHO)$ reacts with semicarbazide $(NH_2NHCONH_2)$ to form acetaldehyde semicarbazone $(CH_3CH=NNHCONH_2)$.This is a nucleophilic addition-elimination reaction.The nitrogen atom of the $-NH_2$ group attached to the $-NH-$ group acts as the nucleophile because its lone pair is not involved in resonance with the carbonyl group,unlike the amide nitrogen.$CH_3CHO + NH_2NHCONH_2 \rightarrow CH_3CH=NNHCONH_2 + H_2O$

Acetaldehyde reacts with semicarbazide,the product will be:

Similar Questions

The correct order of rate of reaction toward nucleophilic addition reaction for the following compounds is:
$(a)$ $p-NO_2-C_6H_4-CHO$
$(b)$ $C_6H_5-CHO$
$(c)$ $p-OMe-C_6H_4-CHO$
$(d)$ $CH_3-CO-CH_2CH_3$
$(e)$ $C_6H_5-CO-C_6H_5$

What is the product obtained in the reaction of acetaldehyde with semicarbazide?

Which one of the following is an acetal?

When an aldehyde is treated with an excess of monohydric alcohol,what is formed?

The major product formed in the following reaction is:

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