This reaction involves the Hofmann bromamide degradation, which converts an amide to an amine with one less carbon atom. The given reactant is a substituted butyramide with a double bond.

Step 1: Identify the Reactant
The reactant is $CH {CH}_{3} CH = C {HCONH}_{2}$ . It is an unsaturated amide.

Step 2: Understand Hofmann Bromamide Reaction
The Hofmann bromamide reaction involves treating an amide with bromine and a strong base (like NaOH or KOH). The general reaction is: $RCONH_{2} + Br_{2} + 4 NaOH \to RNH_{2} + 2 NaBr + Na {CO}_{3} + 2 H_{2} O$ The carbon chain is shortened by one carbon, and the product is a primary amine.

Step 3: Apply to the Given Reactant
The reactant has the structure $CH_{3} CH = C HCONH_{2}$ . The amide group is -CONH₂, so the carbon adjacent to it is the one that will be lost. The double bond is not affected by this reaction.

After Hofmann degradation, the product will be an amine with the formula $CH_{3} CH = C {HNH}_{2}$ , which is CH₃CH=CH–NH₂. However, note that the double bond is between the second and third carbon in the original chain, so the product is an allylic amine: CH₃–CH=CH–NH₂.

Step 4: Compare with Options
Among the options, the correct product is $CH_{3} CH = C {HNH}_{2}$ , which is the third option: CH₃CH=CH–CH₂NH₂ is incorrect because it has an extra carbon; the correct one is CH₃CH=CH–NH₂.

Final Answer: The major product is $CH_{3} CH = C {HNH}_{2}$ .

Related Topics

Hofmann Bromamide Degradation: This is a reaction used to convert primary amides to primary amines with one less carbon atom. It involves the use of bromine and sodium hydroxide. The mechanism includes the formation of an isocyanate intermediate, which then hydrolyzes to give the amine.

Amines and Amides: Amines are derivatives of ammonia where one or more hydrogen atoms are replaced by alkyl or aryl groups. Amides are compounds with a carbonyl group linked to a nitrogen atom. The Hofmann reaction is specific to primary amides.

Formulae

General reaction: $RCONH_{2} + Br_{2} + 4 NaOH \to RNH_{2} + 2 NaBr + Na {CO}_{3} + 2 H_{2} O$