CH_3-CH_2-CH(CH_3)-CONH_2 xrightarrow[ Delta | vedclass

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(A) $(1)$ Hofmann Bromamide Degradation: The amide $CH_3-CH_2-CH(CH_3)-CONH_2$ reacts with $Br_2/KOH$ to form an amine with one less carbon atom,which

Vedclass · Accepted Answer

$CH_3-CH_2-CH=CH_2$

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(A) $(1)$ Hofmann Bromamide Degradation: The amide $CH_3-CH_2-CH(CH_3)-CONH_2$ reacts with $Br_2/KOH$ to form an amine with one less carbon atom,which is $CH_3-CH_2-CH(CH_3)-NH_2$ (sec-butylamine).$(2)$ Hofmann Exhaustive Methylation: The amine reacts with excess $CH_3I$ to form the quaternary ammonium salt $CH_3-CH_2-CH(CH_3)-N^{+}(CH_3)_3 I^{-}$.$(3)$ Hofmann Elimination: Heating the quaternary ammonium hydroxide (formed with $AgOH$) leads to the formation of the least substituted alkene (Hofmann product). In this case,$CH_3-CH_2-CH=CH_2$ ($1$-butene) is the major product.

$CH_3-CH_2-CH(CH_3)-CONH_2$ $\xrightarrow[ \Delta ]{ Br_2/KOH } (A)$ $\xrightarrow[ (2) AgOH/\Delta ]{ (1) CH_3I (excess) } (B)$. The major product $(B)$ is:

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