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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$CH_3-CH_2-CH=CH_2$

Vedclass · Answer

(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:

Similar Questions

Identify the medium required for the formation of $p$-Hydroxyazobenzene from benzene diazonium chloride and phenol.

In the following reaction,the product $(A)$ is

In which of the following reactions is an alcohol formed as the major product?

Identify the reagent $R$ in the following reaction:
$C_6H_5NO_2 \xrightarrow{R} C_6H_5NHOH$

The major product of the reaction is:
$R-CONH_2 \xrightarrow{NaOBr} (A)$
(where $R$ is the chiral group shown in the image)

Vedclass Test Series

Exam Paper Generator

Online Exam Module