Identify substrate A in the following reactio | vedclass

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

Question

(A) The reaction is Hofmann elimination.Substrate $A$ is diethyldimethyl ammonium halide,which reacts with moist $Ag_2O$ (forming $AgOH$) to produce t

Vedclass · Accepted Answer

Diethyldimethyl ammonium halide

Vedclass · Answer

(A) The reaction is Hofmann elimination.Substrate $A$ is diethyldimethyl ammonium halide,which reacts with moist $Ag_2O$ (forming $AgOH$) to produce the quaternary ammonium hydroxide.$(CH_3CH_2)_2N^+(CH_3)_2X^- + AgOH \rightarrow (CH_3CH_2)_2N^+(CH_3)_2OH^- + AgX$Upon heating,the quaternary ammonium hydroxide undergoes Hofmann elimination to form the alkene and the tertiary amine.$(CH_3CH_2)_2N^+(CH_3)_2OH^- \xrightarrow{\Delta} CH_3CH_2N(CH_3)_2 + CH_2=CH_2 + H_2O$

Column-$I$	Column-$II$
$A$. Benzenesulphonyl chloride	$1$. Zwitter ion
$B$. Sulphanilic acid	$2$. Hinsberg reagent
$C$. Alkyl diazonium salts	$3$. Dyes
$D$. Aryl diazonium salts	$4$. Conversion to alcohols

Identify substrate $A$ in the following reaction.
$A + AgOH \xrightarrow[ii) \Delta]{i) \text{moist } Ag_2O} CH_3CH_2N(CH_3)_2 + CH_2=CH_2$

Similar Questions

Match the compounds given in Column-$I$ with the items given in Column-$II$.

Column-$I$ Column-$II$

$A$. Benzenesulphonyl chloride $1$. Zwitter ion

$B$. Sulphanilic acid $2$. Hinsberg reagent

$C$. Alkyl diazonium salts $3$. Dyes

$D$. Aryl diazonium salts $4$. Conversion to alcohols

Among the isomeric amines,select the one with the lowest boiling point.

The final product formed when $Methyl \ amine$ is treated with $NaNO_2$ and $HCl$ is

The increasing order of the $pK_b$ of the following compounds is:

Identify the product of the following reaction:
$\text{Aniline} \xrightarrow[ii) \text{H}_2\text{O}, \Delta]{i) \text{NaNO}_2 + \text{HCl}, 273 \text{ K}} \text{Product}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Identify substrate $A$ in the following reaction.$A + AgOH \xrightarrow[ii) \Delta]{i) \text{moist } Ag_2O} CH_3CH_2N(CH_3)_2 + CH_2=CH_2$