Find out the major product of the following r | vedclass

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

Question

(A) The reaction involves the formation of a Grignard reagent from an allyl bromide followed by a Wurtz-type coupling reaction.$1$. $CH_2=CH-CH_2Br +

Vedclass · Accepted Answer

$CH_2=CH-CH_2-CH_2-CH=CH_2$

Vedclass · Answer

(A) The reaction involves the formation of a Grignard reagent from an allyl bromide followed by a Wurtz-type coupling reaction.$1$. $CH_2=CH-CH_2Br + Mg \xrightarrow{T.H.F.} CH_2=CH-CH_2MgBr$ (Allylmagnesium bromide).$2$. The Grignard reagent reacts with another molecule of allyl bromide: $CH_2=CH-CH_2MgBr + CH_2=CH-CH_2Br \rightarrow CH_2=CH-CH_2-CH_2-CH=CH_2 + MgBr_2$.This is a coupling reaction resulting in the formation of $1,5-hexadiene$.

Find out the major product of the following reaction:
$2 \text{ mole } CH_2=CH-CH_2Br \xrightarrow{Mg, T.H.F.} \text{?}$

Similar Questions

Identify the incorrect order of heat of hydrogenation $(HOH)$.

The reaction of an alkene $X$ with bromine produces a compound $Y$,which has $22.22 \% \ C$,$3.71 \% \ H$ and $74.07 \% \ Br$. The ozonolysis of alkene $X$ gives only one product. The alkene $X$ is,
[Given,atomic mass of $C = 12$; $H = 1$; $Br = 80$ ]

For the reaction: $CH_2=CH-CH=CH_2 + HBr (1 \text{ eq}) \xrightarrow{80^{\circ}C} P_1 + P_2$ (where $P_1$ is the major product),identify $P_1$.

Diimide $(N_2H_2)$ is used to reduce the double bond of:

What is $X$ in the following reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find out the major product of the following reaction:$2 \text{ mole } CH_2=CH-CH_2Br \xrightarrow{Mg, T.H.F.} \text{?}$