In the dehydrohalogenation of 2-bromobutane,w | vedclass

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

Question

(A) The dehydrohalogenation of $2$-bromobutane via the $E2$ mechanism requires an anti-periplanar arrangement of the leaving group $(-Br)$ and the $\b

Vedclass · Accepted Answer

Vedclass · Answer

(A) The dehydrohalogenation of $2$-bromobutane via the $E2$ mechanism requires an anti-periplanar arrangement of the leaving group $(-Br)$ and the $\beta$-hydrogen atom. To form $cis$-but-$2$-ene,the methyl groups on the $C_2$ and $C_3$ carbons must be on the same side of the developing double bond in the transition state. Looking at the Newman projections,the conformation that places the $-Br$ atom anti to the $\beta$-hydrogen while keeping the two methyl groups on the same side is represented by the structure in option $A$. In this conformation,the $C_2$ methyl group and the $C_3$ methyl group are eclipsed or syn-like relative to each other when viewed along the $C_2-C_3$ bond,which leads to the $cis$ isomer upon elimination.

In the dehydrohalogenation of $2$-bromobutane,which conformation leads to the formation of $cis$-but-$2$-ene?

Similar Questions

Identify the chiral molecule from the following.

The product of the reaction between ethylbenzene and $N$-bromosuccinimide $(NBS)$ is:

What product is formed when $1-$chlorobutane reacts with alcoholic $KOH$?

The major product is:

Sodium ethoxide is a specific reagent for

Vedclass Test Series

Exam Paper Generator

Online Exam Module