LDA = text{Lithium di-isopropyl amide}. In th | vedclass

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

Question

(A) The reaction is a base-induced decomposition of a hydrazone derivative,often referred to as a Shapiro-type reaction or related elimination.
$1$.

Vedclass · Accepted Answer

Vedclass · Answer

(A) The reaction is a base-induced decomposition of a hydrazone derivative,often referred to as a Shapiro-type reaction or related elimination.
$1$. $LDA$ (a strong,sterically hindered base) abstracts an $\alpha$-proton from the alkyl chain attached to the hydrazone carbon.
$2$. This leads to the formation of a vinylic carbanion intermediate.
$3$. The carbanion undergoes elimination,releasing $N_2$ gas and the alkene $Ph-CH=CH_2$ as byproducts.
$4$. The final product $A$ is the alkene formed by the double bond between the original hydrazone carbon and the adjacent carbon of the alkyl chain.
$5$. Based on the structure,the product $A$ is $1	ext{-cyclohexylbut-1-ene}$ (or a similar isomer depending on the specific alkyl chain structure provided in the image). Looking at the options,the structure corresponds to the formation of the double bond at the position where the hydrazone was located.

$LDA = \text{Lithium di-isopropyl amide}$. In the given reaction,identify product $A$:

Similar Questions

Reagent $(A)$ in the reaction is

The major product of the following reaction is $:$

How many optical isomers are possible for the compound obtained by the hydrogenation of $(CH_3)_2C = CH-CH_3$?

The number of hydroxyl group$(s)$ in $Q$ is

The major product of the reaction of $1,3$-cyclohexadiene with $1 \text{ mole}$ of $HBr$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module