Which statement is correct for product 'X'?Cl | vedclass

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

Question

(D) The reagent $H_2/Pd/BaSO_4$ is known as Lindlar's catalyst or a poisoned catalyst used in the Rosenmund reduction.It selectively reduces the acid

Vedclass · Accepted Answer

Alkene functional group is present in '$X$'

Vedclass · Answer

(D) The reagent $H_2/Pd/BaSO_4$ is known as Lindlar's catalyst or a poisoned catalyst used in the Rosenmund reduction.It selectively reduces the acid chloride group $(-COCl)$ to an aldehyde group $(-CHO)$ and the alkyne group $(-C \equiv CH)$ to an alkene group $(-CH=CH_2)$.It does not affect the nitrile $(-CN)$ or ketone $(-CO-)$ functional groups under these conditions.Therefore,the product '$X$' contains an alkene functional group.

Which statement is correct for product '$X$'?
$Cl-CO-CH_2-C(C \equiv CH)(CH_2-CN)-CH_2-CH_2-CO-CH_3 \xrightarrow{H_2/Pd/BaSO_4} X$

Similar Questions

Identify the final product in the following reaction sequence: $R-CHO$ $\xrightarrow{HS-(CH_2)_3-SH}$ $\xrightarrow{n-BuLi}$ $\xrightarrow{D_2O}$ $\xrightarrow{HgCl_2/CdCO_3, H_2O} \text{Product}$

Acetaldehyde on treatment with dil. $NaOH$ followed by heating gives:

Which one of the following on treatment with $50 \% \text{ aq. } NaOH$ yields the corresponding alcohol and salt of carboxylic acid?

The reaction of $50\%$ aq $KOH$ on an equimolar mixture of $4-$methylbenzaldehyde and formaldehyde followed by acidification gives

$2CD_2=O + OH^- \xrightarrow{\quad} X + Y$ ($Y$ is an alcohol and $D$ is deuterium). Identify $X$ and $Y$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Which statement is correct for product '$X$'?$Cl-CO-CH_2-C(C \equiv CH)(CH_2-CN)-CH_2-CH_2-CO-CH_3 \xrightarrow{H_2/Pd/BaSO_4} X$