In the given reaction CH_3-CH_2-C(=O)-CH_2-CO | vedclass

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

Question

(B) The starting material contains both a ketone and an ester group.$LiAlH_4$ is a strong reducing agent that reduces both groups.To selectively reduc

Vedclass · Accepted Answer

$HO-CH_2-CH_2-OH + H^{+}$

Vedclass · Answer

(B) The starting material contains both a ketone and an ester group.$LiAlH_4$ is a strong reducing agent that reduces both groups.To selectively reduce the ester group to an alcohol while keeping the ketone group intact,the ketone must first be protected as a cyclic ketal using ethylene glycol $(HO-CH_2-CH_2-OH)$ in the presence of an acid catalyst $(H^{+})$.The subsequent acidic workup $((ii) H_2O/H^{+})$ after reduction removes the protecting group and restores the ketone functionality.

In the given reaction $CH_3-CH_2-C(=O)-CH_2-COOC_2H_5$ $\xrightarrow{[X]} (A)$ $\xrightarrow{(i) LiAlH_4, (ii) H_2O/H^{+}} CH_3-CH_2-C(=O)-CH_2-CH_2OH + C_2H_5OH$,$[X]$ will be:

Similar Questions

Which of the following reducing reagents will be suitable for the conversion shown below?
(Structure: $3-$methylcyclohexanone to methylcyclohexane)

What is the product formed when acetaldehyde reacts with $PCl_5$?

The following reagent converts $C_6H_5COCHO$ to $C_6H_5CHOHCOONa$.

The same product $(A)$ is formed in both of the following reactions:
$1. PhMgBr + CH_3-CN \xrightarrow{H_3O^{+}} (A)$
$2. Ph-COOH \xrightarrow{(1) \text{ excess } CH_3Li, (2) H_3O^{+}} (A)$
Identify product $(A)$.

The compound which does not respond to iodoform test is

Vedclass Test Series

Exam Paper Generator

Online Exam Module