Identify the product A in the following react | vedclass

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

Question

(A) The reaction involves the conjugate addition (also known as $1,4$-addition) of an organometallic reagent to an $\alpha,\beta$-unsaturated carbonyl

Vedclass · Accepted Answer

$2-$phenylcyclohexanone

Vedclass · Answer

(A) The reaction involves the conjugate addition (also known as $1,4$-addition) of an organometallic reagent to an $\alpha,\beta$-unsaturated carbonyl compound.In the presence of copper salts (like $CuCl$),the Grignard reagent $(PhMgBr)$ forms an organocopper species $(PhCu)$,which acts as a 'soft' nucleophile.Soft nucleophiles prefer to attack the $\beta$-carbon of the $\alpha,\beta$-unsaturated system rather than the carbonyl carbon (which is a 'hard' electrophilic site).Therefore,the phenyl group $(Ph)$ adds to the $\beta$-position of the cyclohexenone ring,followed by protonation $(H^+)$ of the resulting enolate to yield $2$-phenylcyclohexanone.

Identify the product $A$ in the following reaction:
$1$-phenylcyclohex-$1$-en-$1$-one $\xrightarrow[(2) H^+]{(1) PhMgBr/CuCl} (A) (C_{19}H_{20}O)$

Similar Questions

In the following reaction:
$\text{Aldehyde} + \text{Alcohol} \xrightarrow{HCl} \text{Acetal}$
$HCHO \qquad ^tBuOH$
$CH_3CHO \qquad MeOH$
The best combination for the formation of acetal is:

Product $(A)$ is

An organic compound $A$ reacts with methyl magnesium iodide to form an addition product,which on hydrolysis gives compound $B$. Compound $B$ gives a blue color in the Victor Meyer test. What are compounds $A$ and $B$ respectively?

What type of compounds is obtained in the first step of the Wolff-Kishner reduction of carbonyl compounds?

Which of the following does not effect the transformation shown below?
$R_2C=O \rightarrow R_2CH_2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Identify the product $A$ in the following reaction:$1$-phenylcyclohex-$1$-en-$1$-one $\xrightarrow[(2) H^+]{(1) PhMgBr/CuCl} (A) (C_{19}H_{20}O)$