The product A in the following reaction is: | vedclass

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

Question

(A) The reaction involves the alkylation of an unsymmetrical ketone,$2$-methylcyclohexanone,using a strong,sterically hindered base,$LDA$ (Lithium dii

Vedclass · Accepted Answer

$2-$ethyl$-6-$methylcyclohexanone

Vedclass · Answer

(A) The reaction involves the alkylation of an unsymmetrical ketone,$2$-methylcyclohexanone,using a strong,sterically hindered base,$LDA$ (Lithium diisopropylamide). $LDA$ is a kinetic base that selectively removes a proton from the less sterically hindered $\alpha$-carbon. In $2$-methylcyclohexanone,there are two types of $\alpha$-carbons: one is the tertiary carbon (substituted with a methyl group) and the other is the secondary carbon ($-CH_2-$ group). $LDA$ removes a proton from the less hindered $-CH_2-$ $\alpha$-carbon to form the kinetic enolate. Subsequent reaction of this kinetic enolate with the alkylating agent,ethyl iodide $(CH_3-CH_2-I)$,leads to the alkylation at the less substituted $\alpha$-carbon. Therefore,the major product $A$ is $2$-ethyl-$6$-methylcyclohexanone.

The product $A$ in the following reaction is:

Similar Questions

If formaldehyde and potassium hydroxide are heated,then we get

Which of the following compounds will not undergo an aldol condensation reaction?

Following compounds are given:
$(i) CH_3-CH_2-OH$
$(ii) CH_3-CO-CH_3$
$(iii) CH_3-CH(OH)-CH_3$
$(iv) CH_3-OH$
Which of the above compound$(s)$,on being warmed with iodine solution and $NaOH$,will give iodoform?

In the given reaction $X + Y \xrightarrow[5^\circ C]{NaOH} CH_3-CH(OH)-CH(CH_3)-CHO$,$(X)$ and $(Y)$ will respectively be:

Which of the following does not give a yellow precipitate with $I_2$ and $NaOH$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module