In the following reaction,the major product X | vedclass

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Question

(A) The reaction proceeds in two steps:Step $1$: Ozonolysis of $1,2$-dimethylcyclopentene followed by reductive workup $(Zn/H_2O)$ leads to the format

Vedclass · Accepted Answer

$3-$methylcyclohex$-2-$en$-1-$one

Vedclass · Answer

(A) The reaction proceeds in two steps:Step $1$: Ozonolysis of $1,2$-dimethylcyclopentene followed by reductive workup $(Zn/H_2O)$ leads to the formation of heptane-$2,6$-dione.Step $2$: The resulting heptane-$2,6$-dione undergoes an intramolecular aldol condensation in the presence of base $(OH^-)$. The enolate formed at the $\alpha$-carbon attacks the other carbonyl group to form a $\beta$-hydroxy ketone intermediate. This intermediate subsequently undergoes dehydration (loss of $H_2O$) to form the stable $\alpha,\beta$-unsaturated carbonyl compound,which is $3$-methylcyclohex-$2$-en-$1$-one.

List-$I$	List-$II$
$A. \ C_6H_5NHNH_2$	$I. \ \text{Schiff base}$
$B. \ NH_2OH$	$II. \ \text{Hydrazone}$
$C. \ C_6H_5NH_2$	$III. \ \text{Oxime}$
	$IV. \ \text{Phenyl hydrazone}$

In the following reaction,the major product $X$ is

Similar Questions

Which of the following will not give the iodoform test?

Identify the reagent $R$ used in the following reaction:
$\text{Ketone} \xrightarrow{R} \text{Semicarbazone}$

Match the following reagents with the products obtained when they react with a ketone.
List-$I$ List-$II$
$A. \ C_6H_5NHNH_2$ $I. \ \text{Schiff base}$
$B. \ NH_2OH$ $II. \ \text{Hydrazone}$
$C. \ C_6H_5NH_2$ $III. \ \text{Oxime}$
$IV. \ \text{Phenyl hydrazone}$

Correct answer is:

Which is the major product of the following reaction?

The $Wolff-Kishner$ reduction is used for which of the following?

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