2,2-dichloropropane stackrel{Aq. KOH}{longrig | vedclass

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

Question

(C) Step $1$: Reaction of $2,2-$dichloropropane with aqueous $KOH$ gives a gem-diol intermediate,which loses a water molecule to form acetone $(CH_3CO

Vedclass · Accepted Answer

propane

Vedclass · Answer

(C) Step $1$: Reaction of $2,2-$dichloropropane with aqueous $KOH$ gives a gem-diol intermediate,which loses a water molecule to form acetone $(CH_3COCH_3)$ as product $A$.Step $2$: Clemmensen reduction of acetone $(CH_3COCH_3)$ using $Zn(Hg)/HCl$ reduces the carbonyl group to a methylene group $(-CH_2-)$,yielding propane $(CH_3CH_2CH_3)$ as product $B$.Therefore,the correct option is $C$.

$2,2-$dichloropropane $\stackrel{Aq. KOH}{\longrightarrow} A$ $\xrightarrow{\text{Clemmensen reduction}} B$. $B$ is

Similar Questions

Which of the following reacts with alcohol in the presence of dry $HCl$ to form an acetal?

If formaldehyde and potassium hydroxide are heated,then we get

What is the final product of this reaction?

$A$ carbonyl compound $X$ $(C_8H_8O)$ undergoes disproportionation with conc. $KOH$ on heating. The product of $X$ with $Zn-Hg/HCl$ is $Y$ and the product of $X$ with $NaBH_4$ is $Z$. What are $Y$ and $Z$ respectively?

The major product $(P)$ in the following reaction is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module