Question

An isomer of \[{C_3}{H_6}C{l_2}\] on boiling with aqueous KOH gives acetone. Hence, the isomer is?​
A. \[2,2\]-dichloropropane
B. \[1,2\]-dichloropropane
C. \[1,1\]-dichloropropane
D. \[1,3\]-dichloropropane

Answer 1

Hint: isomers are compounds with the same molecular formula but different physical and chemical properties. These physical and chemical properties may be due to difference in structure or due to special arrangement of atoms. They are called structural isomer and stereoisomers respectively. A given compound can have various types of isomers- cyclic or acyclic, saturated or unsaturated. These various possibilities can be found using DBE or double bond equivalent.

Complete step-by-step answer:DBE of \[{C_3}{H_6}C{l_2}\] is given by $C + 1 - \left( {\dfrac{{H + X - N}}{2}} \right)$
$=3 + 1 - \left( {\dfrac{{6 + 2 - 0}}{2}} \right)$

Thus, DBE$ = 0$
This means the isomers of \[{C_3}{H_6}C{l_2}\] are only acyclic.
The isomer gives acetone on boiling with $KOH$ aqueous.
Acetone has chemical formula $C{H_3} - CO - C{H_3}$ and it is a ketone. Whereas chlorine converts into
alcohol on boiling with aqueous $KOH$. Thus, the two alcohols should be placed in such a way that the two alcohols should be converted into one ketone group.
This is only possible when both the alcohols are at the same position.

Hence, both the chlorine should be at the same position.
As ketone group in acetone appears at the same carbon, so the correct isomer is \[2,2\]-dichloropropane.

Option ‘A’ is correct

Note: Two $OH$ groups at the same position are unstable and tend to release a water molecule and convert into $C = O$. This is because the energy released during the formation of $C = O$ is far more than energy released by two $C - O$. This shows extreme stability of $C = O$ as compared to $C - O$. This is the driving force of the reaction.