Q.Ethylidene bromide is a/an:

Concept and Intuition

The key idea here is hybridisation of the carbon atom that is directly attached to the group X. X could be any halogen or substituent — the question is about the carbon’s orbital geometry, not about X itself.

Hybridisation is determined by the number of sigma bonds and lone pairs around a carbon.

• A carbon with four sigma bonds (like in an alkane) is $sp^3$ hybridised.
• A carbon with three sigma bonds and one pi bond (like in an alkene or aromatic ring) is $sp^2$ hybridised.
• A carbon with two sigma bonds and two pi bonds (like in an alkyne) is $sp$ hybridised.

So we need to check, in each option, the carbon that holds X: is it making a double bond (or part of an aromatic ring) or only single bonds?

Let’s go through each option one by one.

Step-by-step analysis

1. Option (A): $CH_2=CH-CH_2-X$

   Draw the structure:

   $CH_2=CH-CH_2-X$

   The carbon bonded to X is the last carbon — it is $CH_2-X$. This carbon has four single bonds (to two H atoms, to the adjacent carbon, and to X). No double bond. So it is $sp^3$ hybridised.

   Not the answer.

2. Option (B): $C_6H_5-CH_2-X$

   This is benzyl halide (or benzyl derivative). The carbon attached to X is the $CH_2$ group directly attached to the benzene ring. That carbon is bonded to: the ring carbon, two H atoms, and X — again four sigma bonds, no pi bond. So it is $sp^3$ hybridised.

   Watch out

   A common mistake is to think that because the benzene ring has $sp^2$ carbons, the side-chain carbon is also $sp^2$. It is not — the $CH_2$ group is saturated. Only the ring carbons are $sp^2$.

3. Option (C): $CH_2=CH-X$ …