The question asks which molecule has an sp2 hybridised carbon directly bonded to the leaving group X. Only option (C) CH2=CH−X has the X attached to a vinylic carbon (which is sp2 hybridised). The correct answer is (C).
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 sp3 hybridised.
- A carbon with three sigma bonds and one pi bond (like in an alkene or aromatic ring) is sp2 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
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Option (A): CH2=CH−CH2−X
Draw the structure:
CH2=CH−CH2−X
The carbon bonded to X is the last carbon — it is CH2−X. This carbon has four single bonds (to two H atoms, to the adjacent carbon, and to X). No double bond. So it is sp3 hybridised.
Not the answer.
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Option (B): C6H5−CH2−X
This is benzyl halide (or benzyl derivative). The carbon attached to X is the CH2 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 sp3 hybridised.
A common mistake is to think that because the benzene ring has sp2 carbons, the side-chain carbon is also sp2. It is not — the CH2 group is saturated. Only the ring carbons are sp2.
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Option (C): CH2=CH−X …