Which of the following correctly describes the transition state of the SN2 reaction between CH3Cl and the hydroxide ion?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The carbon atom is bonded to five atoms at the same time, and its three C–H bonds lie in one plane.
In an SN2 reaction the bond to the incoming nucleophile forms while the bond to the leaving group breaks, in a single step with no intermediate:
\( \displaystyle \ce{CH3Cl + OH^- -> CH3OH + Cl^-} \)
Halfway through, the –OH and the Cl are both partly attached to the carbon. Counting these two and the three hydrogens, the carbon is bonded to five atoms, and the three C–H bonds have been pushed into one plane.
Why the other options are incorrect
Option (1) The carbon atom is sp2 hybridised and carries a positive charge, because the C–Cl bond breaks before the –OH group attaches. — Incorrect. This describes the planar carbocation of an SN1 reaction, which forms when the C–X bond breaks first. In SN2 the C–Cl bond is only weakened in the transition state.
Option (2) The carbon atom is bonded to the –OH group, and the chloride ion has already left the carbon. — Incorrect. This is the product stage, methanol and a chloride ion, reached after the transition state.
Option (3) The transition state is a stable species that can be isolated before methanol is formed. — Incorrect. A carbon bonded to both the incoming and the outgoing group is unstable; such a structure cannot be isolated.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “…These two processes take place simultaneously in a single step and no intermediate is formed. … In transition state all the three C-H bonds are in the same plane and the attacking and leaving nucleophiles are partially attached to the carbon.” … “…Thus, in the transition state, carbon is simultaneously bonded to five atoms.”