The reaction \( \ce{CH3Cl + OH- -> CH3OH + Cl-} \) takes place by the SN2 mechanism. At a fixed temperature, the concentrations of both CH3Cl and OH− are doubled. How does the rate of the reaction change?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → It becomes four times
The SN2 reaction of CH3Cl with hydroxide ion follows second-order kinetics: the nucleophile and the alkyl halide take part in the same single step, so the rate depends on the concentration of both reactants.
\( \displaystyle r_1 = k[\ce{CH3Cl}][\ce{OH-}] \)
\( \displaystyle r_2 = k\left(2[\ce{CH3Cl}]\right)\left(2[\ce{OH-}]\right) \)
\( \displaystyle r_2 = 4\,k[\ce{CH3Cl}][\ce{OH-}] = 4r_1 \)
Each doubling doubles the rate, so the rate becomes four times. In an SN1 reaction the rate would depend only on the alkyl halide, and the same change would only double it.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “(a) Substitution nucleophilic bimolecular (SN2) … The reaction between CH3Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics, i.e., the rate depends upon the concentration of both the reactants.”