An SN1 reaction of an alkyl bromide is carried out in a polar protic solvent. How is the energy needed to break the C–Br bond in the slow step obtained?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → From the solvation of the bromide ion with the proton of the protic solvent
An SN1 reaction takes place in two steps. In step I, the polarised C–Br bond breaks slowly into a carbocation and a bromide ion; in step II, the nucleophile attacks the carbocation.
Breaking the bond needs energy, and in a polar protic solvent (water, alcohol, acetic acid) this comes from the solvation of the bromide ion: its attraction to the protons of the solvent stabilises the ion as it forms.
Why the other options are incorrect
Option (2) From the heat released when the carbocation combines with the nucleophile — Incorrect. The carbocation meets the nucleophile only in step II, after the C–Br bond has already broken, so this cannot supply the energy for step I.
Option (3) From the bond formed between the nucleophile and the carbon at the same time as the C–Br bond breaks — Incorrect. Bond making and bond breaking in a single step describe the SN2 mechanism, not SN1.
Option (4) From the absorption of light, which breaks the C–Br bond into free radicals — Incorrect. In SN1 the bond breaks into ions (a carbocation and a bromide ion), not into free radicals; light is not needed.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “Step I is the slowest and reversible. It involves the C–Br bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent.”