Sn1 reactions proceed through a carbocation intermediate.

Solution

Chemistry ma’am · 4 steps

In SN1 the leaving group goes first, all on its own, and that leaves a carbocation behind.

1. Step 1, slow. The C–X bond breaks heterolytically and the halide leaves with the bonding pair. The carbon is left with only six electrons and a positive charge: a carbocation.

   (CH₃)₃C-Br →(slow) (CH₃)₃C⁺ + Br⁻

2. Step 2, fast. The nucleophile then attacks the flat sp² carbocation. It can come in from either face, which is why an optically active substrate gives a largely racemic product.
3. Why unimolecular. Only the substrate takes part in the slow step, so rate = k[R–X]. That is the '1' in SN1. The carbocation is a true intermediate, not just a transition state.
4. So. The statement is true: SN1 reactions go through a carbocation intermediate.

The idea to keep. SN1 means two steps with a carbocation in between, so anything that stabilises a carbocation (3° carbon, polar protic solvent) speeds it up. SN2 is one step with no intermediate.

Answer from the NTA final answer key. Solution by PrepWiser.