AS Chemistry 9701: Why Secondary Halogenoalkanes React by a Mixed SN1/SN2 Mechanism

1-Bromobutane, 2-bromobutane and 2-bromo-2-methylpropane are all reacted separately with aqueous sodium hydroxide.

(a) State which mechanism, SN1 or SN2, is followed almost exclusively by 1-bromobutane, and which is followed almost exclusively by 2-bromo-2-methylpropane. [2]

(b) Explain why 2-bromobutane, unlike the other two compounds, reacts by a mixture of both the SN1 and SN2 mechanisms. Refer to both a steric factor and an electronic (carbocation stability) factor in your answer. [4]

(c) A student increases the concentration of $\text{OH}^-(\text{aq})$ used with 2-bromobutane. Explain the effect this has on the relative proportion of product formed via the SN2 pathway compared with the SN1 pathway. [3]

Part (a): Mechanisms for the primary and tertiary compounds

1-Bromobutane is a primary halogenoalkane. Its C–Br carbon is uncrowded, so $\text{OH}^-$ can approach and attack from the side directly opposite the leaving bromine with little steric hindrance, and a primary carbocation is too unstable to form. It reacts almost entirely by SN2.
2-Bromo-2-methylpropane is a tertiary halogenoalkane. Three alkyl groups crowd the C–Br carbon (blocking backside attack) and strongly stabilise the carbocation that forms when the C–Br bond breaks. It reacts almost entirely by SN1.

Part (b): Why 2-bromobutane reacts by a mixture

2-Bromobutane is a secondary halogenoalkane: its C–Br carbon is bonded to two other carbons.

Steric factor: with two alkyl groups (rather than three) around the C–Br carbon, there is some hindrance to direct backside attack by $\text{OH}^-$ , but not enough to block SN2 completely. SN2 can still occur, just more slowly than for a primary substrate.
Electronic factor: if the C–Br bond does break first, the resulting secondary carbocation is stabilised by the inductive electron donation of its two alkyl groups. This is more stabilisation than a primary carbocation gets (so it can form often enough to matter) but less than a tertiary carbocation gets (so it does not form so readily that it dominates completely).

Because neither pathway is either fully blocked or fully dominant, both the two-step SN1 route (via the moderately stable secondary carbocation) and the one-step SN2 route (via the only partly hindered backside attack) occur simultaneously, giving a mixture of products/pathways for 2-bromobutane.

Part (c): Effect of increasing [OH⁻]

The two pathways have different rate equations: $\text{SN2: rate} = k_2[\text{RBr}][\text{OH}^-] \qquad \text{SN1: rate} = k_1[\text{RBr}]$

Increasing $[\text{OH}^-(\text{aq})]$ directly increases the rate of the SN2 pathway, since $\text{OH}^-$ is a reactant in its rate-determining (and only) step. The SN1 pathway’s rate-determining step is the ionisation of the C–Br bond, which does not involve $\text{OH}^-$ at all, so its rate is unaffected by $[\text{OH}^-]$ .

As a result, raising $[\text{OH}^-]$ increases the proportion of 2-bromobutane that reacts via SN2 relative to SN1, even though the SN1 pathway continues to occur at the same rate as before.

Final answers

(a) 1-bromobutane: SN2. 2-bromo-2-methylpropane: SN1.
(b) Steric hindrance is only partial (two, not three, alkyl groups) so SN2 still proceeds; the secondary carbocation is moderately stabilised (more than primary, less than tertiary) so SN1 can also compete, both pathways therefore occur together.
(c) Increasing $[\text{OH}^-]$ speeds up SN2 (rate depends on $[\text{OH}^-]$ ) but not SN1 (rate is independent of $[\text{OH}^-]$ ), so a greater proportion of product forms via SN2.