Halogenoalkanes: Question 9

Worked solution

Why the mechanism at a stereocentre matters

2-Bromobutane, $\text{CH}_3\text{CH}_2\text{CH}(\text{Br})\text{CH}_3$, has a stereocentre at C2: that carbon is bonded to four different groups (Br, $\text{CH}_2\text{CH}_3$, $\text{CH}_3$, H), so a single enantiomer of it is optically active. Because C2 is secondary, this substrate reacts with aqueous $\text{OH}^-$ through a mixture of the SN1 and SN2 mechanisms, some molecules react by each pathway, not all by the same one. The stereochemical outcome at C2 depends entirely on which pathway a given molecule follows.

The SN2 fraction: inversion

For the molecules that react by SN2, $\text{OH}^-$ attacks C2 directly from the side opposite the leaving $\text{Br}^-$ (“backside attack”), because that is the only direction from which the departing bromide does not block the approaching nucleophile. As the new C–O bond forms and the C–Br bond breaks in one concerted step, the other three groups on C2 are pushed through to the opposite side, like an umbrella turning inside out. This inverts the spatial arrangement of the four groups at C2, so every molecule that reacts by SN2 gives one single, specific configuration of butan-2-ol. The mirror image of what simple retention would give.

The SN1 fraction: racemisation

For the molecules that react by SN1, the C–Br bond breaks first (the rate-determining step) to give a planar carbocation at C2, with the three remaining groups ($\text{CH}_2\text{CH}_3$, $\text{CH}_3$, H) and the empty p-orbital all in one plane. $\text{OH}^-$ can then attack this planar intermediate from either face with equal probability, since both faces are equally exposed. This produces the two possible configurations of butan-2-ol in equal (50:50) amounts, a racemic mixture, from the SN1 fraction alone.

Combining the two fractions

Because 2-bromobutane reacts by both pathways simultaneously (not overwhelmingly by just one, as a primary or tertiary substrate would), the overall product is a combination of:

• a batch of one single configuration, contributed entirely by the SN2 fraction, and
• an even 50:50 split of both configurations, contributed by the SN1 fraction.

Since the SN1 portion adds equal amounts of both configurations, it does not favour either one. All of the net excess of one configuration over the other comes from the SN2 portion. The overall product is therefore still optically active (because the SN2-inverted configuration is present in excess), but its optical purity is lower than it would be if the substrate reacted by SN2 alone, because the SN1 fraction dilutes that excess with racemic material.

Why the other options are wrong

• B wrongly assumes that any SN1 character at all wipes out all optical activity; in fact only the fraction of molecules that actually goes via the planar carbocation is racemised, the SN2 fraction still contributes a genuine excess of one configuration.
• C wrongly treats the SN1 contribution from a secondary substrate as negligible; unlike a primary substrate (almost purely SN2) or a tertiary substrate (almost purely SN1), a secondary substrate has a genuinely significant proportion reacting by each pathway, so the loss of optical purity is real and measurable, not negligible.
• D confuses the chirality of the nucleophile with the stereochemical outcome at the substrate’s stereocentre; $\text{OH}^-$ does not need to be chiral itself to change the spatial arrangement of the four groups already present on C2 when it substitutes for $\text{Br}^-$ there.

Final answer

A. The SN2 fraction inverts configuration to give one specific enantiomer of butan-2-ol, while the SN1 fraction gives a racemic 50:50 mixture via the planar carbocation; combined, the product remains optically active (with the SN2-favoured configuration in excess) but with reduced optical purity compared with a purely SN2 reaction.