Question

In which of the following reactions does hydroxide itself act as a leaving group?

• A. Base-catalysed E2 elimination
• B. Base-catalysed E1 elimination
• C. Base-catalysed aldol condensation
• D. Conversion of alcohol to alkyl halide

Hint:

Ask which of these reactions removes the OH without first protonating it or converting it into another group.

Answer 1

The Correct Option is A

Step 1: Solve this by elimination of the wrong options first, based on what actually leaves as the leaving group in each named reaction. Step 2: Conversion of alcohol to alkyl halide: this reaction always needs the $-OH$ turned into something that leaves more easily, either by protonation with $HX$ so $H_2O$ leaves, not $OH^-$, or by forming an intermediate such as a chlorosulfite ester with $SOCl_2$ so a different, better leaving group departs. Hydroxide ion never leaves directly here, so this option is ruled out. Step 3: Base-catalysed aldol condensation: the mechanism is deprotonation of an alpha-hydrogen to form an enolate, nucleophilic attack of that enolate on a carbonyl carbon, and finally an E1cb-type dehydration of the resulting beta-hydroxy carbonyl. Even in that last dehydration step, the base first removes an alpha-hydrogen to form a stabilized carbanion, and the C-OH bond breaks in a step assisted by conjugation with the carbonyl, so this is not hydroxide leaving directly from a simple alcohol either. Step 4: Base-catalysed E1 elimination is actually a mismatched phrase: E1 mechanisms are defined by a slow, unimolecular ionization step that generates a carbocation before any base gets involved, and that ionization of an alcohol needs acid catalysis, protonation to $-OH_2^+$, because $OH^-$ is too poor a leaving group to depart on its own. So base-catalysed E1 does not correspond to hydroxide leaving directly either. Step 5: That leaves base-catalysed E2 elimination performed directly on an alcohol. Here a strong base abstracts a beta-hydrogen in the very same step that the C-OH bond breaks, forming the new C=C double bond and releasing $OH^-$ directly, with no protonation and no conversion to another leaving group needed beforehand. \[ B^- + H-C-C-OH \rightarrow B-H + C=C + OH^- \] By process of elimination and by the concerted mechanism itself, this is the reaction where hydroxide acts as the leaving group. \[\boxed{\text{Base-catalysed E2 elimination}}\]