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: A leaving group is the species that departs, taking the bonding electron pair with it, when a bond breaks during a reaction. For alcohols, the hydroxide ion, $OH^-$, is normally a poor leaving group because it is a strong base, so most reactions of alcohols first convert $-OH$ into a better leaving group. Step 2: In acid-catalysed dehydration, an E1 process, the alcohol is first protonated to give $-OH_2^+$, and it is water, not hydroxide, that actually leaves in the rate-determining step. Likewise, conversion of an alcohol to an alkyl halide proceeds through protonation or activation by reagents such as $SOCl_2$ or $PX_3$, so again water or another better leaving group departs, not free hydroxide ion. Step 3: Base-catalysed aldol condensation does not involve loss of a hydroxide leaving group from an alcohol at all; it proceeds through enolate formation and nucleophilic addition to a carbonyl, with a final dehydration of the beta-hydroxy carbonyl assisted by prior deprotonation, not a direct alcohol departure. Step 4: Base-promoted E2 elimination carried out directly on an alcohol under strongly basic, high-temperature conditions is the case where no prior protonation activates the hydroxyl group. The base removes a beta-hydrogen while the C-OH bond breaks in a single concerted step, so the hydroxide ion itself departs directly as the leaving group. \[ \text{Base}^- + H-C-C-OH \rightarrow \text{alkene} + \text{Base-H} + OH^- \] Step 5: This is the only listed reaction where hydroxide ion, rather than water or a converted leaving group, is expelled directly. \[\boxed{\text{Base-catalysed E2 elimination}}\]