Hemiacetal Ring Size, Ester Cleavage and Elimination Selectivity

When a hydroxy-aldehyde or hydroxy-ketone cyclises intramolecularly, the -OH oxygen attacks the carbonyl carbon to form a cyclic hemiacetal. Two factors decide whether this happens readily: ring size and carbonyl reactivity. Rings of 5 or 6 atoms form fastest because they combine low angle strain with a favourable entropy cost (the two reacting ends are close enough to meet without excessive loss of conformational freedom) — this is the same preference behind why glucose adopts a 6-membered pyranose ring rather than a 3- or 4-membered alternative. Separately, aldehydes are more electrophilic than ketones (fewer electron-donating alkyl groups on the carbonyl carbon, see the Grignard-reactivity cluster for the full ordering), so a hydroxy-aldehyde cyclises more readily than the corresponding hydroxy-ketone with the same tether length. When a question offers a choice between different tether lengths AND different carbonyl types, apply both filters together: prefer the aldehyde over the ketone, and prefer the tether length that produces a 5- or 6-membered ring over one that would produce a strained 3-/4-membered ring or an entropically unfavourable larger ring.

Ordinary ester acidic hydrolysis proceeds by acyl-oxygen cleavage (water attacks the carbonyl carbon). A tert-butyl ester is the exception: the tert-butyl group can form an unusually stable tertiary carbocation, so under acidic aqueous conditions the alkyl-oxygen bond breaks instead (SN₁-like), releasing the free carboxylic acid directly and generating a tert-butyl cation. That cation does not simply pick up water to give tert-butanol as its only fate — being tertiary, it readily loses a proton (E₁) to form an alkene, here isobutylene ((CH₃)₂C=CH₂), which is the species that decolourises bromine water. Whenever a question flags a tert-butyl (or other very stable carbocation-forming) ester undergoing hydrolysis and then asks about a bromine-water-reactive by-product, think alkyl-oxygen cleavage to a stabilised cation, then E₁ elimination to an alkene — not simple acyl-oxygen hydrolysis.

In E₂ elimination from a tertiary alkyl halide, a small, unhindered base (like methoxide, CH₃O⁻) can reach the more substituted (and usually more accessible) β-hydrogen, favouring the more stable, more substituted (Zaitsev) alkene. A bulky base (like tert-butoxide, (CH₃)₃CO⁻) cannot easily approach a crowded β-hydrogen and instead removes a hydrogen from a less hindered, often less substituted position, favouring the less substituted (Hofmann) alkene. To maximise the yield of an exocyclic alkene like methylenecyclohexane (where the double bond sits outside the ring, at a more hindered position) from a 1-halo-1-methylcyclohexane starting material, pair a good leaving group (bromide, not fluoride — fluoride is a poor leaving group and reacts far too slowly for a clean E₂) with a bulky, sterically hindered base, so the combination of leaving-group ability and base bulk together drives the reaction toward the Hofmann product.