CHE 232-002

When dehydration follows aldol

A lot of people have been asking me, how do you know when an aldol reaction will be followed by elimination of H₂O (dehydration)? Here are the rules of thumb:

1. If the nucleophile is a simple, monocarbonyl compound, and LDA is used to deprotonate it, the reaction stops at the β-hydroxycarbonyl stage — no elimination occurs. That's because all the base is consumed at the outset to deprotonate all of the starting material, and there's no more base around when the electrophile is added and the aldol is obtained. Also, the Li⁺ ion coordinates between the carbonyl O and the β-O^–, making a nice, stable six-membered ring that is reluctant to break apart.

2. If the nucleophile is a simple, monocarbonyl compound, and a weak base such as NaOH or NaOEt is used to deprotonate it, elimination usually occurs (assuming there's a second α-hydrogen on the nucleophile), and the α,β-unsaturated carbonyl compound is usually obtained. That's because there's base present throughout the reaction time, shuttling protons among starting materials and products, and so the aldol product will inevitably be deprotonated some of the time, and when it is, it can expel HO^– to give the α,β-unsaturated carbonyl compound.

3. If the nucleophile is a simple, monocarbonyl compound, and the reaction occurs under acidic conditions, elimination usually occurs (assuming there's a second α-hydrogen on the nucleophile), and the α,β-unsaturated carbonyl compound is usually obtained.

4. If the nucleophile is a 1,3-dicarbonyl compound, elimination always occurs (assuming there's a second α-hydrogen on the nucleophile), and the α,β-unsaturated carbonyl compound is obtained. That's because the deprotonation of the 1,3-dicarbonyl–aldol adduct is so facile, and once it happens, loss of HO^– is inevitable.

Return to the Grossman CHE 232 page.

Return to CHE 232 page

Return to Course Listing

Return to the UK Chemistry Home Page