Organic Chemistry · Chapter 7: Aldehydes & Ketones II

Question

Aldol Condensation: Formation of Aldol Reaction
Step #1: The base removes a proton on the α carbon to form a resonance-stabilized enolate.
Step #2: ____.
Step #3: Protonation of the alkoxide forms the β-hydroxy aldehyde.

Answer

Aldol Condensation: Formation of Aldol Reaction
Step #1: The base removes a proton on the α carbon to form a resonance-stabilized enolate.
Step #2: Nucleophilic attack of the enolate on an electrophilic carbonyl in another molecule of aldehyde forms a new C–C bond.
Step #3: Protonation of the alkoxide forms the β-hydroxy aldehyde.
Aldol Condensation: Formation of Aldol Reaction Step #1: The base removes a proton on the α carbon to form a resonance-…
Full Reaction Examples ↓:
Note ↑: See the conjugated product above, α,ß-unsaturated carbonyl compound, it can be later used in Michael addition reaction: which eventually gives you a 1,4-carbonyl product.
Note: If there are multiple aldehydes or ketones, we cannot easily control which will act as the nucleophile and which will act as the electrophile, and a mixture of products will result. This can be prevented if one of the molecules has no α-hydrogens because the α-carbons are quaternary (like benzaldehyde).

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More cards in Chapter 7: Aldehydes & Ketones II

____ - carbon is adjacent to the carbonyl carbon.

Through ____, oxygen pulls some of the electron density out of these α-Hydrogens bonded to Carbons, weakening them.

Through induction, oxygen pulls some of the ____ out of these α-Hydrogens bonded to Carbons, weakening them.

Through induction, oxygen pulls some of the electron density out of these α-Hydrogens bonded to Carbons, ____ them.

Oxygen pulling on C-H bonds with α-hydrogens, makes it relatively easy to ____ the α -carbon of an aldehyde or ketone.

The ____ of α -hydrogens is augmented by resonance stabilization of the conjugate base.

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