Test: Aldehydes and Ketones - 2 - MCAT MCQ

One mole of aldehyde reacts with one mole of alcohol via a nucleophilic addition reaction to form a product called a hemiacetal. In a hemiacetal, an –OH group, an –OR group, a hydrogen atom, and an –R group are attached to the same carbon atom.

The reactions listed in the answer choices are examples of aldol condensations. In the presence of a base, the α-hydrogen is abstracted from an aldehyde, forming an enolate ion, [CH₃CHCHO]^–. This enolate ion then attacks the carbonyl group of the other aldehyde molecule, CH₃CH₂CHO, forming the pictured aldol.

The aldol condensation is both a dehydration reaction because a molecule of water is lost, and a nucleophilic addition reaction because the nucleophilic enolate attacks and binds to the carbonyl carbon.

At high temperatures and with a weak base like NH3, the thermodynamic enolate will be favored. The reaction proceeds slowly with the weak base, giving the kinetic enolate time to interconvert to the more stable thermodynamic enolate.

When α-carbons are deprotonated, the negative charge is resonance stabilized in part by the electronegative carbonyl oxygen, which is electron-withdrawing. Alkyl groups are actually electron-donating, which destabilizes carbanion intermediates; this invalidates statement II.

When succinaldehyde (or any aldehyde or ketone with α-hydrogens) is treated with a strong base like lithium diisopropylamide (LDA), it forms the more nucleophilic enolate carbanion.

Because benzaldehyde lacks an α-proton, it cannot be reacted with base to form the nucleophilic enolate carbanion. Therefore, acetone will act as our nucleophile, and both choices (A) and (B) can be eliminated. In order to perform this reaction, which is an aldol condensation, acetone will be reacted with a strong base—not a strong acid—in order to extract the α-hydrogen and form the enolate anion, which will act as a nucleophile.

The nomenclature in this question is well above what one needs to be able to draw on the MCAT; however, we can discern that we are forming a ketone and an aldehyde from a single molecule. The hallmark of a reverse aldol reaction is the breakage of a carbon–carbon bond, forming two aldehydes, two ketones, or one of each. In an aldol condensation, choice (A), we would expect to form a single product by combining two aldehydes, two ketones, or one of each. A dehydration reaction, choice (C), should release a water molecule, rather than breaking apart a large organic molecule into two smaller molecules. A nucleophilic attack, choice (D), should feature the formation of a bond between a nucleophile and an electrophile; again, we would not expect to break apart a large organic molecule into two smaller molecules. Note that simply noting how many reactants and products are present in the reaction is sufficient to determine the answer.

Tautomerization is the interconversion of two isomers in which a hydrogen and a double bond are moved. The keto and enol tautomers of aldehydes and ketones are common examples of tautomers seen on Test Day. Note that the equilibrium lies to the left because the keto form is more stable. Esterification, choice (A), is the formation of esters from carboxylic acids and alcohols. Elimination, choice (C), is a reaction in which a part of a reactant is lost and a new multiple bond is introduced. Dehydration, choice (D), is a reaction in which a molecule of water is eliminated.

The keto form generally dominates the equilibrium due to its greater thermodynamic stability. Tautomeric equilibrium tells us about the relative proportion of each tautomer in a given set of conditions. The equilibrium lies far to the keto side because the keto form has lower energy and is more stable.