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MCAT Carbonyl Chemistry and Carboxylic Acid Derivatives Practice Questions

By StudyStore · Published 2026-10-20 · 8 original questions · about 12 min

Eight free MCAT-style practice questions on acyl derivative reactivity, nucleophilic acyl substitution, acetals as protecting groups, selective reduction, the aldol reaction, keto-enol tautomerism and acid strength, with explanations.

Quick concept primer

The carbonyl carbon is electrophilic. Aldehydes and ketones undergo nucleophilic addition, and the tetrahedral intermediate is protonated to give an alcohol, a hemiacetal or an acetal depending on the nucleophile. Carboxylic acid derivatives instead undergo nucleophilic acyl substitution: addition gives a tetrahedral intermediate, then the leaving group is expelled. The reaction goes in the direction of the better leaving group, so reactivity follows acyl chloride > anhydride > thioester > ester > amide > carboxylate.

Reagent choice matters. Sodium borohydride reduces aldehydes and ketones but not esters; lithium aluminum hydride reduces nearly all carbonyls, including esters, acids and amides. Acetals are stable to base and nucleophiles, which makes them standard protecting groups for carbonyls. The aldol reaction requires an alpha hydrogen to form an enolate; dehydration of the aldol product gives an alpha,beta-unsaturated carbonyl. Electron-withdrawing groups near a carboxylic acid increase its acidity by stabilizing the carboxylate.

Traps: a carbonyl without alpha hydrogens cannot act as the enolate component, and amides are the least reactive acid derivatives because of strong resonance.

8 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

Which acid derivative reacts most rapidly with water at room temperature?

  1. Acetyl chloride
  2. Acetamide
  3. Acetic anhydride
  4. Ethyl acetate
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Answer: A. Acetyl chloride

In nucleophilic acyl substitution, reactivity tracks the quality of the leaving group and the electrophilicity of the carbonyl carbon. Chloride is an excellent leaving group (the conjugate base of a strong acid) and chlorine withdraws electron density without donating much by resonance, so acetyl chloride is the most reactive. Anhydrides are next, then esters, and amides are the least reactive.

Question 2

Acetic anhydride is added to ethanol with a small amount of pyridine. What are the products?

  1. Acetaldehyde and ethanol
  2. Acetamide and water
  3. Ethyl acetate and acetic acid
  4. Diethyl ether and acetic acid
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Answer: C. Ethyl acetate and acetic acid

Ethanol attacks one carbonyl of the anhydride to give a tetrahedral intermediate, which collapses and expels acetate as the leaving group. The result is the ester, ethyl acetate, plus acetic acid (neutralized by pyridine). This is a standard way to make esters without needing the acid chloride.

Question 3

A chemist wants to make a Grignard reagent from 4-bromobutan-1-ol so it can be added to a ketone. What must be done first?

  1. Oxidize the alcohol to a carboxylic acid
  2. Convert the hydroxyl group into a tosylate
  3. Protect the hydroxyl group, for example as a silyl ether
  4. Add aqueous acid to the alkyl bromide
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Answer: C. Protect the hydroxyl group, for example as a silyl ether

A Grignard reagent is a very strong base and is destroyed by any acidic O-H bond, so a free alcohol in the same molecule would protonate the carbon-magnesium bond as soon as it formed. Converting the hydroxyl to a silyl ether (stable to strong bases, removed later with fluoride or acid) allows magnesium insertion to proceed. A carboxylic acid is even more acidic, a tosylate is a leaving group that the Grignard reagent would attack rather than a protecting group, and aqueous acid would simply destroy the organometallic reagent.

Question 4

Reduction of N-methylacetamide (CH3CONHCH3) with LiAlH4, followed by aqueous workup, gives mainly which product?

  1. N-ethylmethylamine, CH3CH2NHCH3
  2. Acetic acid and methylamine
  3. Acetaldehyde
  4. Ethanol and methylamine
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Answer: A. N-ethylmethylamine, CH3CH2NHCH3

Amides are unusual among carboxylic acid derivatives because LiAlH4 converts the C=O into CH2 and gives an amine, not an alcohol. After the first hydride addition the oxygen leaves as an aluminate and an iminium ion forms, which accepts a second hydride. Esters and acids give alcohols, so an alcohol product here would be a mistake. Acetic acid and methylamine would arise from hydrolysis, not reduction.

Question 5

Which of the following carbonyl compounds cannot form an enolate and therefore cannot act as the nucleophile in an aldol reaction?

  1. Propanal
  2. Benzaldehyde
  3. Acetone
  4. Acetaldehyde
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Answer: B. Benzaldehyde

An enolate forms by removing a hydrogen on the carbon next to the carbonyl (the alpha carbon). In benzaldehyde, the carbon next to the carbonyl is an aromatic ring carbon with no removable alpha hydrogen, so no enolate can be formed. Benzaldehyde can still serve as the electrophilic partner in a crossed aldol reaction.

Question 6

Two molecules of acetaldehyde react in dilute aqueous base. What is the aldol addition product, and what does heating do to it?

  1. 3-hydroxybutanal; heating causes dehydration to give crotonaldehyde (2-butenal)
  2. Butanoic acid; heating gives butene
  3. 2-hydroxybutanal; heating gives butanal
  4. Ethyl acetate; heating gives acetic acid
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Answer: A. 3-hydroxybutanal; heating causes dehydration to give crotonaldehyde (2-butenal)

Hydroxide forms the enolate of acetaldehyde, which adds to the carbonyl carbon of a second molecule and gives 3-hydroxybutanal (a beta-hydroxy aldehyde). Heating eliminates water through an E1cB mechanism, producing the conjugated alpha,beta-unsaturated aldehyde 2-butenal, which is the condensation product. Conjugation provides the driving force for the dehydration.

Question 7

Which compound has the highest enol content at equilibrium?

  1. Acetaldehyde
  2. Acetone
  3. 2,4-Pentanedione
  4. Cyclohexanone
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Answer: C. 2,4-Pentanedione

For most simple carbonyl compounds the keto form is much more stable than the enol. 2,4-Pentanedione (acetylacetone), a 1,3-dicarbonyl, has a strongly stabilized enol: the C=C is conjugated with the second carbonyl and an intramolecular hydrogen bond forms a six-membered ring. As a result the enol is the major form in many solvents. Simple ketones and aldehydes contain only a tiny fraction of enol.

Question 8

Which ranking of acid strength, from strongest to weakest, is correct?

  1. ClCH2COOH > Cl3CCOOH > CH3COOH
  2. CH3COOH > Cl3CCOOH > ClCH2COOH
  3. CH3COOH > ClCH2COOH > Cl3CCOOH
  4. Cl3CCOOH > ClCH2COOH > CH3COOH
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Answer: D. Cl3CCOOH > ClCH2COOH > CH3COOH

Electronegative substituents near the carboxylate withdraw electron density through the sigma bonds (the inductive effect) and stabilize the conjugate base. More chlorines give greater stabilization, so trichloroacetic acid (pKa about 0.7) is stronger than chloroacetic acid (about 2.9), which is stronger than acetic acid (about 4.8).

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.