Free MCAT-style practice, 5D
MCAT Alcohols and Oxidation-Reduction Practice Questions
Eight free MCAT-style practice questions on alcohol oxidation and reduction, oxidation states, leaving groups, tosylates, phenol acidity and quinones, with full explanations.
Quick concept primer
An alcohol is primary, secondary or tertiary according to how many carbons are attached to the carbon that bears the OH group. Oxidation removes a hydrogen from that carbon, so the class decides the product. A primary alcohol gives an aldehyde with PCC under anhydrous conditions and a carboxylic acid with aqueous chromic acid. A secondary alcohol gives a ketone with either reagent. A tertiary alcohol has no hydrogen on the carbinol carbon and is not oxidized. Reduction runs the other way: hydride reagents turn aldehydes into primary alcohols and ketones into secondary alcohols.
Hydroxide is a poor leaving group. Protonating the OH, or converting it to a tosylate or mesylate, creates a leaving group that is a weak, stable base. Phenols (pKa about 10) are far more acidic than alcohols (pKa about 16) because the phenoxide charge is delocalized into the ring. Phenols with two OH groups in a 1,4 relationship oxidize to quinones, the chemistry behind ubiquinone in the electron transport chain.
Traps: forming a tosylate does not touch the C-O bond, so configuration is retained at that step. PCC stops at the aldehyde, chromic acid in water does not. The 5D Biologically Relevant Molecules and Organic Reactivity pack adds flashcards and more practice across the whole category, and you can try a free sampler first.
8 practice questions
Question 1
A chemist treats 3-methylbutan-1-ol with pyridinium chlorochromate (PCC) in anhydrous dichloromethane. What is the major organic product?
- 3-Methylbutanoic acid
- 3-Methylbutan-2-one
- 3-Methylbutanal
- No reaction occurs
Show answer and explanation
Answer: C. 3-Methylbutanal
3-Methylbutan-1-ol is a primary alcohol, and PCC without water oxidizes a primary alcohol to the aldehyde and stops there, giving 3-methylbutanal. The carboxylic acid forms only when water is present to hydrate the aldehyde, as with chromic acid. A ketone would require a secondary alcohol, and only tertiary alcohols resist oxidation.
Question 2
Three unknown alcohols, each with the formula C5H12O, are treated separately with aqueous chromic acid, which is orange and turns green when chromium is reduced.
| Unknown | Color after 5 minutes | Organic product isolated |
|---|---|---|
| X | Green | A carboxylic acid |
| Y | Green | A ketone |
| Z | Orange | None, starting material recovered |
Which assignment is consistent with the data?
- X is pentan-1-ol, Y is pentan-2-ol, Z is 2-methylbutan-2-ol
- X is pentan-2-ol, Y is pentan-1-ol, Z is 2-methylbutan-2-ol
- X is 2-methylbutan-2-ol, Y is pentan-3-ol, Z is pentan-1-ol
- X is pentan-1-ol, Y is 2-methylbutan-2-ol, Z is pentan-3-ol
Show answer and explanation
Answer: A. X is pentan-1-ol, Y is pentan-2-ol, Z is 2-methylbutan-2-ol
Aqueous chromic acid takes a primary alcohol to a carboxylic acid, a secondary alcohol to a ketone, and leaves a tertiary alcohol unchanged. So X must be primary (pentan-1-ol), Y secondary (pentan-2-ol) and Z tertiary (2-methylbutan-2-ol). Choice B swaps the primary and secondary alcohols, and choices C and D each assign a tertiary alcohol to a sample that reacted.
Question 3
Propan-2-ol is oxidized to propanone (acetone). What is the oxidation state of the central carbon before and after the reaction?
- -1 before and +1 after
- 0 before and +1 after
- +1 before and +2 after
- 0 before and +2 after
Show answer and explanation
Answer: D. 0 before and +2 after
Count +1 for each bond to oxygen, -1 for each bond to hydrogen and 0 for each bond to carbon. In propan-2-ol the central carbon has one C-O bond and one C-H bond, giving 0. In acetone it has two bonds to oxygen and none to hydrogen, giving +2, a two-electron oxidation. Choice A describes a primary alcohol going to an aldehyde.
Question 4
A substituted phenol has a pKa of 10.0. It is dissolved in an aqueous buffer held at pH 8.0. Approximately what percentage of the phenol molecules are deprotonated?
- 0.01 percent
- 1 percent
- 50 percent
- 99 percent
Show answer and explanation
Answer: B. 1 percent
The Henderson-Hasselbalch equation gives a ratio of phenoxide to phenol of 10 raised to (8.0 minus 10.0), which is 0.01. One phenoxide for every 100 phenols means about 1 percent is deprotonated (0.01 divided by 1.01). Fifty percent would require pH equal to pKa, and 99 percent would require a pH two units above it.
Question 5
A student compares the acidity of cyclohexanol, phenol and 4-nitrophenol in water. Which order, from most acidic to least acidic, is correct?
- Cyclohexanol, then phenol, then 4-nitrophenol
- Phenol, then 4-nitrophenol, then cyclohexanol
- 4-Nitrophenol, then phenol, then cyclohexanol
- 4-Nitrophenol, then cyclohexanol, then phenol
Show answer and explanation
Answer: C. 4-Nitrophenol, then phenol, then cyclohexanol
Acidity tracks the stability of the conjugate base. Cyclohexanol gives an alkoxide with the charge fixed on oxygen, so it is the weakest acid. Phenoxide spreads the charge into the ring by resonance. A para nitro group withdraws electrons and extends that delocalization onto its own oxygens, making 4-nitrophenol the strongest acid of the three.
Question 6
(R)-Butan-2-ol is treated with tosyl chloride and pyridine. The purified tosylate is then heated with sodium bromide in a polar aprotic solvent. Which statement describes the outcome?
- Tosylation retains configuration and substitution inverts it, giving (S)-2-bromobutane
- Tosylation inverts configuration and substitution inverts it again, giving (R)-2-bromobutane
- The tosylate ionizes to a planar carbocation, giving racemic 2-bromobutane
- Tosylation retains configuration and so does substitution, giving (R)-2-bromobutane
Show answer and explanation
Answer: A. Tosylation retains configuration and substitution inverts it, giving (S)-2-bromobutane
Tosyl chloride reacts at the oxygen, so the bond between carbon and oxygen is never broken and the stereocenter is unchanged. Bromide then displaces the tosylate by backside attack in an SN2 reaction, favored by a good nucleophile in a polar aprotic solvent, which inverts the center to give (S)-2-bromobutane. A carbocation route would need a polar protic, ionizing solvent.
Question 7
Each of the following converts the hydroxyl group of butan-1-ol into a better leaving group EXCEPT:
- protonation by concentrated HBr.
- reaction with tosyl chloride and pyridine.
- reaction with mesyl chloride and triethylamine.
- deprotonation by sodium hydride.
Show answer and explanation
Answer: D. deprotonation by sodium hydride.
Sodium hydride removes the hydroxyl proton to give an alkoxide, which is a stronger nucleophile but would have to leave as an oxide dianion, an even worse leaving group than hydroxide. Protonation lets the group leave as neutral water, and tosylate and mesylate leave as resonance-stabilized sulfonate anions, the conjugate bases of strong acids.
Question 8
In the electron transport chain, ubiquinone (coenzyme Q) is converted to ubiquinol. Which statement correctly describes this conversion?
- It is an oxidation in which the ring loses two electrons and two protons.
- It is a reduction in which the quinone gains two electrons and two protons to form an aromatic diol.
- It is a hydrolysis in which water adds across one carbonyl group.
- It is a reduction in which one carbonyl group gains two electrons while the ring stays nonaromatic.
Show answer and explanation
Answer: B. It is a reduction in which the quinone gains two electrons and two protons to form an aromatic diol.
A quinone has two ring carbonyl groups. Adding two electrons and two protons converts both carbonyls to hydroxyl groups on a fully aromatic ring, the hydroquinone form called ubiquinol, so the change is a reduction. The reverse reaction, ubiquinol to ubiquinone, is the oxidation described in choice A. No water is added, and both carbonyls change.