Organic Chemistry for the MCAT Exam

Common failure modes

These are the patterns that cause most candidates to lose marks on this topic. Recognising them in advance is half the work.

• !Memorising reaction names without understanding mechanism — MCAT passages present novel substrates that require mechanism reasoning, not name recall
• !Confusing R/S and E/Z designation rules, especially with multiple stereocenters
• !Misreading NMR multiplicity — forgetting n+1 rule or confusing chemical shift ranges for different proton environments
• !Not recognising when a passage uses organic chemistry to set up a biochemistry question (e.g., enzyme mechanism using nucleophilic catalysis)

Study tips

• 1Master the five nucleophilic substitution and elimination patterns: SN1, SN2, E1, E2, and the conditions that favor each (substrate structure, nucleophile strength, solvent, temperature).
• 2Practice drawing all functional groups from their IUPAC name and assigning correct hybridization (sp, sp2, sp3) to every carbon.
• 3For NMR, memorize the chemical shift windows: alkyl ~0–3 ppm, allylic/benzylic ~2–3, adjacent to O or N ~3–5, aromatic ~6–8, aldehyde ~9–10, carboxylic acid ~10–12.
• 4Drill stereochemistry: draw Fischer projections, convert between Newman and wedge-dash, and assign R/S using the priority rules (CIP).

Sample MCAT Organic Chemistry questions

These sample items mirror the format and difficulty of real MCAT questions. Practice with thousands more on the free Koydo question bank.

1. 1

   A secondary alkyl halide is treated with sodium ethoxide in ethanol at room temperature. The major product is the result of which reaction, and what is the predicted stereochemical outcome?

   • ASN2 — complete inversion of configuration
   • BSN1 — racemic mixture
   • CE2 — the more substituted alkene (Zaitsev product)Correct
   • DE1 — the less substituted alkene (Hofmann product)
   Why this answer?

   Sodium ethoxide is a strong, bulky base. Secondary substrates with a strong base favor elimination. E2 proceeds via an anti-periplanar transition state and gives the Zaitsev (more substituted) alkene as the major product. Note: if the nucleophile were the primary consideration, SN2 would compete — but base strength and substrate branching tilt toward elimination. (Illustrative.)

2. 2

   In a ¹H NMR spectrum, a signal appears as a doublet of doublets at 4.2 ppm. This pattern is consistent with a proton that:

   • AHas two adjacent protons that are chemically equivalent
   • BHas two adjacent protons that are chemically non-equivalentCorrect
   • CHas no adjacent protons
   • DIs on an aromatic ring
   Why this answer?

   A doublet of doublets (dd) indicates coupling to two protons with different coupling constants (J values). This occurs when the two adjacent protons are chemically non-equivalent (diastereotopic), producing two separate splitting events. If the two adjacent protons were equivalent, the signal would be a simple triplet.

3. 3

   Which compound has the highest boiling point?

   • APentane (MW 72)
   • B1-Butanol (MW 74)Correct
   • CDiethyl ether (MW 74)
   • DPropanal (MW 58)
   Why this answer?

   1-Butanol is the only compound capable of intermolecular hydrogen bonding through its −OH group. Hydrogen bonds are stronger than dipole-dipole or London dispersion forces, so 1-butanol has the highest boiling point (~118 °C) despite having a molecular weight comparable to diethyl ether (~35 °C).