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MCAT · Chemical and Physical Foundations · Organic Chemistry
Structure, isomerism and reactivity is part of Organic Chemistry in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include sN1, SN2, E1 and E2 selection, alkene additions and carbocation behavior, stereocenters and stereoisomer relationships and stereochemical outcome and conformations.
Common traps: allowing SN2 at a tertiary carbon; making an SN1 rate depend on the nucleophile; favoring SN2 in polar protic solvents without comment.
Question 1
A chemist treats 1-iodo-3-methylbutane with sodium cyanide in dry DMF at room temperature. Which mechanism accounts for most of the product?
SN1
SN2
E1
E2
Answer: B. SN2
The iodide is on a primary carbon, cyanide is a strong nucleophile but a weak base, and DMF is polar aprotic, so backside SN2 dominates (B). SN1 and E1 (A, C) need a stable carbocation, which a primary substrate cannot form, and E2 (D) would need a strong, bulky base rather than cyanide.
Question 2
2-Bromo-2-methylbutane, (CH3)2CBrCH2CH3, is heated with sodium ethoxide in ethanol. The major organic product is:
2-ethoxy-2-methylbutane
3-methyl-1-butene
2-methyl-1-butene
2-methyl-2-butene
Answer: D. 2-methyl-2-butene
A strong base, a tertiary halide, and heat favor E2 elimination, and the more substituted (Zaitsev) alkene predominates: removing H from C3 gives trisubstituted 2-methyl-2-butene, (CH3)2C=CHCH3. Removing H from a methyl group gives the less substituted 2-methyl-1-butene as a minor product, and substitution to form the ether is disfavored at the hindered tertiary carbon under these conditions.
Question 3
Phenol (pKa ≈ 10) is about a million times more acidic than cyclohexanol (pKa ≈ 16) mainly because:
the O–H bond in phenol is much weaker than in cyclohexanol.
the phenoxide ion's negative charge is delocalized into the aromatic ring by resonance.
the benzene ring donates electron density to oxygen, making the proton easier to remove.
cyclohexanol forms stronger hydrogen bonds with water than phenol does.
Answer: B. the phenoxide ion's negative charge is delocalized into the aromatic ring by resonance.
Acid strength reflects the stability of the conjugate base. The phenoxide oxygen's lone pair delocalizes into the ring, spreading the negative charge over the ortho and para carbons, whereas the cyclohexanol alkoxide charge stays localized on oxygen. Bond-strength differences are minor, electron donation into oxygen would destabilize the anion, and solvation of the neutral alcohol is not decisive.
Question 4
2-Bromo-2-methylbutane is heated with potassium tert-butoxide in tert-butanol. What is the major organic product?
2-methylbut-2-ene
2-methylbut-1-ene
2-tert-butoxy-2-methylbutane
2-methylbutan-2-ol
Answer: B. 2-methylbut-1-ene
A bulky strong base with a tertiary halide favors E2, and steric hindrance makes the base remove the more accessible primary hydrogen, giving the less substituted (Hofmann) alkene, 2-methylbut-1-ene (B). A is the Zaitsev product favored by small bases, C would need SN2 at a tertiary carbon, and D requires water as nucleophile.
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