Structure, isomerism and reactivity sample question 2

the phenoxide ion's negative charge is delocalized into the aromatic ring by resonance.

Structure, isomerism and reactivity sample question 5

Lithium aluminum hydride in ether, followed by aqueous acid

MCAT · Chemical and Physical Foundations · Organic Chemistry

Structure, isomerism and reactivity

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.

Sample questions

Question 1
2-Bromo-2-methylbutane, (CH3)2CBrCH2CH3, is heated with sodium ethoxide in ethanol. The major organic product is:
- A.
  2-ethoxy-2-methylbutane
- B.
  3-methyl-1-butene
- C.
  2-methyl-1-butene
- D.
  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 2
Phenol (pKa ≈ 10) is about a million times more acidic than cyclohexanol (pKa ≈ 16) mainly because:
- A.
  the O–H bond in phenol is much weaker than in cyclohexanol.
- B.
  the phenoxide ion's negative charge is delocalized into the aromatic ring by resonance.
- C.
  the benzene ring donates electron density to oxygen, making the proton easier to remove.
- D.
  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 3
A chemist treats 1-iodo-3-methylbutane with sodium cyanide in dry DMF at room temperature. Which mechanism accounts for most of the product?
- A.
  SN1
- B.
  SN2
- C.
  E1
- D.
  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 4
2-Bromo-2-methylbutane is heated with potassium tert-butoxide in tert-butanol. What is the major organic product?
- A.
  2-methylbut-2-ene
- B.
  2-methylbut-1-ene
- C.
  2-tert-butoxy-2-methylbutane
- D.
  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.
Azide Displacement of a Primary Tosylate
A research group prepares 4-phenylbutyl tosylate (TsO–CH₂CH₂CH₂CH₂–C₆H₅) in two steps. Methyl 4-phenylbutanoate is first reduced to 4-phenylbutan-1-ol, and the alcohol is then treated with p-toluenesulfonyl chloride and pyridine. The tosylate is next allowed to react with sodium azide at 25 °C to give 4-phenylbutyl azide and sodium tosylate. The group measures the second-order rate constant in three solvents, each initially 0.10 M in tosylate and 0.20 M in sodium azide.
Table 1 Rate constants for N₃⁻ + R–OTs at 25 °C
Solvent Type k (M⁻¹ s⁻¹)
Dimethylformamide (DMF) Polar aprotic 5.0 × 10⁻³
Methanol Polar protic 5.0 × 10⁻⁶
Water Polar protic 1.5 × 10⁻⁶
Table 2 Approximate pKa values of conjugate acids
Acid pKa
p-Toluenesulfonic acid (TsOH) –2.8
Hydrazoic acid (HN₃) 4.7
Methanol 15.5
In each solvent the rate was found to double when the azide concentration was doubled and to double when the tosylate concentration was doubled. After the DMF reaction the mixture was poured into water and shaken with diethyl ether. The ether layer (density about 0.71 g/mL) was separated and dried, and the product was isolated by evaporating the ether.
Question 5
In the first step, methyl 4-phenylbutanoate is converted to 4-phenylbutan-1-ol. Which reagent system would accomplish this transformation?
- A.
  Lithium aluminum hydride in ether, followed by aqueous acid
- B.
  Sodium borohydride in ethanol at room temperature
- C.
  Pyridinium chlorochromate in CH₂Cl₂ at room temperature
- D.
  H₂ with a palladium catalyst at 1 atm and 25 °C
Answer: A. Lithium aluminum hydride in ether, followed by aqueous acid
LiAlH₄ delivers hydride twice to an ester, giving the primary alcohol after workup. NaBH₄ (B) is too mild to reduce esters. PCC (C) is an oxidant, not a reductant. Pd/H₂ at 1 atm (D) leaves both the ester and the benzene ring intact.

Table 1 Rate constants for N₃⁻ + R–OTs at 25 °C
Solvent	Type	k (M⁻¹ s⁻¹)
Dimethylformamide (DMF)	Polar aprotic	5.0 × 10⁻³
Methanol	Polar protic	5.0 × 10⁻⁶
Water	Polar protic	1.5 × 10⁻⁶

Table 2 Approximate pKa values of conjugate acids
Acid	pKa
p-Toluenesulfonic acid (TsOH)	–2.8
Hydrazoic acid (HN₃)	4.7
Methanol	15.5

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Structure, isomerism and reactivity

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Azide Displacement of a Primary Tosylate

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