Questions & explanations
1. What is the rate law for an E1 elimination reaction?
- Rate = k[substrate][base]
- Rate = k[substrate]
- Rate = k[base]
- Rate = k[substrate]²
Answer: Rate = k[substrate]
E1 elimination is a two-step unimolecular reaction. The rate-determining step is the ionization of the substrate to form a carbocation, which depends only on the concentration of the substrate. Therefore, the rate law is first order in substrate: rate = k[substrate]. The base concentration does not appear in the rate law because the base is involved in the fast second step.
2. 3,3-dimethyl-1-butene is treated separately with (i) H2O/H+, (ii) Hg(OAc)2/H2O then NaBH4, and (iii) BH3 then H2O2/NaOH. Which option correctly matches the major alcohol product?
- (i) 3,3-dimethyl-1-butanol, (ii) 3,3-dimethyl-2-butanol, (iii) 2,3-dimethyl-2-butanol
- (i) 2,3-dimethyl-2-butanol, (ii) 3,3-dimethyl-2-butanol, (iii) 3,3-dimethyl-1-butanol
- (i) 3,3-dimethyl-2-butanol, (ii) 2,3-dimethyl-2-butanol, (iii) 3,3-dimethyl-1-butanol
- (i) 2,3-dimethyl-2-butanol, (ii) 3,3-dimethyl-1-butanol, (iii) 3,3-dimethyl-2-butanol
Answer: (i) 2,3-dimethyl-2-butanol, (ii) 3,3-dimethyl-2-butanol, (iii) 3,3-dimethyl-1-butanol
Acid-catalyzed hydration (i) proceeds via a 2° carbocation that undergoes a methyl shift to a more stable 3° carbocation, yielding 2,3-dimethyl-2-butanol. Oxymercuration (ii) gives Markovnikov addition without rearrangement, yielding 3,3-dimethyl-2-butanol. Hydroboration (iii) gives anti-Markovnikov addition, yielding 3,3-dimethyl-1-butanol. Thus option b is correct.
3. 1-Methylcyclohexene is treated with BH3/THF followed by H2O2/NaOH. What is the major product?
- trans-2-Methylcyclohexanol
- cis-2-Methylcyclohexanol
- 1-Methylcyclohexanol
- 1-Methylcyclohexene oxide
Answer: trans-2-Methylcyclohexanol
Hydroboration-oxidation adds H and OH across the double bond with anti-Markovnikov regiochemistry and syn stereochemistry. BH3 adds to the less substituted carbon (C2) and H to the more substituted carbon (C1) from the same face. Oxidation with retention places OH at C2. The methyl at C1 and OH at C2 end up on opposite faces, giving trans-2-methylcyclohexanol.
4. 2-Bromobutane is treated with ethoxide ion in ethanol. Which alkene is the major product?
- 1-Butene
- 1,3-Butadiene
- Butane
- 2-Butene
Answer: 2-Butene
According to Saytzeff's rule, the more substituted alkene is the major product in E2 elimination with a small base like ethoxide. 2-Bromobutane has β-hydrogens on C1 and C3. Removal of β-H from C3 gives 2-butene (disubstituted), while removal from C1 gives 1-butene (monosubstituted). 2-Butene is more stable due to hyperconjugation and is the major product.
5. Treatment of 2-bromo-2,3-dimethylbutane with KOt-Bu in t-BuOH gives which major alkene?
- 2,3-dimethyl-1-butene (terminal)
- 3,3-dimethyl-1-butene (terminal)
- 2,3-dimethyl-2-butene (tetra-substituted)
- 2,3-dimethyl-1,3-butadiene
Answer: 2,3-dimethyl-2-butene (tetra-substituted)
KOt-Bu is a strong bulky base, but the substrate has only one β-hydrogen (on C3) that is accessible for E2 elimination. The β-hydrogens on C1 are primary but sterically hindered by the bulky base; however, the only available β-hydrogen is on the tertiary carbon, leading to the Zaitsev product 2,3-dimethyl-2-butene as major.
6. Which statement correctly describes the SN1 mechanism?
- It is a one-step concerted process with inversion of configuration.
- It involves a carbocation intermediate and follows first-order kinetics.
- It is a two-step process where the nucleophile attacks first.
- It involves a carbanion intermediate and follows second-order kinetics.
Answer: It involves a carbocation intermediate and follows first-order kinetics.
SN1 proceeds in two steps: first, the leaving group departs to form a planar carbocation (slow, rate-determining); second, the nucleophile attacks. The rate depends only on substrate concentration, giving first-order kinetics. Option b correctly captures the carbocation intermediate and first-order rate law.
7. The peroxide effect (anti-Markovnikov addition) is observed only with HBr because:
- both propagation steps are exothermic only for HBr
- HBr has the weakest H–X bond among hydrogen halides
- Br• is the most stable halogen radical
- HBr is a stronger acid than HCl and HI
Answer: both propagation steps are exothermic only for HBr
For the radical chain to propagate, both steps (Br• addition to alkene and H abstraction from HX) must be exothermic. For HCl, H abstraction is endothermic due to strong H–Cl bond. For HI, Br• addition is endothermic due to weak C–I bond. Only HBr has intermediate bond strength making both steps exothermic.
8. What is the correct order of SN1 reactivity for the following alkyl halides?
- CH3Br > CH3CH2Br > (CH3)2CHBr > (CH3)3CBr
- (CH3)2CHBr > (CH3)3CBr > CH3CH2Br > CH3Br
- (CH3)3CBr > (CH3)2CHBr > CH3CH2Br > CH3Br
- CH3CH2Br > (CH3)2CHBr > (CH3)3CBr > CH3Br
Answer: (CH3)3CBr > (CH3)2CHBr > CH3CH2Br > CH3Br
SN1 reactivity depends on carbocation stability: tertiary > secondary > primary > methyl. (CH3)3CBr forms a tertiary carbocation (most stable), followed by (CH3)2CHBr (secondary), then CH3CH2Br (primary), and CH3Br (methyl, least stable). Thus the correct order is (CH3)3CBr > (CH3)2CHBr > CH3CH2Br > CH3Br.
9. Which of the following statements about E2 elimination in cyclohexane systems is correct?
- Menthyl chloride undergoes E2 faster than neomenthyl chloride because it has more equatorial groups.
- Neither diastereomer undergoes E2 because the leaving group is always equatorial.
- Both diastereomers undergo E2 at the same rate because ring flipping makes the leaving group axial in both.
- Neomenthyl chloride undergoes E2 faster than menthyl chloride because its leaving group is axial in the most stable chair.
Answer: Neomenthyl chloride undergoes E2 faster than menthyl chloride because its leaving group is axial in the most stable chair.
E2 requires anti-periplanar geometry: both H and leaving group must be axial. In neomenthyl chloride, Cl is axial in the most stable chair, allowing fast E2. In menthyl chloride, Cl is equatorial in the most stable chair; ring flipping to axial puts three groups axial, raising energy and slowing reaction.
10. 2-Bromo-2-methylbutane is treated with potassium tert-butoxide (t-BuOK) in t-BuOH. Which alkene is the major product?
- 2-Methyl-2-butene
- 2-Methyl-1-butene
- 3-Methyl-1-butene
- 2-Methylbutane
Answer: 2-Methyl-2-butene
Potassium tert-butoxide is a strong, bulky base. In E2 elimination, it abstracts the most accessible β-hydrogen, which is the one on the more substituted carbon (C3) due to hyperconjugation, leading to the more substituted alkene (Zaitsev product). Thus, 2-methyl-2-butene is the major product.
11. (S)-1-chloro-1-phenylpropane is treated with NaSH in DMSO at room temperature. What is the major product and its stereochemistry?
- (S)-1-mercapto-1-phenylpropane, retention
- (R)-1-mercapto-1-phenylpropane, inversion
- racemic 1-mercapto-1-phenylpropane
- β-methylstyrene (elimination product)
Answer: (R)-1-mercapto-1-phenylpropane, inversion
NaSH is a strong nucleophile and weak base. DMSO is a polar aprotic solvent, which favors SN2. The substrate is secondary benzylic, but SN2 is still dominant with a strong nucleophile in aprotic solvent. SN2 proceeds with inversion of configuration, so (S)-chloride gives (R)-thiol.
12. 2-Bromo-3-methylbutane is treated with KOt-Bu in t-BuOH. The major alkene product is:
- 3-Methyl-1-butene
- 2-Methyl-2-butene
- 2-Methyl-1-butene
- 3-Methyl-2-butene
Answer: 2-Methyl-2-butene
KOt-Bu is a bulky base, but the substrate has only one β-hydrogen on the more substituted carbon (C3), which is tertiary and accessible. The less substituted β-hydrogens are sterically hindered by the tert-butyl group. Thus, the Zaitsev product (2-methyl-2-butene) is major.