Question 1Level 2 · ExamSingle correct
The rate of the reaction R–Br + I− → R–I + Br− is measured for four bromides under identical conditions (NaI in dry acetone, 25 °C, same concentrations): CH3Br, CH3CH2Br, (CH3)2CHBr and (CH3)3CCH2Br. The correct order of rate is ______. Exactly one option is correct.
ACH3Br > CH3CH2Br > (CH3)3CCH2Br > (CH3)2CHBr
BCH3Br > CH3CH2Br > (CH3)2CHBr > (CH3)3CCH2Br
C(CH3)3CCH2Br > CH3CH2Br > CH3Br > (CH3)2CHBr
DCH3CH2Br > CH3Br > (CH3)2CHBr > (CH3)3CCH2Br
Show answer and solution
Answer: B
NaI in dry acetone (polar aprotic, good nucleophile, no carbocation-stabilising solvent) is a classic SN2 (Finkelstein) system, so rate is governed by steric hindrance to backside attack at the carbon bearing Br.
CH3Br (no alkyl groups) > CH3CH2Br (one α-alkyl group) > (CH3)2CHBr (two α-alkyl groups).
(CH3)3CCH2Br (neopentyl bromide) is primary, but the bulky tert-butyl group on the β-carbon blocks the backside of the C–Br carbon; its SN2 rate is lower even than that of the secondary bromide (by roughly three orders of magnitude).
Order: CH3Br > CH3CH2Br > (CH3)2CHBr > (CH3)3CCH2Br.
Trap: 'primary is always faster than secondary in SN2' fails for β-branched neopentyl systems.
Answer: (B).
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
1-Cyclobutylethan-1-ol, CH3CH(OH)–C4H7 (the CH(OH) carbon is attached to one carbon of an unsubstituted cyclobutane ring), is heated with concentrated HBr. The reaction proceeds through carbocations, and every rearrangement that leads to a more stable carbocation takes place. The major organic product is ______. Exactly one option is correct.
A1-bromo-1-cyclobutylethane
B1-bromo-2-methylcyclopentane
C1-bromo-1-methylcyclopentane
Dbromocyclohexane
Show answer and solution
Answer: C
Protonation of OH and loss of water gives the secondary cation CH3CH+–(cyclobutyl).
Step 1 (ring expansion): a ring C–C bond adjacent to the cation migrates to the cationic carbon. The four-membered ring becomes five-membered; the former CH(CH3) carbon is now a ring carbon and the positive charge sits on the old ring carbon next to it: 2-methylcyclopentyl cation (secondary, but the large angle strain of cyclobutane is relieved).
Step 2 (1,2-hydride shift): the H on the CH(CH3) ring carbon shifts to the adjacent cationic carbon, giving the tertiary 1-methylcyclopentyl cation.
Step 3: Br− captures this cation → 1-bromo-1-methylcyclopentane, C6H11Br (formula consistent with C6H12O – OH + Br).
(A) is the unrearranged product; (B) stops after ring expansion; (D) would need a six-membered ring, impossible from a 4-ring plus one exocyclic carbon bearing the cation (the CH3 stays outside the ring).
Answer: (C).
Difficulty: Very hardQuestion 3Level 3 · AdvancedOne or more correct
1-Bromopropane reacts with sodium azide by the SN2 mechanism: CH3CH2CH2Br + N3− → CH3CH2CH2N3 + Br−, carried out in methanol at 25 °C. Starting from this reference experiment and changing only the factor stated (all other conditions the same), which changes increase the initial rate? One or more options may be correct.
AReplacing methanol by dimethyl sulphoxide (DMSO) as solvent
BReplacing 1-bromopropane by 1-chloropropane at the same concentration
CReplacing 1-bromopropane by 1-iodopropane at the same concentration
DDoubling the concentration of N3− while keeping the concentration of the alkyl halide the same
Show answer and solution
Answer: A, C, D
For SN2, rate = k[RX][N3−].
(A) Correct. Methanol (protic) hydrogen-bonds to and stabilises the small anion N3−, lowering its nucleophilicity. DMSO (polar aprotic) solvates Na+ well but leaves N3− poorly solvated ('naked'), so the rate rises sharply.
(B) Incorrect. C–Cl is stronger than C–Br and Cl− is a poorer leaving group, so the rate falls.
(C) Correct. C–I is the weakest C–X bond and I− is the best leaving group among halides: rate increases.
(D) Correct. The reaction is first order in N3−; doubling [N3−] doubles the rate.
Answer: (A), (C), (D).
Difficulty: Very hard