IMO Practice Test — Haloalkanes and Haloarenes
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
A 2° alkyl bromide reacts twice as fast when [Nu] is doubled but its rate is unchanged when only [R–X] is held constant and a better leaving group is used at the same concentration ratio. The reaction most likely proceeds by:
SN2 with inversion
SN1 with racemisation
E1 elimination
free-radical substitution
Explanation: Rate doubling with [Nu] means the nucleophile appears in the rate law (bimolecular), the signature of SN2, which proceeds with inversion.
Question 2 of 14
Which sequence correctly ranks the rate of SN1 hydrolysis: I CH3Br, II (CH3)2CHBr, III (CH3)3CBr, IV C6H5CH2Br?
I > II > III > IV
III > IV > II > I
IV > III > II > I
II > III > IV > I
Explanation: SN1 follows carbocation stability: 3° (III) and resonance-stabilised benzylic (IV) are top, then 2° (II), then methyl (I): III > IV > II > I.
Question 3 of 14
p-Nitrochlorobenzene undergoes nucleophilic substitution far more readily than chlorobenzene because the –NO2 group:
donates electrons by resonance
weakens the ring aromaticity
stabilises the anionic intermediate by withdrawing electrons
increases the electron density at the C–Cl carbon
Explanation: A para –NO2 group withdraws electrons and stabilises the negatively charged intermediate formed on nucleophilic addition, accelerating substitution.
Question 4 of 14
An optically active halide gives a product with the opposite sign of optical rotation and high optical purity. This is best explained by:
SN1 racemisation
retention of configuration
a meso product
complete SN2 inversion
Explanation: A single, clean inversion (SN2) flips the configuration at the stereocentre, often reversing the sign of rotation with high optical purity.
Question 5 of 14
Why does increasing solvent polarity (protic) accelerate the SN1 step but can slow an SN2 reaction with an anionic nucleophile?
It stabilises the developing carbocation/anion (helps SN1) but also solvates and deactivates the anionic nucleophile (hurts SN2)
It raises the activation energy of ionisation
It lowers the temperature
It changes the leaving group
Explanation: Protic solvents stabilise the charges formed on ionisation (favouring SN1) but hydrogen-bond to and shield an anionic nucleophile, reducing its SN2 reactivity.
Question 6 of 14
The major product of dehydrohalogenation of 2-bromo-2,3-dimethylbutane with a strong base is:
the least substituted alkene (Hofmann)
2,3-dimethylbut-2-ene (most substituted, Saytzeff)
an alcohol
a symmetrical ether
Explanation: Saytzeff’s rule favours the most highly substituted, most stable alkene, here the tetrasubstituted 2,3-dimethylbut-2-ene.
Question 7 of 14
Allyl chloride (CH2=CHCH2Cl) is much more reactive in SN1 than 1-chloropropane because:
it has a stronger C–Cl bond
chlorine is more electronegative in it
its carbocation is resonance (allyl) stabilised
it is aromatic
Explanation: Ionisation gives an allyl cation that is delocalised over two carbons (resonance), so it forms easily and speeds up SN1.
Question 8 of 14
Which reagent pair best converts propan-1-ol to 1-iodopropane in good purity?
HI directly only
PCl3 then nothing
AgF only
SOCl2 then NaI in dry acetone (Finkelstein)
Explanation: Convert the alcohol to the chloride with SOCl2 (clean, gaseous by-products), then exchange Cl for I by the Finkelstein reaction with NaI in dry acetone.
Question 9 of 14
Chlorobenzene does not give a Grignard reagent easily in ether but reacts under special conditions; meanwhile its nitration occurs mainly at o/p positions. Both facts arise from:
the strong, resonance-shortened C–Cl bond and lone-pair donation to o/p carbons
the weak C–Cl bond
the sp3 nature of the ring carbon
hydrogen bonding
Explanation: Resonance shortens/strengthens C–Cl (low reactivity) and shifts electron density to the o/p carbons (o/p-directing) — both stem from lone-pair delocalisation into the ring.
Question 10 of 14
Which statement about the SN2 transition state is correct?
The carbon is sp3 and tetrahedral
The carbon is approximately sp2 with the nucleophile and leaving group on opposite sides (trigonal bipyramidal-like)
A free carbocation is present
The nucleophile attacks from the same side as the leaving group
Explanation: In the SN2 TS the central carbon is roughly sp2 with the three other groups planar and Nu and X partially bonded on opposite sides (five-coordinate).
Question 11 of 14
Why does anti-Markovnikov addition occur only with HBr (not HCl or HI) in the presence of peroxides?
Only HBr is ionic
HBr is the strongest acid
Both chain-propagation steps are exothermic and feasible only for HBr; for HCl/HI one step is too endothermic or too slow
Peroxides only dissolve HBr
Explanation: For the radical chain both propagation steps are energetically favourable only with HBr; with HCl one step is too endothermic and with HI the addition of I radical is too slow, so the peroxide effect is unique to HBr.
Question 12 of 14
A symmetrical alkane R–R is best made from R–X by:
Finkelstein reaction
Swarts reaction
Sandmeyer reaction
Wurtz reaction (Na/dry ether)
Explanation: The Wurtz reaction couples two identical alkyl halides with sodium in dry ether to give a symmetrical alkane R–R.
Question 13 of 14
CFCs damage the ozone layer because, on reaching the stratosphere, UV light:
cleaves a C–Cl bond to release Cl radicals that catalytically destroy O3
converts them to CO2
makes them dissolve in clouds
turns them into DDT
Explanation: UV homolysis of the C–Cl bond releases Cl radicals, each of which destroys many ozone molecules in a chain (Cl + O3 → ClO + O2).
Question 14 of 14
Which combination would give the cleanest single alkylarene by a Wurtz–Fittig reaction?
two different aryl halides
an aryl halide and an alkyl halide with Na/dry ether
two different alkyl halides
an alcohol and an aryl halide
Explanation: Wurtz–Fittig couples an aryl halide with an alkyl halide (Na/dry ether) to give an alkylarene such as toluene from chlorobenzene + chloromethane.