IMO Practice Test — Hydrocarbons
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
An alkene of formula C5H10 on ozonolysis gives propanone and ethanal. The alkene is:
pent-1-ene
2-methylbut-2-ene
pent-2-ene
2-methylbut-1-ene
Explanation: Propanone (CH3)2C=O and ethanal CH3CHO recombine to (CH3)2C=CHCH3, i.e. 2-methylbut-2-ene.
Question 2 of 12
Which carbocation forms preferentially when H+ adds to 2-methylbut-2-ene?
primary
secondary
tertiary
methyl
Explanation: Protonation gives the most stable (3°) carbocation, which directs the Markovnikov product.
Question 3 of 12
Of the following, the species that is NOT aromatic is:
benzene
cyclopentadienyl anion
cyclobutadiene
tropylium cation
Explanation: Cyclobutadiene has 4 π electrons (4n), so it is antiaromatic; the others have (4n+2).
Question 4 of 12
In the nitration of chlorobenzene, the nitro group enters mainly at:
ortho and para
meta only
the ipso position
no reaction occurs
Explanation: Halogens are deactivating but o/p directing (lone-pair donation), so o- and p-products dominate.
Question 5 of 12
The order of acidity of the C-H bond is correctly given by:
ethane > ethene > ethyne
ethyne > ethene > ethane
ethene > ethyne > ethane
all equal
Explanation: Acidity rises with s-character of the C: sp (ethyne) > sp2 (ethene) > sp3 (ethane).
Question 6 of 12
Addition of one mole of HBr to buta-1,3-diene can give a 1,4-product because of:
steric hindrance
antiaromaticity
allylic carbocation resonance
free rotation
Explanation: The allylic cation is resonance-delocalised, so Br- can add at C-2 (1,2) or C-4 (1,4).
Question 7 of 12
Friedel-Crafts alkylation of benzene with 1-chloropropane/AlCl3 often gives cumene (isopropylbenzene) because:
benzene rearranges
AlCl3 is reduced
propene forms
the 1° cation rearranges to a 2° cation
Explanation: The primary propyl cation rearranges to the more stable secondary (isopropyl) cation before attack.
Question 8 of 12
Number of monochlorination products (structural isomers) of 2-methylbutane is:
2
3
4
5
Explanation: Four distinct types of H give four isomeric monochlorides.
Question 9 of 12
Markovnikov addition of water (dil. H2SO4) to but-1-ene gives:
butan-2-ol
butan-1-ol
butanal
butanone
Explanation: OH goes to the more substituted carbon (C-2) via the 2° cation: butan-2-ol.
Question 10 of 12
Which set of conditions reduces but-2-yne to trans-but-2-ene?
H2/Lindlar
H2/Pt
HBr/peroxide
Na in liquid NH3
Explanation: Dissolving-metal reduction (Na/liq. NH3) delivers the trans (E) alkene; Lindlar gives cis.
Question 11 of 12
Sulphonation of benzene is described as reversible because:
SO3H is a strong base
the ring loses aromaticity
dilute acid/steam removes the –SO3H group
FeCl3 decomposes
Explanation: Heating benzenesulphonic acid with dilute acid/steam reverses the substitution (desulphonation).
Question 12 of 12
Aniline (C6H5NH2) on bromination in water gives 2,4,6-tribromoaniline rapidly because –NH2 is:
a weak meta director
a strong activating o/p director
deactivating
non-directing
Explanation: The lone pair on N strongly activates the ring (o/p directing), so all three free o/p positions react.