Online Test — Hydrocarbons
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
The general formula of an alkyne is:
CnH2n+2
CnH2n
CnH2n-2
CnHn
Explanation: Alkynes contain one triple bond and have the formula CnH2n-2.
Question 2 of 20
The Wurtz reaction is used to prepare:
symmetrical alkanes
alkenes
alkynes
arenes
Explanation: 2 R-X + 2 Na → R-R + 2 NaX gives symmetrical alkanes.
Question 3 of 20
The most stable conformation of ethane is:
eclipsed
staggered
skew
all equal
Explanation: Staggered ethane has minimum torsional strain (H atoms 60° apart).
Question 4 of 20
Propene reacting with HBr (no peroxide) gives mainly:
1-bromopropane
2-bromopropane
1,2-dibromopropane
propan-1-ol
Explanation: Markovnikov addition puts Br on the more substituted carbon: 2-bromopropane.
Question 5 of 20
The peroxide (anti-Markovnikov) effect is observed only with:
HCl
HBr
HI
H2O
Explanation: Only HBr gives anti-Markovnikov addition via a viable free-radical chain (Kharasch effect).
Question 6 of 20
Cold dilute alkaline KMnO4 used to detect unsaturation is called:
Baeyer's reagent
Tollens' reagent
Fehling's solution
Lucas reagent
Explanation: Baeyer's reagent (cold dilute alkaline KMnO4) is decolourised by alkenes/alkynes.
Question 7 of 20
A terminal alkyne gives a white precipitate with:
bromine water
FeCl3
NaHCO3
ammoniacal AgNO3
Explanation: The acidic ≡C-H reacts with ammoniacal AgNO3 to give silver acetylide.
Question 8 of 20
Ozonolysis of CH3CH=CHCH3 (then Zn/H2O) gives:
only HCHO
only CH3CHO
CH3CHO + HCHO
CH3COCH3
Explanation: Both halves are identical, so cleavage gives two molecules of CH3CHO.
Question 9 of 20
Benzene contains how many delocalised π electrons?
2
4
6
12
Explanation: Six p-electrons are delocalised over the ring, satisfying Huckel's 4n+2 rule with n=1.
Question 10 of 20
By Huckel's rule, a species is aromatic if it has:
(4n+2) π electrons
4n π electrons
(2n+1) π electrons
n π electrons
Explanation: Aromaticity requires a cyclic, planar, conjugated ring with (4n+2) π electrons.
Question 11 of 20
The electrophile in the chlorination of benzene (Cl2/FeCl3) is:
Cl-
Cl+
Cl·
HCl
Explanation: FeCl3 polarises Cl2 to give Cl+, the attacking electrophile.
Question 12 of 20
Which group is a meta director?
–OH
–CH3
–NH2
–COOH
Explanation: –COOH is electron-withdrawing and deactivating, directing meta.
Question 13 of 20
Nitration of toluene gives mainly:
only meta product
ortho and para products
only the ortho product
no reaction
Explanation: The activating methyl group is an o/p director, so o- and p-nitrotoluene dominate.
Question 14 of 20
The intermediate in electrophilic aromatic substitution is the:
carbanion
free radical
arenium ion
carbene
Explanation: Addition of E+ gives the resonance-stabilised arenium ion (sigma complex).
Question 15 of 20
Heating sodium benzoate with soda-lime gives:
benzene
toluene
phenol
benzaldehyde
Explanation: Decarboxylation removes COONa as carbonate, leaving benzene.
Question 16 of 20
Hydroboration-oxidation of propene gives:
propan-2-ol
propanone
propanal only
propan-1-ol
Explanation: Boron adds to the terminal carbon (anti-Markovnikov), so OH ends up on C-1: propan-1-ol.
Question 17 of 20
Which hydrocarbon is a known carcinogen?
methane
ethene
benzene
propane
Explanation: Benzene and polynuclear aromatics are carcinogenic; chronic exposure can cause leukaemia.
Question 18 of 20
Cyclic polymerisation of three molecules of ethyne over a red-hot tube gives:
cyclohexane
toluene
benzene
naphthalene
Explanation: 3 HC≡CH ⟶[873 K, Cu] C6H6 (benzene).
Question 19 of 20
But-1-ene shows geometrical isomerism:
yes, two forms
no, the =CH2 carbon has two identical H
only at high T
only with a catalyst
Explanation: Cis-trans isomerism needs each C of the double bond to bear two different groups; =CH2 has two identical H.
Question 20 of 20
The hybridisation of carbon in benzene is:
sp
sp2
sp3
sp3d
Explanation: Each ring carbon is sp2 hybridised and planar, leaving a p-orbital for the π system.