Online Test — Haloalkanes and Haloarenes
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
Which carbon is bonded to chlorine in vinyl chloride (CH2=CHCl)?
sp2
sp3
sp
sp3d
Explanation: In CH2=CHCl the C–Cl carbon is part of the C=C and is sp2 hybridised.
Question 2 of 20
The IUPAC name of (CH3)3CBr is:
1-bromobutane
2-bromo-2-methylpropane
2-bromobutane
1-bromo-2-methylpropane
Explanation: The central carbon bears Br and three methyls: 2-bromo-2-methylpropane (tert-butyl bromide).
Question 3 of 20
In the Swarts reaction an alkyl chloride is converted to an alkyl fluoride using:
NaI in acetone
Cu2Cl2
AgF or SbF3
PCl5
Explanation: Metallic fluorides such as AgF, Hg2F2, CoF2 or SbF3 supply fluorine in the Swarts reaction.
Question 4 of 20
The Finkelstein reaction is driven to completion because:
NaI is a strong oxidiser
iodide is a weaker nucleophile
acetone reacts with R–Cl
NaCl is insoluble in acetone and precipitates
Explanation: NaCl precipitates from dry acetone, removing product and shifting the equilibrium toward R–I.
Question 5 of 20
Addition of HBr to propene with peroxide gives predominantly:
1-bromopropane
2-bromopropane
1,2-dibromopropane
propan-2-ol
Explanation: The peroxide (free-radical) effect gives anti-Markovnikov addition, so Br adds to the terminal carbon: 1-bromopropane.
Question 6 of 20
The boiling points of CH3Cl, CH3Br and CH3I follow the order:
CH3I < CH3Br < CH3Cl
CH3Cl < CH3Br < CH3I
all equal
CH3Br < CH3Cl < CH3I
Explanation: Boiling point rises with molar mass/polarisability: CH3Cl < CH3Br < CH3I.
Question 7 of 20
The rate law for an SN2 reaction is:
Rate = k[R–X]
Rate = k[Nu]
Rate = k[R–X][Nu]
Rate = k[R–X]2
Explanation: SN2 is bimolecular; both the substrate and the nucleophile appear in the rate law (second order overall).
Question 8 of 20
SN1 reactivity of alkyl halides follows the order:
methyl > 1° > 2° > 3°
1° > 2° > 3° > methyl
all react equally
3° > 2° > 1° > methyl
Explanation: SN1 rate depends on carbocation stability, which increases 3° > 2° > 1° > methyl.
Question 9 of 20
Walden inversion (inversion of configuration) is observed in:
SN2
SN1
E1
free-radical halogenation
Explanation: SN2 proceeds by backside attack, inverting the configuration at the reacting carbon.
Question 10 of 20
Hydrolysis of an optically active halide giving a racemic (optically inactive) product indicates:
SN2 with inversion
SN1 via a planar carbocation
no reaction
pure retention
Explanation: A planar carbocation (SN1) is attacked from both faces, giving roughly equal enantiomers, i.e. racemisation.
Question 11 of 20
Polar protic solvents (e.g. water) favour:
SN2 by exposing the nucleophile
elimination only
SN1 by stabilising the carbocation
no substitution
Explanation: Polar protic solvents solvate and stabilise the ions formed on ionisation, favouring the SN1 pathway.
Question 12 of 20
When 2-bromobutane reacts with alcoholic KOH, the major product (Saytzeff) is:
but-1-ene
butan-2-ol
butane
but-2-ene
Explanation: Saytzeff’s rule predicts the more substituted, more stable alkene, but-2-ene, as the major elimination product.
Question 13 of 20
A Grignard reagent R–MgX must be made under anhydrous conditions because:
water protonates the nucleophilic carbon to give R–H
Mg dissolves in water
water oxidises R–X
ether reacts with water
Explanation: The carbanion-like carbon of R–MgX abstracts an acidic H from water, giving the alkane R–H and destroying the reagent.
Question 14 of 20
The Wurtz reaction of two molecules of CH3CH2Cl gives:
ethane
n-butane
propane
ethene
Explanation: 2C2H5Cl + 2Na → C4H10 (n-butane) + 2NaCl; two ethyl groups couple.
Question 15 of 20
Aryl halides are much less reactive than alkyl halides toward nucleophiles mainly due to:
weaker C–X bond
lower electronegativity of the halogen
partial double-bond character of C–X from resonance
smaller molecular size
Explanation: Resonance donation of the halogen lone pair gives the C–X bond partial double-bond character, making it short, strong and hard to break.
Question 16 of 20
In electrophilic substitution, a halogen substituent on benzene is:
activating, ortho/para-directing
deactivating, meta-directing
activating, meta-directing
deactivating, ortho/para-directing
Explanation: The −I effect deactivates the ring, but the +R effect directs the electrophile to the ortho and para positions.
Question 17 of 20
Chloroform is stored in dark bottles with a little ethanol to prevent formation of:
phosgene
carbon tetrachloride
methane
chlorine
Explanation: Light and air oxidise CHCl3 to poisonous phosgene (COCl2); ethanol and darkness prevent and destroy it.
Question 18 of 20
Which polyhalogen compound is a persistent, bioaccumulating insecticide now widely banned?
Dichloromethane
DDT
Freon-12
Iodoform
Explanation: DDT is chemically stable and fat-soluble, so it accumulates up the food chain; it is banned or restricted in many countries.
Question 19 of 20
Coupling of two molecules of chlorobenzene with Na in dry ether (Fittig reaction) gives:
toluene
benzene
biphenyl
phenol
Explanation: 2C6H5Cl + 2Na → C6H5–C6H5 (biphenyl) + 2NaCl.
Question 20 of 20
Thionyl chloride is preferred for R–OH → R–Cl because:
it is cheap
it works without heating
it gives an alkene
both by-products (SO2, HCl) are gases, giving a pure product
Explanation: SO2 and HCl escape as gases, so the alkyl chloride is obtained pure, unlike PCl5 which leaves liquid POCl3.