Chemistry · Haloalkanes and haloarenes · chapter 6 of 10

Class 12 Chemistry Chapter 6: Haloalkanes and Haloarenes MCQs with answers

10 multiple-choice questions on Haloalkanes and Haloarenes, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
The halide (CH₃CH₂)₃C–Cl is classified as
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Answer: (c) a tertiary alkyl halide

The carbon carrying chlorine is bonded to three ethyl groups, i.e. to three other carbon atoms, so the halide is tertiary (3°). The carbon is sp³, so it is not vinylic.

Question 2
Which of the following is an allylic halide?
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Answer: (b) CH₃–CH=CH–CH₂Br

In CH₃CH=CH–CH₂Br the bromine is on an sp³ carbon next to the C=C: allylic. In (a) the CH₂Cl is two carbons away from the double bond; (c) has Br on an sp² carbon of the C=C (vinylic); in (d) the CH₂Cl is not attached to the ring, so it is not benzylic either.

Question 3
The IUPAC name of CH₃CH₂CH(CH₃)CH₂CH₂Cl is
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Answer: (a) 1-chloro-3-methylpentane

The longest chain has five carbons; numbering from the CH₂Cl end gives Cl locant 1 and CH₃ locant 3 (lower set than 5 and 3). Prefixes are cited alphabetically: chloro before methyl. (d) does not use the longest chain.

Question 4
CH₃CH₂CH₂CHCl₂ is best described as
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Answer: (b) a gem-dihalide, 1,1-dichlorobutane

Both chlorine atoms sit on the same (terminal) carbon, so the compound is geminal (an alkylidene halide, butylidene chloride); its IUPAC name is 1,1-dichlorobutane. Vicinal would need the two halogens on adjacent carbons.

Question 5
How many structurally different monochloro products can form when 2,3-dimethylbutane, (CH₃)₂CH–CH(CH₃)₂, undergoes free radical chlorination?
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Answer: (b) 2

The molecule is symmetrical: all twelve CH₃ hydrogens are equivalent and the two CH hydrogens are equivalent. Two types of hydrogen give two monochlorides: ClCH₂CH(CH₃)CH(CH₃)₂ and (CH₃)₂CClCH(CH₃)₂.

Question 6
Butan-1-ol is converted to 1-chlorobutane by thionyl chloride. The method is preferred because the by-products are
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Answer: (c) SO₂ and HCl, both gases

R–OH + SOCl₂ → R–Cl + SO₂ + HCl. Both by-products are gases that escape, leaving the alkyl chloride pure. H₃PO₃ comes from PCl₃, POCl₃ from PCl₅.

Question 7
Which alcohol is converted to its chloride simply by shaking with concentrated HCl at room temperature, with no ZnCl₂?
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Answer: (c) 2-methylbutan-2-ol

Tertiary alcohols react with concentrated HCl on shaking at room temperature; primary and secondary alcohols need ZnCl₂. Only 2-methylbutan-2-ol, (CH₃)₂C(OH)CH₂CH₃, is tertiary; (a) and (d) are primary, (b) secondary.

Question 8
Which carbon–halogen bond has the lowest bond enthalpy?
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Answer: (d) CH₃–I

Down the group the halogen grows and the C–X bond lengthens, from 139 pm (C–F) to 214 pm (C–I); the bond enthalpy falls accordingly, 452, 351, 293 and 234 kJ mol⁻¹. CH₃–I has the longest bond and the lowest enthalpy.

Question 9
1-Bromopropane is best converted to 1-fluoropropane by heating it with
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Answer: (b) Hg₂F₂

Alkyl fluorides are best made by heating an alkyl bromide or chloride with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃: the Swarts reaction. NaI gives the iodide; fluorine is too reactive for controlled halogenation.

Question 10
1-Methylcyclohexene reacts with HCl. The main product is
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Answer: (a) 1-chloro-1-methylcyclohexane

By Markovnikov's rule H adds to the ring CH of the double bond (the carbon with more hydrogens) and Cl to the carbon carrying the methyl group, giving the tertiary chloride 1-chloro-1-methylcyclohexane. (d) would need Cl₂, not HCl.

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