Match the methods of preparing halogen compounds in List-I with their features in List-II.
| List-I (Method) | List-II (Feature) |
|---|---|
| (A) Free radical chlorination of an alkane | (I) Two products are possible, but one predominates as per Markovnikov’s rule |
| (B) Addition of HBr to propene | (II) The reaction is reversible and needs an oxidising agent such as HNO3 or HIO4 |
| (C) Addition of bromine in CCl4 to an alkene | (III) A complex mixture of isomeric products forms, so the yield of any single compound is low |
| (D) Iodination of an arene with iodine | (IV) Used in the laboratory to detect a double bond; the product is colourless |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-I, C-IV, D-II
(A) Free radical chlorination of an alkane — This matches (III) A complex mixture of isomeric products forms. Chlorine can replace any of the hydrogen atoms, and more than one of them, so isomeric mono- and polyhaloalkanes form together. They are hard to separate, and the yield of any one compound is low. For butane, the monochloride C4H9Cl is a mixture of 1-chlorobutane and 2-chlorobutane:
\( \displaystyle \ce{C4H10 + Cl2} \xrightarrow{h\nu} \ce{C4H9Cl + HCl} \)
(B) Addition of HBr to propene — This matches (I) Two products are possible, but one predominates as per Markovnikov’s rule. Br can add to either carbon of the double bond; the major product is 2-bromopropane, and 1-bromopropane is the minor one.
\( \displaystyle \ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3} \) (major)
(C) Addition of bromine in CCl4 to an alkene — This matches (IV) Used in the laboratory to detect a double bond; the product is colourless. The reddish brown colour of bromine is discharged as it adds across the C=C, and the vic-dibromide formed is colourless.
\( \displaystyle \ce{CH3-CH=CH2 + Br2} \xrightarrow{\ce{CCl4}} \ce{CH3-CHBr-CH2Br} \)
(D) Iodination of an arene with iodine — This matches (II) The reaction is reversible and needs an oxidising agent. The HI formed would take the reaction back, so an oxidising agent (HNO3 or HIO4) is added to oxidise the HI.
\( \displaystyle \ce{C6H6 + I2 <=> C6H5I + HI} \)
NCERT (2026-27 edition):
6.4.2 From Hydrocarbons – “(I) From alkanes by free radical halogenation … Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any single compound is low.”
“Propene yields two products, however only one predominates as per Markovnikov’s rule.”
“(ii) Addition of halogens: In the laboratory, addition of bromine in CCl4 to an alkene resulting in discharge of reddish brown colour of bromine constitutes an important method for the detection of double bond in a molecule. The addition results in the synthesis of vic-dibromides, which are colourless (Unit 9, Class XI).”
6.5 Preparation of Haloarenes – “…Reactions with iodine are reversible in nature and require the presence of an oxidising agent (HNO3, HIO4) to oxidise the HI formed during iodination.”