JEE Chemistry Haloalkanes and Haloarenes MCQs Set 05

Answer: (b) 73.4

Explanation:
1. Molar mass of \( \ce{C2H5Br} \) = 24 + 5 + 80 = 109.
2. Percentage Br = 80/109 x 100 = 73.4.

Teacher's Note:
1. Classify the halide by the type of carbon holding the halogen: 1, 2 or 3 degree.
2. Write the structural formula first, then count or classify; do not trust the name alone.

Answer: (b) \( \ce{CH3CBr2CH3} \)

Explanation:
1. Gem-dihalides have both halogen atoms on the same carbon atom, as in \( \ce{CH3CBr2CH3} \)
2. The others are vicinal or 1,3 dibromides.

Teacher's Note:
1. Write the structural formula first, then count or classify; do not trust the name alone.
2. Learn the common names of halogen derivatives (gem, vicinal, benzylidene, haloform) with their structures.

Answer: (c) \( \ce{CH3CHBr2} \)

Explanation:
1. Ethylidene dihalides are gem-dihalides of ethane: both Br atoms on the same carbon, \( \ce{CH3CHBr2} \)
2. \( \ce{BrCH2CH2Br} \) is ethylene dibromide.

Teacher's Note:
1. Learn the common names of halogen derivatives (gem, vicinal, benzylidene, haloform) with their structures.
2. Classify the halide by the type of carbon holding the halogen: 1, 2 or 3 degree.

Answer: (a) Isopropyl chloride

Explanation:
1. In isopropyl chloride, \( \ce{(CH3)2CHCl} \), the C bearing Cl is joined to two other carbons, so it is secondary.

Teacher's Note:
1. Write the structural formula first, then count or classify; do not trust the name alone.
2. Learn the common names of halogen derivatives (gem, vicinal, benzylidene, haloform) with their structures.

Answer: (b) Methane

Explanation:
1. Haloforms have the formula \( \ce{CHX3} \) and are derived from methane (chloroform, bromoform, iodoform).

Teacher's Note:
1. The iodoform test is given by compounds with a CH3CO- or CH3CH(OH)- group; methanol, formaldehyde and diethyl ketone do not respond.
2. Learn the common names of halogen derivatives (gem, vicinal, benzylidene, haloform) with their structures.

Answer: (a) 1,2,3,4,5,6-hexachlorocyclohexane

Explanation:
1. BHC is formed by addition of three \( \ce{Cl2} \) to benzene, giving a saturated ring with one Cl on each carbon: 1,2,3,4,5,6-hexachlorocyclohexane.

Teacher's Note:
1. BHC is formed by free-radical addition of chlorine to benzene in sunlight; only the gamma isomer is a strong insecticide.
2. Classify the halide by the type of carbon holding the halogen: 1, 2 or 3 degree.

Answer: (a) \( C_nH_{2n+1}X \)

Explanation:
1. An alkyl halide is an alkane in which one H is replaced by X: \( C_nH_{2n+1}X \).

Teacher's Note:
1. Learn the common names of halogen derivatives (gem, vicinal, benzylidene, haloform) with their structures.
2. Classify the halide by the type of carbon holding the halogen: 1, 2 or 3 degree.

Answer: (d) Neohexyl chloride

Explanation:
1. Neohexyl chloride, \( \ce{(CH3)3CCH2CH2Cl} \), has Cl on a carbon attached to only one other carbon.

Teacher's Note:
1. Classify the halide by the type of carbon holding the halogen: 1, 2 or 3 degree.
2. Write the structural formula first, then count or classify; do not trust the name alone.

Answer: (a) 1,1,1-trichloro-2,2-bis(p-chlorophenyl)ethane

Explanation:
1. DDT is \( \ce{(ClC6H4)2CH-CCl3} \): a CCl3 carbon next to a CH carrying two p-chlorophenyl groups.

Teacher's Note:
1. DDT is made from chloral and chlorobenzene with conc. sulphuric acid; it is a persistent insecticide.
2. Write the structural formula first, then count or classify; do not trust the name alone.

Answer: (b) Darzens procedure

Explanation:
1. Converting an alcohol to the alkyl chloride with thionyl chloride in pyridine is Darzens procedure.
2. The by-products SO2 and HCl are gases, so the product is easy to isolate.

Teacher's Note:
1. Know the named halogenation methods: Hunsdiecker (silver salt + Br2), Finkelstein (NaI/acetone), Swarts (metal fluoride) and Darzens (SOCl2).
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (d) 2-Bromo-2-methylpropane

Explanation:
1. By Markovnikov's rule H adds to the CH2 carbon, giving the tertiary carbocation, which captures Br-: \( \ce{(CH3)3CBr} \).

Teacher's Note:
1. Markovnikov addition puts H on the carbon with more hydrogens; the peroxide effect reverses this for HBr only.
2. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.

Answer: (b) An oxidative process

Explanation:
1. Replacing C-H by C-X raises the oxidation state of carbon, so it is an oxidation; it is also exothermic.

Teacher's Note:
1. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.
2. Link every preparation with its reagent: SOCl2 (Darzens), Cl2/hv, HX with ZnCl2, Sandmeyer, Hunsdiecker, Finkelstein, Swarts.

Answer: (a) \( \ce{CH3Br} \)

Explanation:
1. \( \ce{CH3COOAg + Br2 -> CH3Br + CO2 + AgBr} \); the product has one carbon less than the acid.

Teacher's Note:
1. Know the named halogenation methods: Hunsdiecker (silver salt + Br2), Finkelstein (NaI/acetone), Swarts (metal fluoride) and Darzens (SOCl2).
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (b) Sandmeyer's reaction

Explanation:
1. Replacing the diazonium group by Cl, Br or CN using the cuprous salt is the Sandmeyer reaction.

Teacher's Note:
1. Diazonium salts give Cl, Br, CN with cuprous salts (Sandmeyer), F with HBF4 and heat, and H with hypophosphorous acid.
2. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.

Answer: (d) \( \ce{CH3CH2Cl + HCl + SO2} \)

Explanation:
1. \( \ce{C2H5OH + SOCl2 -> C2H5Cl + SO2 + HCl} \)
2. Only option (d) lists all three products.

Teacher's Note:
1. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.
2. Link every preparation with its reagent: SOCl2 (Darzens), Cl2/hv, HX with ZnCl2, Sandmeyer, Hunsdiecker, Finkelstein, Swarts.

Answer: (a) n-Butane

Explanation:
1. n-Butane has two kinds of hydrogen (terminal CH3 and internal CH2), giving 1-chlorobutane and 2-chlorobutane.
2. Isobutane gives 1-chloro-2-methylpropane and tert-butyl chloride; neopentane and cyclobutane give one product each.

Teacher's Note:
1. Note the by-products as well; gaseous by-products explain why a method gives a pure product.
2. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.

Answer: (a) \( \ce{Cu2Cl2} \)

Explanation:
1. Sandmeyer reaction: the diazonium salt with \( \ce{Cu2Cl2} \)/HCl gives \( \ce{C6H5Cl} \).

Teacher's Note:
1. Diazonium salts give Cl, Br, CN with cuprous salts (Sandmeyer), F with HBF4 and heat, and H with hypophosphorous acid.
2. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.

Answer: (a) HI > HBr > HCl > HF

Explanation:
1. Reactivity follows acid strength and nucleophilicity of X-: HI > HBr > HCl > HF.

Teacher's Note:
1. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.
2. Link every preparation with its reagent: SOCl2 (Darzens), Cl2/hv, HX with ZnCl2, Sandmeyer, Hunsdiecker, Finkelstein, Swarts.

Answer: (a) Raschig process

Explanation:
1. The Raschig process passes benzene, HCl and air over a copper chloride catalyst to give chlorobenzene.

Teacher's Note:
1. Aryl and vinyl halides are inert in ordinary nucleophilic substitution because the C-X bond has partial double-bond character.
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (a) The bromonium ion formed initially may react with \( \ce{Br-} \) or \( \ce{CH3OH} \)

Explanation:
1. The cyclic bromonium ion can be opened by either Br- or the solvent CH3OH, giving two products.

Teacher's Note:
1. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.
2. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.

Answer: (a) Substitution

Explanation:
1. A hydrogen atom of the alkane is replaced by a chlorine atom.

Teacher's Note:
1. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.
2. Link every preparation with its reagent: SOCl2 (Darzens), Cl2/hv, HX with ZnCl2, Sandmeyer, Hunsdiecker, Finkelstein, Swarts.

Answer: (a) Tertiary > Secondary > Primary

Explanation:
1. The rate depends on carbocation stability (SN1 pathway): 3° > 2° > 1°.

Teacher's Note:
1. SN1 follows carbocation stability (3 degree > 2 degree > 1 degree), while SN2 is fastest for the least crowded carbon.
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (b) \( \ce{C6H6Cl6} \)

Explanation:
1. In UV light benzene adds three molecules of \( \ce{Cl2} \) (free-radical addition) to give benzene hexachloride, \( \ce{C6H6Cl6} \).

Teacher's Note:
1. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.
2. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.

Answer: (c) Bright sunlight

Explanation:
1. Addition (not substitution) occurs under light; \( \ce{AlCl3} \) would give chlorobenzene instead.

Teacher's Note:
1. Aryl and vinyl halides are inert in ordinary nucleophilic substitution because the C-X bond has partial double-bond character.
2. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.

Answer: (a) \( \ce{CHI3} \)

Explanation:
1. This is the iodoform reaction; iodoform is a yellow solid.

Teacher's Note:
1. The iodoform test is given by compounds with a CH3CO- or CH3CH(OH)- group; methanol, formaldehyde and diethyl ketone do not respond.
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (c) Both (a) and (b)

Explanation:
1. Bleaching powder with water gives \( \ce{Ca(OH)2} \) and \( \ce{Cl2} \); chlorine oxidises and chlorinates, and the base hydrolyses chloral.

Teacher's Note:
1. DDT is made from chloral and chlorobenzene with conc. sulphuric acid; it is a persistent insecticide.
2. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.

Answer: (c) Reduction

Explanation:
1. The steps are oxidation of ethanol, chlorination of ethanal and hydrolysis of chloral; no reduction takes place.

Teacher's Note:
1. DDT is made from chloral and chlorobenzene with conc. sulphuric acid; it is a persistent insecticide.
2. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.

Answer: (d) Propan-2-ol

Explanation:
1. The haloform reaction needs \( \ce{CH3CH(OH)-} \) or \( \ce{CH3CO-} \) groups.
2. Propan-2-ol is oxidised to acetone and then gives chloroform; the others do not have this group.

Teacher's Note:
1. The iodoform test is given by compounds with a CH3CO- or CH3CH(OH)- group; methanol, formaldehyde and diethyl ketone do not respond.
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.