JEE Chemistry Haloalkanes and Haloarenes MCQs Set 06

Answer: (d) Chloral

Explanation:
1. Ethanol is oxidised to ethanal and then exhaustively chlorinated to chloral, \( \ce{CCl3CHO} \).

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) Ethyl iodide

Explanation:
1. \( \ce{C2H5OC2H5 + 2HI -> 2C2H5I + H2O} \) (excess HI converts the alcohol also to the iodide).

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: (c) Diethyl ketone

Explanation:
1. Iodoform needs a \( \ce{CH3CO-} \) or \( \ce{CH3CH(OH)-} \) group.
2. Diethyl ketone, \( \ce{CH3CH2COCH2CH3} \), has neither.

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: (d) Acetamide

Explanation:
1. Propan-2-ol, acetophenone and butanone all contain the \( \ce{CH3CO-} \) or \( \ce{CH3CH(OH)-} \) unit.
2. Acetamide, \( \ce{CH3CONH2} \), does not give the reaction under these conditions.

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. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.

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

Explanation:
1. Haloform reaction of \( \ce{CH3COCH3} \) gives iodoform and sodium acetate.

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. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.

Answer: (d) \( \ce{HNO2} \) and then heating with \( \ce{Cu2Cl2} \)

Explanation:
1. Aniline is diazotised with \( \ce{NaNO2/HCl} \) (nitrous acid) and the diazonium salt is warmed with \( \ce{Cu2Cl2} \) (Sandmeyer).

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. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (a) \( \ce{CH3OH} \)

Explanation:
1. Methanol with HI (or red P and iodine) gives methyl iodide.

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) Ethyl iodide

Explanation:
1. Grignard reagents are prepared from Mg and an alkyl halide in dry ether, e.g.
2. \( \ce{C2H5MgI} \).

Teacher's Note:
1. Grignard reagents need dry ether and are destroyed by water, alcohols and any acidic hydrogen.
2. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.

Answer: (c) Chloroform

Explanation:
1. Chlorine liberated at the anode and alkali at the cathode act like bleaching powder, giving chloroform.

Teacher's Note:
1. Chloroform is oxidised by air and light to poisonous phosgene, so it is stored in dark, full bottles with a little ethanol.
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (b) NaCl

Explanation:
1. NaCl cannot convert an alcohol to an alkyl halide because the OH group is a poor leaving group and Cl- alone cannot displace it; the other three reagents all work.

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: (b) \( \ce{C2H5Cl + C6H5COCl + POCl3} \)

Explanation:
1. The ester O-C bonds are cleaved: \( \ce{C6H5COOC2H5 + PCl5 -> C6H5COCl + C2H5Cl + POCl3} \).

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: (c) Gammexane

Explanation:
1. In sunlight benzene adds three Cl2 molecules (free-radical addition) to give 1,2,3,4,5,6-hexachlorocyclohexane, which is gammexane (lindane, the gamma isomer of BHC).

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. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (c) HBr

Explanation:
1. The peroxide effect (free-radical addition) works only for HBr.

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: (a) Isopropyl bromide

Explanation:
1. By Markovnikov's rule H adds to the CH2 carbon: \( \ce{CH3CHBrCH3} \).

Teacher's Note:
1. Markovnikov addition puts H on the carbon with more hydrogens; the peroxide effect reverses this for HBr only.
2. For alcohols with HX remember the order 3 degree > 2 degree > 1 degree and HI > HBr > HCl.

Answer: (b) DDT

Explanation:
1. Two molecules of chlorobenzene condense with \( \ce{CCl3CHO} \) in conc.
2. \( \ce{H2SO4} \) to give DDT.

Teacher's Note:
1. DDT is made from chloral and chlorobenzene with conc. sulphuric acid; it is a persistent insecticide.
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (a) Chloroform

Explanation:
1. Haloform reaction of acetone with bleaching powder gives chloroform.

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. Anti-Markovnikov addition applies to HBr only, and only in the presence of peroxides.

Answer: (c) Ethanol

Explanation:
1. Ethanol is oxidised and chlorinated to chloral, which is hydrolysed to chloroform.

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: (b) Ethylene dibromide

Explanation:
1. Electrophilic addition of \( \ce{Br2} \) across the double bond gives 1,2-dibromoethane.

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) \( \ce{ROH + SOCl2} \)

Explanation:
1. With \( \ce{SOCl2} \) the by-products \( \ce{SO2} \) and HCl escape, so the alkyl chloride is obtained pure and in good yield.

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: (b) \( \ce{CCl3-CHO} \)

Explanation:
1. Chloral, \( \ce{CCl3CHO} \), condenses with two molecules of chlorobenzene in conc. sulphuric acid.

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) Trichloroacetaldehyde

Explanation:
1. The reagent is trichloroacetaldehyde (chloral), \( \ce{CCl3CHO} \), which condenses with two molecules of chlorobenzene.

Teacher's Note:
1. DDT is made from chloral and chlorobenzene with conc. sulphuric acid; it is a persistent insecticide.
2. Note the by-products as well; gaseous by-products explain why a method gives a pure product.

Answer: (a) 1-Bromo-3-(trichloromethyl)benzene (meta)

Explanation:
1. The strongly electron-withdrawing \( \ce{CCl3} \) group deactivates the ring and directs the incoming electrophile to the meta position; one equivalent of Br2 gives the mono-bromo product.

Teacher's Note:
1. Chlorobenzene and vinyl chloride are unreactive because the C-Cl bond has partial double-bond character.
2. Chloroform, DDT, BHC and chloropicrin are asked repeatedly: memorise their preparations and uses.

Answer: (c) The action of moist silver oxide

Explanation:
1. Moist \( \ce{Ag2O} \) acts as AgOH and substitutes Br by OH. Alcoholic KOH would give ethene instead.

Teacher's Note:
1. Alcoholic KOH eliminates HX to give the more substituted alkene (Saytzeff rule); aqueous KOH substitutes.
2. Chloroform, DDT, BHC and chloropicrin are asked repeatedly: memorise their preparations and uses.

Answer: (c) Tetraethylammonium chloride

Explanation:
1. With excess alkyl halide ammonia is alkylated repeatedly (1, 2, 3° amines) and finally forms the quaternary salt \( \ce{(C2H5)4N+Cl-} \).

Teacher's Note:
1. Haloform and carbylamine reactions are tests; know the reagent, the product and the observation.
2. Separate substitution (aqueous KOH, AgCN, KCN) from elimination (alcoholic KOH) before reading the options.

Answer: (c) Light and air

Explanation:
1. Chloroform is slowly oxidised to phosgene by air in the presence of light, which is why it is stored in dark bottles.

Teacher's Note:
1. Chloroform is oxidised by air and light to poisonous phosgene, so it is stored in dark, full bottles with a little ethanol.
2. Separate substitution (aqueous KOH, AgCN, KCN) from elimination (alcoholic KOH) before reading the options.

Answer: (c) Carbonyl chloride

Explanation:
1. Phosgene is \( \ce{COCl2} \), carbonyl chloride.

Teacher's Note:
1. Chlorobenzene and vinyl chloride are unreactive because the C-Cl bond has partial double-bond character.
2. Chloroform, DDT, BHC and chloropicrin are asked repeatedly: memorise their preparations and uses.

Answer: (c) Isopropyl chloride and methyl chloride

Explanation:
1. Wurtz coupling of isopropyl (3 carbons) and methyl (1 carbon) gives \( \ce{(CH3)2CH-CH3} \), isobutane (2-methylpropane).

Teacher's Note:
1. Wurtz coupling joins two alkyl groups with sodium in dry ether; fluorides do not react and mixed halides give a mixture of alkanes.
2. Remember the ambident nucleophiles: KCN gives nitriles, AgCN gives isocyanides, KNO2 gives nitrites, AgNO2 gives nitroalkanes.

Answer: (b) Warming them with magnesium powder in dry ether

Explanation:
1. \( \ce{RX + Mg -> RMgX} \) in dry (anhydrous) ether; any trace of water or alcohol destroys the reagent.

Teacher's Note:
1. Grignard reagents need dry ether and are destroyed by water, alcohols and any acidic hydrogen.
2. Haloform and carbylamine reactions are tests; know the reagent, the product and the observation.