JEE Chemistry Haloalkanes and Haloarenes MCQs Set 30

Answer: (a) C₂H₆ (excess) + Cl₂, UV light

Explanation:
1. Free-radical chlorination needs light.
2. Using ethane in excess makes it likely that a chlorine radical meets ethane and not C₂H₅Cl, which limits further substitution.
3. Answer: (a)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (c) C₆H₅CH₂C(CH₃)₂Cl

Explanation:
1. C is benzaldehyde, which has no alpha hydrogen, and D is propanone, which has alpha hydrogens but no CHO.
2. So B is C₆H₅CH=C(CH₃)₂.
3. B arises from the tertiary chloride C₆H₅CH₂C(CH₃)₂Cl, which also gives an immediate precipitate with AgNO₃.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (a) CHCl₃ does not ionise in water

Explanation:
1. Chloroform is covalent and does not give free Cl⁻ ions, so Ag⁺ finds nothing to precipitate as AgCl.
2. Answer: (a)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (c) 1-Bromobicyclo[2.2.2]octane

Explanation:
1. In S_N2 the nucleophile must attack from the back of the C–Br bond.
2. At a bridgehead carbon this side is blocked by the cage, and inversion is impossible.
3. The other bromides allow back-side attack.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (c) 1-Chloro-2,4-dinitrobenzene

Explanation:
1. Hydrazine displaces chloride from 1-chloro-2,4-dinitrobenzene by nucleophilic aromatic substitution.
2. The nitro groups at the ortho and para positions activate the ring.
3. The product is 2,4-dinitrophenylhydrazine.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (b) 2,2-Bis(4-chlorophenyl)-1,1,1-trichloroethane (DDT)

Explanation:
1. Chloral is protonated and attacks two molecules of chlorobenzene at the para positions.
2. The product is p,p'-dichlorodiphenyltrichloroethane, DDT.
3. Answer: (b)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (d) Propionic acid

Explanation:
1. Three OH groups replace the three Cl atoms to give CH₃CH₂C(OH)₃.
2. This is unstable and loses water to give propanoic acid (as its potassium salt).
3. Answer: (d)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (b) NCCH₂CH₂CN

Explanation:
1. Alcoholic KOH gives ethene (X).
2. Bromine gives 1,2-dibromoethane (Y).
3. KCN replaces both bromines to give succinonitrile (Z).
4. Answer: (b)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (d) NO₂ withdraws electrons from the ortho and para positions

Explanation:
1. Nitro groups at the ortho and para positions withdraw electron density by delocalisation and induction.
2. They stabilise the negatively charged intermediate of nucleophilic aromatic substitution.
3. Answer: (d)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (c) Cyclopentene

Explanation:
1. The Grignard reagent has MgBr next to the remaining Br.
2. On heating, Mg and Br leave from adjacent carbons by a β-elimination.
3. This gives cyclopentene.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (d) 4-Nitrobenzyl chloride

Explanation:
1. All are primary benzylic chlorides, which react by S_N2 with an open transition state.
2. The electron-withdrawing nitro group makes the benzylic carbon more electrophilic and stabilises the transition state with the incoming nucleophile.
3. Answer: (d)

Teacher's Note:
1. Compare the electron demand of the four substituents.

Answer: (b) b only

Explanation:
1. S_N2 rate falls with steric crowding, so the order in (a) is right, and (c) follows leaving group ability.
2. In S_N1, the more substituted benzylic halide gives the more stable cation, so the correct order is PhCBrMePh > PhCBrMe₂ > PhCHBrMe > PhCH₂Br.
3. The order in (b) is reversed.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (b) CH₃OCH₂Cl

Explanation:
1. Loss of Cl⁻ from CH₃OCH₂Cl gives CH₃O⁺=CH₂, stabilised by the oxygen lone pair.
2. The tert-butoxy compound is similar but more hindered, and the other two give poor cations.
3. Answer: (b)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (a) 1-Chloro-2,4,6-trinitrobenzene

Explanation:
1. Three nitro groups at the ortho and para positions stabilise the Meisenheimer intermediate the most.
2. An electron-donating NMe₂ group would slow the reaction.
3. Answer: (a)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (d) B with aqueous NaOH (S_N2) gives CH₃CH(OH)CH₂CH₂Br and not CH₃CHBrCH₂CH₂OH

Explanation:
1. At the higher temperature the more stable 1,4-adduct A is formed, and Markovnikov addition of HBr gives B.
2. Mg couples the two C–Br bonds intramolecularly to give methylcyclopropane.
3. In S_N2 the primary C–Br is attacked faster than the secondary one, so the product is CH₃CHBrCH₂CH₂OH, which makes (d) wrong.
4. Answer: (d)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (c) 1-Bromo-4-(deuteriomethyl)benzene

Explanation:
1. The benzylic C–Cl bond reacts with Mg more readily than the aryl C–Br bond.
2. The Grignard reagent is BrC₆H₄CH₂MgCl, and D₂O replaces MgCl with D.
3. The product is BrC₆H₄CH₂D.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

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

Explanation:
1. The CCl₃ group is electron withdrawing, so HCl adds anti-Markovnikov to give Cl₃CCH₂CH₂Cl.
2. Aqueous KOH converts CCl₃ to COOK and CH₂Cl to CH₂OH.
3. B is Cl₃C–CHBr–CH₂Br, which has one stereocentre, so it can be resolved.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (b) Benzyne

Explanation:
1. The strong base removes a proton ortho to Cl, and Cl⁻ leaves to give benzyne.
2. NH₂⁻ then adds to benzyne, and protonation gives aniline.
3. This is the elimination–addition mechanism.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (b) CH₃CH(OH)CH₂Cl

Explanation:
1. X is propan-1-ol and dehydration gives propene (Y).
2. Chlorine water forms a chloronium ion, which water opens at the more substituted carbon.
3. Z is the chlorohydrin CH₃CH(OH)CH₂Cl.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (d) Bromobenzene

Explanation:
1. This is the Hunsdiecker reaction, which replaces COOAg with Br.
2. The product is bromobenzene, C₆H₅Br.
3. Answer: (d)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (b) PhCH(OH)COOH

Explanation:
1. OH⁻ hydrolyses the gem-dibromide to the hydrate, which gives phenylglyoxal, PhCOCHO.
2. An intramolecular Cannizzaro reaction then gives the mandelate.
3. Acidification gives mandelic acid, PhCH(OH)COOH.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (c) A mixture of both phenols, with ¹⁴C at C–OH and at the adjacent carbon

Explanation:
1. At this temperature elimination–addition takes place through benzyne.
2. The nucleophile can attack either carbon of the triple bond, so the label ends up at both positions.
3. A mixture of the two phenols forms.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.

Answer: (d) meso-2,3-Dibromobutane

Explanation:
1. Bromine adds in an anti manner through a cyclic bromonium ion.
2. Anti addition to the trans alkene gives the meso compound.
3. The cis alkene would give the racemic mixture.
4. Answer: (d)

Teacher's Note:
1. Decide first whether the halide reacts by S_N1, S_N2, elimination or a radical path, then apply that rule to the options.