JEE Chemistry Haloalkanes and Haloarenes MCQs Set 25

Answer: (c) C₆H₅CH₂C(CH₃)₂Cl

Explanation:
1. An immediate white precipitate with AgNO₃ points to a tertiary chloride, which forms a stable carbocation: C₆H₅CH₂C(CH₃)₂Cl.
2. Alcoholic KOH gives the conjugated, more substituted alkene B = C₆H₅CH=C(CH₃)₂, and ozonolysis gives benzaldehyde C (no α-hydrogen, Cannizzaro) and acetone D (α-hydrogens, aldol, no Cannizzaro).
3. The primary chloride (a) would not precipitate immediately, (b) would give two aldehydes with α-hydrogens, and (d) is a benzylic chloride with no β-hydrogen for elimination.
4. Answer: (c)

Teacher's Note:
1. Tertiary halide: immediate AgCl.
2. Aldehyde without α-H: Cannizzaro; with α-H: aldol.

Answer: (a) CHCl₃ does not ionise in water

Explanation:
1. Chlorine in chloroform is covalently bonded and no free Cl⁻ ions are present, so no AgCl is formed.
2. Organic halides do not give the test unless Cl⁻ is released.
3. Answer: (a)

Teacher's Note:
1. Precipitation needs free halide ions.
2. Covalent C–Cl bonds do not ionise.

Answer: (b) CH₃CH₂CHBrCH₃

Explanation:
1. Bromination is a free-radical reaction in which bromine radicals abstract the weaker secondary C–H hydrogen preferentially.
2. The main product is therefore 2-bromobutane, CH₃CH₂CHBrCH₃.
3. Answer: (b)

Teacher's Note:
1. Radical stability: 3° > 2° > 1°.
2. Bromination is highly selective.

Answer: (c) D > C > A > B

Explanation:
1. For nucleophiles with the same attacking atom (oxygen), nucleophilicity follows basicity.
2. Basicity decreases in the order CH₃O⁻ > HO⁻ > C₆H₅O⁻ > CH₃COO⁻, because the conjugate acids are in the order of increasing acidity: methanol < water < phenol < acetic acid.
3. Answer: (c)

Teacher's Note:
1. Same attacking atom: stronger base is the better nucleophile.
2. Resonance delocalisation lowers nucleophilicity.

Answer: (b) The reaction of methylmagnesium bromide with acetone gives butan-2-ol

Explanation:
1. Methylmagnesium bromide adds to acetone to give, after hydrolysis, 2-methylpropan-2-ol (tertiary butyl alcohol), not butan-2-ol.
2. The other statements are correct: reduction gives ethane, HX reactivity is HI > HBr > HCl, and C₂H₄Cl₂ exists as 1,1- and 1,2-dichloroethane.
3. Answer: (b)

Teacher's Note:
1. Grignard reagent + ketone gives a tertiary alcohol.
2. Acetone + CH₃MgBr gives (CH₃)₃COH.

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

Explanation:
1. 2,4-Dinitrophenylhydrazine is made by nucleophilic aromatic substitution of the chlorine of 1-chloro-2,4-dinitrobenzene by hydrazine.
2. The nitro groups ortho and para to chlorine activate the ring for this substitution.
3. Answer: (c)

Teacher's Note:
1. Brady's reagent is 2,4-dinitrophenylhydrazine.
2. SNAr needs ortho/para nitro groups.

Answer: (c) (p-ClC₆H₄)₂CH–CCl₃ (DDT)

Explanation:
1. Chloral condenses with two molecules of chlorobenzene (electrophilic substitution at the para position) with loss of water.
2. The product is 1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane, DDT: (p-ClC₆H₄)₂CH–CCl₃.
3. Answer: (c)

Teacher's Note:
1. DDT is made from chlorobenzene and chloral.
2. Chlorine directs para.

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

Explanation:
1. The nitro group withdraws electrons by −M and −I effects from the ortho and para positions, making these carbons electron poor.
2. The carbanion intermediate formed by attack of OH⁻ is stabilised by the nitro groups at the ortho and para positions, so substitution is easy.
3. Answer: (d)

Teacher's Note:
1. Electron withdrawing groups at o/p favour nucleophilic aromatic substitution.
2. Meta nitro groups give no resonance stabilisation.

Answer: (b) racemic mixture

Explanation:
1. Free radical chlorination at the secondary carbon gives a planar radical, which is attacked by chlorine equally from both faces.
2. The chiral product 2-chlorobutane is therefore formed as an equal mixture of d and l forms, a racemic mixture.
3. Answer: (b)

Teacher's Note:
1. Planar radical intermediate gives both enantiomers.
2. Product is optically inactive.

Answer: (d) trans-pent-2-ene

Explanation:
1. Strong base in ethanol favours E2 elimination over substitution.
2. By Saytzeff's rule the more substituted alkene, pent-2-ene, is the major product, and the trans isomer is more stable than the cis isomer.
3. Answer: (d)

Teacher's Note:
1. Saytzeff: more substituted alkene.
2. Trans alkenes are more stable than cis.

Answer: (a) CH₃CH₂CH₂NHCOCH₃

Explanation:
1. CH₃CH₂Cl + NaCN → CH₃CH₂CN (X).
2. Reduction gives propylamine, CH₃CH₂CH₂NH₂ (Y).
3. Acetylation by acetic anhydride gives N-propylacetamide, CH₃CH₂CH₂NHCOCH₃ (Z).
4. Answer: (a)

Teacher's Note:
1. Nitrile reduction adds one carbon to the amine.
2. Amine + anhydride gives amide.

Answer: (c) 1-fluoro-2,4,6-trinitrobenzene

Explanation:
1. The reaction is nucleophilic aromatic substitution (addition–elimination).
2. The rate rises with the number of electron-withdrawing nitro groups at ortho and para positions, which stabilise the negatively charged intermediate.
3. The trinitro compound has three such groups and reacts fastest.
4. Answer: (c)

Teacher's Note:
1. More o/p nitro groups, faster S_NAr.
2. Fluorine is a good group in S_NAr because it activates the ipso carbon.

Answer: (d) CH₃CCl₂CH₃ (2,2-dichloropropane)

Explanation:
1. Immediate turbidity in the Lucas test means III is a tertiary alcohol.
2. Tertiary alcohol III = (CH₃)₃COH, formed from acetone (II) and CH₃MgBr.
3. Acetone arises from hydrolysis of the gem-dihalide 2,2-dichloropropane.
4. Answer: (d)

Teacher's Note:
1. Lucas test: tertiary alcohol reacts immediately.
2. Hydrolysis of gem-dihalide gives a ketone or aldehyde.

Answer: (b) C₆H₅CH₂Br > C₆H₅CHBrCH₃ > C₆H₅CBr(CH₃)₂ > C₆H₅CBr(CH₃)C₆H₅ (S_N1)

Explanation:
1. For S_N1 the rate depends on carbocation stability.
2. The cation from C₆H₅CBr(CH₃)C₆H₅ is tertiary and stabilised by two phenyl groups, so it forms fastest, and the primary benzyl bromide is slowest: the order in (b) is reversed.
3. (a) and (c) are correct for S_N2.
4. Answer: (b)

Teacher's Note:
1. S_N1 rate: stability of carbocation.
2. S_N2 rate: less hindrance and better leaving group.

Answer: (b) C₆H₅CH=C(CH₃)CH₂CH₃

Explanation:
1. A tertiary halide with a strong base undergoes E2 elimination.
2. Removal of a benzylic hydrogen gives C₆H₅CH=C(CH₃)CH₂CH₃, in which the C=C is conjugated with the benzene ring and is the most substituted and most stable alkene, so it is the major product.
3. Answer: (b)

Teacher's Note:
1. Elimination, not substitution, with strong base and tertiary halide.
2. Conjugation with the ring stabilises the alkene.

Answer: (a) 2,4,6-trinitrochlorobenzene

Explanation:
1. Hydrolysis proceeds by nucleophilic aromatic substitution, which is fastest when strongly electron withdrawing groups sit ortho and para to chlorine.
2. 2,4,6-Trinitrochlorobenzene has three such nitro groups and reacts under mild conditions.
3. Answer: (a)

Teacher's Note:
1. Nitro groups at o/p activate; amino groups deactivate.
2. More nitro groups mean faster reaction.

Answer: (d) B on S_N2 with aqueous NaOH gives CH₃CHOHCH₂CH₂Br and not CH₃CHBrCH₂CH₂OH

Explanation:
1. A is the 1,4-addition product, B is 1,3-dibromobutane, and Mg gives intramolecular coupling to methylcyclopropane (C).
2. In an S_N2 reaction the unhindered primary bromine is replaced faster than the secondary one, so the product is CH₃CHBrCH₂CH₂OH; statement (d) is the reverse.
3. Answer: (d)

Teacher's Note:
1. S_N2 prefers primary carbon.
2. 1,3-Dihalide with Mg gives a cyclopropane.

Answer: (c) benzyl chloride

Explanation:
1. In benzyl chloride the chlorine is on an sp³ carbon, and C–Cl cleavage gives the resonance stabilised benzyl cation.
2. In the chlorotoluenes the C–Cl bond is on an sp² carbon with partial double-bond character from resonance, so it is stronger.
3. Answer: (c)

Teacher's Note:
1. sp³ C–Cl is weaker than sp² C–Cl.
2. Benzylic cation is stable.

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

Explanation:
1. The CCl₃ group is electron withdrawing, so HCl adds against Markovnikov's rule to give A = Cl₃CCH₂CH₂Cl.
2. Aqueous KOH hydrolyses CCl₃ to COOK and CH₂Cl to CH₂OH, giving HOCH₂CH₂COOK.
3. B = Cl₃CCHBrCH₂Br has a chiral carbon (CHBr), so it can be resolved into enantiomers.
4. Answer: (c)

Teacher's Note:
1. Strongly electron withdrawing group reverses orientation.
2. A carbon with four different groups is chiral.

Answer: (d) B and C

Explanation:
1. The intermediate (Meisenheimer complex) is formed by addition of CH₃O⁻ to the ring carbon bearing fluorine, which becomes sp³, so aromaticity is lost and A is false.
2. The negative charge is delocalised over the ring and onto the nitro group (B true), and electron-withdrawing groups stabilise it (C true).
3. Answer: (d)

Teacher's Note:
1. Addition step breaks aromaticity.
2. Nitro group delocalises negative charge.

Answer: (b) CH₃CHOHCH₂Cl

Explanation:
1. CH₃CH₂CH₂Br → propan-1-ol (X) → propene (Y, dehydration).
2. Chlorine water adds HOCl across the double bond (chlorohydrin formation): the Cl goes to the terminal carbon and OH to the more substituted carbon, giving CH₃CH(OH)CH₂Cl.
3. Answer: (b)

Teacher's Note:
1. Cl₂/H₂O gives a halohydrin.
2. OH adds to the carbon that better stabilises positive charge.

Answer: (b) C₆H₅CH(OH)COOH

Explanation:
1. Hydrolysis of the gem-dibromide gives phenylglyoxal, C₆H₅COCHO (A).
2. With hydroxide, phenylglyoxal undergoes an intramolecular Cannizzaro rearrangement to the mandelate ion (B), C₆H₅CH(OH)COO⁻.
3. Acidification gives mandelic acid, C₆H₅CH(OH)COOH (C).
4. Answer: (b)

Teacher's Note:
1. Aldehyde with no α-H: Cannizzaro.
2. Intramolecular Cannizzaro gives a hydroxy acid.

Answer: (c) on both the labelled carbon and the adjacent carbon (two products)

Explanation:
1. At high temperature with a strong base the reaction proceeds through benzyne (elimination–addition).
2. OH⁻ can add to either carbon of the triple bond, so the OH ends up on the labelled carbon in one product and on the adjacent carbon in the other, in roughly equal amounts.
3. Answer: (c)

Teacher's Note:
1. Benzyne has two equivalent ends.
2. Labelling confirms the benzyne mechanism.

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

Explanation:
1. Bromine adds anti through a cyclic bromonium ion.
2. Anti addition to the trans alkene gives the meso compound (the two stereocentres are mirror images of each other, giving an internal plane of symmetry).
3. Answer: (d)

Teacher's Note:
1. Anti addition on trans-alkene gives meso.
2. Anti addition on cis-alkene gives racemic mixture.

Answer: (d) o- and p-chlorotoluene

Explanation:
1. FeCl₃ generates Cl⁺ equivalent, an electrophile, giving ring substitution (not side chain).
2. The methyl group is ortho/para directing, so the products are o- and p-chlorotoluene.
3. Answer: (d)

Teacher's Note:
1. Lewis acid: ring substitution; light or heat: side chain.
2. CH₃ is activating and o/p directing.

Answer: (c) 2 > 3 > 1

Explanation:
1. Solvolysis proceeds through a carbocation (S_N1).
2. The tertiary halide (2) gives the most stable cation and is fastest, the secondary halide (3) is next, and the vinylic halide (1) does not ionise readily because the vinyl cation is unstable and the C–Br bond is strengthened by resonance.
3. Answer: (c)

Teacher's Note:
1. Rate follows carbocation stability.
2. Vinylic halides are the least reactive.