JEE Chemistry Haloalkanes and Haloarenes MCQs Set 13

Answer: (c) III < II < I

Explanation:
1. Methyl groups donate electrons and increase the electron density of the ring, which opposes attack by a nucleophile.
2. The more methyl groups, the lower the reactivity: III < II < I.

Teacher's Note:
1. Electron donating groups deactivate towards nucleophilic substitution.

Answer: (a) butane < 1-chlorobutane < 1-bromobutane < 1-iodobutane

Explanation:
1. Boiling point increases with the molar mass and polarisability of the halogen.
2. Butane (-0.5 degrees C) < 1-chlorobutane (78) < 1-bromobutane (102) < 1-iodobutane (131).

Teacher's Note:
1. Larger halogen means stronger dispersion forces.

Answer: (b) 1-bromoethane < 1-bromopropane < 1-bromobutane < bromobenzene

Explanation:
1. Within the series the boiling point rises with molar mass: bromoethane (38 degrees C) < 1-bromopropane (71) < 1-bromobutane (102).
2. Bromobenzene (156 degrees C) has the highest molar mass and the largest surface area, so it boils highest.

Teacher's Note:
1. Compare molar masses and surface areas.

Answer: (d) \( \ce{CH3Cl} \)

Explanation:
1. \(S_N2\) reactions are fastest when steric hindrance at the carbon is minimal.
2. Methyl chloride is the least hindered, so it reacts fastest.

Teacher's Note:
1. \(S_N2\) order: methyl > primary > secondary > tertiary.

Answer: (d) 3-phenylcyclopentene

Explanation:
1. E2 elimination needs a hydrogen anti-periplanar to Br.
2. In the trans isomer the hydrogen on C2 is cis to Br, so it cannot be removed.
3. The H on C5 is anti, so the double bond forms between C1 and C5, which gives 3-phenylcyclopentene.

Teacher's Note:
1. E2 needs an anti hydrogen; Zaitsev product is not always possible.

Answer: (a) IV > III > I > II

Explanation:
1. The rate follows the nucleophilicity, which follows the basicity of the attacking oxygen.
2. Basicity: methoxide > hydroxide > phenoxide (delocalised charge) > acetate (more delocalised).

Teacher's Note:
1. Weaker conjugate acids give stronger nucleophiles for the same attacking atom.

Answer: (c) 2-methyl-2-propanol

Explanation:
1. The first CH\(_3\)MgI gives acetone by displacing bromide.
2. Acetone then reacts with a second equivalent to give the alkoxide, which on hydrolysis with NH\(_4\)Cl gives the tertiary alcohol (CH\(_3\))\(_3\)COH, 2-methyl-2-propanol.

Teacher's Note:
1. Acid halides react twice with Grignard reagents.

Answer: (b) DDT

Explanation:
1. Two molecules of chlorobenzene condense with chloral (\( \ce{CCl3CHO} \)) in the presence of conc. H\(_2\)SO\(_4\).
2. The product is 1,1,1-trichloro-2,2-bis(4-chlorophenyl)ethane, i.e. DDT.

Teacher's Note:
1. DDT stands for dichlorodiphenyltrichloroethane.

Answer: (d) \( \ce{CH3C6H4NC} \)

Explanation:
1. This is the carbylamine (isocyanide) test for primary amines.
2. The primary amine group of p-toluidine is converted into an isocyanide group, giving p-tolyl isocyanide.

Teacher's Note:
1. Carbylamine reaction: R-NH\(_2\) + CHCl\(_3\) + 3KOH gives R-NC.

Answer: (a) tert-butyl chloride

Explanation:
1. Wurtz coupling of tert-butyl chloride gives 2,2,3,3-tetramethylbutane, \( \ce{(CH3)3C-C(CH3)3} \).
2. All the 18 hydrogens are equivalent, so monochlorination gives only one product.

Teacher's Note:
1. A hydrocarbon with all H atoms equivalent gives only one monochloro compound.

Answer: (a) 2,3-dimethylpentane

Explanation:
1. In 2,3-dimethylpentane, C3 carries H, CH\(_3\), C\(_2\)H\(_5\) and CH(CH\(_3\))\(_2\), four different groups, so it is chiral.
2. 2,2-Dimethylpentane and 2-methylhexane have no chiral carbon.

Teacher's Note:
1. Check each carbon for four different substituents.

Answer: (a) 1, 2 and 4

Explanation:
1. A carbon with four different groups is chiral.
2. In 1 the C2 bears H, D, CH\(_3\), CH\(_2\)Cl; in 2 the C bears H, D, Cl, C\(_2\)H\(_5\); in 4 the C bears H, Cl, CH\(_3\), CH\(_2\)D.
3. In 3 no carbon has four different groups.

Teacher's Note:
1. Isotopes make carbons chiral.

Answer: (b) \( \ce{NO2} \) withdraws electrons from the ortho and para positions

Explanation:
1. The nitro groups at the ortho and para positions withdraw electrons by the resonance effect.
2. This stabilises the intermediate carbanion and makes the C-Cl carbon more electron deficient, so the chlorine is replaced easily.

Teacher's Note:
1. Activated nucleophilic aromatic substitution needs o/p electron withdrawing groups.

Answer: (b) calcium hypochlorite

Explanation:
1. Bleaching powder (calcium hypochlorite) acts as a source of chlorine and as a base.
2. It chlorinates acetone to trichloroacetone, which is then cleaved by Ca(OH)\(_2\) to give chloroform.

Teacher's Note:
1. Haloform reaction of acetone with bleaching powder.

Answer: (b) benzene

Explanation:
1. Chlorobenzene gives the Grignard reagent \( \ce{C6H5MgCl} \).
2. The Grignard reagent is protonated by the acidic hydrogen of ethanol to give benzene.

Teacher's Note:
1. Alcohols decompose Grignard reagents to alkanes or arenes.

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

Explanation:
1. Hydrolysis proceeds through nucleophilic substitution, activated by o/p electron withdrawing groups.
2. Three nitro groups (2,4,6-) give the strongest activation, while the dimethylamino group is electron donating and deactivates.

Teacher's Note:
1. More o/p nitro groups give faster hydrolysis.

Answer: (d) carbonyl chloride

Explanation:
1. Phosgene is \( \ce{COCl2} \), carbonyl chloride.
2. It is formed by oxidation of chloroform in air and light, and is highly poisonous.

Teacher's Note:
1. Chloroform is stored in dark bottles to prevent phosgene formation.

Answer: (a) Phenetole (ethyl phenyl ether)

Explanation:
1. This is the Williamson synthesis: \( \ce{C6H5ONa + C2H5I -> C6H5OC2H5 + NaI} \).
2. The product is phenetole.

Teacher's Note:
1. Phenoxide acts as a nucleophile towards primary alkyl halides.

Answer: (d) all of these

Explanation:
1. The tertiary halide with a strongly basic methoxide mainly undergoes E2 elimination, giving the alkenes (b) and (c) (Zaitsev product (c) is major).
2. A small amount of the ether (a) forms by substitution, so all three can form.

Teacher's Note:
1. Tertiary halides with a strong base give mostly elimination.

Answer: (c) 3

Explanation:
1. The first RMgX replaces Cl (giving an ester RCOOEt).
2. The second RMgX adds to the ester carbonyl and expels ethoxide to give a ketone, and the third adds to the ketone.
3. After work-up the product is \( \ce{R3C-OH} \), a tertiary alcohol, so Y = 3.

Teacher's Note:
1. Chloroformate esters need three equivalents of Grignard reagent.