JEE Chemistry Haloalkanes and Haloarenes MCQs Set 24

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Chapter-wise Objective Questions: Haloalkanes and Haloarenes

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Multiple Choice Questions

Question 1: Ethanol is treated with \( \ce{SOCl2}\)/pyridine to give A, and A reacts with sodium acetylide (\( \ce{HC#CNa} \)) to give B. B is:
(a) But-2-yne
(b) Propyne
(c) Butane
(d) But-1-yne
Show Answer & Explanation

Answer: (d) But-1-yne

Explanation:
1. A = chloroethane.
2. Acetylide ion displaces Cl- (\( S_N2 \)) giving \( \ce{HC#C-CH2CH3} \), but-1-yne.

Teacher's Note:
1. Acetylide is a good nucleophile towards primary halides.

Question 2: Propene is treated with (i) \( \ce{BH3}\), THF and (ii) \( \ce{H2O2/OH-} \) to give X, which on treatment with \( \ce{PCl5} \) gives Y. Y is:
(a) 2-Chloropropane
(b) 1,1-Dichloropropane
(c) 2,2-Dichloropropane
(d) 1-Chloropropane
Show Answer & Explanation

Answer: (d) 1-Chloropropane

Explanation:
1. Hydroboration-oxidation gives the anti-Markovnikov alcohol, propan-1-ol (X).
2. \( \ce{PCl5} \) replaces the OH with Cl: 1-chloropropane.

Teacher's Note:
1. Hydroboration gives the primary alcohol.

Question 3: Ethene is converted to P by Baeyer's reagent, P to Q by \( \ce{PCl5} \), and Q to R by 1 mole of \( \ce{NaNH2} \). Which statement is true for R?
(a) It has a larger dipole moment than ethyl chloride
(b) It has a longer C-Cl bond than ethyl chloride
(c) It is less reactive than ethyl chloride in \( S_N \) reactions
(d) All of these
Show Answer & Explanation

Answer: (c) It is less reactive than ethyl chloride in \( S_N \) reactions

Explanation:
1. P = ethane-1,2-diol, Q = 1,2-dichloroethane, R = chloroethene (vinyl chloride).
2. In vinyl chloride the C-Cl bond has partial double bond character (shorter, stronger) and the dipole moment is lower, so it is much less reactive in substitution.

Teacher's Note:
1. Vinyl halides are unreactive towards nucleophilic substitution.

Question 4: 2-Chloropropane is converted by alcoholic KOH into A, A by HBr with peroxide into B, and B by aqueous KOH into C. C is:
(a) Propan-2-ol
(b) Propene
(c) Propan-1-ol
(d) 1-Bromopropane
Show Answer & Explanation

Answer: (c) Propan-1-ol

Explanation:
1. A = propene; B = 1-bromopropane (anti-Markovnikov addition).
2. Aqueous KOH replaces Br with OH, giving propan-1-ol.

Teacher's Note:
1. Alcoholic KOH eliminates; aqueous KOH substitutes.

Question 5: tert-Butyl bromide is heated with alcoholic KOH to give A, and A reacts with HBr in the presence of peroxide to give B. B is:
(a) 2-Bromo-2-methylpropane
(b) 1-Bromobutane
(c) 1-Bromo-2-methylpropane
(d) 2-Bromobutane
Show Answer & Explanation

Answer: (c) 1-Bromo-2-methylpropane

Explanation:
1. Dehydrohalogenation gives 2-methylpropene (A).
2. The peroxide effect gives anti-Markovnikov addition: Br adds to the terminal carbon, giving \( \ce{(CH3)2CHCH2Br} \).

Teacher's Note:
1. Peroxide with HBr gives the anti-Markovnikov product.

Question 6: 2-Iodopropane (P) is converted by alcoholic KOH to X, HBr with peroxide gives Y, and KI in acetone gives Z. P and Z are:
(a) chain isomers
(b) homologues
(c) position isomers
(d) identical
Show Answer & Explanation

Answer: (c) position isomers

Explanation:
1. X = propene; Y = 1-bromopropane (anti-Markovnikov); Z = 1-iodopropane.
2. P (2-iodopropane) and Z (1-iodopropane) differ in the position of iodine.

Teacher's Note:
1. Peroxide effect gives the anti-Markovnikov product.

Question 7: Ethane reacts with \( \ce{Br2}/h\nu \) to give P. P with KCN gives Q, which on acid hydrolysis gives S; P with AgCN gives R, which on acid hydrolysis gives T. S and T can be obtained by ozonolysis of:
(a) But-1-yne
(b) Hex-3-yne
(c) Propyne
(d) But-2-yne
Show Answer & Explanation

Answer: (a) But-1-yne

Explanation:
1. P = bromoethane. Q = propanenitrile, S = propanoic acid. R = ethyl isocyanide, T = ethylamine + formic acid (the formic acid being the other product).
2. Ozonolysis of but-1-yne \( \ce{CH3CH2C#CH} \) gives propanoic acid and formic acid.

Teacher's Note:
1. Match the acid fragments with the alkyne ends.

Question 8: But-2-ene undergoes ozonolysis to give X, X is treated with \( \ce{PCl5} \) to give Y, and Y with (i) alcoholic KOH and (ii) \( \ce{NaNH2} \) gives Z. Z is:
(a) Chloroethene
(b) Ethyne
(c) Ethanal
(d) 1,1-Dichloroethane
Show Answer & Explanation

Answer: (b) Ethyne

Explanation:
1. X = ethanal; Y = 1,1-dichloroethane (PCl5 replaces C=O with CCl2).
2. Double dehydrohalogenation by alcoholic KOH and \( \ce{NaNH2} \) gives ethyne.

Teacher's Note:
1. PCl5 converts a carbonyl group into a gem-dichloride.

Question 9: 1-Bromo-4-chlorobenzene is treated with 1 mole of Mg in THF to give A, A is treated with \( \ce{CO2} \) to give B and B with \( \ce{H+} \) to give C. C is:
(a) Benzene-1,4-dicarboxylic acid
(b) 4-Bromobenzoic acid
(c) Chlorobenzene
(d) 4-Chlorobenzoic acid
Show Answer & Explanation

Answer: (d) 4-Chlorobenzoic acid

Explanation:
1. Mg inserts at the weaker, more reactive C-Br bond.
2. The Grignard reagent adds to \( \ce{CO2} \) and, after acidification, gives 4-chlorobenzoic acid.

Teacher's Note:
1. C-Br reacts before C-Cl with magnesium.

Question 10: Propyne is treated successively with HBr (X), aqueous KOH (Y) and Zn-Hg/HCl (Z). Z is:
(a) Propane
(b) Propene
(c) Propan-2-ol
(d) Propanone
Show Answer & Explanation

Answer: (a) Propane

Explanation:
1. HBr adds to give 2-bromopropene (X).
2. Aqueous KOH converts it to the enol, which tautomerises to acetone (Y).
3. Clemmensen reduction of acetone gives propane (Z).

Teacher's Note:
1. Follow each step: addition, hydrolysis, reduction.

Question 11: Ethyne is passed through a red hot iron tube to give A, A is brominated with \( \ce{Br2/Fe} \) in the dark to give B, and B is heated with sodium in dry ether to give C. C is:
(a) Diphenylmethane
(b) Bromobenzene
(c) Toluene
(d) Biphenyl
Show Answer & Explanation

Answer: (d) Biphenyl

Explanation:
1. A = benzene; B = bromobenzene.
2. Sodium with an aryl halide in dry ether couples two aryl groups (Wurtz-Fittig/Fittig reaction), giving biphenyl.

Teacher's Note:
1. Aryl halide plus sodium gives a biaryl.

Question 12: Which of the following substitutions is feasible? (1) R-Br with \( \ce{Hg2F2} \); (2) R-Br with NaI in acetone; (3) R-Cl with AgF
(a) All of these
(b) (1) only
(c) (2) only
(d) (1) and (2) only
Show Answer & Explanation

Answer: (a) All of these

Explanation:
1. Swarts reaction: heavy metal fluorides (\( \ce{Hg2F2}\), AgF) convert alkyl chlorides or bromides to alkyl fluorides.
2. Finkelstein reaction: NaI in dry acetone converts R-Br to R-I because NaBr precipitates.

Teacher's Note:
1. Know the Swarts and Finkelstein reactions.

Question 13: Methane reacts with excess chlorine in the presence of light. Which statement is true for the major product?
(a) It reacts with aqueous \( \ce{AgNO3} \) to give a white precipitate of AgCl
(b) It is used as an anaesthetic
(c) All of these
(d) It has zero dipole moment
Show Answer & Explanation

Answer: (d) It has zero dipole moment

Explanation:
1. Excess chlorine gives carbon tetrachloride, \( \ce{CCl4} \).
2. It is tetrahedral and symmetrical, so the dipole moment is zero.
3. It is a covalent compound (no AgCl precipitate) and is not an anaesthetic (chloroform was).

Teacher's Note:
1. Excess halogen replaces all hydrogen atoms.

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Haloalkanes and Haloarenes Objective Questions & Solutions for JEE Chemistry

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