Haloalkanes and Haloarenes — JEE Main practice

36 questions

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Sample questions with solutions

Q1 · 2026

The correct order of reactivity of the following benzyl halides towards reaction with KCN is:

  • A.

    a > b > c > d

  • B.

    b > a > d > c

  • C.

    b > a > c > d

  • D.

    a > b > d > c

Answer: B

  1. Identify the reaction mechanism. Since benzyl halides reacting with KCN typically proceed via SN1S_N1 (especially when substituents can stabilize a benzylic carbocation),

  2. Determine what controls the rate. Given the rate depends on the stability of the carbocation intermediate formed — more electron-donating substituents on the ring (which stabilize the developing positive charge via resonance/+M effect) increase reactivity, while electron-withdrawing substituents decrease it.

  3. Rank the four given structures (a-d) by the electron-donating/withdrawing character of their substituents.

[IMAGE: structures a-d — not extracted]

Hence, based on this stability analysis, the order is b > a > d > c — the answer is option B.

Q2 · 2026

The correct order of the rate of the reaction for the following reaction with respect to nucleophiles is:

CH3Br+Nu−⟶CH3Nu+Br−\mathrm{CH}_3\mathrm{Br} + \mathrm{Nu}^- \longrightarrow \mathrm{CH}_3\mathrm{Nu} + \mathrm{Br}^-

  • A.

    OH−>PhO−>CH3COO−>ClO4−\mathrm{OH}^- > \mathrm{PhO}^- > \mathrm{CH}_3\mathrm{COO}^- > \mathrm{ClO}_4^-

  • B.

    ClO4−>CH3COO−>OH−>PhO−\mathrm{ClO}_4^- > \mathrm{CH}_3\mathrm{COO}^- > \mathrm{OH}^- > \mathrm{PhO}^-

  • C.

    CH3COO−>PhO−>OH−>ClO4−\mathrm{CH}_3\mathrm{COO}^- > \mathrm{PhO}^- > \mathrm{OH}^- > \mathrm{ClO}_4^-

  • D.

    PhO−>OH−>CH3COO−>ClO4−\mathrm{PhO}^- > \mathrm{OH}^- > \mathrm{CH}_3\mathrm{COO}^- > \mathrm{ClO}_4^-

Answer: A

  1. Recognize the mechanism. Since CH3Br\mathrm{CH}_3\mathrm{Br} reacts via SN2S_N2, the rate increases with nucleophilicity of Nu−\mathrm{Nu}^-.

  2. Rank oxygen nucleophiles by resonance delocalization. Given more resonance delocalization of the negative charge reduces nucleophilicity,

  • OH−\mathrm{OH}^-: charge fully localized on one oxygen — strongest
  • PhO−\mathrm{PhO}^- (phenoxide): charge delocalized into the ring — weaker than OH−\mathrm{OH}^-
  • CH3COO−\mathrm{CH}_3\mathrm{COO}^- (acetate): charge delocalized over two oxygens — even weaker
  • ClO4−\mathrm{ClO}_4^-: charge extensively delocalized over four oxygens — essentially non-nucleophilic
  1. Combine into the overall order:
OH−>PhO−>CH3COO−>ClO4−\mathrm{OH}^- > \mathrm{PhO}^- > \mathrm{CH}_3\mathrm{COO}^- > \mathrm{ClO}_4^-

Hence, the answer is option A.

Q3 · 2026

The correct order of reactivity of CH3Br\mathrm{CH}_3\mathrm{Br} in methanol with the following nucleophiles is

F−,I−,C2H5O−\mathrm{F}^-, \mathrm{I}^-, \mathrm{C}_2\mathrm{H}_5\mathrm{O}^- and C6H5O−\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-

  • A.

    I−>C2H5O−>C6H5O−>F−\mathrm{I}^- > \mathrm{C}_2\mathrm{H}_5\mathrm{O}^- > \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- > \mathrm{F}^-

  • B.

    I−>F−>C6H5O−>C2H5O−\mathrm{I}^- > \mathrm{F}^- > \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- > \mathrm{C}_2\mathrm{H}_5\mathrm{O}^-

  • C.

    I−>C6H5O−>F−>C2H5O−\mathrm{I}^- > \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- > \mathrm{F}^- > \mathrm{C}_2\mathrm{H}_5\mathrm{O}^-

  • D.

    I−>C2H5O−>F−>C6H5O−\mathrm{I}^- > \mathrm{C}_2\mathrm{H}_5\mathrm{O}^- > \mathrm{F}^- > \mathrm{C}_6\mathrm{H}_5\mathrm{O}^-

Answer: A

  1. Recall that CH₃Br reacts via SN2S_N2, so the rate depends on the nucleophilicity of the attacking species in the protic solvent methanol.

  2. Consider solvation effects in a protic solvent. Since smaller, more charge-dense anions like F−\mathrm{F}^- are strongly solvated by methanol (hydrogen bonding), making them less available to attack, F−\mathrm{F}^- is the weakest nucleophile here. Larger, less charge-dense anions like I−\mathrm{I}^- are less solvated, making I−\mathrm{I}^- the strongest.

  3. Compare the two oxygen nucleophiles. Given C6H5O−\mathrm{C}_6\mathrm{H}_5\mathrm{O}^- (phenoxide) has its negative charge delocalized into the aromatic ring via resonance, reducing its nucleophilicity, while C2H5O−\mathrm{C}_2\mathrm{H}_5\mathrm{O}^- (ethoxide) has a more localized charge, making it more nucleophilic,

C2H5O−>C6H5O−\mathrm{C}_2\mathrm{H}_5\mathrm{O}^- > \mathrm{C}_6\mathrm{H}_5\mathrm{O}^-
  1. Combine into the overall order:
I−>C2H5O−>C6H5O−>F−\mathrm{I}^- > \mathrm{C}_2\mathrm{H}_5\mathrm{O}^- > \mathrm{C}_6\mathrm{H}_5\mathrm{O}^- > \mathrm{F}^-

Hence, the answer is option A.

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