Haloalkanes and Haloarenes Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on haloalkanes and haloarenes with step-by-step solutions covering named reactions, SN1 reactivity, carbocation stability, free-radical chlorination and boiling-point trends.
A curated set of JEE Main 2026 previous-year questions on haloalkanes and haloarenes, each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Recall each named reaction and its characteristic reagent:
- A. Finkelstein — halide exchange $R\text{-}X + \text{NaI} \xrightarrow{\text{dry acetone}} R\text{-}I + \text{NaX}$. Reagent: NaI (III).
- B. Swarts — replacement of Cl/Br by F using metallic fluorides: $R\text{-}X + \text{SbF}_3 \rightarrow R\text{-}F$. Reagent: $\text{SbF}_3$ (I).
- C. Sandmeyer — diazonium to aryl halide: $\text{ArN}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2} \text{Ar-Cl}$. Reagent: $\text{Cu}_2\text{Cl}_2$ (IV).
- D. Fittig — coupling of aryl halides with sodium in dry ether: $2\,\text{Ar-X} + 2\text{Na} \rightarrow \text{Ar-Ar}$. Reagent: Na, dry ether (II).
So A-III, B-I, C-IV, D-II.
Answer: B
Solution
Statement I: 3-phenylpropene is allylbenzene, $\text{C}_6\text{H}_5\text{-}\text{CH}_2\text{-}\text{CH}=\text{CH}_2$. Markovnikov addition of HBr places Br on the more substituted internal carbon:
$$\text{C}_6\text{H}_5\text{-CH}_2\text{-CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{C}_6\text{H}_5\text{-CH}_2\text{-}\underset{\text{Br}}{\text{CH}}\text{-CH}_3$$The Br-bearing carbon carries four different groups ($\text{-Br}$, $\text{-H}$, $\text{-CH}_3$, $\text{-CH}_2\text{C}_6\text{H}_5$), so it is a chiral secondary bromide. True.
Statement II: Both routes start from a diazonium salt. Sandmeyer uses cuprous salts ($\text{Cu}_2\text{Cl}_2$ for Ar-Cl, CuCN for Ar-CN); Gattermann uses copper powder with HCl (Ar-Cl) or with KCN/Cu (Ar-CN). Both give aryl chlorides and aryl cyanides. True.
Both statements are true.
Answer: A
Solution
The rate-determining step of $S_N1$ is carbocation formation, so the more stable the cation, the faster the reaction.
- Benzyl cation $\text{C}_6\text{H}_5\text{-}\overset{+}{\text{CH}}_2$ is stabilized by resonance — the positive charge is delocalized into the aromatic ring (three additional resonance structures).
- Ethyl cation $\text{CH}_3\text{-}\overset{+}{\text{CH}}_2$ is a primary cation stabilized only by weak hyperconjugation.
Since resonance stabilization $\gg$ hyperconjugation, the benzyl cation is far more stable, so benzyl chloride reacts faster in $S_N1$. Statement I is true. Statement II correctly states the reason and is also true, and it is the correct explanation of Statement I.
Answer: A
Solution
Evaluate each statement:
- A. True. Alcoholic KOH (a less polar medium) removes $\beta$-hydrogens and promotes elimination ($E2$), whereas aqueous KOH favours substitution.
- B. False. $S_N1$ rate follows cation stability. The benzhydryl cation $(\text{C}_6\text{H}_5)_2\overset{+}{\text{CH}}$ from $\text{C}_6\text{H}_5\text{-CHCl-C}_6\text{H}_5$ is stabilized by two rings and is more stable than the benzyl cation, so the order is reversed: $\text{C}_6\text{H}_5\text{-CHCl-C}_6\text{H}_5 > \text{C}_6\text{H}_5\text{-CH}_2\text{-Cl}$.
- C. False. Aryl halides ($\text{C-X}$ has partial double-bond character, resonance) are far less reactive than alkyl halides; their properties are not similar.
- D. False. Allyl chloride $\text{CH}_2=\text{CH-CH}_2\text{-Cl}$ is a haloalkene, not a haloalkyne.
- E. True. Aliphatic alcohols give R-Cl with $\text{SOCl}_2$, but phenols (Ar-OH) do not give Ar-Cl with $\text{SOCl}_2$ because of the strong, partial-double-bond $\text{C(aryl)-O}$ linkage.
Correct statements: A and E only.
Answer: C
Solution
Identify P. Monochlorination of n-butane $\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_3$ gives 1-chlorobutane and 2-chlorobutane. Only 2-chlorobutane $\text{CH}_3\text{-}\underset{\text{Cl}}{\text{CH}}\text{-CH}_2\text{-CH}_3$ has a chiral carbon, so it is the optically active product P.
Second chlorination of 2-chlorobutane. Label the carbons $\text{C}^1\text{H}_3\text{-}\text{C}^2\text{HCl-}\text{C}^3\text{H}_2\text{-}\text{C}^4\text{H}_3$. A second Cl can replace an H at any of the four carbons:
- New Cl at $\text{C}^1$ → 1,2-dichlorobutane
- New Cl at $\text{C}^2$ → 2,2-dichlorobutane
- New Cl at $\text{C}^3$ → 2,3-dichlorobutane
- New Cl at $\text{C}^4$ → 1,3-dichlorobutane (renumbering $\text{ClCH}_2\text{-CH}_2\text{-CHCl-CH}_3$)
Ignoring stereoisomers, these are 4 distinct dichloro compounds.
Answer: 4 (option B)
Solution
Statement I: Within a homologous series, increasing molecular size (surface area) increases van der Waals forces and hence boiling point. Since chain length increases $\text{CH}_3\text{I} < \text{CH}_3\text{CH}_2\text{I} < \text{CH}_3\text{CH}_2\text{CH}_2\text{I}$, the boiling-point order stated is correct. True.
Statement II: The para isomer packs into a more symmetric crystal lattice, giving a higher melting point than the ortho isomer (standard NCERT trend). For boiling point, the ortho isomer has a small net dipole moment while the para isomer’s opposing dipoles cancel; thus the para boiling point is slightly lower than the ortho. True.
Both statements are true.
Answer: A