JEE (Advanced) · Chemistry, Class 12
Haloalkanes and Haloarenes: JEE Advanced PYQ, 2007–2026
Haloalkanes and Haloarenes (Chemistry, Class 12) has been asked 10 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.3 questions a paper, 1% of all Chemistry questions. It is the 23rd most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 1in 2 papers |
| 2009 | 1in 2 papers |
| 2010 | 0in 2 papers |
| 2011 | 1in 2 papers |
| 2012 | 0in 2 papers |
| 2013 | 1in 2 papers |
| 2014 | 0in 2 papers |
| 2015 | 0in 2 papers |
| 2016 | 0in 2 papers |
| 2017 | 1in 2 papers |
| 2018 | 1in 2 papers |
| 2019 | 0in 2 papers |
| 2020 | 0in 2 papers |
| 2021 | 1in 2 papers |
| 2022 | 0in 2 papers |
| 2023 | 2in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 0in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- SN1 and SN2 mechanisms7
- Nature of the C–X bond2
- Elimination reactions1
Try it
10 Haloalkanes and Haloarenes questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Haloalkanes and Haloarenes
The C–Cl bond length in chloromethane is 178 pm and its dipole moment is 1.86 D. Treat the entire dipole moment as arising from the C–Cl bond, and assume the bond consists of charges +δ on carbon and –δ on chlorine separated by 178 pm. What is δ, as a multiple of the electronic charge e? (1 D = 3.336 × 10⁻³⁰ C m, e = 1.602 × 10⁻¹⁹ C)
Answer: (A)
Dipole moment μ = charge × distance = δe × d. Convert μ to SI units: 1.86 D = 1.86 × 3.336 × 10⁻³⁰ C m = 6.21 × 10⁻³⁰ C m. Then e·d = 1.602 × 10⁻¹⁹ C × 1.78 × 10⁻¹⁰ m = 2.85 × 10⁻²⁹ C m. So δ = 6.21 × 10⁻³⁰ / 2.85 × 10⁻²⁹ ≈ 0.22. The C–Cl bond is therefore polar covalent and not ionic, with about 0.22 e of charge displaced from C to Cl.
Q2 · Chemistry · Haloalkanes and Haloarenes
2-Bromo-1-phenylpropane (C₆H₅CH₂CHBrCH₃) is heated with alcoholic KOH. Which statement describes the main product correctly?
Answer: (A)
Dehydrohalogenation removes HBr, with H lost from a β-carbon. Here there are two β-carbons. The CH₃ carbon gives C₆H₅CH₂CH=CH₂, which is non-conjugated and monosubstituted. The benzylic CH₂ carbon gives C₆H₅CH=CHCH₃, which is disubstituted and conjugated with the aromatic ring. By Zaitsev's rule, with extra stabilisation from conjugation, the second alkene is the major product (mainly the trans isomer). Alcoholic KOH is not a bulky base, so the Hofmann product is not favoured, and the medium favours elimination over substitution.
Q3 · Chemistry · Haloalkanes and Haloarenes
Laboratory chloroform is stored in dark brown bottles that are filled to the brim and closed tightly. Which of the following gives the correct reason for this practice?
Answer: (A)
Principle: chloroform is slowly oxidised by atmospheric oxygen in the presence of light to carbonyl chloride (COCl2, phosgene), 2CHCl3 + O2 → 2COCl2 + 2HCl. Phosgene is a highly poisonous gas. Application: to stop this reaction, both light and oxygen must be removed. Dark brown bottles cut off light, and filling the bottle to the brim leaves almost no air. The other options either confuse chloroform with CCl4, treat the practice as purely physical, or propose a hydrolysis reaction that is not the hazard.
Q4 · Chemistry · Haloalkanes and Haloarenes
A sample of 11.95 g of chloroform (M = 119.5 g/mol) is kept in a clear, partly filled bottle in sunlight. Over time 30% of the chloroform is oxidised by atmospheric oxygen according to 2CHCl3 + O2 → 2COCl2 + 2HCl. What mass of phosgene (COCl2, M = 99 g/mol) is formed, and how many moles of O2 are consumed?
Answer: (A)
Principle: use the balanced equation 2CHCl3 + O2 → 2COCl2 + 2HCl and convert mass to moles and back with the right molar mass. Application: n(CHCl3) = 11.95/119.5 = 0.100 mol, and 30% of it is 0.030 mol. The ratio CHCl3 : COCl2 is 1:1, so n(COCl2) = 0.030 mol and its mass is 0.030 × 99 = 2.97 g. The ratio CHCl3 : O2 is 2:1, so n(O2) = 0.015 mol. This slow air oxidation in light, which also releases HCl, is why chloroform is stored in dark, completely filled bottles with no air space.
Q5 · Chemistry · Haloalkanes and Haloarenes
Chlorobenzene is nitrated with a mixture of concentrated HNO3 and concentrated H2SO4. Which statement correctly describes the reactivity of the ring and the orientation of the product?
Answer: (A)
A halogen on a benzene ring has two opposing electronic effects. The –I effect, from the high electronegativity of Cl, withdraws electron density from the ring and makes it less reactive than benzene toward electrophiles. The +R effect, from donation of a lone pair into the ring, increases electron density at the ortho and para positions. The orientation is controlled by +R, because the carbocation intermediate from o/p attack has an extra resonance form with the positive charge on the carbon bearing Cl, which is stabilised by its lone pair. The result is that haloarenes are deactivating but ortho/para-directing.
Q6 · Chemistry · Haloalkanes and Haloarenes
Chloromethane has a dipole moment of 1.86 D and a C–Cl bond length of 178 pm. Assume the whole dipole moment arises from the C–Cl bond, treated as point charges +δe and −δe separated by the bond length. Use 1 D = 3.34 × 10⁻³⁰ C m and e = 1.60 × 10⁻¹⁹ C. What is the magnitude of the fractional charge δ on the chlorine atom?
Answer: (A)
A bond dipole moment is μ = q × d, where q is the charge on each end and d is the separation between the charges. Here q = δe. Convert the units first: μ = 1.86 × 3.34 × 10⁻³⁰ = 6.21 × 10⁻³⁰ C m, and d = 178 pm = 1.78 × 10⁻¹⁰ m. Then q = 6.21 × 10⁻³⁰ / 1.78 × 10⁻¹⁰ = 3.49 × 10⁻²⁰ C, so δ = 3.49 × 10⁻²⁰ / 1.60 × 10⁻¹⁹ ≈ 0.22. The C–Cl bond is therefore about 22% ionic, polar covalent and nowhere near fully ionic. The chlorine carries a partial negative charge and the carbon a partial positive charge. The common errors are halving the charge, halving the distance, and slipping a power of ten in the picometre conversion.
Q7 · Chemistry · Haloalkanes and Haloarenes
The hydrolysis of 2-chloro-2-methylpropane (initial concentration 0.010 M) by the SN1 mechanism is carried out at a fixed temperature in aqueous acetone containing 0.050 M NaOH. The half-life of the haloalkane is 30 minutes. The experiment is repeated at the same temperature with 0.50 M NaOH, and the reaction still proceeds entirely by SN1. What fraction of the original haloalkane remains after 90 minutes?
Answer: (C)
In an SN1 reaction the rate law is rate = k[RX]. The nucleophile concentration does not appear, so the first-order half-life (ln 2/k) is unchanged when [OH⁻] is raised from 0.050 M to 0.50 M. It remains 30 min. In 90 min, n = 90/30 = 3 half-lives have passed, so the fraction left is (1/2)³ = 1/8, or 12.5%.
Q8 · Chemistry · Haloalkanes and Haloarenes
A volatile halomethane contains only carbon and chlorine, with 92.2% chlorine by mass (C = 12, Cl = 35.5). Acute exposure to its vapour causes dizziness, nausea and an irregular heartbeat. Which option correctly identifies the compound and a consequence of its release?
Answer: (C)
Principle: the composition clue fixes the formula, and the hazards then follow from known properties. Application: for CCl4, % Cl = (4 × 35.5)/(12 + 142) × 100 = 142/154 × 100 = 92.2%. CHCl3 gives 106.5/119.5 = 89.1% and CH2Cl2 gives 71/85 = 83.5%, and both contain hydrogen. CCl2F2 contains fluorine and has 58.7% Cl. Only CCl4 fits. Its documented effects include dizziness, nausea, irregular heartbeat, damage to nerve cells and liver cancer. On release it rises to the stratosphere and contributes to ozone depletion.
Q9 · Chemistry · Haloalkanes and Haloarenes
1,2-Dibromoethane is heated with excess sodamide (NaNH₂) in liquid ammonia, and the mixture is then treated with water. The final organic product is:
Answer: (B)
A vicinal dihalide loses one HBr to give a vinyl halide (CH₂=CHBr). With excess of a very strong base such as NaNH₂, a second HBr is lost to give ethyne, which is deprotonated to sodium acetylide. Water then liberates ethyne. Stopping at the vinyl bromide, forming the diol by substitution, or forming ethene are the typical wrong answers.
Q10 · Chemistry · Haloalkanes and Haloarenes
The dipole moment of chlorobenzene is 1.5 D. Assume the dipole moment of a polychlorobenzene is the vector sum of the individual C–Cl moments, each equal to that of chlorobenzene and directed along the line from the ring centre to the substituted carbon, and ignore any interaction between substituents. What is the dipole moment of 1,2,4-trichlorobenzene?
Answer: (B)
Each C–Cl bond contributes a vector of 1.5 D pointing from the ring towards chlorine. Number the ring carbons so that the angular positions of carbons 1, 2 and 4 are 60°, 120° and 240°. Carbons 1 and 4 are para to each other, so their moments are equal and opposite and cancel. Only the moment of the Cl at carbon 2 remains, which is 1.5 D. A full vector sum gives the same result: (0.5, 0.866) + (−0.5, 0.866) + (−0.5, −0.866) = (−0.5, 0.866), with magnitude 1. In units of 1.5 D, the answer is 1.5 D. The answer is 0 only for the 1,3,5-isomer, and 3.0 D belongs to the 1,2,3-isomer. This works because the C–Cl bond is polar, so each bond has a direction-dependent moment that adds like a vector.
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All 10 real JEE Advanced questions from Haloalkanes and Haloarenes.
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About Haloalkanes and Haloarenes in JEE Advanced.
How many questions come from Haloalkanes and Haloarenes in JEE Advanced?
Haloalkanes and Haloarenes (Chemistry, Class 12) has been asked 10 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.3 questions a paper, 1% of all Chemistry questions. It is the 23rd most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it.
Which topics of Haloalkanes and Haloarenes are asked most in JEE Advanced?
SN1 and SN2 mechanisms (7), Nature of the C–X bond (2), Elimination reactions (1), counted over the papers from 2007–2026.
Is Haloalkanes and Haloarenes important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 23 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Haloalkanes and Haloarenes JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Haloalkanes and Haloarenes with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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