Solved Examples (Dedicated Problem Set)

How to Use This Problem Set

This is your full workout for Haloalkanes and Haloarenes, grouped by theme: classification and nomenclature, preparation and named reactions, SN1 vs SN2 mechanism and reactivity, stereochemistry, elimination and Saytzeff, haloarene reactivity, and polyhalogen uses.

Quick reminders:

• Finkelstein (NaI/dry acetone) makes iodides; Swarts (AgF) makes fluorides; Sandmeyer (CuX/HX) makes aryl halides.
• SN2: rate = k[RX][Nu], inversion, CH$_3$ > 1 > 2 > 3. SN1: rate = k[RX], racemisation, 3 > 2 > 1 > CH$_3$.
• aq. KOH → substitution (alcohol); alc. KOH → elimination (alkene, Saytzeff).

Solved Examples - Classification & Naming

Example 1. Classify (CH$_3$)$_3$CCl as 1/2/3 degree.

Solution: Tertiary (3 degree) - the C-Cl carbon has three carbons attached.

Example 2. IUPAC name of (CH$_3$)$_2$CHBr?

Solution: 2-bromopropane.

Example 3. Classify CH$_2$=CH-CH$_2$-Br by C-X type.

Solution: Allylic halide (sp$^3$ C-Br adjacent to C=C).

Solved Examples - Preparation

Example 4. Convert 1-chlorobutane to 1-iodobutane.

Solution: Finkelstein reaction: NaI in dry acetone.

Example 5. How to make an alkyl fluoride from an alkyl bromide?

Solution: Swarts reaction: heat with AgF (or Hg$_2$F$_2$, CoF$_2$, SbF$_3$).

Example 6. Product of propene + HBr with peroxide?

Solution: Anti-Markovnikov: 1-bromopropane.

Solved Examples - SN1/SN2

Example 7. SN2 reactivity: rank CH$_3$Br, (CH$_3$)$_2$CHBr, (CH$_3$)$_3$CBr.

Solution: CH$_3$Br > (CH$_3$)$_2$CHBr > (CH$_3$)$_3$CBr (steric hindrance increases).

Example 8. SN1 reactivity: rank the same three.

Solution: (CH$_3$)$_3$CBr > (CH$_3$)$_2$CHBr > CH$_3$Br (carbocation stability).

Example 9. A first-order substitution (rate = k[RX]) is which mechanism?

Solution: SN1.

Solved Examples - Stereochemistry

Example 10. SN2 at a chiral carbon gives what stereochemistry?

Solution: Inversion (Walden inversion).

Example 11. SN1 of an optically active halide gives what?

Solution: A racemic mixture (racemisation).

Example 12. Is CHFClBr chiral?

Solution: Yes - four different groups on the carbon.

Solved Examples - Elimination

Example 13. 2-bromopropane + alcoholic KOH gives?

Solution: Propene (elimination).

Example 14. 2-bromopropane + aqueous KOH gives?

Solution: Propan-2-ol (substitution).

Example 15. Major alkene from 2-bromobutane elimination?

Solution: but-2-ene (Saytzeff - more substituted).

Solved Examples - Metals & Haloarenes

Example 16. 2 CH$_3$CH$_2$Br + 2 Na/dry ether gives?

Solution: n-Butane (Wurtz reaction).

Example 17. CH$_3$CH$_2$Br + Mg/dry ether gives?

Solution: Ethylmagnesium bromide (Grignard reagent).

Example 18. Why is chlorobenzene less reactive than chloroethane?

Solution: Resonance (partial double-bond C-Cl), sp$^2$ carbon, unstable phenyl cation.

Solved Examples - Haloarenes & Uses

Example 19. Where does nitration of chlorobenzene occur?

Solution: Mainly ortho and para (halogen is o/p-directing).

Example 20. Toluene from chlorobenzene + CH$_3$Cl + Na?

Solution: Wurtz-Fittig reaction.

Example 21. Why store chloroform in dark bottles with ethanol?

Solution: To prevent oxidation to poisonous phosgene (ethanol destroys any phosgene).

Solved Examples - Mixed

Example 22. Best leaving group: F-, Cl-, Br-, I-?

Solution: I- (weakest C-I bond).

Example 23. Which gives a racemic product: SN1 or SN2?

Solution: SN1.

Example 24. Product of ethene + Br$_2$?

Solution: 1,2-dibromoethane (vicinal dihalide).

Example 25. Freon-12 formula and use?

Solution: CCl$_2$F$_2$, used as a refrigerant.

Solved Examples - Mixed (continued)

Example 26. Reactivity of alcohols with HX: 1, 2, 3 degree order?

Solution: 3 > 2 > 1 degree.

Example 27. Which HX shows the peroxide (anti-Markovnikov) effect?

Solution: Only HBr.

Example 28. Solvent used in Finkelstein and why?

Solution: Dry acetone - NaI soluble, NaCl/NaBr insoluble (precipitate, drives reaction).

Example 29. Reagent: aqueous KOH gives substitution or elimination?

Solution: Substitution (alcohol).

Example 30. What is iodoform's antiseptic action due to?

Solution: The liberated free iodine.

JEE Main & Advanced Level Solved Examples

These problems demand multi-step reasoning, stereochemical analysis and mechanism work that go beyond direct recall — the kind that separate top scorers. Identify the substrate type and the operative mechanism first, then reason carefully.

Example 31: Assign R/S configuration [JEE Main]

Rank the four groups on the chiral carbon of 2-bromobutane (CH$_3$-CHBr-CH$_2$-CH$_3$) by CIP priority, and state how R or S is then assigned.

Solution: The chiral carbon (C2) carries Br, CH$_2$CH$_3$ (ethyl), CH$_3$ and H.

• Priority is fixed first by the atomic number of the directly bonded atom: Br (Z = 35) is highest, H is lowest.
• Ethyl vs methyl: both join through carbon, so compare the next atoms — ethyl's carbon carries (C, H, H) while methyl's carbon carries (H, H, H). Since C > H, ethyl outranks methyl.
• Final order: Br > CH$_2$CH$_3$ > CH$_3$ > H.

Now point the lowest priority (H) away from you. If Br → ethyl → methyl traces a clockwise arc the centre is R; anticlockwise is S. (If H happens to point toward you, read the direction and then flip the label.)

Example 32: Count the stereoisomers of 2,3-dibromobutane [JEE]

How many stereoisomers does CH$_3$-CHBr-CHBr-CH$_3$ have?

Solution: There are two chiral carbons (C2 and C3) bearing the same set of groups, so the naive count $2^2 = 4$ overcounts.

• (2R,3R) and (2S,3S) are non-superimposable mirror images → one pair of enantiomers (optically active).
• (2R,3S) has an internal mirror plane, so it is superimposable on its mirror image — a single meso form (optically inactive).

Total = three stereoisomers (a d/l pair plus one meso). Identical ends collapse two of the four into one meso isomer.

Example 33: Stereoisomers of 2-bromo-3-chlorobutane [JEE]

How many stereoisomers does CH$_3$-CHBr-CHCl-CH$_3$ have?

Solution: Again two chiral carbons, but C2 (Br) and C3 (Cl) now carry different groups, so there is no internal symmetry and no meso form. All $2^2 = 4$ combinations are distinct: (2R,3R), (2S,3S), (2R,3S), (2S,3R) — two pairs of enantiomers, four stereoisomers. Contrast with Example 32: a meso form appears only when the two stereocentres are constitutionally identical.

JEE/Advanced — Mechanisms & Rearrangements

Example 34: Rearrangement during solvolysis [JEE Advanced]

Neopentyl bromide, (CH$_3$)$_3$C-CH$_2$-Br, is a primary halide, yet it gives almost no normal SN2 product and on solvolysis yields rearranged products. Explain.

Solution: SN2 is throttled because the bulky tert-butyl group shields the backside of the C-Br carbon (severe steric hindrance). Ionisation instead gives an unstable primary neopentyl cation, which at once undergoes a 1,2-methyl shift: a methyl migrates from the adjacent quaternary carbon to produce the far more stable tertiary cation (CH$_3$)$_2$C$^+$-CH$_2$-CH$_3$ (the tert-amyl cation). This cation is then trapped, giving rearranged products such as 2-methylbutan-2-ol (substitution) or 2-methylbut-2-ene (elimination). A change in the carbon skeleton is the fingerprint of a carbocation intermediate.

Example 35: Saytzeff vs Hofmann elimination [JEE Main]

2-Bromo-2-methylbutane is treated with (a) alcoholic KOH and (b) potassium tert-butoxide. Predict the major alkene in each case.

Solution: Two alkenes are possible — 2-methylbut-2-ene (more substituted, Saytzeff) and 2-methylbut-1-ene (less substituted, Hofmann).

• (a) Alcoholic KOH (small base) follows Saytzeff's rule → the more stable, more substituted 2-methylbut-2-ene dominates.
• (b) Potassium tert-butoxide (bulky base) cannot easily reach the hindered internal hydrogens, so it removes a more accessible terminal H → the Hofmann product, 2-methylbut-1-ene, dominates.

Takeaway: a bulky base switches the selectivity from Saytzeff to Hofmann.

JEE/Advanced — Haloarenes: Benzyne & SNAr

Example 36: The benzyne mechanism [JEE Advanced]

Chlorobenzene resists NaOH except under drastic conditions, yet with the stronger base sodamide (NaNH$_2$) in liquid NH$_3$ it gives aniline. When the C bearing Cl is labelled with $^{14}$C, the product has the -NH$_2$ on the labelled carbon and on the carbon next to it in nearly equal amounts. Explain.

Solution: This is elimination-addition (the benzyne mechanism), not direct substitution. The strong base abstracts a hydrogen ortho to chlorine; loss of Cl$^-$ then generates a highly reactive benzyne intermediate carrying an extra, strained π bond between two adjacent ring carbons. The amide ion (NH$_2^-$) adds across this symmetrical unit at either carbon with about equal probability, so the $^{14}$C label is split roughly 50:50 between the original and the neighbouring position. This isotope-scrambling result is the classic proof of the benzyne pathway.

Example 37: Reactivity in nucleophilic aromatic substitution [JEE]

Arrange chlorobenzene, p-nitrochlorobenzene, 2,4-dinitrochlorobenzene and 2,4,6-trinitrochlorobenzene (picryl chloride) in increasing order of reactivity toward aqueous NaOH, and explain.

Solution: When the nucleophile adds, it forms a negatively charged (Meisenheimer) intermediate; electron-withdrawing -NO$_2$ groups at the ortho and para positions delocalise and stabilise that charge, so each o/p-NO$_2$ speeds the reaction. Increasing reactivity:

chlorobenzene < p-nitrochlorobenzene < 2,4-dinitrochlorobenzene < 2,4,6-trinitrochlorobenzene.

Chlorobenzene needs ~623 K and ~300 atm, whereas picryl chloride (three o/p-NO$_2$) hydrolyses very easily. A meta-NO$_2$ would not help — it cannot delocalise the charge onto nitrogen.

JEE/Advanced — Synthesis, Reactivity & Numericals

Example 38: Multi-step conversion [JEE Main]

Convert propan-2-ol into propan-1-ol using only reactions from this chapter.

Solution: A direct one-step swap is impossible, so relocate the functional group through an alkene:

1. Dehydrate propan-2-ol with hot concentrated H$_2$SO$_4$ → propene (CH$_3$-CH=CH$_2$).
2. Add HBr with an organic peroxide (anti-Markovnikov, Kharasch effect) → 1-bromopropane (CH$_3$-CH$_2$-CH$_2$Br).
3. Hydrolyse with aqueous KOH (SN2) → propan-1-ol.

The peroxide effect is the crux: it places Br on the terminal carbon, moving -OH from C2 to C1.

Example 39: Ranking SN1 reactivity across halide types [JEE]

Arrange toward SN1 hydrolysis: chlorobenzene, vinyl chloride (CH$_2$=CH-Cl), n-propyl chloride, allyl chloride (CH$_2$=CH-CH$_2$-Cl) and benzyl chloride (C$_6$H$_5$-CH$_2$-Cl).

Solution: SN1 rate tracks the stability of the carbocation formed on ionisation.

• Benzyl and allyl cations are resonance-stabilised → fastest.
• n-Propyl gives an ordinary, unstabilised primary cation → slow but possible.
• Vinyl and aryl (chlorobenzene) would need an unstable sp$^2$ cation and have partial double-bond C-Cl character → effectively no SN1.

Order: vinyl chloride ≈ chlorobenzene << n-propyl chloride < benzyl chloride ≈ allyl chloride (benzyl is usually placed marginally above allyl owing to aromatic resonance).

Example 40: Optical purity numerical [JEE Advanced]

The pure (+)-enantiomer of 2-bromobutane has specific rotation +23.1°. A sample shows an observed specific rotation of +9.2°. Find its optical purity (enantiomeric excess) and the percentage of each enantiomer.

Solution: Optical purity (ee) = (observed rotation / rotation of pure enantiomer) × 100 = (9.2 / 23.1) × 100 ≈ 40%.

An ee of 40% means the (+) form is in 40% excess over the (-). Let (+) = x% and (-) = (100 - x)%. Then x - (100 - x) = 40, giving x = 70. So the sample is 70% (+) and 30% (-) 2-bromobutane. Takeaway: ee is the excess of one enantiomer; convert to composition with 'excess = difference'.