Alkyl halides are best prepared from alcohols, which are easily accessible.
-- NCERT Class 12 Chemistry, Ch. 6, p. 163Preparation Properties
8 MCQs9-step worked example
Source: NCERT Organic Compounds Containing HalogensPYQ coverage: NEET 2020, 2021, 2022, 2024Official key: NTA-verifiedLast updated: 26 Sep 2026
Preparation Properties, explained for NEET
The costly habit in this topic is treating "eliminate HX from a haloalkane" as if the alkene were forced. It is not. When the halogen sits on a carbon with two different β-carbons, two alkenes are possible, and Zaitsev's rule says the major product is the more substituted one. 2-Bromopentane with alcoholic KOH gives pent-2-ene as the major product, not pent-1-ene. Students who reach for the terminal alkene because it "looks simpler" lose the mark outright.
Two routes matter for preparation. From alcohols, R–OH plus HX, PX₃, or SOCl₂ gives the chloride or bromide; SOCl₂ is preferred because SO₂ and HCl leave as gases and the alkyl chloride is obtained pure (NCERT Class 12 Chemistry Part 2, Chapter 6, printed page 164). Alkenes take up HX by Markovnikov addition, and halogen addition across the double bond gives the vicinal dihalide.
Fluorides and iodides do not come cleanly from the direct alcohol route. Iodoalkanes are made by Finkelstein: R–Cl or R–Br heated with NaI in dry acetone, where NaCl or NaBr precipitates and drives the equilibrium. Fluoroalkanes come from Swarts — the bromo- or chloroalkane is heated with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃ (printed page 180). Both are halogen-exchange reactions on an existing haloalkane, not substitutions on an alcohol.
On properties: haloalkanes are polar but form no hydrogen bonds with water, so they are only very slightly soluble in it, while dissolving readily in organic solvents. Boiling points for a given alkyl group rise RI > RBr > RCl > RF, tracking molecular mass and van der Waals forces, and fall as the chain branches (printed page 189).
Watch-out: named-reagent questions here are recall-cheap and trap-rich. Pair each conversion with its reagent and its by-product before exam day.
Can you answer these Preparation Properties MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which reagent converts an alcohol to an alkyl chloride while giving only gaseous by-products, leaving the haloalkane pure?
Show answer and why every option is right or wrong
Answer: C. SOCl₂ converts R–OH to R–Cl with SO₂ and HCl released as gases, so the alkyl chloride is obtained in a pure state — this is why NCERT names it the preferred reagent (Class 12 Chemistry Part 2, Chapter 6, printed page 163).
Why A is wrong: A is wrong because PCl₅ leaves POCl₃ behind in the flask; the by-product is not gaseous, so purification is still needed.
Why B is wrong: B is wrong because the Lucas reagent route works but leaves water and the zinc salt in the mixture rather than escaping as gas.
Why D is wrong: D is wrong because NaI in dry acetone is the Finkelstein reagent acting on a haloalkane, not on an alcohol.
MCQ 2Easy RecallPractice
Fluoroalkanes are prepared by heating a chloro- or bromoalkane with a metallic fluoride such as AgF or SbF₃. This reaction is named after:
Show answer and why every option is right or wrong
Answer: A. Swarts reaction is the halogen exchange that installs fluorine using a metallic fluoride on an existing bromo- or chloroalkane (NCERT Class 12 Chemistry Part 2, Chapter 6, printed page 166).
Why B is wrong: B is wrong because Finkelstein is the NaI-in-dry-acetone exchange that installs iodine, not fluorine.
Why C is wrong: C is wrong because Wurtz couples two alkyl halides with sodium to give an alkane; no halogen ends up in the product.
Why D is wrong: D is wrong because Sandmeyer converts a diazonium salt to an aryl halide using a copper(I) halide, which is a different starting material entirely.
MCQ 3Direct ApplicationPractice
2-Bromopentane is heated with alcoholic KOH. Applying Zaitsev's rule, the major alkene is:
Show answer and why every option is right or wrong
Answer: A. Zaitsev's rule selects the more substituted alkene, so β-hydrogen removal from C-3 gives pent-2-ene as the major product rather than the terminal isomer.
Why B is wrong: B is wrong because pent-1-ene is the less substituted (Hofmann) minor product; picking it is the standard Zaitsev trap.
Why C is wrong: C is wrong because the carbon skeleton is unbranched and no rearrangement occurs in this elimination, so a methylbutene cannot form.
Why D is wrong: D is wrong because a single dehydrohalogenation removes one HBr and creates one double bond, not two.
MCQ 4Easy RecallPractice
Why are haloalkanes only very slightly soluble in water despite being polar molecules?
Show answer and why every option is right or wrong
Answer: B. Dissolving a haloalkane requires breaking hydrogen bonds between water molecules, and the new haloalkane–water attractions are too weak to compensate because no hydrogen bonding is possible (NCERT Class 12 Chemistry Part 2, Chapter 6, printed page 169).
Why A is wrong: A is wrong because a monohaloalkane has a single C–X dipole with nothing to cancel it; the molecule is definitely polar.
Why C is wrong: C is wrong because haloalkanes are not hydrolysed by plain water at room temperature; they need aqueous alkali.
Why D is wrong: D is wrong because density governs whether an insoluble liquid sinks or floats, not whether it dissolves.
MCQ 5Direct ApplicationPractice
1-Chlorobutane is heated with sodium iodide in dry acetone. Which observation confirms the reaction is proceeding?
Show answer and why every option is right or wrong
Answer: C. NaI is soluble in dry acetone but NaCl is not, so the sodium chloride precipitates and its removal drives the Finkelstein equilibrium towards 1-iodobutane (NCERT Class 12 Chemistry Part 2, Chapter 6, printed page 165).
Why A is wrong: A is wrong because brown vapour indicates free bromine or nitrogen dioxide; no free halogen is liberated in this exchange.
Why B is wrong: B is wrong because a deep blue colour is the starch–iodine test, which requires free I₂ and starch, neither of which is present.
Why D is wrong: D is wrong because SO₂ belongs to the SOCl₂ preparation from an alcohol, not to a halogen-exchange reaction.
MCQ 6Direct ApplicationPractice
Propene is treated with HBr in the absence of peroxides. The major product is:
Show answer and why every option is right or wrong
Answer: D. Markovnikov addition places the halogen on the more substituted carbon, so propene plus HBr gives 2-bromopropane as the major product.
Why A is wrong: A is wrong because 1-bromopropane is the anti-Markovnikov product, formed only when peroxides are present, and the stem excludes them.
Why B is wrong: B is wrong because a vicinal dihalide requires addition of Br₂ across the double bond, not of HBr.
Why C is wrong: C is wrong because allylic substitution needs a halogen under free-radical conditions and leaves the double bond intact; here the double bond is consumed.
MCQ 7CalculationPractice
A student needs 1-iodopropane but has only propan-1-ol. Which two-stage route is sound?
Show answer and why every option is right or wrong
Answer: D. The alcohol route reliably delivers the chloride, and Finkelstein exchange on that chloride then installs iodine — the two stages together are the standard preparation of an iodoalkane from an alcohol.
Why A is wrong: A is wrong because there is no direct I₂-plus-alcohol substitution; iodides are not obtained cleanly from the direct alcohol route.
Why B is wrong: B is wrong because the Swarts product is the least reactive haloalkane towards further exchange, and the sequence needlessly installs fluorine before removing it.
Why C is wrong: C is wrong because eliminating to propene and re-adding HI would work only for the Markovnikov product 2-iodopropane, giving the wrong isomer.
MCQ 8Concept TrapPractice
Two statements are made about preparing haloalkanes. Statement I: heating an alcohol with a metallic fluoride is the standard route to a fluoroalkane. Statement II: heating an alcohol with NaI in dry acetone is the standard route to an iodoalkane. Which is correct?
Show answer and why every option is right or wrong
Answer: D. Swarts and Finkelstein are both halogen-exchange reactions carried out on an existing bromo- or chloroalkane, so neither starts from an alcohol (NCERT Class 12 Chemistry Part 2, Chapter 6, printed page 165).
Why A is wrong: A is wrong because accepting both statements ignores that fluorides and iodides are not obtained from the direct alcohol route at all.
Why B is wrong: B is wrong because Swarts acts on a haloalkane; an alcohol plus a metallic fluoride is not the named preparation.
Why C is wrong: C is wrong because Finkelstein likewise needs an alkyl chloride or bromide as substrate, since the driving force is precipitation of NaCl or NaBr.
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How do you solve a Preparation Properties question? A worked example
- 1
Given
2-Bromopentane, CH₃–CHBr–CH₂–CH₂–CH₃, is heated with alcoholic KOH.
- 2
Required
The major alkene, and the reason the minor alkene is minor.
- 3
Concept
Alcoholic KOH promotes β-elimination (dehydrohalogenation). A β-hydrogen is one on a carbon adjacent to the C–Br carbon. Where two different β-carbons exist, Zaitsev's rule predicts the more substituted alkene as the major product because it is the more stable one.
- 4
Formula/rule
Major product = alkene with the greater number of alkyl groups attached to the doubly bonded carbons.
- 5
Substitution
Number the chain: C1 (CH₃), C2 (CHBr), C3 (CH₂), C4 (CH₂), C5 (CH₃). Bromine sits on C2. The two β-carbons are C1 and C3. Removing H from C1 with the Br from C2 gives pent-1-ene, a C1=C2 double bond bearing one alkyl group. Removing H from C3 instead gives pent-2-ene, a C2=C3 double bond bearing two alkyl groups.
- 6
Calculation
Count substituents on the double bond: pent-1-ene is monosubstituted; pent-2-ene is disubstituted. There is no arithmetic and no measured quantity here — the chain positions C1 to C5 are counting integers (exact) and contribute no significant figures.
- 7
Final answer
Pent-2-ene is the major product; pent-1-ene is the minor product.
- 8
Common trap
Choosing pent-1-ene because the terminal position "looks simpler" or because it is the product of removing the first hydrogen you meet when reading the structure left to right. Zaitsev does not care about reading direction. Locate every β-carbon first, then compare substitution on the resulting double bonds.
- 9
Similar NEET-style question
2-Bromo-2-methylbutane is heated with alcoholic KOH. Identify the major alkene and state how many distinct β-carbons the substrate has. (Expected reasoning: three β-carbons, two of which are equivalent methyls; the major product is 2-methylbut-2-ene, the trisubstituted alkene.)
What to remember before solving Preparation Properties questions
Aryl halides are prepared by electrophilic substitution to arenes. Fluorides and iodides are best prepared by halogen exchange method.
-- NCERT Class 12 Chemistry, Ch. 6, p. 189Since b-hydrogen atom is involved in elimination, it is often called b-elimination. If there is possibility of formation of more than one alkene due to the availability of more than one b-hydrogen atoms, usually one alkene is formed as the major product. These form part of a pattern first observed by Russian chemist, Alexander Zaitsev (also pronounced as Saytzeff) who in 1875 formulated a rule which can be summarised as ‘in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.’ Thus, 2-bromopentane gives pent-2-ene as the major product.
-- NCERT Class 12 Chemistry, Ch. 6, p. 180Key Fact
Grignard reagents
Most organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds. Such compounds are known as organo-metallic compounds. An important class of organo-metallic compounds discovered by Victor Grignard in 1900 is alkyl magnesium halide, RMgX, referred as Grignard Reagents.
-- NCERT Class 12 Chemistry, Ch. 6, p. 180The above methods are not applicable for the preparation of aryl halides because the carbon-oxygen bond in phenols has a partial double bond character and is difficult to break being stronger than a single bond.
-- NCERT Class 12 Chemistry, Ch. 6, p. 164(i) From hydrocarbons by electrophilic substitution Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts like iron or iron(III) chloride. The ortho and para isomers can be easily separated due to large difference in their melting points. Reactions with iodine are reversible in nature and require the presence of an oxidising agent (HNO3, HIO4) to oxidise the HI formed during iodination. Fluoro compounds are not prepared by this method due to high reactivity of fluorine. (ii) From amines by Sandmeyer's reaction When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br. Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.
-- NCERT Class 12 Chemistry, Ch. 6, p. 166Preparation Properties questions from past NEET papers
4 questions from NEET 2020, 2021, 2022, 2024. Answers verified against NTA official keys.
NEET 2024Revised key
1POCl₃ and H₃PO₃
2POCl₃ and H₃PO₄
3H₃PO₄ and POCl₃
4H₃PO₃ and POCl₃
NTA Answer: Option 4(revised_final)
NEET 2022
Which of the following sequence of reactions is suitable to synthesize chlorobenzene?
1Aniline, HCl, Heating
2Benzene, Cl2, anhydrous FeCl3
3Phenol, NaNO2, HCl, CuCl
4Benzene, HCl
NTA Answer: Option 2(final)
NEET 2021
1Huckel's Rule
2Saytzeff's Rule
3Hund's Rule
4Hofmann Rule
NTA Answer: Option 2(final)
NEET 2020
1(a), (b), (d)
2(a), (b), (c)
3(a), (c), (d)
4(b), (c), (d)
NTA Answer: Option 2(final)
All 7 past-paper questions from Organic Compounds Containing Halogens →
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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