Haloalkanes are alkyl groups with one or more halogen atoms. CBSE Class 12 and NEET cover them thoroughly. Expect two questions a year on SN1/SN2 mechanisms and preparation.
Core Concepts
Preparation
From alcohols with HX, PX3 or SOCl2. From alkenes by HX addition (Markovnikov). From alkanes by free radical halogenation. From silver salts of acids (Hunsdiecker).
Let us compare the three methods of converting alcohols to haloalkanes:
| Reagent | Reaction | Advantage |
|---|---|---|
| HX (HCl, HBr, HI) | Simple, cheap | |
| PX3 (PBr3, PCl3) | Clean, no rearrangement | |
| SOCl2 (thionyl chloride) | Byproducts are gases — easy purification |
SOCl2 is the best laboratory method for making chloroalkanes because both byproducts (SO2 and HCl) are gases that escape the reaction mixture, making purification trivial. This is a favourite NEET reasoning question.
Physical properties
Higher boiling points than corresponding alkanes due to higher molecular mass and polarity. Insoluble in water, soluble in organic solvents.
Boiling point trends: for the same alkyl group, because molecular mass increases down the group, and London forces dominate. For the same halogen, longer chains give higher boiling points.
SN1 and SN2
SN1 — two steps, carbocation intermediate, favoured by tertiary substrates and polar protic solvents. Rate depends only on substrate. SN2 — one step, backside attack, inversion of configuration, favoured by primary substrates and polar aprotic solvents. Rate depends on both substrate and nucleophile.
Only the substrate appears in the rate expression because the rate-determining step is the slow ionisation to form the carbocation. The nucleophile attacks in a fast second step.
Both reactants are involved in the single concerted step. The nucleophile attacks the carbon from the back side, pushing out the leaving group. This causes Walden inversion — the configuration at the carbon flips.
Here is the decision tree that handles most NEET questions:
Primary substrate → SN2 (strong nucleophile) or E2 (strong base + heat)
Secondary substrate → SN2 with strong nucleophile in polar aprotic solvent, or SN1/E1 in polar protic solvent
Tertiary substrate → SN1 (weak nucleophile, polar protic) or E2 (strong base)
Elimination reactions
E1 and E2. Give alkenes. Compete with substitution. Zaitsev’s rule — major product is the more substituted alkene.
E1 mechanism: Two steps, similar to SN1. Carbocation forms first, then a base removes a proton from the adjacent carbon. Favoured by tertiary substrates and polar protic solvents.
E2 mechanism: One step, concerted. Base removes proton and leaving group departs simultaneously. Requires anti-periplanar geometry (H and X on opposite sides). Favoured by strong bases and elevated temperature.
The competition between substitution and elimination is a favourite JEE topic. The general rule: raising temperature favours elimination because elimination has a higher activation energy (more sensitive to temperature changes). Using a bulky base (like tert-butoxide) also favours elimination because it cannot easily approach the carbon for substitution.
Uses and dangers
CFCs were widely used but are now banned. DDT is a pesticide that bioaccumulates. Chloroform is used as a solvent and was once an anaesthetic. Carbon tetrachloride is carcinogenic.
Worked Examples
The carbocation formed is stabilised by three electron-donating alkyl groups. Primary carbocations are unstable, so primary substrates go SN2.
The C-X bond in haloarenes has partial double bond character due to resonance, making it stronger and harder to break. Haloarenes need harsh conditions for nucleophilic substitution.
NaOH is both a nucleophile and a base. With a secondary substrate:
- In aqueous NaOH (polar protic) → SN1 gives 2-butanol (racemic mixture)
- In alcoholic NaOH (less protic, higher temp) → E2 gives but-2-ene (Zaitsev product, trans major)
The solvent determines the pathway. This is why NEET questions always specify “aqueous” or “alcoholic” NaOH.
When (R)-2-bromobutane reacts with NaOH via SN2, the product is (S)-2-butanol. The nucleophile attacks from the back side, flipping the configuration. If the question says “with inversion of configuration,” it is SN2.
In both cases, the C-X bond has partial double bond character:
- In vinyl halides (), the lone pair on X conjugates with the C=C pi bond
- In aryl halides (), the lone pair conjugates with the aromatic ring
This shortens and strengthens the C-X bond. Also, in aryl halides, the sp2 carbon is more electronegative than sp3, holding the electrons tighter.
Comparison of SN1, SN2, E1, E2
| Feature | SN1 | SN2 | E1 | E2 |
|---|---|---|---|---|
| Steps | 2 | 1 | 2 | 1 |
| Intermediate | Carbocation | None (transition state) | Carbocation | None |
| Rate law | First order | Second order | First order | Second order |
| Substrate preference | 3° > 2° | Methyl > 1° > 2° | 3° > 2° | 3° > 2° |
| Solvent | Polar protic | Polar aprotic | Polar protic | Any |
| Stereochemistry | Racemisation | Inversion | - | Anti-periplanar |
| Product | Substituted | Substituted | Alkene | Alkene |
Common Mistakes
Saying all substitutions are SN2. Tertiary substrates prefer SN1.
Confusing E1 and E2. E1 has a carbocation intermediate; E2 is concerted.
Writing that haloarenes react easily. They are much less reactive than haloalkanes.
Forgetting that SN2 causes inversion. If the question gives a specific configuration (R or S) and says SN2, the product must have the opposite configuration.
Not specifying the solvent when predicting the reaction pathway. Aqueous NaOH favours substitution; alcoholic NaOH favours elimination. The solvent is not optional information.
Exam Weightage and Revision
Haloalkanes and Haloarenes carry 2-3 questions in NEET every year and about 8 marks in CBSE Class 12 boards. JEE asks more mechanistic questions (stereochemistry of SN2, competition between SN and E). This chapter is high-weightage and predictable.
| Question Type | NEET Frequency | Difficulty |
|---|---|---|
| SN1 vs SN2 identification | Every year | Medium |
| Preparation from alcohol | Most years | Easy |
| Haloarene reactivity | Every 2 years | Medium |
| Product prediction | Every year | Medium |
| Elimination vs substitution | JEE focus | Hard |
The most reliable NEET question from this chapter: “Arrange the following in order of SN1 or SN2 reactivity.” For SN1, order by carbocation stability (3° > 2° > 1°). For SN2, reverse order (methyl > 1° > 2°, no 3°).
Practice Questions
Q1. Why is thionyl chloride preferred over PCl5 for preparing chloroalkanes in the laboratory?
With SOCl2, the byproducts are SO2 and HCl — both gases that escape the reaction mixture. This makes purification of the product very easy. With PCl5, the byproduct is POCl3, which is a liquid that stays in the mixture and requires additional separation steps.
Q2. Predict the major product when 2-bromo-2-methylpropane reacts with (a) aqueous NaOH, (b) alcoholic KOH.
(a) Aqueous NaOH: SN1 gives 2-methyl-2-propanol (tert-butyl alcohol). The tertiary substrate forms a stable carbocation. (b) Alcoholic KOH: E2 gives 2-methylpropene (isobutylene). The strong base and heat favour elimination. The substrate is tertiary, so SN2 is impossible.
Q3. An optically active haloalkane gives a racemic product on reaction with aqueous acetone. What is the mechanism?
SN1. The optically active substrate ionises to form a planar carbocation (which is achiral). The nucleophile can attack from either face equally, giving a 50:50 mixture of R and S products — a racemic mixture. SN2 would give inversion, not racemisation.
Q4. Chlorobenzene does not undergo nucleophilic substitution under normal conditions but does so with NaOH at 300°C and 200 atm. Explain.
The C-Cl bond in chlorobenzene has partial double bond character due to resonance between the chlorine lone pair and the aromatic ring. This makes it stronger than a typical C-Cl bond. Extreme conditions (high temperature and pressure) provide enough energy to break this bond. The reaction proceeds via a benzyne intermediate.
FAQs
Why does Markovnikov addition give a specific product with HBr but anti-Markovnikov with HBr + peroxide?
Without peroxide: ionic mechanism. H+ adds first to the carbon with more H atoms (more stable carbocation intermediate). With peroxide: radical mechanism. Br radical adds first to the less substituted carbon (more stable radical intermediate on the more substituted carbon). Only HBr shows this peroxide effect — not HCl or HI.
What is a good leaving group?
A good leaving group is a stable species after departure — it should be able to hold the electron pair it takes with it. Halide ions (I- > Br- > Cl-) are good because they are stable anions. OH- is a poor leaving group, which is why alcohols are first converted to tosylates or protonated before substitution.
Why do we study both SN and E together?
Because they compete with each other in real reactions. The same substrate, nucleophile, and solvent can give either substitution or elimination products depending on conditions. Understanding when each wins is essential for predicting real reaction outcomes.
Memorise substrate preferences — primary goes SN2, tertiary goes SN1, secondary either. That decision tree handles most PYQs.
Grignard Reagent from Haloalkanes
Haloalkanes react with magnesium in dry ether to form Grignard reagents — one of the most versatile synthetic tools:
The Grignard reagent () acts as a source of carbanion (), which is a powerful nucleophile. It reacts with:
- Formaldehyde → primary alcohol ()
- Other aldehydes → secondary alcohol ()
- Ketones → tertiary alcohol ()
- CO2 → carboxylic acid () after acidification
- Epoxides → alcohol with 2 extra carbons ()
- Water → alkane () — this is why anhydrous conditions are essential
The absolute condition for Grignard reactions is anhydrous (dry) ether. Even trace moisture destroys the Grignard reagent: . This condition is tested in almost every Grignard question.
Environmental Concerns — CFCs and DDT
Two haloalkane derivatives have had enormous environmental impact:
CFCs (chlorofluorocarbons): Once widely used as refrigerants (Freon-12, CCl2F2) and aerosol propellants. In the stratosphere, UV light breaks the C-Cl bond, releasing Cl radicals that catalytically destroy ozone. One Cl radical can destroy 100,000 ozone molecules. The Montreal Protocol (1987) banned CFC production — one of the most successful environmental treaties ever.
DDT (dichlorodiphenyltrichloroethane): A powerful insecticide that helped eradicate malaria in many countries. But DDT is persistent (half-life of 15 years in soil), lipophilic (accumulates in fat), and biomagnifies up the food chain. Birds at the top (eagles, pelicans) accumulated so much DDT that their eggshells became too thin to survive. DDT is now banned in agriculture but still used in limited quantities for malaria control in some countries.
Both examples show how chemical stability (strong C-Cl and C-F bonds) can be a problem when these compounds enter the environment.
Haloalkanes are the gateway to carbocation chemistry. Understand SN1 vs SN2 and you are ready for most organic mechanism problems.