Chemistry · Haloalkanes And Haloarenes · NEET
Take an alkane and mix it with a halogen like Cl2 or Br2. Give it sunlight, UV light, or heat (573-773 K). One hydrogen atom of the alkane is replaced by a halogen atom. Example: CH4 + Cl2 in sunlight gives CH3Cl + HCl. This is called free radical substitution because it goes through halogen free radicals. This is a common NEET reaction type, so know it well.
Light breaks the halogen molecule (Cl2 -> 2 Cl radicals). This is the initiation step. Without this energy the reaction does not start, because alkanes are very unreactive (only strong single bonds, no easy site for attack). The light gives the energy to make the first free radicals that then keep the chain going.
Once one H is replaced, the product still has other H atoms that can also be replaced. So CH4 gives CH3Cl, then CH2Cl2, then CHCl3, then CCl4. You get a mixture. This is why alkane halogenation is a poor lab method for making a pure single haloalkane. NEET often tests this weakness.
1) Initiation: light splits Cl2 into two Cl radicals. 2) Propagation: a Cl radical takes an H from the alkane making an alkyl radical + HCl; the alkyl radical then grabs a Cl from Cl2 making the haloalkane + a new Cl radical (chain keeps going). 3) Termination: two radicals join together and the chain stops. Remember the order: Initiation, Propagation, Termination.
Alkene + HX is an ADDITION reaction: HX adds across the C=C double bond, no atom is lost, and it needs no light. Alkane + X2 is a SUBSTITUTION reaction: an H is swapped for X, HX is released, and it needs light or heat. Alkene addition is clean and gives one main product; alkane substitution gives a mixture.
Normally HBr adds by Markovnikov's rule (Br goes to the carbon with fewer H atoms). But in the presence of a peroxide, HBr adds the opposite way (Br goes to the carbon with more H atoms). This is the anti-Markovnikov or peroxide effect, and it happens only with HBr, not HCl or HI. It goes through a free radical mechanism.
Reactivity order is F2 > Cl2 > Br2 > I2. Fluorination is too violent (explosive) and iodination is very slow and reversible, so it needs an oxidising agent. That is why chlorination and bromination are the useful ones for NEET.
The major product of the following reaction is: (CH3)2CH-CH=CH2 + HBr, in the presence of benzoyl peroxide (C6H5CO)2O2 ?
What is the change in oxidation number of carbon in the reaction CH4(g) + 4Cl2(g) -> CCl4(l) + 4HCl(g)?
For the reactions: (a) CH3CH2CH2Br + KOH(alc) -> CH3CH=CH2 ; (b) CH3CHBrCH2CH3 + KOH(aq) -> CH3CH(OH)CH2CH3 ; (c) cyclohexene + Br2 -> trans-1,2-dibromocyclohexane. Which statement is correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. It gives a mixture of mono-, di-, tri- and tetra-substituted products, so it is hard to get one pure product. For a clean single haloalkane, preparation from alcohols is far better.
No. Addition of HX to an alkene happens easily at room temperature and needs no light. Only free radical halogenation of alkanes needs sunlight, UV, or high heat.
The peroxide effect works only with HBr. The H-Cl bond is too strong to break into radicals easily, and the H-I bond breaks so easily that the radicals recombine. So only HBr gives the clean anti-Markovnikov product with a peroxide.
F2 > Cl2 > Br2 > I2. Fluorination is explosive and iodination is slow and reversible, so only chlorination and bromination are useful in practice.
Stability order is 3 degree > 2 degree > 1 degree > methyl. A more substituted (tertiary) carbon loses its H more easily, so halogenation prefers that position, giving more of that product.
