Chemistry · Haloalkanes And Haloarenes · NEET
When you add an acid HX (like HBr, HCl, HI) across a C=C double bond of an unsymmetrical alkene, the hydrogen (H) of HX joins the carbon that ALREADY has more hydrogen atoms. The halogen (X) goes to the other carbon (the one with fewer H). Example: propene CH3-CH=CH2 + HBr gives CH3-CHBr-CH3 (2-bromopropane). The H went to the =CH2 end (2 H's) and Br went to the middle carbon. Easy phrase to remember: 'the rich get richer'.
The real reason is carbocation stability, not just the counting trick. HX adds in two steps. First, H+ attaches to one double-bond carbon and makes a positive carbocation on the other carbon. Nature picks the path that gives the MORE STABLE carbocation (3° more stable than 2°, 2° more stable than 1°). For propene, adding H+ to the terminal CH2 makes a secondary (2°) cation on the middle carbon, which is stable. Then Br- attaches there. So Markovnikov's rule is really just 'form the most stable carbocation'.
If a peroxide (like benzoyl peroxide) is added along with HBr, the addition reverses. Now Br goes to the carbon with MORE hydrogens (the terminal carbon), and H goes to the other. Propene + HBr + peroxide gives CH3-CH2-CH2Br (1-bromopropane), NOT 2-bromopropane. This is also called anti-Markovnikov addition or the Kharasch effect. It happens because the peroxide starts a free-radical chain (not an ionic path), and the reaction goes through the MORE STABLE free radical.
This is a favourite NEET trap. The peroxide effect needs a free-radical chain that keeps going. For HCl, the H-Cl bond is too strong, so the chlorine radical step is too slow (endothermic) and the chain breaks. For HI, the H-I bond is weak but the iodine radicals just recombine to I2 instead of adding, so the chain also fails. Only HBr has bond energies that make BOTH chain steps energetically favourable. So: anti-Markovnikov happens for HBr only.
They are for opposite reactions, so do not mix them. Markovnikov's rule is for ADDITION of HX to an alkene (making a haloalkane); it decides which carbon gets the H and which gets the X. Saytzeff's (Zaitsev's) rule is for ELIMINATION (removing HX from a haloalkane to make an alkene); it says the alkene with more alkyl groups on the double bond (more substituted) is the major product. One builds the C-X bond, the other breaks it.
No. Markovnikov's rule and the peroxide effect only matter for UNSYMMETRICAL alkenes, where the two double-bond carbons are different. For a symmetrical alkene (like CH3-CH=CH-CH3, but-2-ene), both carbons have the same number of hydrogens, so H and Br can add either way and give the SAME product. Peroxide makes no difference there.
The major product of the following reaction is: (CH3)2CH-CH=CH2 + HBr, in the presence of benzoyl peroxide (C6H5CO)2O2
In the sequence, X and Z respectively are: CH3CH2CH2OH + PCl3 -> CH3CH2CH2Cl + X + HCl ; CH3CH2CH2Cl --alc.KOH,heat--> Y ; Y --HBr/peroxide--> Z
Try the real previous-year questions from this chapter — each with the answer and a full solution.
When HX adds to an unsymmetrical alkene, the negative part (X) goes to the carbon carrying the fewer hydrogen atoms, and H goes to the carbon with more hydrogens.
Only HBr shows the peroxide (anti-Markovnikov) effect. HCl and HI do not, because their chain steps are not energetically favourable.
A free radical intermediate. The reaction picks the path that forms the more stable free radical, which is why the final product looks 'reversed'.
Yes. Ionic HX addition forms the most stable carbocation first (3° over 2° over 1°), and that is what actually decides the product.
Reactions of HX with alkenes appear almost every year, often hidden inside multi-step sequences. Spotting the 'peroxide' keyword and applying the flip correctly is a quick, guaranteed mark.
