Chemistry · Hydrocarbons · NEET
Only HBr. This is the single most tested point in NEET. With HCl the H-Cl bond is too strong (430.5 kJ/mol), so the free radical cannot break it. With HI the H-I bond is weak, but iodine radicals just join together to form I2 instead of adding to the double bond. Only HBr has the 'just right' bond strength (363.7 kJ/mol) to run the chain reaction. So if you see HCl or HI with peroxide, the product is still the normal Markovnikov product.
Br goes to the carbon that has MORE hydrogen atoms (the less substituted, terminal carbon). This is the exact opposite of Markovnikov's rule. Example: CH3-CH=CH2 + HBr (peroxide) gives CH3-CH2-CH2Br (1-bromopropane), where Br sits on the end CH2 carbon. Without peroxide you would get 2-bromopropane instead.
Because the mechanism changes from ionic (using H+ and a carbocation) to a free-radical chain. The Br radical adds FIRST to the double bond. It adds to the carbon that gives the more stable carbon radical. For CH3-CH=CH2, Br adds to the terminal CH2 so the radical forms on the middle carbon, which is a more stable secondary (2°) radical. Then that radical grabs H from HBr. Result: Br ends up on the terminal carbon - anti-Markovnikov.
In Markovnikov addition (no peroxide, ionic mechanism), the H of the acid goes to the carbon with more H, so Br goes to the carbon with FEWER H (more substituted). In anti-Markovnikov / peroxide effect (with peroxide, radical mechanism, HBr only), Br goes to the carbon with MORE H. Same alkene, same HBr, but the presence of peroxide flips which carbon gets the Br.
Benzoyl peroxide, (C6H5CO)2O2, is the initiator. On heating it breaks into free radicals. These radicals pull a Br atom off HBr to make a bromine radical, which starts the free-radical chain. So 'benzoyl peroxide' in a NEET question is a signal word: it tells you to expect the anti-Markovnikov product (but only if the acid is HBr).
The major product of the following reaction is: (CH3)2CH-CH=CH2 + HBr, in the presence of (C6H5CO)2O2 (benzoyl peroxide) →
Ph-C≡C-Me is treated: with Na/liq. NH3 → L; with H2, Lindlar's catalyst → K. Then L + HBr/benzoyl peroxide → N, and K + HBr → M. Choose the correct option.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. The peroxide effect, the Kharasch effect, and anti-Markovnikov addition of HBr all describe the same reaction. It was reported by M.S. Kharasch and F.R. Mayo in 1933.
Only HBr. HCl and HI do not show the peroxide effect, so with them you always get the normal Markovnikov product even if peroxide is present.
A free-radical chain mechanism with three steps: initiation (peroxide makes radicals), propagation (Br radical adds to the alkene, then the carbon radical takes H from HBr), and termination. This is different from Markovnikov addition, which is ionic.
1-bromopropane, CH3-CH2-CH2Br. Br goes to the terminal carbon. Without peroxide, you get 2-bromopropane (Markovnikov).
Yes, hydroboration-oxidation (BH3 then H2O2/OH-) adds water anti-Markovnikov to give a primary alcohol, but its mechanism is different (not a radical chain). For NEET, treat both as 'anti-Markovnikov', but only the HBr/peroxide reaction is called the Kharasch/peroxide effect.
