Chemistry · Hydrocarbons · NEET
The H adds to the double-bond carbon that already has MORE hydrogen atoms. The negative part (Br, Cl, OH) adds to the carbon with FEWER hydrogen atoms. Example: in propene CH3-CH=CH2, the =CH2 carbon has 2 H's, so H adds there and Br adds to the middle carbon, giving 2-bromopropane as the major product.
Because the reaction goes through a carbocation (a carbon with a + charge). The alkene first grabs H+, and this can form two possible carbocations. The molecule always makes the MORE STABLE carbocation (3 degree > 2 degree > 1 degree, because more alkyl groups push electrons and spread the + charge). Then the negative part attacks that stable carbocation carbon. So Markovnikov's rule is really just carbocation stability in disguise.
Step 1: break the double bond and add H+ to each carbon one at a time. Step 2: see which side gives the more stable carbocation (3 degree beats 2 degree beats 1 degree). Step 3: put the H on the carbon you did NOT charge, and put Cl/Br/OH on the charged carbon. The product from the more stable carbocation is the MAJOR product.
Yes. HCl, HBr, HI, cold conc. H2SO4, and water (H2O with H2SO4 catalyst) all add to alkenes by Markovnikov's rule through a carbocation. The ONLY exception is HBr WITH a peroxide - that reverses the rule (anti-Markovnikov). HCl and HI do NOT reverse even with peroxide.
Look at the reagent line. Plain HBr, HCl, HI, H2O/H2SO4 = Markovnikov (halogen on more substituted carbon). But if you see 'peroxide', 'benzoyl peroxide', '(C6H5CO)2O2', or 'Kharasch effect' WITH HBr, then it is anti-Markovnikov and the Br goes to the LESS substituted carbon. This peroxide reversal happens only with HBr.
An alkene 'A' on reaction with O3 and Zn-H2O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene 'A' gives 'B' as the major product. The structure of product 'B' is:
The major product of the following reaction is: (CH3)2CH-CH=CH2, HBr, (C6H5CO)2O2 (benzoyl peroxide) → ?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
When an unsymmetrical reagent (like H-Br) adds to an unsymmetrical alkene, the negative part attaches to the double-bond carbon that has fewer hydrogen atoms, because that path gives the more stable carbocation.
Almost every year NEET asks the 'major product' of adding HX or water to an alkene or alkyne. You cannot pick the right option without Markovnikov's rule plus carbocation stability. It also links directly to ozonolysis and anti-Markovnikov questions, so one idea unlocks several marks.
Yes. Unsymmetrical alkynes add HX and water by Markovnikov's rule through a vinylic cation. For example, propyne + H2O/H2SO4/HgSO4 gives acetone (Markovnikov hydration), not propanal.
The alkene first takes H+ to form a carbocation. Two carbocations are possible; the molecule forms the MORE stable one (more alkyl groups = more stable). The nucleophile (Br-, Cl-, OH-) then attacks that stable carbocation carbon, which is the more substituted carbon.
Only when HBr is added in the presence of a peroxide (benzoyl peroxide) - the Kharasch or peroxide effect. Then a free-radical chain forms the more stable radical, so Br ends up on the less substituted carbon (anti-Markovnikov). HCl and HI never reverse this way.
