Q.Explain Markownikoff's rule with suitable example.

Step 1. Statement of the rule. When an unsymmetrical alkene reacts with a hydrogen halide (H-X), the hydrogen adds to the double-bond carbon that ALREADY carries more hydrogens, while the halogen adds to the carbon carrying FEWER hydrogens; equivalently, the most electronegative part of the reagent ends up on the more-substituted (least-hydrogen-bearing) double-bond carbon.

Step 2. Worked example. Propene, CH3-CH=CH2, is unsymmetrical: the terminal carbon (=CH2) carries two H's, while the internal carbon (=CH-) carries only one H (plus the methyl group). Adding HBr: the electrophile H+ attacks preferentially so as to generate the MORE stable carbocation -- protonating the terminal CH2 gives a secondary carbocation at the internal carbon (CH3-CH+-CH3), which is more stable than the primary carbocation that would form from protonating the internal carbon instead.

Step 3. Br- then attacks this secondary carbocation, giving 2-bromopropane (CH3-CHBr-CH3) as the MAJOR product; 1-bromopropane forms only as a minor product, via the less-favoured primary-carbocation pathway. …

State the rule in both its classical form (H to the carbon with more H's) AND its mechanistic form (the more stable carbocation forms and decides the outcome) -- the second form also correctly predict …

• Stating the rule as 'halogen adds to the carbon with more hydrogens' -- this is backwards; it is the HYDROGEN that adds to the carbon that already has more hydrogens, and the halogen goes to the other, less-hydrogenated (more substituted) carbon. …