Question

Markovnikov's rule for the addition of hydrogen halides to alkenes states that the incoming hydrogen bonds to the :

• A. $s{p}^2$ carbon with the most hydrogens already
• B. $s{p}^2$ carbon with the fewest hydrogens already
• C. $s{p}^3$ carbon with the most hydrogens already
• D. $s{p}^3$ carbon with the few hydrogens already

Answer 1

The correct option is A $s{p}^2$ carbon with the most hydrogens already

When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the reagent gets attached to that unsaturated carbon atom which carries a lesser number of hydrogen atoms.

The addition of hydrogen bromide to propene gives isopropyl bromide as a major product according to Markownikov's rule.

$C{H}_3-CH=C{H}_2+H-Br\to C{H}_3-CH\left(Br\right)-C{H}_3$

Here, the Hydrogen atom attaches to the unsaturated Carbon atom with most hydrogens already.

Hence the correct option is A.