Question

$3\text{-Phenyl propene}$ on reaction with $HBr$ gives (as a major product):

• A. $C_6{H}_{5}C{H}_{2}CH\left(Br\right)C{H}_3$
• B. $C_6{H}_{5}CH\left(Br\right)C{H}_{2}C{H}_3$
• C. $C_6{H}_{5}C{H}_{2}C{H}_{2}C{H}_{2}Br$
• D. $C_6{H}_{5}CH\left(Br\right)CH=C{H}_2$

Answer 1

The correct option is B $C_6{H}_{5}CH\left(Br\right)C{H}_{2}C{H}_3$

According to Markovnikov's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.