Question

The addition of HBr to 1-butene gives a mixture of products (I),(II) and (III):
The mixture consists of :

• A. (I) and (II) as major and (III) as minor products
• B. (II) as major, (I) and (III) as minor products
• C. (II) as minor, (I) and (III) as major products
• D. (I) and (II) as minor and (III) as major products.

Answer 1

Answer: (A)

(I) and (II) as major and (III) as minor products

$C{H}_{3}C{H}_2-CH=C{H}_2\underset{\text{Markovnikov's addition}}{\overset{HBr}{\to }}\underset{\text{Major product}}{\underset{\text{exist in 2 enantiomers I and II}}{C{H}_{3}C{H}_2-\underset{\stackrel{|}{Br}}{\stackrel{\ast }{CH}}-C{H}_3}}+\underset{\text{III (Minor product)}}{C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br}$

Hence, the mixture consists of (I) and (II) as major and as (III) as minor products.