Question

$C{H}_{3}C{H}_{2}CH=C{H}_2+CB{r}_4\stackrel{Peroxides}{\to }?$

What is the product of the reaction?

• A. $C{H}_{3}C{H}_2\underset{CB{r}_3}{\underset{|}{C}H}C{H}_{2}Br$
• B. $C{H}_{3}C{H}_{2}CH=CHCB{r}_3$
• C. $C{H}_{3}C{H}_2\underset{Br}{\underset{|}{C}}HC{H}_{2}CB{r}_3$
• D. $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}CB{r}_3$

Answer 1

The correct option is A $C{H}_{3}C{H}_2\underset{CB{r}_3}{\underset{|}{C}H}C{H}_{2}Br$

$1-butene$ reacts with tetra Bromo methane to form anti-Markownikoff's product. Peroxides used for the preparation of anti-Markownikoff's product.

$C{H}_{3}C{H}_{2}CH=C{H}_2+CB{r}_4\stackrel{peroxides}{\to }C{H}_{3}C{H}_{2}CH\left(CB{r}_3\right)C{H}_{2}Br$

So the answer is A.