Question

Which of the following reactions will result in the formation of a chiral center in the product?

• A. ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2\mathrm{HBr}\to$
• B. ${\mathrm{CH}}_2={\mathrm{CH}}_2+\mathrm{HOBr}\to$
• C. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to$
• D. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2}{\to }$

Answer 1

The correct option is C

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to$

Explanation for the correct option:

(c ) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to$

• The reaction of $1-\mathrm{butene}$ with $\mathrm{HBr}$ results in the production of a chiral center in the output.
• The addition takes place according to Markovnikov's rule.
• $\mathrm{Br}$ is added to a $\mathrm{C}$atom that has fewer $\mathrm{H}$ atoms.
• ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to {\mathrm{CH}}_3{\mathrm{CH}}_2\stackrel{*}{\mathrm{C}}{\mathrm{HBrCH}}_3$
• Asterisk indicates the presence of chiral carbon in the product. ${\mathrm{CH}}_3{\mathrm{CH}}_2\stackrel{*}{\mathrm{C}}{\mathrm{HBrCH}}_3$
• Hence this option is correct

Explanation for the incorrect options:

(a) ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to$

• ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$
• No chiral center is produced
• Thus, the option is incorrect.

(b)${\mathrm{CH}}_2={\mathrm{CH}}_2+\mathrm{HOBr}\to$

• ${\mathrm{CH}}_2={\mathrm{CH}}_2+\mathrm{HOBr}\to {\mathrm{CH}}_2{\mathrm{BrCH}}_2\mathrm{OH}$
• No chiral center is produced
• Thus, the option is incorrect.

(d) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2}{\to }$

• ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$
• No chiral center is produced
• Thus, the option is incorrect.

Hence the correct option is (c ) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to$