Ncert Exercises & Solutions

Question 13.16:-

Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanisms.

Addition of HBr to propene is an example of an electrophilic substitution reaction.
Hydrogen bromide provides an electrophile, H+. This electrophile attacks the double bond to form 1° and 2° carbocations as shown:

H_{3} C^{3} - C^{2} H = \overset{1}{C} H_{2} + H - B r

↓ H^{+}

H_{3} C - C H_{2} - \overset{+}{C H_{2} + B \bar{r}}

(L e s s s t a b l e)

p r i m a r y c a r v b o c a t i o n

H_{3} C - \overset{+}{C H - C H_{3}} + B \bar{r}

(M o v e s t a b l e)

s e c o n d a r y c a r b o c a t u i o n

Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will format a faster rate. Thus, in the next step, Br– attacks the carbocation to form 2 – bromopropane as the major product

H_{3} C - \overset{+}{C H} - C H_{3} + B r \to C H_{3} - \underset{B r}{C H} - C H_{3}

2-Bromopropane

This reaction follows Markovnikov’s rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms. In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov’s rule. The reaction follows a free radical chain mechanism as:

C_{6} H_{5} - \underset{\underset{O}{∥}}{C} - O - O - \underset{\underset{O}{∥}}{C}

↓ H o m k l y s i s

2 C_{6} H_{S} - \underset{\underset{O}{∥}}{C} - ⊛ \to {\overset{*}{2 C}}_{6} H_{5} + 2 C O_{2}

C_{6} H_{5} + H - B r \overset{H o m o l y s i s}{\to} C_{6} H_{5} + B r

C H_{3} - C H = C H_{2} + B r

C H_{3} - \underset{|}{C H} - C H_{2}

B r

(L e s s s t a b l e)

p r i m a r y f r e e r a d i c a l

C H_{3} - C H - C H_{2} - B r

(M o r e s t a b l e)

s e c o n d a r y f r e e r a d i c a l

Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1 – bromopropane is obtained as the major product.

C H_{3} - \overset{*}{C H} - C H_{2} B r + H - B r

C H_{3} - C H_{2} - C H_{2} B r + B r \cdot

1 - B r o m o p r o a n e

M a j o r p r o d u c t

In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained on addition of HBr to propene in the absence and presence of peroxide

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