
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Answer
575.4k+ views
Hint: Addition of HBr to propene is an example of an electrophilic substitution reaction.Hydrogen bromide provides an electrophile, hydron.
Complete step by step answer:
-Addition of HBr to propene is an example of an electrophilic substitution reaction.Hydrogen bromide provides an electrophile, hydron. This electrophile attacks the double bond to form $1^{\circ}$ and $2^{\circ}$ carbocations as shown:
-Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step, Br- attacks the carbocation to form 2 - bromopropane as the major product.
-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 additional reaction takes place anti to Markovnikov's rule. The reaction follows a free radical chain mechanism as:
-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.
-In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained in addition to HBr to propene in the absence and presence of peroxide.
Note: Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. In the presence of peroxide, Br free radical acts as an electrophile.
Complete step by step answer:
-Addition of HBr to propene is an example of an electrophilic substitution reaction.Hydrogen bromide provides an electrophile, hydron. This electrophile attacks the double bond to form $1^{\circ}$ and $2^{\circ}$ carbocations as shown:
-Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step, Br- attacks the carbocation to form 2 - bromopropane as the major product.
-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 additional reaction takes place anti to Markovnikov's rule. The reaction follows a free radical chain mechanism as:
-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.
-In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained in addition to HBr to propene in the absence and presence of peroxide.
Note: Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. In the presence of peroxide, Br free radical acts as an electrophile.
Recently Updated Pages
Class 12 Question and Answer - Your Ultimate Solutions Guide

Two men on either side of the cliff 90m height observe class 10 maths CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

How is gypsum formed class 10 chemistry CBSE

If the line 3x + 4y 24 0 intersects the xaxis at t-class-10-maths-CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

State the principle of an ac generator and explain class 12 physics CBSE

Give 10 examples of unisexual and bisexual flowers

