# Which one of the following reactions would be the best for the formation of 2-bromobutane(1) $CH_3CH=CHCH_2CH_3 + HBr \rightarrow$ (2) $CH_3CH_2CH=CH_2 + HBr \rightarrow$(3) $CH_3CH=CHCH_3 + {Br}_2 \rightarrow$(4) $CH_3CH_2CH=CH_2 + HBr \ Peroxide \rightarrow$A. (1)B. (2)C. (3) D. (4)

Last updated date: 27th Jan 2023
Total views: 19.8k
Views today: 0.13k
Verified
19.8k+ views
Hint: Addition reaction of alkene has a characteristic feature it goes through a different type of mechanism:
(a) Classical carbocation mechanism and (b) non-classical carbocation mechanism
In classical carbocation, the mechanism charge is completely developed on the carbon atom and in non-classical carbocation, the charge is partially developed on the carbon atom.

According to the addition mechanism of the alkene as the formation of carbocation takes place first the electrophile gets attached to the carbon which contains more no. of hydrogen atoms attached to the carbon and then the nucleophile gets attached to the carbon that contains less number of hydrogen atoms, because according to markovnikov’s rule the nucleophile will attach with the carbon which contains less number of hydrogen.

Now according to the addition reaction the product of the given reactions are:
(1) $CH_3CH=CHCH_2CH_3 + HBr \rightarrow \\ CH_3CHBr-CH_2CH_2CH_3+CH_3CH_2-CHBrCH_2CH_3$
In the given reaction two products will be formed 2-bromopentane (minor) and 3- bromopentane (major).

(2) $CH_3CH_2CH=CH_2 + HBr \rightarrow C H_3CH_2CHBr-CH_3+CH_3CH_2CH_2CH_2Br$
In the given reaction two products will be formed 2-bromobutane (major) and 1- bromobutane (minor).

(3) $CH_3CH=CHCH_3 + {Br}_2 \rightarrow C H_3CHBr CHBrCH_3$
In the given reaction one product will be formed which is dihalide 2,3-dibromobutane.

(4) $CH_3CH_2CH=C{Hr}_2 + \ HBr + Peroxide \rightarrow C H_3CH_2CH_2CH_2Br$
In the given reaction the product formed will be 1- bromobutane because in presence of peroxide the mechanism goes through anti-markovnikov’s rule.
Thus, we can see that only $CH_3CH_2CH=CH_2$ on reaction with HBr produces 2-bromobutane so it is best for the formation of 2-bromobutane.
Thus, Option (B) is correct

Note: Not all reactions follow markovnikov’s rule. In the second reaction 2- bromobutane contains a chiral carbon which means it will have stereoisomers. It will have two stereoisomers; they will be enantiomers that means mirror image of each other.