The radical halogenation of 2-methylpropane gives two products: (CH3)2CHCH2X (minor) and (CH3)3CX (major) Chlorination gives a larger amount of the minor product than does bromination, Why?

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TuteeHUB · Accepted Answer

The methyl groups are more hindered to attack by the larger bromine atom.

TuteeHUB · Answer

Bromine is more reactive than chlorine and is able to attack the less reactive 3º C-H.

TuteeHUB · Answer

Bromine atoms are less reactive (more selective) than chlorine, and preferentially attack the weaker 3º C-H bond

TuteeHUB · Answer

Bromination is reversible and the more stable 3º-alkyl bromide is formed exclusively.

1.	The radical halogenation of 2-methylpropane gives two products: (CH3)2CHCH2X (minor) and (CH3)3CX (major) Chlorination gives a larger amount of the minor product than does bromination, Why?
A.	Bromine is more reactive than chlorine and is able to attack the less reactive 3º C-H.
B.	Bromine atoms are less reactive (more selective) than chlorine, and preferentially attack the weaker 3º C-H bond
C.	The methyl groups are more hindered to attack by the larger bromine atom.
D.	Bromination is reversible and the more stable 3º-alkyl bromide is formed exclusively.
Answer» C. The methyl groups are more hindered to attack by the larger bromine atom.

The radical halogenation of 2-methylpropane gives two products: (CH3)2CHCH2X (minor) and (CH3)3CX (major) Chlorination gives a larger amount of the minor product than does bromination, Why?

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