Question

Consider the reaction

CH₃ - CH - CH₃ + Br₂ □(→┴Light, 400K )
|
CH₃
CH₃
|
CH₃ - CH - CH₂Br + CH₃ - C - CH₃
| |
CH₃ Br
(I) (II)

(a) Compounds I and II are formed in nearly equal amounts
(b) Compounds I is formed in larger proportions
(c) Compounds II is formed in larger proportions
(d) Relative amount of II cannot be predicted

Answer 1

Answer:

I think that the answer is option b it will help you

Answer 2

Compound 2 is formed majorly.

Explanation:

• The addition of a halogen to an alkane in presence of sunlight is a free radicle reaction.

• It means, the intermediate products are free radicle in nature.

• The bromine molecule is broken into bromine radicles by sunlight.

• This bromine radicle takes a hydrogen from the alkane forming alkyl free radicle.

• These free radicles then join in between each other giving multiple products.

• Now, stability of an alkyl free radicle depends as 3°>2°>1°.

• So the major product formed in the reaction will be product 2.

For more information about free radicle substitution,

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