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A 10g mixture of iso-butane and iso-bute...

A 10g mixture of iso-butane and iso-butene requires 20g of `Br_2`(in `CCl_4`) for complete addition. If 10g of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at `127^(@)`C, how much of it would be formed?
(Atom weight of bromine=80)

A

24.21g

B

20.0g

C

30.0g

D

12g

Text Solution

Verified by Experts

The correct Answer is:
A
Let iso-butane be a g and iso-butene b g.
Thus a + b = 10g
`CH_3-underset((56))underset("iso-butene")underset(CH_3)underset(|)C=CH_2+underset((80xx2))(Br_2)toCH_2-underset(CH_3)underset(|)overset(Br)overset(|)C-CH_3Br`
`CH_3-underset("iso-butane")underset(CH_3)underset(|)(CH)-CH_3+Br_2to` No addition reaction
Now 160 g of `Br_2` reacts with 56 g of iso-butene
`:.` 20g of `Br_2`reacts with `(56xx20)/160=7.25` g of iso-butane
Total amount of iso-butane available for 10g mixture
`=(7.25+3)=10.25 g`
`CH_3-underset((58))underset(CH_3)underset(|) (CH)-CH_3+Br_2tounderset("tert-butyl bromide(137)")(CH_3-underset(CH_3)underset(|)overset(Br)overset(|)C-CH_3)`
Now 58g of iso-butane gives 137 fo tert-butyl bromide
`:.` 10.25g of iso-butane gives `(137xx10.25)/58=24.21g` of tert-butyl bromide.
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