A 10 g mixture of isobutane and isobutene requires 20 g of `Br_(2) ("in "C CI_(4))` for complete addition . If 10 g of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the pressence of light at 393K which exclusive product and how much of it would be formed ?
(Atomic weight of bromine =80)

Explanation

`CH_(3)-underset((56))underset("Isobutene")underset(CH_(3))underset(|)C=CH_(2) underset((80xx2))(Br_(2)) to CH_(3)-overset(Br)overset(|)underset(CH_(3))underset(|)C-CH_(2)Br`
`CH_(3)-underset("Isobutane")underset(CH_(3))underset(|)CH-CH_(3)+Br to "No addition reactin "`
Now 160 g of `Br_(2)` reacts with 56 g of isobutene
`:.` 20 g of `Br_(2)` reacts with `(56xx 20)/(160) =7` g of isobutene (b)
Thus , amount of isobutane (a) is (10 -7) =3 g
`CH_(3)-underset((56))underset(CH_(3))underset(|)C=CH_(2)+H_(2) to CH_(3)-underset((58))underset(CH_(3))underset(|)CH-CH_(3)`
Now 56 g of isobutene gives 58 g of isobutane
`:.` 7 g of isobutene gives `(58xx7)/(56) =7.25` g of isobutane
Total amount of isobutane available for 10 g mixture
`=(7.25 +3) g =10.25 g`
`CH_(3)-underset((58))underset(CH_(3))underset(|)CH-CH_(3)+Br_(2)overset(400K)underset("Light")(to) " "CH_(3)-underset("tert-Butyl bromide(137)")underset(CH_(3))underset(|)overset(Br)overset(|)C-CH_(3)`
Now 58 g of isobutane gives 137 g of ter-butyl bromide
`:.` 10.25 g of isobutane gives `(13.7xx 10.25)/(58) =24.21` g of tert-butyl brominde.