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 10g mixture of iso-butane and iso-butene requires 20g of Br_2(in CCl

1.	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)
Answer» 24.21g 20.0g 30.0g 12g Solution :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.