InterviewSolution
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InterviewSolution
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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. 24.21gB. 20.0gC. 30.0gD. 12g |
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Answer» Correct Answer - 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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