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An organic compound (X) containing C, H and O has a vapour density 37. 0.2750g of (X) produced 0.6540g of CO_(2) and 0.3375 of H_(2)O. The compound (X) on dehydration gave a hydrocarbon (Y) containing 85.71% of C. (Y) on treatment wit HI followed by hydrolysis, gave (Z) which was isomeric with (X). Give structural formulae of (X), (Y) and (Z) |
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Answer» SOLUTION :Moles of C in (X) `=1 xx` moles of `C_(2)` `=1 xx (0.6540)/(44)=0.01486` Moles of H in (X) `=2xx` moles of `H_(2)O` `=2 xx (0.3375)/(18)=0.0375` Wt. of oxygen in (X) = 0.2750 - (wt. of C+ wt. of H) `=0.2750- (0.01486 xx 12 + 0.0375 xx 1)` =0.0592g `THEREFORE` MOLE of O in (X) `=(0.0592)/(16)= 0.0037` `therefore` mole of `C:H: O= 0.01486 : 0.0375: 0.0037` `=4:10:1` `therefore` empirical formula of (X) is `C_(4)H_(10)O` (74) As molecular WEIGHT of (X) `=2xx VD` `=2 xx 37= 74`, `therefore` molecular formula of X is `C_(4)H_(10)O` Now for the compound (Y) Moles of `C: H = (85.71)/(12): (14.29)/(1)` `=7.143: 14.29` `=1: 2` Empirical formula of (Y) is `CH_(2)` Since (Y) is produced from (X) by dehydration, (Y) must be `C_(4)H_(10)O- H_(2)O`, i.e., `C_(4)H_(8)` Further, as `C_(4)H_(8)` is an unsaturated hydrocarbon, (X), i.e., `C_(4)H_(10)O` must be an alcohol `therefore` (X) may be `CH_(3).CH_(2).CH_(2).CH_(2).OH` (primary) or `CH_(3).CH_(2).CH(OH).CH_(3)` (secondary) and (Y) may be `CH_(3).CH_(2).CH=CH_(2)` or `CH_(3).CH=CH.CH_(3)` But as given in the question, `underset((X))(CH_(3).CH_(2).CH_(2).CH_(2)OH) underset(-H_(2)O)overset(HI)rarr CH_(3).CH_(2).CHI.CH_(3)overset("hydrolysis")rarr underset((Z))(CH_(3).CH_(2).CH(OH).CH_(3))` [ISOMERIC to `CH_(3).CH_(2).CH_(2).CH_(2)OH` (X)] Thus, (X) is `CH_(3).CH_(2).CH_(2).CH_(2)OH` (Y) is `CH_(3).CH_(2).CH=CH_(2)` (Z) is `CH_(3).CH_(2).underset(underset(OH)(|))(CH).CH_(3)` |
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