An unknown compound I was found to have a molecular weight of 137and an analysis

Question

An unknown compound I was found to have a molecular weight of 137and an analysis of C 35.03%; H 6.57% Br ; 58.4% . Upon treatmentwith sodium ethoxide in hot ethanol, the compound wasconverted into a new compound II which, analyzed as C 85.71%; H14.28%. Upon treatment with HBr compound II was converted tocompound III which analyzed the same as compound I. However theboiling point of III was 13o lower than that ofcompound I, clearly indicating that the two compounds weredifferent.

Calculate the molecular formulas of compounds I, II, III.

Explanation / Answer

137*.3503=weight of carbon=48g *1 atom/12 g ---> 4 carbonatoms 137*.0584=weight of bromine=80g--->1 Br atom 137*.0657=weight of hydrogen=9g--->9 H atoms formula i C4H9Br when treated with sodium ethoxide in hot ethanol, the bromine groupleaves and an elimination occurs, losing an H atom as well. the newformula ii is C4H8 (this can be verified by doing the math) formula iii is the same as formula i C4H9Br, the difference is thatin formula i the bromine is on a terminal carbon and in formal iiithe bromine is on an internal carbon, accounting for the boilingpoint differnces

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