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A gaseous hydrocarbon decolourise bromin...

A gaseous hydrocarbon decolourise bromine in `C""Cl_4`. One molecule of acetone and one molecule of acetaldehyde are obtained as a result of its ozonolysis.
Determine the structural formula of the compound and write its IUPAC name. Give all reactions involved.

Text Solution

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1. The given hydrocarbon decolurises bromine solution. Hence, it is an unsaturated hydrocarbon.
2. The hydrocarbon, on ozonolysis, produces 1 molecule of acetone and 1 molecule of acetaldehyde. So, the structure of the unsaturated hydrocarbon many be obtained by writing the Formula of the two carbonyl compounds side by side facing their carbonyl groups with respect to each other followed by removal of the oxygen atoms and joining the two carbonyl carbons by a double bond. Therefore, the IUPAC name of the hydrocarbon is 2-methylbut-2-ene
`underset("Acetone")(CH_(3) - overset(overset(CH_(3))(|))(C) = O) + underset("Acetaldehyde")(O = overset(overset(H)(|))(C) - CH_(3)) rArr underset("2-methylbut-2-ene")(CH_(3) - overset(overset(CH_(3))(|))(C) - CHCH_(3))`
The reactions are as follows:
`underset(underset("(colourless)")("2-methylbut-2-ene"))((CH_(3))_(2) C = CHC_(3)) + underset(("red"))(Br_(2)) rarr underset(underset(("colourless"))("2,3-dibromo-2-methylbutane"))((CH_(3))_(2) CBrCHBrCH_(3))`
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