
How do alkenes decolorize bromine water?
Answer
546.3k+ views
Hint: When the hydrocarbon decolorizes the Bromine water, hydrocarbons are likely to be unsaturated one's that is alkene as well as alkyne because of double or triple bond break with those valence of carbon is fulfill by bromine. When bromine reacts with them it decolorizes. This reaction is used as the confirmed test for unsaturated compounds.
Complete step-by-step answer:
As we know Hydrocarbon compound decolorizes bromine water it is unsaturated hydrocarbon as well as hydrocarbon is an alkane because bromide and it undergoes an addition reaction along with unsaturated compound with containing double or a triple bond became and which in results become too much unstable. Thus bromine water is broken down in result to release $B{{r}^{-}}$ as well as $H$ for substitution in the carbon-carbon chain. Whenever bromine water is added to an unsaturated compound it gets added to latter and hence turning them into colorless from its original reddish brown color from a product of ozonolysis. Therefore we can predict the structure of alkene and we know that on ozonolysis subsequent one gives \[3-methyl\text{ }butanal\] with formaldehyde. The following reaction is given:
\[C{{H}_{3}}CH\left( C{{H}_{3}} \right)C{{H}_{2}}CH=C{{H}_{2}}+O\text{ }\to \text{ }C{{H}_{3}}CH\left( C{{H}_{3}} \right)C{{H}_{2}}CH=O+HCH{{O}_{2}}\]
Thus, name of compound is given by \[4-methyl-1-pentene\] and following formula is
\[C{{H}_{3}}CH\left( C{{H}_{3}} \right)C{{H}_{2}}CH=C{{H}_{2}}\]
Note: Also note that ozonolysis is a organic chemical reaction where in ozone \[\left( {{O}_{3}} \right)\] is employed to cleave unsaturated bond of alkene, alkyne, as well as $AZO$ compound which is with \[diazinyl\] functional group. Ozonolysis is organic redox reaction and the equivalent oxidation of alkene is with help of ozone and provides alcohol, aldehyde, ketone or carboxylic acid as well as ozonolysis of alkyne gives acid anhydride or \[diketone\].
Complete step-by-step answer:
As we know Hydrocarbon compound decolorizes bromine water it is unsaturated hydrocarbon as well as hydrocarbon is an alkane because bromide and it undergoes an addition reaction along with unsaturated compound with containing double or a triple bond became and which in results become too much unstable. Thus bromine water is broken down in result to release $B{{r}^{-}}$ as well as $H$ for substitution in the carbon-carbon chain. Whenever bromine water is added to an unsaturated compound it gets added to latter and hence turning them into colorless from its original reddish brown color from a product of ozonolysis. Therefore we can predict the structure of alkene and we know that on ozonolysis subsequent one gives \[3-methyl\text{ }butanal\] with formaldehyde. The following reaction is given:
\[C{{H}_{3}}CH\left( C{{H}_{3}} \right)C{{H}_{2}}CH=C{{H}_{2}}+O\text{ }\to \text{ }C{{H}_{3}}CH\left( C{{H}_{3}} \right)C{{H}_{2}}CH=O+HCH{{O}_{2}}\]
Thus, name of compound is given by \[4-methyl-1-pentene\] and following formula is
\[C{{H}_{3}}CH\left( C{{H}_{3}} \right)C{{H}_{2}}CH=C{{H}_{2}}\]
Note: Also note that ozonolysis is a organic chemical reaction where in ozone \[\left( {{O}_{3}} \right)\] is employed to cleave unsaturated bond of alkene, alkyne, as well as $AZO$ compound which is with \[diazinyl\] functional group. Ozonolysis is organic redox reaction and the equivalent oxidation of alkene is with help of ozone and provides alcohol, aldehyde, ketone or carboxylic acid as well as ozonolysis of alkyne gives acid anhydride or \[diketone\].
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