
n-Propyl bromide on treatment with ethanolic potassium hydroxide produces;
(a) propane
(b) propene
(c) propyne
(d) propanol
Answer
566.7k+ views
Hint: When alkyl halide is made to undergo reaction with the ethanolic potassium hydroxide, to result in the formation of the unsaturated compound and there is elimination of the bromine. Now you can easily solve the reaction and identify the product formed.
Complete step by step answer:
First of all, let’s discuss the n-propyl bromide. N-propyl bromide is an alkyl halide and is formed by the reaction of the alkane with the halogen’
Alkanes are the simplest hydrocarbons and consist of the carbon-carbon single bond. N-propyl bromide means that there are three carbon atoms and the hydrogen of first carbon is replaced by the halogen i.e. the bromine and results in the formation of the alkyl halide.
Now considering the statement,
When this alkyl halide i.e. n-propyl bromide is made to undergo reaction with the ethanolic potassium hydroxide i.e. $KOH$, there is formation of the carbon- carbon double bond and removal of bromine and the product formed is the unsaturated compound which consists of the double i.e. the compound is alkene.
Besides this, there is formation of potassium bromide and water as well.
The reaction occurs as;
$\begin{align}
& C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{Ethanolic\text{ }KOH}C{{H}_{3}}CH=C{{H}_{2}}+KBr+{{H}_{2}}O \\
& n-propyl\text{ }propene \\
& bromide \\
\end{align}$
So, thus n-Propyl bromide on treatment with ethanolic potassium hydroxide produces propene.
Hence, option (b) is correct.
Note: Alkanes and alkenes both come under the category of the hydrocarbons. But the difference is that the alkanes consists of carbon-carbon single bond and is saturated hydrocarbon whereas on the other hand, alkenes consist of the carbon-carbon double bond and are, thus, unsaturated hydrocarbons.
Complete step by step answer:
First of all, let’s discuss the n-propyl bromide. N-propyl bromide is an alkyl halide and is formed by the reaction of the alkane with the halogen’
Alkanes are the simplest hydrocarbons and consist of the carbon-carbon single bond. N-propyl bromide means that there are three carbon atoms and the hydrogen of first carbon is replaced by the halogen i.e. the bromine and results in the formation of the alkyl halide.
Now considering the statement,
When this alkyl halide i.e. n-propyl bromide is made to undergo reaction with the ethanolic potassium hydroxide i.e. $KOH$, there is formation of the carbon- carbon double bond and removal of bromine and the product formed is the unsaturated compound which consists of the double i.e. the compound is alkene.
Besides this, there is formation of potassium bromide and water as well.
The reaction occurs as;
$\begin{align}
& C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{Ethanolic\text{ }KOH}C{{H}_{3}}CH=C{{H}_{2}}+KBr+{{H}_{2}}O \\
& n-propyl\text{ }propene \\
& bromide \\
\end{align}$
So, thus n-Propyl bromide on treatment with ethanolic potassium hydroxide produces propene.
Hence, option (b) is correct.
Note: Alkanes and alkenes both come under the category of the hydrocarbons. But the difference is that the alkanes consists of carbon-carbon single bond and is saturated hydrocarbon whereas on the other hand, alkenes consist of the carbon-carbon double bond and are, thus, unsaturated hydrocarbons.
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