
Write the structure of alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic $KOH$.
Answer
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Hint: We have to draw the structure of \[1 - bromo - 1 - methylcyclohexane\] then only we can find the structure of alkene and understand the meaning of the word dehydrohalogenation.
Complete step by step solution:
Let's start with the formal definition, in dehydrohalogenation de stands for removal or elimination, hydro stands for hydrogen and halogen stands for halogen itself. So, dehydrohalogenation is the process of elimination of hydrogen and halogen. Therefore, we can understand that hydrogen bromide is being removed.
Now, we are given a substrate which is 1-Bromo-1-Methylcyclohexane whose molecular formula is $C_7H_13Br$. From the name of the substrate, we can see that the bromine, as well as the methyl group both, is present at the same position. So, the structure of the substrate is as follows
Now, in the presence of alcoholic KOH, bromine (halogen) and hydrogen from an adjacent carbon is being removed which leads to the formation of an alkene. The formation of alkene is shown below
Therefore we can conclude that the alkene formed is\[1 - methylcyclohex - 1 - ene\]. This nomenclature is due to the presence of a methyl group at 1st position and a double bond present between the 1st and 2nd position. The other products of this reaction are KBr and $H_2O$.
Additional information:
- Halogen groups represent all the atoms present in Group 17 of periodic table i.e. \[Fluorine{\text{ }}\left( F \right),{\text{ }}Chlorine{\text{ }}\left( {Cl} \right),{\text{ }}Bromine{\text{ }}\left( {Br} \right),{\text{ }}Iodine{\text{ }}\left( I \right),{\text{ }}and{\text{ }}Astatine{\text{ }}\left( {At} \right).\]
- Alcoholic KOH is being used so that KOH acts as a base not as a nucleophile, if KOH acts as a nucleophile it will lead to the formation of alcohol rather than an alkene.
Note: We must understand that quicklime $CaO$ is used in daily household works and with the demand increasing in its production; industrial revolution is taking place with proper enumeration using it.
Complete step by step solution:
Let's start with the formal definition, in dehydrohalogenation de stands for removal or elimination, hydro stands for hydrogen and halogen stands for halogen itself. So, dehydrohalogenation is the process of elimination of hydrogen and halogen. Therefore, we can understand that hydrogen bromide is being removed.
Now, we are given a substrate which is 1-Bromo-1-Methylcyclohexane whose molecular formula is $C_7H_13Br$. From the name of the substrate, we can see that the bromine, as well as the methyl group both, is present at the same position. So, the structure of the substrate is as follows
Now, in the presence of alcoholic KOH, bromine (halogen) and hydrogen from an adjacent carbon is being removed which leads to the formation of an alkene. The formation of alkene is shown below
Therefore we can conclude that the alkene formed is\[1 - methylcyclohex - 1 - ene\]. This nomenclature is due to the presence of a methyl group at 1st position and a double bond present between the 1st and 2nd position. The other products of this reaction are KBr and $H_2O$.
Additional information:
- Halogen groups represent all the atoms present in Group 17 of periodic table i.e. \[Fluorine{\text{ }}\left( F \right),{\text{ }}Chlorine{\text{ }}\left( {Cl} \right),{\text{ }}Bromine{\text{ }}\left( {Br} \right),{\text{ }}Iodine{\text{ }}\left( I \right),{\text{ }}and{\text{ }}Astatine{\text{ }}\left( {At} \right).\]
- Alcoholic KOH is being used so that KOH acts as a base not as a nucleophile, if KOH acts as a nucleophile it will lead to the formation of alcohol rather than an alkene.
Note: We must understand that quicklime $CaO$ is used in daily household works and with the demand increasing in its production; industrial revolution is taking place with proper enumeration using it.
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