
A hydrocarbon reacts with hypochlorous acid to give 1-chloro-2-hydroxyethane. The hydrocarbon is
A. Ethylene
B. Methane
C. Ethane
D. Acetylene
Answer
164.1k+ views
Hint: The hypochlorous acid has the chemical formula of HClO or HOCl. It is a weak acid formed due to the reaction of chlorine with water. Here, we will try to find out the hydrocarbon that will give 1-chloro-2-hydroxyethane.
Complete Step by Step Answer:
Let’s write the given reaction first. It is given that the reaction of hypochlorous acid with a hydrocarbon gives 1-chloro-2-hydroxyethane.
\[{\rm{Hydrocarbon}} + {\rm{HOCl}} \to {\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{OH}}\]
Let’s understand how the chlorine atom is bonded to the hydrocarbon. In an acidic medium, the proton consumes the OH of HOCl and gives water. And also the formation of an electrophile of chlorine happens. The chemical reaction is as follows:
\[{\rm{HOCl}} \overset{H^{+}}{\rightarrow} {{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{C}}{{\rm{l}}^ + }\]
Now, an electron-rich species attacks the chlorine electrophile. The electron-rich species can be a double bond. So, the hydrocarbon may be ethene or ethylene because the product 1-chloro-2-hydroxyethane has two atoms of carbon. So, we will take ethylene as the double-bonded hydrocarbon.
\[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{C}}{{\rm{l}}^ + } \to {\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{C}}{{\rm{H}}_{\rm{2}}}\]
So, we have seen that the electron-deficient carbon is attacked by a molecule of water.
\[{\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{C}}{{\rm{H}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{O}}{{\rm{H}}_{\rm{2}}}\]
Now, a proton of the water molecule leaves the compound and we get an alcohol compound.
\[{\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{O}}{{\rm{H}}_{\rm{2}}} \overset{-H^{+}}{\rightarrow}{\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{OH}}\]
So, now we get the product given in the question, that is, 1-chloro-2-hydroxyethane. Therefore, the hydrocarbon required is ethylene.
Hence, option A is right.
Note: There are many useful applications of hypochlorous acid in chemistry such as it is useful in the conversion of alkenes to chlorohydrins. It is an oxidising agent whose nature is strong. It also possesses many antibacterial properties.
Complete Step by Step Answer:
Let’s write the given reaction first. It is given that the reaction of hypochlorous acid with a hydrocarbon gives 1-chloro-2-hydroxyethane.
\[{\rm{Hydrocarbon}} + {\rm{HOCl}} \to {\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{OH}}\]
Let’s understand how the chlorine atom is bonded to the hydrocarbon. In an acidic medium, the proton consumes the OH of HOCl and gives water. And also the formation of an electrophile of chlorine happens. The chemical reaction is as follows:
\[{\rm{HOCl}} \overset{H^{+}}{\rightarrow} {{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{C}}{{\rm{l}}^ + }\]
Now, an electron-rich species attacks the chlorine electrophile. The electron-rich species can be a double bond. So, the hydrocarbon may be ethene or ethylene because the product 1-chloro-2-hydroxyethane has two atoms of carbon. So, we will take ethylene as the double-bonded hydrocarbon.
\[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{C}}{{\rm{l}}^ + } \to {\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{C}}{{\rm{H}}_{\rm{2}}}\]
So, we have seen that the electron-deficient carbon is attacked by a molecule of water.
\[{\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{C}}{{\rm{H}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{O}}{{\rm{H}}_{\rm{2}}}\]
Now, a proton of the water molecule leaves the compound and we get an alcohol compound.
\[{\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {}^ + {\rm{O}}{{\rm{H}}_{\rm{2}}} \overset{-H^{+}}{\rightarrow}{\rm{Cl}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{OH}}\]
So, now we get the product given in the question, that is, 1-chloro-2-hydroxyethane. Therefore, the hydrocarbon required is ethylene.
Hence, option A is right.
Note: There are many useful applications of hypochlorous acid in chemistry such as it is useful in the conversion of alkenes to chlorohydrins. It is an oxidising agent whose nature is strong. It also possesses many antibacterial properties.
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