Question

A hydrocarbon reacts with hypochlorous acid and gives 2 chloroethanol. The hydrocarbon is:
(a) Ethylene
(b) Methane
(c) Ethane
(d) Acetylene

Answer 1

First of all, convert the reaction statement into molecular formulas and then try to solve further. Try to observe that hypochlorous acid would break and tend to attack the hydrocarbon we consider in the beginning and then match it with the product given and the options as well.

Complete step-by-step answer: Let’s start with the conversion of reaction statement into molecular formulas:-
Suppose the hydrocarbon be R.
Hypochlorous acid= HOCl 2-chloroethanol =$Cl-C{{H}_{2}}-C{{H}_{2}}-OH$
Now the reaction would have taken place as follows:-
$R+HOCl\to Cl-C{{H}_{2}}-C{{H}_{2}}-OH$
Let’s check the reactions taking place in between:-
-We will first see how this Cl atom would have joined the hydrocarbon. In acidic medium, ${{H}^{+}}$ consumes $O{{H}^{-}}$ of HOCl and forms${{H}_{2}}O$. Now the chlorine ion formed is positively charged (cation) as when ${{H}^{+}}$arrived, hydroxide ion breaks itself from Cl and takes the shared paired of electron and becomes negatively charged to attack${{H}^{+}}$. It is shown below:-
$HOCl\xrightarrow{{{H}^{+}}}{{H}_{2}}O+C{{l}^{+}}$
-Now this $C{{l}^{+}}$ act as electrophile and can provoke any electron rich specie to attack on itself such as double bond. So, the hydrocarbon we supposed has a double bond and now R is ${{R}_{1}}={{R}_{2}}$ . The reaction occurs as follows:-
${{R}_{1}}={{R}_{2}}+C{{l}^{+}}\to Cl-{{R}_{1}}-{{R}_{2}}^{+}$
-Since, the other part of hydrocarbon has become electron deficient, it is attacked by the ${{H}_{2}}O$ molecule (which may be reaction produced ${{H}_{2}}O$ or solvent).
$Cl-{{R}_{1}}-{{R}_{2}}^{+}+{{H}_{2}}O\to Cl-{{R}_{1}}-{{R}_{2}}{{-}^{+}}O{{H}_{2}}$
-Now, O has a positive charge on itself which it cannot tolerate being an electronegative element. Therefore, the H attached beside it, leaves itself as ${{H}^{+}}$ and we get an alcohol group beside the hydrocarbon.
$Cl-{{R}_{1}}-{{R}_{2}}{{-}^{+}}O{{H}_{2}}\xrightarrow{-{{H}^{+}}}Cl-{{R}_{1}}-{{R}_{2}}-OH$

On comparing the following 2 molecules of product we get:
$Cl-C{{H}_{2}}-C{{H}_{2}}-OH$and $Cl-{{R}_{1}}-{{R}_{2}}-OH$

${{R}_{1}}=C{{H}_{2}}\text{ and }{{R}_{2}}=C{{H}_{2}}$

The hydrocarbon which reacted with hypochlorous acid to give 2-chloroethanol is $C{{H}_{2}}=C{{H}_{2}}$

Therefore the right option is: (a) Ethylene.

Note: For such kind of questions, always go through the mechanism part, so as to know which specie is breaking and which ion formation is taking place because without the formation of correct specie we would not be able to show attack on each other and formation of the product would be different than expected.