Question

Explain the mechanism of the following reaction:
$C{H_3} - C{H_2} - OH\,\xrightarrow[{443\,K}]{{{H^ + }}}\,C{H_2} = C{H_2}\, + \,{H_2}O$

Answer 1

Hint:As we see here that water molecule is being released so we will name this reaction as the dehydration reaction. A dehydration reaction is the one in which a water molecule is lost from alcohol and as a result an unsaturated compound like alkene is formed. It is an example of elimination reaction.

Complete step-by-step answer:The mechanism of the reaction will proceed as:-
In the first step we will see that the ${H^ + }$ ion which is coming from an acid will attract the oxygen atom of ethanol towards itself and will get attached to the oxygen atom thereby forming an ethyl oxonium ion.
$C{H_3} - C{H_2} - OH + {H^ + }\,\, \rightleftarrows \,\,C{H_3} - C{H_2} - O{H_2}^ + $
In the second step we see that an intermediate is being formed. The intermediate formed is known as the carbocation. A carbocation is the one in which a positive charge is present on the carbon and is very stable. This step is usually very slow that is why it is known as the rate determining step.
$C{H_3} - C{H_2} - O{H_2}^ + \,\, \rightleftarrows \,\,C{H_3} - C{H_2}^ + \, + \,{H_2}O$
In the last step we see that a proton i.e. the ${H^ + }$ ion will be eliminated from the carbocation and hence will lead to the formation of ethene. The acid which was consumed in the first step is released in the last step. After the formation of ethene, the ${H^ + }$ ion will be removed so as to shift the equilibrium in the forward direction.
$C{H_3} - C{H_2}^ + \,\, \rightleftarrows \,\,{H_2}C = C{H_2}\, + \,{H^ + }$.

Note:As we know that ethanol is a primary alcohol so it undergoes the ${E_2}$ mechanism. Also the rate of dehydration depends upon the stability of the intermediate formed i.e the carbocation. Tertiary carbocations are the most stable and therefore the rate of dehydration is also the highest for tertiary carbocations, then comes the secondary carbocation and the last comes the primary.