Question

Write the structures of alkenes obtained on dehydration of:
A.propan - 1 - ol
B.propan - 2 - ol
C.2- methylpropan -1 -ol
D.2- methylpropan -2 -ol

Answer 1

Hint:We know that one way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo E1 or E2 mechanisms to lose water and form a double bond. The dehydration reaction of alcohols to produce alkene proceeds by heating the alcohols at high temperatures in the presence of a strong acid, such as sulphuric or phosphoric acid.

Complete answer:
There are different types of alcohols that may dehydrate through a slightly different mechanism pathway. The dehydration reaction occurs when the $-OH$ group in the alcohol donates two electrons to ${H^ + }$ from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid i.e., the nucleophile then attacks the hydrogen adjacent to the carbocation and form a double bond.
A.propan - 1 – ol
Dehydration of propan - 1 – ol gives prop-1-ene:
$C{H_3}C{H_2}C{H_2} - OH + \Delta \to C{H_3}CH = C{H_2}$
B.​ propan - 2 – ol
Dehydration of propan - 2 – ol gives prop-1-ene:
$C{H_3} - CH\left( {OH} \right) - CH{_3} + \Delta \to C{H_3}CH = C{H_2}$
C.2- methylpropan -1 –ol
Dehydration of 2- methylpropan -1 –ol gives 2- methylprop -1 – ene:
$C{H_3} - CH\left( {C{H_3}} \right) - CH{_2} - OH + \Delta \to C{H_3} - C\left( {C{H_3}} \right) = C{H_2}$
D.2- methylpropan -2 -ol
Dehydration of 2- methylpropan -2 –ol gives 2- methylprop -1 – ene:
$C{H_3} - C\left( {C{H_3}} \right)\left( {OH} \right) - CH{_3} + \Delta \to C{H_3} - C\left( {C{H_3}} \right) = C{H_2}$

Note:
It should be noted that the primary alcohols undergo bimolecular elimination while the secondary alcohols and tertiary alcohols undergo unimolecular elimination. The relative reactivity of alcohols in dehydration reaction is ranked as the following way: Methanol < primary < secondary < tertiary.