Question: In the below shown reaction, X is: \[C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{conc.{{H}_{2}}S{{O}_{4}},{...

Question

In the below shown reaction, X is:
$C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{conc.{{H}_{2}}S{{O}_{4}},{{170}^{{}^\circ }}C}X$
(A) ethene
(B) isopropyl alcohol
(C) neopentyl alcohol
(D) 2-methyl-2-propanol

Answer 1

Solution

Elimination reaction, any of a class of organic chemical reactions in which a pair of atoms or groups of atoms are removed from a molecule, usually through the action of acids, bases, or metals and, in some cases, by heating to a high temperature.

Complete step by step solution: When conc. ${{H}_{2}}S{{O}_{4}}$ is added to an alcohol at such a high temperature, it undergoes elimination and thus, gives an alkene.
Thus the corresponding alkene formed will be ethene.
The reaction is given below:
$C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{conc.{{H}_{2}}S{{O}_{4}},{{170}^{{}^\circ }}C}C{{H}_{2}}=C{{H}_{2}}$
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo ${{E}_{1}}$ or ${{E}_{2}}$ mechanisms to lose water and form a double bond. The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures.
Different types of alcohols may dehydrate through a slightly different mechanism pathway. However, the general idea behind each dehydration reaction is that 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 (the base) then reacts with the hydrogen adjacent to the carbocation and forms a double bond.

Hence the correct option is A option.

Note: The ${{E}_{2}}$ elimination of tertiary alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorus oxychloride ( $POC{{l}_{3}}$ ) in pyridine. This procedure is also effective with hindered secondary alcohols, but for unhindered and primary alcohols an $S{{N}_{2}}$ chloride ion substitution of the chlorophosphate intermediate competes with elimination.