Question

How many structures of X is possible?

(A) 4
(B) 5
(C) 6
(D) 3

Answer 1

Hint: The intermediate structures for any given reaction give some basic reaction rules that would be vital and sensible.
Some of these rules are resonance effect, inductive effect, etc.

Complete step by step solution:
The given reaction is about the dehydrogenation of Butan-2-ol. Let us see the mechanism and conditions for the reaction.
There are 2 steps for the given reaction i.e.
1. The positive ion removes a water molecule from the butan-2-ol and carbocation is formed.
2. The carbocation loses the hydrogen ion to form a double bond.
The structures arising from the second step will be the answer for the given problem.
When the carbocation loses the hydrogen ion (from adjacent carbon atom of the carbon atom already having positive charge) to form double bond, it has two choices i.e.
1. Lose the hydrogen ion from the adjacent $C{{H}_{3}}$ group.
-Due to this move, But-1-ene ($C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}$) is formed.
2. Loss the hydrogen ion from the adjacent $C{{H}_{2}}$ group.
-Due to this move, But-2-ene ($C{{H}_{3}}-CH=CH-C{{H}_{3}}$) is formed.
-But, here the effect of resonance will take place as But-2-ene has a geometrical isomer. We get a mixture of two isomers i.e. cis-but-2-ene and trans-but-2-ene.
Thus,
Dehydration of Butan-2-ol leads to mixture containing,
a. But-1-ene
b. cis-but-2-ene
c. trans-but-2-ene
Therefore, three types of intermediates or structures would be formed for the given dehydration of the But-2-ol.

Hence, option (D) is correct.

Note: Do note to consider the phenomenon of resonance when we see the possibilities of having geometrical isomers.The loss of hydrogen ion from a carbocation always takes place from the alpha carbon atom of carbon holding the positive charge. Thus, the number of alpha carbons decides the basic number of intermediate structures having double bonds within themselves.