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Question: \({{(C{{H}_{3}})}_{2}}C=CHCOC{{H}_{3}}\xrightarrow{{{(C{{H}_{3}})}_{2}}CO}'X'\). Here, X is: A.Mes...

(CH3)2C=CHCOCH3→(CH3)2CO′X′{{(C{{H}_{3}})}_{2}}C=CHCOC{{H}_{3}}\xrightarrow{{{(C{{H}_{3}})}_{2}}CO}'X'. Here, X is:
A.Mesityl oxide
B. Phorone
C .Acetic acid
D. Mesitylene

Explanation

Solution

The concept of aldol condensation is required in this question. We can observe that the reactant already contains alpha hydrogen hence, it will undergo aldol and obtain the desired product by losing water molecules.

Complete step by step answer:
In order to solve this question, we need to learn about alcohol condensation reaction. Carbon atom of the carbonyl group is sp2s{{p}^{2}} hybridised having triangular planar geometry. Carbon atoms form three sigma bonds and one n(pi) bond, out of which 3 sigma bonds are located in the same plane whereas the fourth bond, which is formed by lateral or sideways overlapping, is situated above and below the plane. Bond angles are approximately 1200{{120}^{0}}. Carbonyl group is polarised due to the difference in electronegativity between carbon and oxygen. Oxygen being more electronegative pulls the shared pair of electrons more towards itself making oxygen as a nucleophilic centre and carbon as an electrophilic centre. Hence carbonyl compounds have substantial dipole moments and their polarity can be expressed on the basis of resonance. Aldehydes and ketones having at least one a-Hydrogen atom take part in Aldol condensation in the presence of dilute alkali [e.g. NaOHNaOH, Na2CO3N{{a}_{2}}C{{O}_{3}} or Ba(OH)2Ba{{(OH)}_{2}}. Two molecules of carbonyl compound condense to form a B-hydroxy aldehyde or B-hydroxy ketone which gets dehydrated on heating into α,β\alpha ,\beta -unsaturated aldehyde or ketone. Acidity of a-H atoms of carbonyl compounds is due to strong e withdrawing effect of carbonyl group and resonance stabilisation of the conjugate base. Now, let us come to our question and see the reaction:
(CH3)2C=CHCOCH3→(CH3)2O(CH3)2C=CHOH=C(CH3)2+H2O{{(C{{H}_{3}})}_{2}}C=CHCOC{{H}_{3}}\xrightarrow{{{(C{{H}_{3}})}_{2}}O}{{(C{{H}_{3}})}_{2}}C=CHOH=C{{(C{{H}_{3}})}_{2}}+{{H}_{2}}O
The alpha hydrogens get out as water and we get the final product as phorone.

Hence, we obtain our answer as option B.

Note: Aldol has two functional groups together i.e., aldehyde and alcohol. Similarly ketols have ketonic and alcoholic groups. Ketol and aldol both lose water readily to give α,β\alpha, \beta unsaturated carbonyl compounds, hence this reaction is called Aldol condensation.