Question

Compound (X) ${{C}_{4}}{{H}_{8}}O$ decolourises Baeyer's reagent. It undergoes hydrolysis on reaction with dil.${{H}_{2}}S{{O}_{4}}$, and produces (Y) and (Z). Both (Y) and (Z) give positive iodoform tests. But, only (Y) gives a positive Tollens test. Choose the correct statements.

A.
$(Y)+aqNaOH,\Delta \to C{{H}_{3}}-CH=CH-CHO$
B.
$(Z)+\text{Pyridinium chlorochromate(PCC)} \to (Y)$
C.
$(X)+B{{r}_{2}}/CC{{l}_{4}}\to Br-C{{H}_{2}}-CH-Br-O-C{{H}_{2}}-C{{H}_{3}}$
D.
$(X)+{{O}_{3}},Zn/{{H}_{2}}O\to (Y)+ \text{ other product}$

Answer 1

Hint Tollens test is only given by aldehydes, whereas iodoform test proves the presence of a carbonyl group. PCC is used for oxidation mainly and during reductive ozonolysis, cleavage of the bonds takes place.

Complete step by step solution:
In order to answer our question, we have to proceed step by step and find out the reagents and products of the reaction. Now, we have been given the compound ${{C}_{4}}{{H}_{8}}O$, and it is said that it decolourises Bayer’s reagent. Baeyer's test is a test for unsaturation. Since the compound reacts to the Baeyer's test, that means a double bond or unsaturation is present in the compound. Now, let us check the degree of unsaturation for the compound. The formula for degree of unsaturation is $C-\dfrac{(H+X+N)}{2}+1$, where the letters represent the number of atoms of the respective elements. So, the degree of unsaturation of ${{C}_{4}}{{H}_{8}}O$ is 4+1-4 which comes out to be 1. Now a d.u of 1 means that it can be either a ring or a double bond. However, based on the conditions given in the question, we can assume the compound to be a straight chain double bonded species. As we have been given that compound Y and Z respond to iodoform test, that means that an aldehydic or ketonic group is present there. Precisely, the $CO-C{{H}_{3}}$ or $CH(OH)-C{{H}_{3}}$ group is present. Moreover, as Y gives the Tollens test, that means Y is definitely an aldehyde. So, we can represent the reaction of X given in the question as:
${{H}_{2}}C=CHO-C{{H}_{2}}C{{H}_{3}}\xrightarrow{{{H}_{2}}S{{O}_{4}}}{{H}_{3}}C-CHO(Y)+C{{H}_{3}}C{{H}_{2}}OH(Z)$
So now verifying the options, we have
A.
$C{{H}_{3}}C{{H}_{2}}OH+aqNaOH\to C{{H}_{3}}C{{H}_{2}}CHO$
It is aldol condensation
B.
$C{{H}_{3}}C{{H}_{2}}OH+PCC\to C{{H}_{3}}CHO$
It is oxidation by PCC
C.
$C{{H}_{2}}CHOC{{H}_{2}}C{{H}_{3}}+B{{r}_{2}}/CC{{l}_{4}}\to C{{H}_{2}}BrCHBrOC{{H}_{2}}C{{H}_{3}}$

However, option D is not true. So we obtain our correct answer as options A, B and C.

NOTE: During aldol condensation, an enolate ion is formed which takes part in reaction. It requests NaOH diluted whereas Cannizaro requires concentrated alkaline bases. For aldol, alpha hydrogen needs to be present.