Question

When benzene-1,3-dicarbaldehyde reacts with 50% of $NaOH$, the possible products formed is/are:
A.
B.
C.
D.

Answer 1

It is a Cannizzaro reaction, when a self oxidation reduction reaction in which aldehydes that do not have any $\alpha$-hydrogen atom undergo disproportionation reaction (i.e., self-redox reaction) in the presence of 50% aqueous or ethanolic solution of alkali in which one of the molecules being reduced to alcohol and other being oxidised to the salt of the corresponding acid.

Complete answer:
We are given benzene-1,3-dialdehyde which has a structure like this.

At both $\alpha$-position of the compound that is 1 and 3, we don’t have a hydrogen atom. So the condition for Cannizzaro reaction is true and we have a 50% of $NaOH$. Hence the Cannizzaro reaction will occur here. Also since there are two carbonyl groups it undergoes intramolecular Cannizzaro reaction.
As we know that it is a disproportionation reaction, both oxidation and reduction happen in the molecule. When the carbonyl compound is attached to 1st and 3rd position of the benzene it gets oxidised to form $- CO{O^ - }$. So we understood that at position 1 and 3 at some point of time we will have $- CO{O^ - }$ attached, hence option (a) is a possible product.
Also when benzene-1,3-dialdehyde gets reduced to form alcohols at position 1 and 3 we get $- C{H_2}OH$. Hence option (b) is also a possible product.
And on the intermediate of this, it may also occur at some point of time one $- CO{O^ - }$ and one $- C{H_2}OH$ is formed. Therefore, option (c) is also a possible product.
Now, coming to option(d) we can see that the position of $- CO{O^ - }$ and $- C{H_2}OH$are at 1 and 4 respectively, which is never going to happen as the carbonyl group are present and 1 and 3 position of the compound. Hence this product never forms.

**Hence, options A,B and C are correct.

Note:**
In the presence of $\alpha$-hydrogen the reaction occurred is known as aldol condensation. In this reaction, two molecules of aldehydes or ketones having $\alpha$-hydrogen condense together in presence of a base like $NaOH$ to form $\beta$-hydroxy aldehyde or $\beta$-hydroxy ketone respectively which are collectively known as aldol.