Question

Describe the following:
i) Cross aldol condensation
ii) Decarboxylation

Answer 1

The cross aldol condensation reaction is performed by using two different molecules of aldehyde or ketone. One can elaborate on this reaction by using any two different aldehydes or ketones. The decarboxylation reaction can be said as the reaction which removes the carbon dioxide molecule from a starting compound.

Complete step by step answer:
i) Cross aldol condensation:
First of all, we will understand the concept behind the reaction of a cross aldol condensation. Cross aldol condensation reaction can be defined as the reaction of two different kinds of aldehyde or ketone molecules in presence of a strong base which gives the $\beta -$hydroxy aldehyde or $\beta -$hydroxy ketone as a product.
Now let's see the reaction where we will use two different aldehydes as acetaldehyde and propionaldehyde as below,
Self aldol products,
$C{H_3}CHO + C{H_3}C{H_2}CHO\xrightarrow[\Delta ]{{NaOH}}C{H_3} - CH = CH - CHO + C{H_3} - C{H_2} - CH = C - CHO - C{H_3}$
Cross aldol products,
$C{H_3}CHO + C{H_3}C{H_2}CHO\xrightarrow[\Delta ]{{NaOH}}C{H_3} - CH = C(C{H_2}) - CHO + C{H_3} - C{H_2} - CH = CHCHO$
In the above reaction, the different aldehydes used as acetaldehyde and propionaldehyde can undergo self-condensation reactions with the same types of molecules present in the reaction. The acetaldehyde will produce But${\text{ - 2 - }}$enal and propionaldehyde will produce ${\text{2 - }}$Methylpent${\text{ - 2 - }}$enal as self-condensation products.
In the cross aldol product, the products formed are ${\text{2 - }}$Methylbut${\text{ - 2 - }}$enal formed from acetaldehyde and Pent${\text{ - 2 - }}$enal from propionaldehyde.
ii) Decarboxylation:
First of all, let's understand what exactly is the decarboxylation reaction. It is a chemical reaction that causes the removal of a carboxyl group and releases carbon dioxide which produces hydrocarbons as a product. Now let's see the reaction for the same as below,
$C{H_3}COONa + NaOH\xrightarrow[\Delta ]{{CaO}}C{H_4} + N{a_2}C{O_3}$
In the above reaction, the sodium ethanoate is reacted with the sodium hydroxide base in presence of calcium oxide and heat which produces methane and sodium carbonate.

Note:
It is important to remember that in the cross aldol condensation reaction there are always Four products formed of which two are self aldol and two are cross aldol products. The reaction of decarboxylation is about the removal of a carbon atom from a carbon chain which forms a hydrocarbon that has one less carbon than the initial reactant.