Question

Question: What is the composition of the soap prepared by hydrolysis of the following triacylglycerol?

Short answer

Answer

Saponification of given glyceraldehyde

Therefore, the soap will contain two carboxylate ions:

$$\begin{gathered} {{\mathrm{Na}}^{+}}^{-}\mathrm{OOC}{\left({\mathrm{CH}}_2\right)}_{15}{\mathrm{CH}}_3 \\ {{\mathrm{Na}}^{+}}^{-}\mathrm{OOC}{\left({\mathrm{CH}}_2\right)}_7\mathrm{CH}=\mathrm{CH}{\left({\mathrm{CH}}_2\right)}_7{\mathrm{CH}}_3 \end{gathered}$$

Step-by-step solution

Step-by-Step Solution Step 1: Synthesis of soap

Soap is synthesized from triacylglycerol by hydrolysis. The reaction is carried out in the presence of a base. The reaction is also called saponification.

The reaction produces glycerol and salt of three fatty acids. The sodium salt is formed if sodium hydroxide is used and potassium salt is formed if potassium hydroxide is used.

Carboxylate salt is a soap that has two structurally different regions making it capable of cleaning dirt.

The triacylglycerols are generally obtained from vegetable oil or animal fat.

Hydrolysis of the ester

Esters are produced by the condensation of acids and alcohols. This reaction is called esterification.

The reverse reaction of esterification that cleaves the ester bond to form acids and alcohols is called ester hydrolysis.

If the hydrolysis is catalyzed by acids, alcohols and acids are obtained. If the catalyst is a base, alcohols and carboxylate salts are obtained as products.

Synthesis of soap from given triacylglyceride

Triacylglycerides contain three ester groups.

In the presence of a base like sodium hydroxide, all the ester groups undergo hydrolysis to form glycerol.

The acid part of each ester remains as the sodium salt. The product will contain sodium salt of acids from all three esters. This mixture is called soap.

Hydrolysis of triacylglyceride

Therefore, the soap will contain two carboxylate ions:

$$\begin{gathered} {{\mathrm{Na}}^{+}}^{-}\mathrm{OOC}{\left({\mathrm{CH}}_2\right)}_{15}{\mathrm{CH}}_3 \\ {{\mathrm{Na}}^{+}}^{-}\mathrm{OOC}{\left({\mathrm{CH}}_2\right)}_7\mathrm{CH}=\mathrm{CH}{\left({\mathrm{CH}}_2\right)}_7{\mathrm{CH}}_3 \end{gathered}$$