Question 31.3 - Organic Chemistry

Question

Draw the products formed when triacylglycerol A is treated with each reagent. Rank compounds A, B and C in order of increasing melting point.

a. ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ , ${\mathbf{H}}^{\mathbf{+}}$

b. ${\mathbf{H}}_{\mathbf{2}}$(excess), $\mathbf{Pd}\mathbf{.}\mathbf{C}\mathbf{\to }\mathbf{B}$

c. ${\mathbf{H}}_{\mathbf{2}}$(1 equiv), $\mathbf{Pd}\mathbf{.}\mathbf{C}\mathbf{\to }\mathbf{C}$

Step-by-Step Solution

Answer

Answer

a.

b.

c.

                             $$\begin{gathered} \mathrm{A} < \mathrm{C} < \mathrm{B} \\ 2 \mathrm{double} \mathrm{bonds} 1 \mathrm{double} \mathrm{bond} 0 \mathrm{double} \mathrm{bond} \\ \mathrm{lowest} \mathrm{melting} \mathrm{intermediate} \mathrm{highest} \mathrm{melting} \\ \mathrm{point} \mathrm{melting} \mathrm{point} \mathrm{point} \end{gathered}$$

Step-by-Step Solution Step 1: Triacylglycerol

Triacylglycerols comprise esters acquired from glycerol and three fatty acids. They form the primary part of body fat and vegetable fat. Unsaturated fats constitute a class of triglycerides and possess a low melting point.

Step 2: Melting point

The phase transformation from solid to liquid happens at a particular temperature termed the melting point. Pressure is one factor that impacts the melting point, and it is indicated at standard pressure.

Step 3: Treatment of triacylglycerol A with different reagents

a. The treatment of compound A with ${\mathrm{H}}_2\mathrm{O}$, ${\mathrm{H}}^{+}$ can be given as:

                              Treatment of compound A with ${\mathrm{H}}_2\mathrm{O}$, ${\mathrm{H}}^{+}$

b. The product obtained on the treatment of compound A with ${\mathrm{H}}_2$ (excess) and Pd-C can be given as:

                         Treatment of compound A with ${\mathrm{H}}_2$ (excess) and Pd-C

c.The product obtained on the treatment of compound A with ${\mathrm{H}}_2$ (1 equiv) and Pd-C can be given as:

                       Treatment of compound A with ${\mathrm{H}}_2$ (1 equiv) and Pd-C

The compound comprising more significant double bonds possesses a minor melting point. So, compound A, comprising two double bonds, has the lowest melting point. 

Compound C will have an intermediate melting point, and compound B with zero double bonds has the highest melting point.

The compounds can be arranged in the order of increasing melting point as:

                      $$\begin{gathered} \mathrm{A} < \mathrm{C} < \mathrm{B} \\ 2 \mathrm{double} \mathrm{bonds} 1 \mathrm{double} \mathrm{bond} 0 \mathrm{double} \mathrm{bond} \\ \mathrm{lowest} \mathrm{melting} \mathrm{intermediate} \mathrm{highest} \mathrm{melting} \\ \mathrm{point} \mathrm{melting} \mathrm{point} \mathrm{point} \end{gathered}$$