Q 14-14-59 Eb - Organic Chemistry (Mcmurry)

Question

Hydrocarbon A, C₁₀H₁₄ , has a UV absorption at ${\mathrm{\lambda }}_{\max }=236\mathrm{nm}$and gives hydrocarbon B, C₁₀H₁₈, on catalytic hydrogenation. Ozonolysis of A, followed by zinc/acetic acid treatment, yields the following diketo dialdehyde:

b) Hydrocarbon A reacts with maleic anhydride to yield a Diels-Alder adduct. Which of your structures for A is correct?

Step-by-Step Solution

Answer

Structure II for Hydrocarbon A reacts with maleic anhydride to yield a Diels-Alder adduct. 

Finding degree of unsaturation

In the analysis of  the molecular formula of organic molecules, the degree of unsaturation is a calculation that determines the total numbers of rings and pi bonds in the system. For finding out the DOU, you can use formula,

$$\begin{gathered} \mathrm{Degree} \mathrm{of} \mathrm{unsaturation}=\frac{1}{2}\times (2+2\times \mathrm{C}-\mathrm{H}+\mathrm{N}-\mathrm{X}) \\ \mathrm{Degree} \mathrm{of} \mathrm{unsaturation}=\frac{1}{2}\times (2+2\times 10-14+0-0) \\ \mathrm{Degree} \mathrm{of} \mathrm{unsaturation}=\frac{1}{2}\times (2+2\times 10-14+0-0) \\ \mathrm{Degree} \mathrm{of} \mathrm{unsaturation}=\frac{1}{2}\times (22-14) \\ \mathrm{Degree} \mathrm{of} \mathrm{unsaturation}=4 \end{gathered}$$

From the formula , degree of saturation   for A comes out to be 4 and from the ozonolysis product contains all the carbons and a di-keto dialdehyde one can  say, there are 2 double bonds and 2 rings in A.

Prediction of two structures

The structures which can be in agreement with the information drawn above come out to be,

Explanation

Structure II is the structure which can give Diels-Alder adduct on reaction with maleic anhydride through Diels-Alder reaction. And Diels-Alder reaction is possible in structure II only because in that structure it is possible for the double bonds to assume s-cis conformation through free rotation about the single bond ,which is an essential condition for Diels-Alder reaction to occur.