Q8-65 E - Organic Chemistry (Mcmurry)

Question

Compound A has the formula ${\mathit{C}}_{\mathbf{8}}{\mathit{H}}_{\mathbf{8}}$. It reacts rapidly with $\mathit{K}\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{4}}$to give $\mathit{C}{\mathit{O}}_{\mathbf{2}}$ and a carboxylic acid, B $\left({\mathit{C}}_7{\mathit{H}}_6{\mathit{O}}_2\right)$, but reacts with only 1 molar equivalent of ${\mathbf{H}}_{\mathbf{2}}$on catalytic hydrogenation over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4 equivalents of ${\mathbf{H}}_{\mathbf{2}}$ are taken up and hydrocarbon C $\left({\mathit{C}}_8{\mathit{H}}_{16}\right)$is produced. What are the structures of A, B, and C? Write the reactions.

Step-by-Step Solution

Answer

Answer:

Step1: Structure of compound A ( C 8 H 8 )

Compound A has four double bonds in which three bands are present in the aromatic ring, and the one is a c=c double bond. On reaction with ${\mathrm{KMnO}}_4$which is a strong oxidizing agent leads to carboxylic and carbon dioxide formation.

$C_8{H}_8 C_6{H}_{5}C{O}_{2}H + C{O}_2$

Step2: Reactions of compound A ( C 8 H 8 ) to form compound B

When compound A reacts with ${\mathrm{KMnO}}_4$ which is a strong oxidizing agent leads to the formation of carboxylic and carbon dioxide by breaking the double bond, followed by 1,2 diol formation.

Step3: Reactions of compound A ( C 8 H 8 ) to form compound C

One double bond of compound A reacts with lead to the alkene hydrogenation. Only the double bond in the ring remains the same.

The three double bonds in the around ring undergo hydrogenation in strong conditions by taking four equivalents, leading to hydrogenation of all the double bonds in the compound. 

Step 4: Reaction of compound A

The formation of the products and the reaction with compound A have been shown below