Q8-59 E - Organic Chemistry (Mcmurry)

Question

An unknown hydrocarbon A with the formula ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}$ reacts with 1 molar equivalent of ${\mathbf{H}}_{\mathbf{2}}$ over a palladium catalyst. Hydrocarbon A also reacts with ${\mathbf{OsO}}_{\mathbf{4}}$ to give diol B. When oxidized with ${\mathbf{KMnO}}_{\mathbf{4}}$ in acidic solution, A gives two fragments. One fragment is propanoic acid, ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}$, and the other fragment is ketone C. What are the structures of A, B, and C? Write all reactions, and show your reasoning.

Step-by-Step Solution

Answer

Answer

Structure of compound A

Structure of compound B

Structure of compound C

Step 1: Catalytic hydrogenation

Catalytic hydrogenation is the reaction of alkene or alkyne with hydrogen in the presence of a metal catalyst like nickel, palladium, or platinum. Hydrogenation reduces double and triple bonds in hydrocarbons and adds hydrogen.

Step 2: OsO 4 reactions

Inorganic chemistry, OsO₄ is used for the oxidation of alkenes to the vicinal diols. It adds two hydroxyl groups at the same site known as syn addition.

Step 3: KMnO 4 reactions

The acidified KMnO₄ solution oxidizes alkenes by breaking the double bond between carbons and replacing it with two carbonyl groups.

In this case, the product formed has hydrogen attached to the carbonyl group. It forms aldehyde and aldehydes are readily oxidized to give carboxylic acids.

Step 4: Reaction interpretation and reasoning

In the reaction as only 1 mole of dihydrogen is used, it indicates only one bond is broken and only two hydrogens are attached to 1 mole of ${\mathrm{C}}_6{\mathrm{H}}_{12}$. 

• As here ${\mathrm{OsO}}_4$is reacting with hydrocarbon A this indicated presence of one alkene group. 
• As on reaction with ${\mathrm{KMnO}}_4$ one ketone is generated here it indicates the presence of a side group on one of the carbons of the double bond.
• As the product formed is carboxylic acids so one double bond carbon must have a hydrogen attached to the carbonyl group which readily oxidized to give carboxylic acid.

Step 5: Structure interpretation

Using the steps outlined above, it is possible to identify the compounds. The following are probable compounds based on the given information.

Structure of compound A

Structure of compound B

Structure of compound C

Step 6: Catalytic hydrogenation reaction

The reduction of the double bond in the compound can be done using palladium as a catalyst and dihydrogen gas.

The reaction can be shown as:

Catalytic hydrogenation

Step 7: OsO 4 reaction

The double bond present in the given compound can be converted into a dihydroxyl compound.

The reaction can be shown as:

${\mathrm{OsO}}_4$Reaction

Step 8: KMnO 4 reaction

The double bond present in the given compound can be converted into the carbonyl compound by using ${\mathrm{KMnO}}_4$in acidic medium.

The reaction can be shown as:

${\mathrm{KMnO}}_4$reaction