Q8-61 E - Organic Chemistry (Mcmurry)

Question

Compound A, C₁₀H₁₈O, undergoes reaction with dilute H₂SO₄ at 50 °C to yield a mixture of two alkenes, C₁₀H₁₆. The major alkene product, B, gives only cyclopentanone after ozone treatment followed by reduction with zinc in acetic acid. Identify A and B, and write the reactions. 

Cyclopentanone

Step-by-Step Solution

Answer

Answer

Structure of compound A

Structure of compound B

Step 1: Reaction with H 2 SO 4

When an alkene is treated with sulphuric acid and heat, it results in dehydration and forms an alkene.

Step 2: O 3 and Zn reactions

When an alkene is treated with ozone and subjected to reductive workup with zinc, the carbon-carbon double bond is cleaved to give ketones or aldehydes, depending on the alkene structure.

Oxidative cleavage reaction can be performed by ${\mathbf{O}}_{\mathbf{3}}$

Step 3: Reaction interpretation and reasoning

• As C₁₀H₁₈O gives C₁₀H₁₆ on reaction with dilute H₂SO₄ in presence of heat so it is losing one H₂O so the structure must have an OH group.
• The water removal must have created a double bond.
• As after ozone treatment, only one product is obtained, which is Cyclopentanone, so the alkene product must be having two cyclopentane rings joined by the double bond. 

Step 4: Structure of A and B

From the steps mentioned earlier, we can identify the compounds. So the possible compound, according to the given information, are mentioned below here.

Structure of A

Structure of B

Step 5: Reaction with H 2 SO 4

The reaction of compound A with sulphuric acids leads to the formation of the two products with double bonds present in different positions. 

The reaction can be shown as:

Reaction with dilute H₂SO₄

Step 6: Reaction with O 3 and Zn

The compound B with and Zn reaction leads to the breaking of the two double and results in a formation of two symmetrical carbonyl compounds.

The reaction can be shown as:

O₃ and Zn reaction