1.Describe the ozonolysis of alkenes (of 5 marks question)

2.one mole of a hydrocarbon(A) reacts with one mole of beomine giving a dibromo compound C5div0Br2.Substance A on treatment with cold dilute kMnO4 solution forms a compound C5C12O2(C5div2O2) on ozonolysis A,gives equimolar quantities of propanone and ethanol.Deduce the structure of substance A.

The ozonolysis of alkenes is an important reaction in organic chemistry, particularly useful for breaking double bonds to produce carbonyl compounds. This reaction involves the treatment of alkenes with ozone (O3), leading to the formation of ozonides, which can be further hydrolyzed to yield aldehydes, ketones, or carboxylic acids. Let's break this down step-by-step.

Understanding Ozonolysis

During ozonolysis, the alkene is reacted with ozone in a two-step process:

• Formation of Ozonide: The alkene reacts with ozone to form a cyclic intermediate called an ozonide. This step typically occurs at low temperatures to minimize side reactions.
• Hydrolysis: The ozonide is then treated with a reducing agent, often water or zinc, which leads to the formation of carbonyl compounds (aldehydes or ketones) as the ozonide is cleaved.

Example of Ozonolysis

For instance, if we take 1-butene as our alkene, ozonolysis will yield acetaldehyde and butan-2-one upon hydrolysis. This illustrates how the process effectively cleaves the double bond and transforms it into carbonyl functionalities.

Analyzing the Given Hydrocarbon

Now, let’s analyze the second part of your question regarding substance A. We know that one mole of hydrocarbon A reacts with one mole of bromine to yield a dibromo compound (C5H10Br2). After treatment with cold dilute KMnO4, it forms C5H12O2, indicating the presence of alcohols or carbonyl compounds in the product.

Deduction of Structure for Substance A

Given that ozonolysis of substance A yields equimolar quantities of propanone and ethanol, we can deduce some important structural features:

• Propanone (acetone) has the structure CH3COCH3, indicating that one of the products is a ketone.
• Ethanol (CH3CH2OH) suggests that there is an alcohol group present in the product.

To form these products upon ozonolysis, substance A must have a double bond that allows for the formation of both a ketone and an alcohol. Considering the molecular formula of A (C5H10), we can contemplate a structure like 4-pentene:

• The double bond between the 4th and 5th carbon atoms allows for the ozonolysis to produce the required carbonyl and alcohol products.

Therefore, the structure of substance A can be represented as:

• CH3-CH2-CH=CH-CH3 (4-pentene)

After ozonolysis, this molecule can yield propanone and ethanol, fitting the criteria provided in the question. The production of C5H12O2 upon oxidation with KMnO4 further supports this hypothesis, as it would generate alcohols or ketones consistent with the expected transformations.

In summary, the ozonolysis of alkenes is a powerful method for synthesizing carbonyl compounds, and through logical deductions based on the provided reaction data, we can infer that substance A is likely 4-pentene, leading to the detailed products upon ozonolysis.

1)The first question has a lengthy mechanism so Id recommend you to read it in some standard the textbook.. Basically,the final products after performing ozonolysis are 1)aldehydes and ketones (in O3+Zn+water medium) 2) carboxylic acids and ketones( in O3+water medium) 3) alcohols (in O3 + LiAlH4 medium)

2) I guess A is

The final products after test 3 are propanol and enthanal...not ethanol!