The products of oxidation of 2 -pentanone with concentrated nitric acid are (a) propanoic acid and ethanoic acid (b) butanoic acid and methanoic acid (c) propanoic acid and methanoic acid (d) 2 -methyl propanoic acid and methanoic acid

Short Answer

Expert verified
Answer: The products of oxidation of 2-pentanone with concentrated nitric acid are propanoic acid and ethanoic acid.

Step by step solution

01

Write the structure of 2-pentanone

Firstly, write the structure of 2-pentanone. The structure is as follows: CH3 - CH2 - CO - CH2 - CH3
02

Oxidation of 2-pentanone with concentrated nitric acid

When 2-pentanone reacts with concentrated nitric acid, it undergoes an oxidation reaction by breaking carbon-carbon bonds adjacent to the carbonyl group. The carbonyl carbon (C=O) will oxidize to a carboxylic acid group (-COOH). For 2-pentanone, it will break the bond between the 2nd and 3rd carbon after oxidation. Now, we will have the following products: 1. A 2-carbon chain with a carboxylic acid group (-COOH) attached to the carbonyl carbon. 2. A 3-carbon chain with a carboxylic acid group (-COOH) attached to the carbonyl carbon.
03

Write the IUPAC names of the products

Write the IUPAC names for the obtained compounds in step 2: 1. The 2-carbon compound with a carboxylic acid group is called ethanoic acid (CH3-COOH). 2. The 3-carbon compound with a carboxylic acid group is called propanoic acid (CH3-CH2-COOH).
04

Match the obtained products with the given choices

Now, let's match our products with the given choices: (a) propanoic acid and ethanoic acid (b) butanoic acid and methanoic acid (c) propanoic acid and methanoic acid (d) 2 -methyl propanoic acid and methanoic acid Our products, propanoic acid and ethanoic acid, match with choice (a). Therefore, the correct answer is (a) propanoic acid and ethanoic acid.

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