Question

Which one of the following reactions will not yield propionic acid ?

• A. CH\(_3\)CH\(_2\)CH\(_3\) + KMnO\(_4\)(Heat), OH\(^-\)/H\(_3\)O\(^+\)
• B. CH\(_3\)CH\(_2\)CCl\(_3\) + OH\(^-\)/H\(_3\)O\(^+\)
• C. CH\(_3\)CH\(_2\)COCH\(_3\) + OI\(^-\)/H\(_3\)O\(^+\)
• D. CH\(_3\)CH\(_2\)CH\(_2\)Br + Mg, CO\(_2\) dry ether/H\(_3\)O\(^+\)

Hint:

Remember the standard methods for synthesizing carboxylic acids: oxidation of primary alcohols and aldehydes, hydrolysis of nitriles or esters, hydrolysis of gem-trihalides, Grignard carboxylation, and the haloform reaction for methyl ketones. Knowing the carbon count change in each reaction is crucial (e.g., Grignard carboxylation adds one carbon).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to analyze four different reactions and determine which one will not produce propionic acid (CH\(_3\)CH\(_2\)COOH) as the major product.
Step 2: Detailed Explanation:
- (A) CH\(_3\)CH\(_2\)CH\(_3\) + KMnO\(_4\)(Heat), OH\(^-\)/H\(_3\)O\(^+\): This represents the vigorous oxidation of propane. Simple alkanes are very resistant to oxidation. Under such harsh conditions, C-C bond cleavage occurs, leading to a mixture of smaller carboxylic acids (like acetic and formic acid), ketones, and complete combustion to CO\(_2\) and H\(_2\)O. This is not a synthetic method to produce propionic acid. Therefore, this reaction will not yield propionic acid.
- (B) CH\(_3\)CH\(_2\)CCl\(_3\) + OH\(^-\)/H\(_3\)O\(^+\): This is the alkaline hydrolysis of a gem-trihalide (1,1,1-trichloropropane). The three chlorine atoms on the terminal carbon are replaced by hydroxyl groups to form an unstable gem-triol, CH\(_3\)CH\(_2\)C(OH)\(_3\). This intermediate readily loses a water molecule to form propionic acid. The final acidification step ensures the product is in its acidic form. This reaction yields propionic acid.
- (C) CH\(_3\)CH\(_2\)COCH\(_3\) + OI\(^-\)/H\(_3\)O\(^+\): This is the haloform reaction. The substrate, butan-2-one, is a methyl ketone (a compound with a CH\(_3\)CO- group). Methyl ketones react with hypoiodite (OI\(^-\)) to form iodoform (CHI\(_3\)) and the salt of a carboxylic acid with one less carbon atom. In this case, butan-2-one will give iodoform and sodium propionate (CH\(_3\)CH\(_2\)COO\(^-\)Na\(^+\)). Subsequent acidification yields propionic acid. This reaction yields propionic acid.
- (D) CH\(_3\)CH\(_2\)CH\(_2\)Br + Mg, CO\(_2\) dry ether/H\(_3\)O\(^+\): This is a Grignard synthesis of a carboxylic acid. Propyl bromide (a 3-carbon compound) first forms propylmagnesium bromide (CH\(_3\)CH\(_2\)CH\(_2\)MgBr). This Grignard reagent then reacts with carbon dioxide (CO\(_2\)) to add a carboxyl group. The final product after hydrolysis is butyric acid (CH\(_3\)CH\(_2\)CH\(_2\)COOH), which is a 4-carbon acid. This reaction does not yield propionic acid.
Step 3: Final Answer:
The oxidation of propane with KMnO\(_4\) will not yield propionic acid.