The molecular formula of second homologue in the homologous series of monocarboxylic acid is
- \( \text{C}_2\text{H}_2\text{O}_2 \)
- \( \text{CH}_2\text{O} \)
- \( \text{C}_2\text{H}_4\text{O}_2 \)
- \( \text{C}_3\text{H}_6\text{O}_2 \)
The Correct Option is A
Approach Solution - 1
To determine the second homologue in the homologous series of monocarboxylic acids, it's essential to understand the general formula for monocarboxylic acids. The general formula for a monocarboxylic acid is \(\text{C}_n\text{H}_{2n+1}\text{COOH}\) where \(n\) is a positive integer.
Monocarboxylic acids, also known as aliphatic carboxylic acids, consist of a carboxyl group (\(\text{COOH}\)) attached to an alkyl chain. The series starts with formic acid (HCOOH) where \(n = 0\).
To find the second homologue:
- Formic acid (first compound in this series): \(\text{HCOOH}\) (equivalently written as \(\text{CH}_2\text{O}_2\)) corresponds to \(n = 0\).
- Acetic acid (second compound in the series), substituting \(n = 1\) in the formula: \(\text{C}_1\text{H}_{2(1)+1}\text{COOH} = \text{C}_2\text{H}_4\text{O}_2\).
Thus, the correct molecular formula of the second homologue in the homologous series of monocarboxylic acids is \(\text{C}_2\text{H}_4\text{O}_2\).
Correction: The initial provided correct answer seems to have been misunderstood as \(\text{C}_2\text{H}_2\text{O}_2\). Upon re-evaluation, using the formula explained above, \(\text{C}_2\text{H}_4\text{O}_2\) is indeed the correct formula for acetic acid, the second homologue.
- Option 1 (\(\text{C}_2\text{H}_2\text{O}_2\)) does not fit the general formula for monocarboxylic acids.
- Option 2 (\(\text{CH}_2\text{O}\)) lacks the carboxyl group.
- Option 3 (\(\text{C}_2\text{H}_4\text{O}_2\)) is the correct molecular formula for acetic acid.
- Option 4 (\(\text{C}_3\text{H}_6\text{O}_2\)) corresponds to propionic acid, the third homologue.
Approach Solution -2
The first member of the homologous series of mono carboxylic acids is formic acid (\(\text{HCOOH}\)).
The second member is acetic acid (\(\text{CH}_3\text{COOH}\)).
The molecular formula for acetic acid is \(\text{C}_2\text{H}_4\text{O}_2\).
Hence, the correct answer is \(\text{C}_2\text{H}_4\text{O}_2\).
The correct option is (A) : \( \text{C}_2\text{H}_2\text{O}_2 \)
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Concepts Used:
Preparation - Aldehydes, Ketones and Carboxylic Acids
Aldehydes, ketones, and carboxylic acids are organic compounds that can be prepared by a variety of methods.
Preparation of Aldehydes:
Aldehydes can be prepared by the oxidation of primary alcohols using mild oxidizing agents such as pyridinium chlorochromate (PCC) or by the ozonolysis of alkenes followed by reductive workup. Another method involves the dehydrogenation of alcohols in the presence of a dehydrogenating agent like pyridinium chlorochromate (PCC).
Preparation of Ketones:
Ketones can be prepared by the oxidation of secondary alcohols using oxidizing agents such as potassium permanganate (KMnO4), chromium trioxide (CrO3), or by the ozonolysis of alkenes followed by oxidative workup. Another method involves the Friedel-Crafts acylation of aromatic compounds.
Preparation of Carboxylic Acids:
Carboxylic acids can be prepared by the oxidation of primary alcohols or aldehydes using strong oxidizing agents such as potassium permanganate (KMnO4) or chromic acid (H2CrO4). Another method involves the hydrolysis of nitriles, which produces carboxylic acids and ammonia. Additionally, carboxylic acids can be prepared from Grignard reagents by reacting with carbon dioxide followed by acidic hydrolysis.
In summary, aldehydes, ketones, and carboxylic acids can be prepared by various methods, including oxidation, ozonolysis, and hydrolysis. Understanding the preparation methods of these compounds is important for predicting their behavior and understanding their potential applications in various fields, including chemistry, biology, and industry.