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1-propanol and 2 -propanol can be best distinguished by (a) oxidation with \(\mathrm{KMnO}_{4}\) followed by reaction with Fehling solution (b) oxidation with acidic dichromate followed by reaction with Fehling solution (c) oxidation by heating with copper followed by reaction with Fehling solution (d) oxidation with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) followed by reaction

Short Answer

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Option (c) oxidation by heating with copper distinguishes them.

Step by step solution

01

Understand the Alcohol Types

1-Propanol is a primary alcohol, while 2-propanol is a secondary alcohol. The distinction between these types is crucial for determining how they react with oxidizing agents. Primary alcohols can be oxidized to aldehydes and further to carboxylic acids, whereas secondary alcohols can be oxidized to ketones.
02

Consider the Oxidizing Agent Reactions

Fehling’s solution reacts with aldehydes to form a red precipitate of cuprous oxide. It does not react with ketones. This means, for the Fehling’s test to differentiate between 1-propanol and 2-propanol, the 1-propanol must first be oxidized to an aldehyde.
03

Analyze Each Option

- Option (a): KMnO₄ is a strong oxidizing agent that will convert 1-propanol to a carboxylic acid, not just an aldehyde. - Option (b): Acidic dichromate is also a strong oxidizing agent, similar to KMnO₄, and will take primary alcohols to carboxylic acids. - Option (c): Heating with copper can gently oxidize primary alcohols to aldehydes, suitable for Fehling's test. - Option (d): H₂SO₄ is not typically used as an oxidizing agent alone in such a reaction.
04

Choose the Appropriate Method

Heating with copper for option (c) allows the primary alcohol (1-propanol) to be oxidized to an aldehyde, which will react with Fehling’s solution. The secondary alcohol (2-propanol) will be converted to a ketone, which does not react with Fehling's solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Primary Alcohols
Primary alcohols are a type of alcohol where the hydroxyl group (-OH) is attached to a carbon atom that is only connected to one other carbon atom. This makes the carbon attached to the -OH group end the carbon chain. An example is 1-propanol, where the hydroxyl group is on the first carbon.

Primary alcohols can be oxidized in a two-step process to form aldehydes and carboxylic acids. First, the primary alcohol is oxidized to an aldehyde. With further oxidation, the aldehyde can be turned into a carboxylic acid.

During oxidation, an alcohol loses electrons, often by gaining an oxygen or losing hydrogen. This concept is crucial when considering reactions with different oxidizing agents since the degree of oxidation affects the final product of the reaction.
Secondary Alcohols
Secondary alcohols have their hydroxyl group (-OH) attached to a carbon atom that is connected to two other carbon atoms. This structure creates more branching compared to primary alcohols. An example is 2-propanol, often known as isopropanol.

When oxidized, secondary alcohols form ketones, which do not undergo further oxidation to carboxylic acids. This is because the carbon atom carrying the -OH group does not have an additional hydrogen atom required for further oxidation beyond a ketone.

Unlike primary alcohols that can become more acidic through further oxidation, secondary alcohols reach their terminus at the ketone stage. This is useful in distinguishing them from primary alcohols, especially in tests that target aldehydes or carboxylic acids, such as the Fehling’s test.
Oxidizing Agents
Oxidizing agents play a fundamental role in converting alcohols to their oxidized forms. Different oxidizing agents can achieve varying levels of oxidation depending on their strength.

  • Potassium permanganate (\( \mathrm{KMnO}_{4} \)) is a powerful oxidizing agent, turning primary alcohols into carboxylic acids directly.
  • Acidic dichromate is another strong oxidizer, also creating carboxylic acids from primary alcohols.
  • Copper, when used at elevated temperatures with primary alcohols, gently oxidizes them to aldehydes, perfect for attempts to identify aldehydes using the Fehling’s test.


It's important to choose the right oxidizing agent for the desired oxidation product whether an aldehyde, ketone, or carboxylic acid is needed. This is especially important when distinguishing between primary and secondary alcohols.
Fehling's Test
Fehling's test is a chemical reaction specifically for detecting aldehyde functional groups. This classical method uses Fehling’s solution, consisting of copper(II) sulfate and potassium sodium tartrate in an alkaline solution.

The test works as follows: if an aldehyde is present, the solution will react, resulting in the reduction of blue copper(II) ions to a red cuprous oxide precipitate. This change is visible, making it a useful qualitative test for aldehydes.

However, this test is not applicable for ketones, such as those derived from secondary alcohol oxidation, since ketones do not reduce the copper(II) ions. This characteristic is what makes Fehling's test useful in distinguishing between primary and secondary alcohols, where primary alcohols (through their aldehyde intermediates) will give a positive result, while secondary alcohols will not.

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Most popular questions from this chapter

The ionization constant of phenol is higher than that of ethanol because (a) phenoxide ion is a stronger base than ethoxide ion (b) phenoxide ion is stabilized through delocalization electron (c) phenoxide ion is less stable than ethoxide ion (d) phenoxide ion is bulkier than ethoxide ion

In the reaction, \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH} \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\), the reagent used would include (a) \(\mathrm{Hg}^{2+} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (b) Na in liquid \(\mathrm{NH}_{3}\), followed by \(\mathrm{CH}_{3} \mathrm{Cl}\) (c) \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} / \mathrm{Pt}\), followed by \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{H}_{2} \mathrm{O}\) (d) \(2 \mathrm{~mol}\) of \(\mathrm{H}_{2} / \mathrm{Pt}\), followed by aq. \(\mathrm{NaOH}\)

The molecule which the highest boiling point is (a) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) (d) \(\mathrm{CH}_{3}-\mathrm{CHOH}-\mathrm{CH}_{3}\)

Which of the following is/are correct? (a) Phenol gives paraquinol with \(\mathrm{S}_{2} \mathrm{O}_{8}^{-2} / \mathrm{OH}^{-}\)as a major product. (b) Phenol can give electrophilic aromatic substitution reactions in absence of lewis acid. (c) Salicylic acid can give white/yellow solution with \(\mathrm{Br}_{2} / \mathrm{CS}_{2}\) (d) Ipso substitution is possible in phenol derivative if it contains \(-\mathrm{COOH}\) or \(-\mathrm{SO}_{3} \mathrm{H}\) group at ortho or para position with respect to OH group.

\(\mathrm{CH}_{2}=\mathrm{CH}_{2} \frac{\mathrm{X}}{\mathrm{Ag}} \rightarrow \mathrm{X} \frac{\text { steam }}{473 \mathrm{~K}}-\mathrm{Y}\) The compound \(\mathrm{Y}\) is (a) ethylene glycol (b) epoxyethane (c) ethanal (d) ethanol

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