Question

The most useful reagent to distinguish ethanol from propan-1-ol is (a) \(\mathrm{HCl}\) and \(\mathrm{ZnCl}_{2}\) (b) \(\mathrm{I}_{2}\) and \(\mathrm{NaOH}\) (c) Cu at \(570 \mathrm{~K}\) (d) \(\mathrm{KMnO}_{4}\)

Short answer

Answer: The most useful reagent to distinguish ethanol from propan-1-ol is reagent (b), the iodine-forming reagent with \(\mathrm{I}_{2}\) and \(\mathrm{NaOH}\).

Step-by-step solution

Reagent (a): \(\mathrm{HCl}\) and \(\mathrm{ZnCl}_{2}\)

The combination of hydrochloric acid (\(\mathrm{HCl}\)) and zinc chloride (\(\mathrm{ZnCl}_{2}\)) is known as the Lucas reagent. This reagent is used to distinguish between primary, secondary, and tertiary alcohols. Both ethanol and propan-1-ol are primary alcohols, which means that the Lucas reagent will not be helpful in distinguishing them.

Reagent (b): \(\mathrm{I}_{2}\) and \(\mathrm{NaOH}\)

Iodine (\(\mathrm{I}_{2}\)) and sodium hydroxide (\(\mathrm{NaOH}\)) combine to form the iodine-forming reagent, which is used to test for the presence of alcohols. The reaction produces a different compound depending on the type of alcohol. For ethanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\)), iodofrom (\(\mathrm{CHI}_{3}\)) is produced whereas for propan-1-ol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\)), the reaction produces iodoalkane (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{I}\)). The iodofrom will appear as a solid yellow precipitate, while the iodoalkane will be a colorless liquid. Thus, reagent (b) is useful for distinguishing between ethanol and propan-1-ol.

Reagent (c): Copper at \(570 \mathrm{~K}\)

When alcohols are heated with copper at \(570 \mathrm{~K}\), they undergo dehydrogenation and form aldehydes or ketones depending on the type of alcohol. However, both ethanol and propan-1-ol will form aldehydes upon dehydrogenation, so this reagent will not be able to differentiate between the two alcohols.

Reagent (d): \(\mathrm{KMnO}_{4}\)

Potassium permanganate (\(\mathrm{KMnO}_{4}\)) is a strong oxidizing agent that can oxidize alcohols to carboxylic acids. However, both ethanol and propan-1-ol will be oxidized to their respective carboxylic acids, making them indistinguishable using this reagent.

Conclusion

The most useful reagent to distinguish ethanol from propan-1-ol is reagent (b), the iodine-forming reagent with \(\mathrm{I}_{2}\) and \(\mathrm{NaOH}\). This reagent causes the formation of different compounds for the two alcohols, making them easily distinguishable.

Key concepts

Lucas Reagent Application

The Lucas reagent, a combination of hydrochloric acid (\textsc{HCl}) and zinc chloride (\textsc{ZnCl}\(_2\)), is particularly useful in differentiating between primary, secondary, and tertiary alcohols through a nucleophilic substitution reaction.

The test traditionally involves adding an alcohol to the Lucas reagent at room temperature and observing the rate at which the alcohol reacts to form an alkyl chloride. Tertiary alcohols react almost immediately, resulting in a cloudy solution due to the formation of an insoluble alkyl chloride. Secondary alcohols usually take a few minutes to form the cloudy mixture, whereas primary alcohols, including ethanol and propan-1-ol, react very slowly, typically requiring elevated temperatures to react. This slow reaction rate indicates that the Lucas test on its own isn't an effective method to distinguish between ethanol and propan-1-ol, as both would not react significantly at room temperature, thus providing similar test results.

Visual Indicators

When an alcohol reacts quickly with the Lucas reagent, the clarity of the solution changes as the alkyl chloride forms. However, the Lucas reagent, with its distinctive behavior with different classes of alcohols, is certainly valuable in categorizing alcohols based on their reactivity, even though it may not serve our specific comparison between ethanol and propan-1-ol.

Iodoform Test for Alcohols

The iodoform test is an analytical reaction used to identify the presence of methyl ketones or secondary alcohols with at least one methyl group attached to the carbon bearing the hydroxyl group. In our context, it's an exceptional test to differentiate between ethanol and propan-1-ol.

In the presence of iodine (\textsc{I}\(_2\)) and sodium hydroxide (\textsc{NaOH}), ethanol (\textsc{CH}\(_3\)CH\(_2\)OH) undergoes halogenation and then deprotonation to produce iodoform (\textsc{CHI}\(_3\)), a yellow crystalline solid with a characteristic odor. This reaction occurs because ethanol has a methyl group adjacent to the hydroxyl group. Propan-1-ol (\textsc{CH}\(_3\)CH\(_2\)CH\(_2\)OH), on the other hand, does not have this arrangement and will not form the same distinct yellow precipitate, effectively allowing us to distinguish the two.

Finding the Yellow Precipitate

When carrying out this test, one would look for a yellow precipitate to form. If such a precipitate appears, iodoform has been created, confirming the presence of an alcohol like ethanol that can undergo this specific reaction. This diagnostic test is both qualitative, visually confirming the presence of the methyl group next to the hydroxyl group, and semi-quantitative, as the amount of precipitate can give a rough indication of the concentration of the functional group in question.

Alcohol Oxidation Reactions

Alcohol oxidation reactions involve the conversion of an alcohol to a ketone, an aldehyde, or a carboxylic acid through the loss of hydrogen atoms. These reactions are critical for differentiating alcohols by examining the products they form upon oxidation.

For primary alcohols like ethanol and propan-1-ol, oxidation can yield aldehydes and further oxidation leads to carboxylic acids. Strong oxidizing agents, such as potassium permanganate (\textsc{KMnO}\(_4\)), are used to carry out these transformations. In the case of ethanol, oxidation yields acetaldehyde, which can be further oxidized to acetic acid. For propan-1-ol, oxidation gives propanal, which can then be converted to propanoic acid.

Oxidation Levels

Understanding the products of these oxidations lets us appreciate that while the end products (carboxylic acids) don't differ markedly for the two alcohols and thus don't serve as good distinguishing features, the immediate products (aldehydes) can be distinct for different primary alcohols. Nevertheless, it's important to note that for differentiation purposes, we usually avoid complete oxidation to the carboxylic acid stage, as this prevents us from identifying unique intermediate products that could be more telling of the original alcohol's structure.