Question

\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{+}, \Delta\) is a strong oxidising agent which can oxidise \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) into \(\mathrm{CH}_{3}-\mathrm{COOH}\) but still \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) is obtained by oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}_{3}-\mathrm{OH}\) by using distillation flask at temperature \(55^{\circ} \mathrm{C}_{\text {with }}\) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{H}^{+}, \Delta\). It is possible because (A) Acetaldehyde distills out at this temperature and escape further oxidation (B) At this temperature, acetaldehyde is very stable and escape further oxidation (C) Oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) is slower than oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) (D) None

Short answer

(A) Acetaldehyde distills out at this temperature and escapes further oxidation.

Step-by-step solution

Option A

Acetaldehyde distills out at this temperature and escapes further oxidation. In this case, as \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) is oxidized, acetaldehyde (\(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\)) is formed. The boiling point of acetaldehyde is around 20.8°C, which is much lower than the given temperature (55°C). Due to its lower boiling point, acetaldehyde will be distilling out at this temperature, preventing further oxidation into \(\mathrm{CH}_{3}-\mathrm{COOH}\). This allows only the formation of acetaldehyde, justifying Option A. Therefore, Option A is plausible.

Option B

At this temperature, acetaldehyde is very stable and escapes further oxidation. The stability of a chemical compound is not directly related to the temperature of the reaction. As the temperature in the given question is 55°C, it does not significantly increase the stability of acetaldehyde. Moreover, the temperature will in fact provide the necessary heat for the oxidation reactions to occur. Hence, Option B is not a correct explanation.

Option C

Oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) is slower than the oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\). This statement implies that, within the given reaction, \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) (ethanol) will oxidize at a slower rate compared to acetaldehyde (\(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\)). However, this is not true. Ethanol is the primary product, and it is easier to oxidize ethanol than acetaldehyde. Consequently, Option C is not a valid explanation.

Option D

None. Considering that we've verified the plausibility of Option A, Option D is automatically ruled out as an answer. In conclusion, the correct option that explains the mentioned observations in the exercise is: (A) Acetaldehyde distills out at this temperature and escapes further oxidation.

Key concepts

Alcohol Oxidation

Alcohol oxidation is a fundamental chemical reaction where alcohols are transformed into aldehydes, ketones, or carboxylic acids. This process is essential in organic chemistry for converting simple alcohol substances into more complex compounds. Typically, primary alcohols like ethanol (\(\mathrm{CH}_3\mathrm{-CH}_2\mathrm{-OH}\)) are oxidized to produce aldehydes or carboxylic acids, depending on the reaction conditions.

• In the presence of a strong oxidizing agent, such as potassium dichromate \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\) and acid (\(\mathrm{H}^+\)), alcohol can be fully oxidized to carboxylic acid (\(\mathrm{CH}_3\mathrm{-COOH}\)).
• However, by controlling the reaction conditions, such as temperature, you can limit the oxidation to form acetaldehyde (\(\mathrm{CH}_3\mathrm{-CH} = \mathrm{O}\)) instead of acetic acid.

This control is achieved by using specific reaction setups like a distillation flask, which helps in removing the product as it forms, thus preventing further oxidation.

Acetaldehyde Formation

Acetaldehyde (\(\mathrm{CH}_3\mathrm{-CHO}\)) is a key molecule formed as an intermediate during the oxidation of ethanol. In the step-by-step solution, we see how acetaldehyde is initially formed from ethanol in the presence of heat and an oxidizing agent.

• The process involves the removal of hydrogen from the alcohol group in ethanol, transforming it into a carbonyl group characteristic of aldehydes.
• This very simple transformation is crucial in many industrial and biological processes where acetaldehyde acts as a precursor to various chemical compounds.

What is fascinating about acetaldehyde is its volatility, which plays a significant role in its behavior during reactions like oxidation, as covered in the exercise.

Boiling Point Effects

Boiling point is the temperature at which a substance turns from a liquid into a vapor. Understanding boiling points is critical in reactions where separation of compounds is required.

• Acetaldehyde has a low boiling point of about 20.8°C, which makes it volatile. In the context of the exercise, maintaining a temperature of 55°C is well above this boiling point, causing acetaldehyde to distill out of the reaction mixture.
• This property is key because it enables the continuous removal of acetaldehyde from the reaction mixture, preventing it from being further oxidized to acetic acid, which has a much higher boiling point.

Thus, controlling the boiling point is a practical way to ensure only the desired oxidation product is formed.

Chemical Stability Analysis

Chemical stability refers to the tendency of a chemical compound to resist decomposition or unwanted reactions. In the oxidation reaction of ethanol to acetaldehyde, stability plays a role in ensuring that acetaldehyde does not undergo further reactions to convert into acetic acid.

• In the step-by-step solution, it's highlighted that the stability of acetaldehyde does not increase significantly due to the reaction temperature.
• However, by distilling acetaldehyde as it forms, you effectively reduce its presence in the reaction mixture, steering clear of conditions that would promote further oxidation.

This approach demonstrates an external control over chemical stability, rather than relying on inherent stability properties to dictate the outcome of the reaction.