Question

Identify the incorrect statement among the following. (a) Both pyruvic acid and lactic acid answer the haloform test (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) can be reduced to \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) using red \(\mathrm{P}\) and \(\mathrm{HI}\) (c) Kolbe's electrolysis of potassium succinate gives ethylene and \(\mathrm{CO}_{2}\) (d) Strong heating of ammonium acetate gives methylcyanide

Short answer

a) Both pyruvic acid and lactic acid answer the haloform test. b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) can be reduced to \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) using red \(\mathrm{P}\) and \(\mathrm{HI}\). c) Kolbe's electrolysis of potassium succinate gives ethylene and \(\mathrm{CO}_{2}\). d) Strong heating of ammonium acetate gives methylcyanide. Answer: a) Both pyruvic acid and lactic acid answer the haloform test.

Step-by-step solution

Statement A

Haloform test is generally given by compounds containing a \(\ce{C-CH3-CO-}\) group, such as methyl ketones. Neither pyruvic acid nor lactic acid contains this group. So, this statement is incorrect.

Statement B

Ethanoic acid (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\)) can be reduced to ethane (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\)) using red phosphorus and hydrogen iodide. This reaction is a reduction process where the carboxyl group is replaced with a hydrogen atom which results in a complete reduction of the ethanoic acid to ethane. So, this statement is correct.

Statement C

Kolbe's electrolysis is a process where a carboxylate salt undergoes electrolysis to form a radical, which further decomposes to give alkane radicals, alkene, and carbon dioxide. Potassium succinate, when subjected to Kolbe's electrolysis, would give ethylene (\(\ce{C2H4}\)) and carbon dioxide (\(\ce{CO2}\)) as products. So, this statement is correct.

Statement D

When ammonium acetate is subjected to strong heating, it undergoes thermal decomposition to produce methylcyanide (\(\ce{CH3CN}\)) as the major product. This reaction proceeds via the formation of an amide as an intermediate. So, this statement is correct. Based on the above analysis, the incorrect statement is (a) Both pyruvic acid and lactic acid answer the haloform test.

Key concepts

Carboxylic Acids

Carboxylic acids are fascinating because they contain a functional group known as the carboxyl group. This group is written as \(- ext{COOH}\) and includes both a carbonyl (\(- ext{C=O}\)) and a hydroxyl group (\(- ext{OH}\)).
Natural carboxylic acids like acetic acid are found in vinegar, and they have interesting chemical properties.

• They can act as acids because they release hydrogen ions (H⁺) in solutions.
• Carboxylic acids can form hydrogen bonds, leading to relatively high boiling points.
• They are known for their acidic smell and sour taste.

In reactions, carboxylic acids can be reduced to alcohols but can also participate in various elimination and substitution reactions. This conversion is particularly significant in organic chemistry and forms the basis for broader applications and transformations. Understanding this concept is essential for performing reduction reactions, another core topic discussed here.

Haloform Reaction

The haloform reaction is an interesting chemical reaction where a methyl ketone is transformed into a haloform. The reaction requires a compound with a \(- ext{C-CH}_3- ext{CO}-\) grouping. Interestingly, neither pyruvic acid nor lactic acid has this structural requirement, and this explains why they do not undergo the haloform reaction.
The haloform reaction proceeds in the following steps:

• Halogenation: The hydrogen on the methyl group is replaced by a halogen atom.
• Formation of Trihalomethyl Ketone: All hydrogens are replaced by halogens, resulting in trihalomethyl ketones.
• Cleavage: The trihalomethyl ketone reacts with a base, leading to the formation of a haloform and a carboxylate ion.

This reaction is historically significant in detecting methyl ketones through the production of haloforms such as chloroform or bromoform, which are easily recognizable by their distinctive odors and properties.

Kolbe Electrolysis

Kolbe electrolysis is a nifty process utilized in the electrochemical decarboxylation of carboxylic acids. Essentially, it is a way to modify carboxylate salts via electrolysis. The process generates alkane radicals that are further transformed into alkenes and carbon dioxide.
When potassium succinate is subjected to Kolbe electrolysis, it decomposes to yield ethylene and carbon dioxide. The overall process involves the following steps:

• Formation of Carboxylate Radical: The electrolysis of the carboxylate salt forms radicals at the electrode.
• Decarboxylation: These radicals release carbon dioxide, forming alkane radicals.
• Coupling of Radicals: The alkane radicals couple to form alkanes or further rearrange to form alkenes.

This method is utilized in the laboratory to produce alkanes and alkenes efficiently, providing a sustainable approach to these transformations.

Reduction Reactions

Reduction reactions are fundamental in organic chemistry, describing how molecules gain electrons or hydrogen atoms. An organic compound is said to be reduced if it loses oxygen or gains hydrogen. Carboxylic acids, for instance, can be reduced to alcohols through reduction reactions.
In the context of the problem, ethanoic acid (acetic acid) can be fully reduced to ethane by using red phosphorus and hydrogen iodide, illustrating this reduction. The reduction process entails:

• Reduction of the Carbonyl Group: The carbonyl group of the carboxylic acid receives electrons or hydrogen atoms.
• Formation of Alcohol: Initially, this leads to the formation of an aldehyde or alcohol.
• Further Reduction: The alcohol might be reduced further to an alkane completely depending on the reagents used.

Understanding these mechanisms allows chemists to control and guide how molecules are transformed, essential for creating desired products in both research and industry.