Question

Compound A of molecular formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{~N}_{2} \mathrm{O}_{2}\) reacts with aqueous nitrous acid to give compound \(B\) of formula \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\). Compound B readily loses water on heating to give \(\mathrm{C}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{3}\). Compound A can also react with a solution of bromine and sodium hydroxide in water to give \(\mathrm{D}, \mathrm{C}_{4} \mathrm{H}_{12} \mathrm{~N}_{2}\), which on treatment with nitrous acid and perchloric acid gives methyl ethyl ketone. Write structures of all compounds and the equations involved.

Short answer

Compound A is a dipeptide Gly-Ala with the molecular formula \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{N}_2\mathrm{O}_2\), and its structure is: \(\mathrm{H}_{2}\mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NH}-\mathrm{CH}_{3}-\mathrm{COOH}\). A reacts with aqueous nitrous acid to form B, losing two hydrogen atoms and gaining two oxygen atoms: \(\mathrm{H}_{2}\mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NO}-\mathrm{CH}_{3}-\mathrm{COOH}\). B loses water upon heating to form C: \(\mathrm{H}_{2}\mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NO}-\mathrm{CH}_{3}-\mathrm{CO}\). A reacts with bromine and sodium hydroxide to form D: \(\mathrm{H}_{2}\mathrm{N}-\mathrm{CH}_{2}\mathrm{NH}_{2}\) and \(\mathrm{H}_{2}\mathrm{N}-\mathrm{CH}_{3}\). D reacts with nitrous and perchloric acids to produce methyl ethyl ketone: \(\mathrm{CH}_{3}\mathrm{COCH}_{2}\mathrm{CH}_{3}\).

Step-by-step solution

Analyzing the Reactions

First, we need to analyze the given information and understand what happens during each chemical reaction. 1. Compound A \((\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2})\) reacts with aqueous nitrous acid to form compound B \((\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4})\). 2. Compound B loses water on heating to form compound C \((\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{3})\). 3. Compound A also reacts with a solution of bromine and sodium hydroxide in water to form compound D \((\mathrm{C}_{4} \mathrm{H}_{12} \mathrm{N}_{2})\), which upon treatment with nitrous acid and perchloric acid gives methyl ethyl ketone. Let's move on to identifying the structures of each compound, starting with compound A.

Identifying the Structure of Compound A

The molecular formula of compound A is \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\). The nitrogen and oxygen atoms in the molecular formula suggest that it could be a dipeptide. This is supported by the fact that compound A reacts with bromine and sodium hydroxide, which is the reaction to determine amino acid content in protein analysis. The reaction between nitrous and perchloric acids suggests that the amino groups produce methyl ethyl ketone, so we can identify the amino acids as glycine (Gly) and alanine (Ala). Thus, the structure of compound A can be represented as Gly-Ala: \[ \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NH}-\mathrm{CH}_{3}-\mathrm{COOH} \]

Writing the Reaction Equation for Compound B

Compound A reacts with aqueous nitrous acid, which causes the loss of two hydrogen atoms and gaining two oxygen atoms, creating compound B with the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\). The reaction can be written as: \[ \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NH}-\mathrm{CH}_{3}-\mathrm{COOH} + 2\mathrm{HNO}_2 \rightarrow \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NO}-\mathrm{CH}_{3}-\mathrm{COOH} + 2\mathrm{H}_2\mathrm{O} \]

Writing the Reaction Equation for Compound C

Compound B loses water upon heating, which causes the molecular formula to change to \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{3}\), forming compound C. The reaction can be written as: \[ \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NO}-\mathrm{CH}_{3}-\mathrm{COOH} \xrightarrow{\triangle} \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NO}-\mathrm{CH}_{3}-\mathrm{CO} + \mathrm{H}_{2}\mathrm{O} \]

Writing the Reaction Equation for Compound D

Compound A reacts with a solution of bromine and sodium hydroxide in water to form compound D \((\mathrm{C}_{4} \mathrm{H}_{12} \mathrm{N}_{2})\). The reaction can be written as: \[ \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{CO}-\mathrm{NH}-\mathrm{CH}_{3}-\mathrm{COOH} + 2\mathrm{NaBr} + 2\mathrm{NaOH} \rightarrow \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{NH}_{2} + \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{3} + 2\mathrm{Na}_{2}\mathrm{CO}_{3} + 2\mathrm{H}_{2}\mathrm{O} \]

Writing the Reaction Equation for Methyl Ethyl Ketone formation

Compound D, upon treatment with nitrous acid and perchloric acid, results in the formation of methyl ethyl ketone. The reaction can be written as: \[ \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}\mathrm{NH}_{2} + \mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{3} + 2\mathrm{HNO}_2 + 2\mathrm{HClO}_4 \rightarrow \mathrm{CH}_{3}\mathrm{COCH}_{2}\mathrm{CH}_{3} + 2\mathrm{N}_{2} + 4\mathrm{H}_{2}\mathrm{O} \]

Key concepts

Molecular Formula Analysis

Understanding molecular formulas is a crucial step in organic chemistry as it provides insights into the types and numbers of atoms in a compound. For instance, with Compound A having the formula \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{N}_2 \mathrm{O}_2 \), we can determine it contains 6 carbon atoms, 12 hydrogen atoms, 2 nitrogen atoms, and 2 oxygen atoms. This combination suggests a complex structure, probably indicative of organic molecules like peptides.

By comparing molecular formulas before and after a reaction, students can identify changes such as the loss or addition of atoms, like the transformation from Compound A to Compound B, where nitrous acid reactions are involved. This change implies processes like oxidation or reduction have occurred, which are key in reconstructing the structural identities of unknown compounds.

Dipeptide Structure

A dipeptide is a molecule composed of two amino acids linked by a peptide bond. In the given problem, Compound A is identified by its formula \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{N}_2 \mathrm{O}_2 \), hinting at a dipeptide structure. Amino acids like glycine and alanine form simple dipeptides.

The peptide bond is formed by a condensation reaction where the carboxyl group of one amino acid (Glycine) joins the amino group of the other (Alanine), releasing a molecule of water. The resulting structure has an amide linkage:

• \( \mathrm{H}_2 \mathrm{N} - \mathrm{CH}_2\mathrm{CO} - \mathrm{NH} - \mathrm{CH}_3 \mathrm{COOH} \).

Identifying dipeptides is important in protein chemistry, as these are foundational building blocks of proteins.

Nitrous Acid Reaction

The reaction with nitrous acid is a staple in organic chemistry, especially in the formation and transformation of nitrogen-containing compounds. In this scenario, Compound A reacts with nitrous acid to transform into Compound B, indicating a change in its chemical structure, possibly through the formation of a diazo group.

Typically, reactions with nitrous acid involve the replacement of an amino group with a nitrogen group, which can lead to the formation of alcohols, alkenes, or other functional groups depending on the nature of the reacting compound. Recognizing these changes can help in predicting the structure of resulting compounds (like Compound B), which can further undergo dehydrogenation reactions upon heating.

Ketone Formation

Ketone formation is an essential mechanism in organic reaction pathways, particularly in the context of molecular synthesis. In this problem, the transformation of a dipeptide to form an intermediate compound (Compound D), which further reacts to form methyl ethyl ketone, showcases the complex reaction sequences characteristic of organic chemistry.

Ketones, characterized by a carbonyl group bonded to two alkyl groups, serve various roles in biology and industry as solvents and precursors. In the problem's context, knowing how the initial compound converts into an intermediary before eventually forming a ketone is crucial. This knowledge helps in understanding synthesis pathways, structural changes, and underlying chemistry. The reaction, here, utilizes nitrous acid and perchloric acid, leading to deamination and eventual formation of the ketone structure known as methyl ethyl ketone \( \mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_3 \).