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Compound A is chiral and is a liquid with the formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{O}_{2} \mathrm{~N}\). A is insoluble in water and dilute acid but dissolves in sodium hydroxide solution. Acidification of a sodium hydroxide solution of chiral A gives racemic A. Reduction of chiral A with hydrogen over nickel produces chiral compound B of formula \(\mathrm{C}_{5} \mathrm{H}_{13} \mathrm{~N}\). Treatment of chiral B with nitrous acid gives a mixture containing some chiral alcohol \(\mathrm{C}\) and some 2 -methyl-2-butanol. Write structures for compounds \(\mathrm{A}, \mathrm{B}\), and C that agree with all the given facts. Write balanced equations for all the reactions involved. Show your reasoning. In this type of problem, one should work backward from the structures of the final products, analyzing each reaction for the structural information it gives. The key questions to be inferred in the preceding problem are (a) What kind of chiral compound or compounds could give 2-methyl-2-butanol and a chiral alcohol with nitrous acid? (b) What kinds of compounds could give \(\mathrm{B}\) on reduction? (c) What does the solubility behavior of A indicate about the type of compound that it is? (d) Why does chiral A racemize when dissolved in alkali?

Short Answer

Expert verified
Compound A is an N-acyl derivative, B is N-methyl-2-butanamine, and C is a chiral alcohol.

Step by step solution

01

Analyze the Solubility of Compound A

Compound A is insoluble in water and dilute acid but dissolves in sodium hydroxide. This indicates that A is likely a carboxylic acid or a phenol because these functional groups typically become soluble salts in basic solutions like sodium hydroxide. Given the formula \(\mathrm{C}_5\mathrm{H}_{11}\mathrm{O}_2\mathrm{N}\), it might be an N-substituted amino acid or ester.
02

Determine the Racemization of Compound A

The fact that chiral compound A racemizes (becomes a 50/50 mix of enantiomers) upon acidification of its solution suggests the presence of an enol or a tautomerizable center which can lead to a racemic mixture. This supports the idea that A might possess a labile chiral center adjacent to a carbonyl group, making it susceptible to racemization in basic medium.
03

Examine the Reduction Reaction to Obtain Compound B

The reduction of A with hydrogen over nickel results in the removal or transformation of an oxygen-containing functional group to provide chiral B with the formula \(\mathrm{C}_5\mathrm{H}_{13}\mathrm{N}\). This suggests a reduction of a carbonyl group such as an ester or amide, leading to an amine product. Thus, B might be a chiral amine like a secondary or tertiary amine derived from A.
04

Determine the Products of Treatment with Nitrous Acid

When B is treated with nitrous acid, one product is a chiral alcohol (C), and another is 2-methyl-2-butanol. The presence of 2-methyl-2-butanol suggests B could be a secondary amine. A common reaction is the formation of diazo compounds that rearrange to alcohols, so B could be a primary aromatic amine or secondary amine, suggesting its structure derivation from A involved secondary amine functionalities.
05

Propose the Structures and Reactions

Based on the analysis, we propose: - Compound A is likely an N-acyl derivative like an amino acid with a carboxyl group that reduces to an amine (B) having the structure of a secondary amine. - Compound B, based on reduction product structure, can be N-methyl-2-butanamine (a secondary amine). - Compound C, following reaction with nitrous acid, is predicted to be a stereoisomeric alcohol similar in structure to the derived amine. Balanced equations for reactions are dependent on specific structures of the proposed components in these transformations.

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Key Concepts

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

Racemization
Racemization is a fascinating process where a chiral compound transforms into a racemic mixture - a 50/50 mix of two enantiomers. Imagine it as balance in chaos. This process is often triggered by an external influence, like a change in pH, temperature, or chemical environment. In the context of compound A, racemization occurs when it is acidified after being dissolved in sodium hydroxide. This suggests that compound A must have delicate chiral centers next to reactive groups, capable of switching between different configurations under certain conditions.
  • Compound A has a labile chiral center.
  • This chiral center is sensitive to basic environments due to its neighboring group, likely a carbonyl group.
  • In a basic medium, this neighboring group could transform, promoting the temporary formation of an enol, which can lead to racemization upon re-equilibration to a racemic product.
Racemization is key in many pharmaceutical and scientific processes and understanding it helps us control the stereochemical properties of a compound.
Nitrous Acid Reactions
Nitrous acid reactions are particularly interesting for their ability to transform amines into alcohols, sometimes with exciting rearrangements. When compound B, which is a secondary amine, is treated with nitrous acid, it leads to the production of 2-methyl-2-butanol and another chiral alcohol.
  • One outcome of nitrous acid reacting with an amine is the formation of diazonium salts. However, for aliphatic amines like those derived from compound A to B, stable diazonium ions are not formed.
  • Instead, these reactions can push rearrangements or simple substitutions, turning the amine into an alcohol.
  • In our case, the generation of 2-methyl-2-butanol indicates a rearrangement or sidenote reaction, showing the side-chain flexibility and adaptability.
This highlights how nitrous acid can initiate profound transformations in organic compounds, giving us valuable tools in synthetic chemistry.
Solubility Behavior
The solubility behavior of a compound can tell us a lot about its structure. When we say that compound A is insoluble in water and dilute acids but dissolves in sodium hydroxide, we're getting critical information about its chemical nature.
  • Insolubility in water and dilute acids suggests that compound A remains neutral under these conditions - indicating the absence of free ionic groups in mild conditions.
  • However, its solubility in sodium hydroxide highlights the presence of acidic protons in the structure. These may come from functional groups like carboxylic acids or phenols, which lose their hydrogen and form water-soluble salts in basic conditions.
  • This behavior helps us pinpoint compound A's potential structure, supporting the idea that it involves an acid-based functional group.
Understanding solubility is crucial in predicting how a compound behaves in different environments - a key consideration in both formulary and pharmacology.
Functional Group Transformations
In organic chemistry, functional group transformations are the cornerstone of synthesis and mechanistic studies. These transformations enable the conversion of simple precursors into complex molecules. Let's break down the transformations in the case of compounds A, B, and C.
  • Beginning with compound A, it exhibits a functional group that allows for reduction by hydrogen over nickel - likely reducing a carbonyl (from an ester or amide) to amine, thereby converting to compound B (a secondary amine).
  • This transformation can be seen as the removal of an oxygen-containing group - a tell-tale sign of successful reduction.
  • Next, compound B undergoes transformation when treated with nitrous acid: its transformation into 2-methyl-2-butanol and a secondary chiral alcohol exemplifies nitrous acid's transformative power.
Each transformation not only changes the functionality but also provides insights into the reactivity and possible rearrangements occurring within the molecule. Mastering these transformations allows chemists to design intricate synthetic routes and understand complex reaction mechanisms.

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Most popular questions from this chapter

Benzenediazonium chloride solvolyzes in water to give a mixture of benzenol and chlorobenzene. Some of the facts known about this and related reactions are 1\. The ratio \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl} / \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) increases markedly with \(\mathrm{Cl}^{\ominus}\) concentration but the rate hardly changes at all. 2\. There is no rearrangement observed with 4-substituted benzenediazonium ions, and when the solvolysis is carried out in \(\mathrm{D}_{2} \mathrm{O}\), instead of \(\mathrm{H}_{2} \mathrm{O}\), no \(\mathrm{C}-\mathrm{D}\) bonds are formed to the benzene ring. 3\. 4-Methoxybenzenediazonium chloride solvolyzes about 30 times faster than 4-nitrobenzenediazonium chloride. 4\. Benzenediazonium salts solvolyze in \(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) at almost the same rate as in \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) and, in these solutions, the effective \(\mathrm{H}_{2} \mathrm{O}\) concentration differs by a factor of 1000 . Show how these observations support an \(S_{\mathrm{N}} 1\) reaction of benzenediazonium chloride, and can be used to argue against a benzyne-type elimination-addition reaction with water acting as the \(E 2\) base (Section \(14-6 \mathrm{C}\) ) or an \(S_{\mathrm{N}} 2\) reaction with water as the nucleophile (Section 8-4, Mechanism B, and Section 14-6).

The point of this exercise is to show that reactions of known stereospecificity can be used to establish configuration at chiral centers. A carboxylic acid of \((+)\) optical rotation was converted to an amide by way of the acyl chloride. The amide in turn was converted to a primary amine of one less carbon atom than the starting carboxylic acid. The primary amine was identified as 2\(S\) -aminobutane. What was the structure and configuration of the \((+)\) -carboxylic acid? Indicate the reagents you would need to carry out each step in the overall sequence \(\mathrm{RCO}_{2} \mathrm{H} \rightarrow \mathrm{RCOCl} \rightarrow \mathrm{RCONH}_{2} \rightarrow \mathrm{RNH}_{2}\).

a. Explain why 1,3-diazacyclopentadiene (imidazole) is a much stronger acid than azacyclopentadiene (pyrrole). b. Would you expect benzenamine to be a stronger or weaker acid than cyclohexanamine? Give your reasoning.

Assess the possibility of O-alkylation in the reaction of \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NH}^{\ominus} \mathrm{Na}^{\oplus} .\) Give your reasoning.

Explain why triphenylamine is a much weaker base than benzenamine and why its electronic absorption spectrum is shifted to longer wavelengths compared with the spectrum of benzenamine. Would you expect \(\mathrm{N}\) -phenylcarbazole to be a stronger, or weaker, base than triphenylamine? Explain.

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