a) (-) - Erythrose, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}\), gives tests with Tollens' reagent and Benedict's solution, and is oxidized by bromine water to an optically active acid, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{5}\). Treatment with acetic anhydride yields \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O}_{7}\). Erythrose consumes three moles of \(\mathrm{HlO}_{4}\), and yields three moles of formic acid and one mole of formaldehyde. Oxidation of erythrose by nitric acid yields an optically inactive compound of formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{6}\) (-) - Threose, an isomer of erythrose, shows similar chemical behavior except that nitric acid oxidation yields an optically active compound of formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{6}\). On the basis of this evidence what structure or structures are possible for (-) - erythrose? For (-) -threose? (b) When \(\mathrm{R}-\) glyceraldehyde, \(\mathrm{CH}_{2} \mathrm{OHCHOHCHO}\), is treated with cyanide and the resulting product is hydrolyzed, two monocarboxylic acids are formed. These acids are identical with the acids obtained by oxidation with bromine water of (-) - threose and (-) - erythrose. These acids are identical with the acids obtained by oxidation with bromine water of \((-)\) - threose and \((-)\) - erythrose. Assign a single structure to (-) -erythrose and to \((-)-\) threose.

Step by step solution

01

Analyzing the given reactions

Analyze the given reactions and properties of Erythrose and Threose. This will help us get the structural information of these molecules. Erythrose (C4H8O4) - Tollens' reagent test and Benedict's solution test - Oxidized by bromine water to an optically active acid (C4H8O5) - Treatment with acetic anhydride yields (C10H14O7) - Consumes three moles of HIO4, and yields three moles of formic acid and one mole of formaldehyde - Oxidation by nitric acid yields optically inactive compound (C4H6O6) Threose (C4H8O4, isomer of Erythrose) - Similar chemical behavior as Erythrose - Oxidation by nitric acid yields an optically active compound (C4H6O6)

02

Determining possible structures

Using the information from the reactions: - Since erythrose can be oxidized to an optically active acid by bromine water, it must have at least one chiral carbon. - The reaction with HIO4 shows erythrose contains 3 vicinal diols (-OH groups on adjacent carbons) resulting in the formation of 3 moles of formic acid and 1 mole of formaldehyde. Considering these properties, the possible structure of erythrose can be: \( \mathrm{CH}_{2} \mathrm{OHCHOHCHOHCHO} \) By a similar analysis, the possible structure for threose will also be the same: \( \mathrm{CH}_{2} \mathrm{OHCHOHCHOHCHO} \) ##Part B## R-glyceraldehyde reacts with cyanide to form a product that, upon hydrolysis, forms two monocarboxylic acids identical to the products obtained by oxidation of Erythrose and Threose with bromine water.

03

Assigning a single structure to Erythrose and Threose

We have found out that both erythrose and threose possibly share the same structure \( \mathrm{CH}_{2} \mathrm{OHCHOHCHOHCHO} \). As products of R-glyceraldehyde-cyanide reaction are said to be identical with the monocarboxylic acids obtained upon oxidation of erythrose and threose (both possess bromine water oxidation products), we can single out the structures as follows: Erythrose structure: \( \mathrm{CH}_{2} \mathrm{OHCHOHCHOHCHO} \) Threose structure: \( \mathrm{CH}_{2} \mathrm{OHCHOHCHOHCHO} \) Since erythrose and threose are isomers and have same molecular formula, their difference must be position/internal arrangement of atoms, here chiral carbons are different. Their structures will be: Erythrose structure: \( \mathrm{HOCH}_{2} \mathrm{CHOHCHOHCHO} \) Threose structure: \( \mathrm{HOCH}_{2} \mathrm{CHOHCHOHCH}_{2} \mathrm{OH} \)

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