A chemist possesses succinic anhydride and wants to make \(\beta\) -alanine ( \(\beta\) -aminopropionic acid). How should the synthesis be performed?

Short Answer

Expert verified
To synthesize β-alanine (β-aminopropionic acid) from succinic anhydride, perform the following two steps: 1. Hydrolyze the anhydride using water: \[ \ce{C2H2(CO)2O + H2O -> HOOC-CH2-COOH} \] 2. Perform reductive amination using ammonia and sodium cyanoborohydride (\(\ce{NaBH3CN}\)): \[ \ce{HOOC-CH2-COOH + NH3 + NaBH3CN -> NH2-CH2-CH2-COOH + NaBH3CN_{spent}} \] This synthesis route will yield β-alanine (β-aminopropionic acid).

Step by step solution

01

Analyze chemical structures

Succinic anhydride has the following structure: \[ \ce{C2H2(CO)2O} \] β-alanine (β-aminopropionic acid) has the following structure: \[ \ce{NH2-CH2-CH2-COOH} \] Now we identify the functional groups and the required reactions to convert succinic anhydride into β-alanine.
02

Identify functional groups and required reactions

Succinic anhydride contains a cyclic anhydride functional group \((\ce{-CO-O-CO-})\) and β-alanine contains an amine \((\ce{-NH2})\) and a carboxylic acid \((\ce{-COOH})\) functional group. The key steps to synthesize β-alanine from succinic anhydride will involve the opening of the anhydride followed by the introduction of an amine group. Let's break down these steps.
03

Opening the anhydride

To open the anhydride, we can hydrolyze it using water. This will break the cyclic anhydride ring to form two carboxylic acid groups. The reaction can be written as follows: \[ \ce{C2H2(CO)2O + H2O -> HOOC-CH2-COOH} \]
04

Introducing an amine group

Now, we have one carboxylic acid group that needs to be converted into an amine group. We can achieve this by performing a reductive amination using ammonia and a reducing agent such as sodium cyanoborohydride (\(\ce{NaBH3CN}\)). The reaction can be written as follows: \[ \ce{HOOC-CH2-COOH + NH3 + NaBH3CN -> NH2-CH2-CH2-COOH + NaBH3CN_{spent}} \]
05

Final synthesis of β-alanine

Putting the steps together, we have the following synthesis route to obtain β-alanine from succinic anhydride: 1. Hydrolysis of the anhydride: \[ \ce{C2H2(CO)2O + H2O -> HOOC-CH2-COOH} \] 2. Reductive amination: \[ \ce{HOOC-CH2-COOH + NH3 + NaBH3CN -> NH2-CH2-CH2-COOH + NaBH3CN_{spent}} \] Following this synthesis route, the chemist can obtain β-alanine (β-aminopropionic acid) from succinic anhydride.

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Most popular questions from this chapter

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.

A chemist added water and acid to an unknown ester of formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\) to produce an acid \(\mathrm{X}\) and an alcohol \(\mathrm{Y}\). Oxidation of \(\mathrm{Y}\) with chromic acid produced \(\mathrm{X}\). Determine the structure of the original ester. Write equations for all reactions that occur.

Indicating the mechanism, show how ethyl propionate may be prepared from propionic acid.

Methyl ethyl ketone can undergo Claisen condensation with a given ester, for example, ethyl benzoate, to yield either of two products, depending upon experimental conditions. What are these two products? How can one tell which product one has obtained?

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.

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