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Chapter 17: Problem 28

Draw a structure for each of the following molecules: (a) 2 -aminopropanoic acid (alanine) (b) 2-hydroxypropanoic acid (lactic acid) (c) \((Z)-2\)-methyl-2-butenoic acid (d) 2 -hydroxy \(-1,2,3\)-propanetricarboxylic acid (citric acid) (e) 2-hydroxybenzoic acid (o-hydroxybenzoic acid, or salicylic acid) (f) 2 -oxopropanoic acid (pyruvic acid)

Short Answer

Expert verified
Alanine: CH3-CH(NH2)-COOH, Lactic acid: CH3-CH(OH)-COOH, (Z)-2-methyl-2-butenoic acid: CH3-C(CH3)=CH-COOH, Citric acid: HO-C-(COOH)-CH2-COOH-CH2-COOH, Salicylic acid: OH-C6H4-COOH, Pyruvic acid: CH3-CO-COOH

Step by step solution

01

Structure of 2-aminopropanoic acid (alanine)

For alanine, start with the backbone of the molecule which is a propanoic acid. Then, attach an amino group (-NH2) to the second carbon. The structure is: CH3-CH(NH2)-COOH
02

Structure of 2-hydroxypropanoic acid (lactic acid)

Lactic acid has a propanoic acid backbone with a hydroxyl group (-OH) attached to the second carbon. The structure is: CH3-CH(OH)-COOH
03

Structure of (Z)-2-methyl-2-butenoic acid

This molecule has a butenoic acid backbone with a methyl group on the second carbon and the double bond in the Z (cis) configuration. The structure is: CH3-C(CH3)=CH-COOH
04

Structure of 2-hydroxy-1,2,3-propanetricarboxylic acid (citric acid)

Citric acid has a propane backbone with three carboxylic acid groups (-COOH) and one hydroxyl group (-OH). The structure is: HO-C-(COOH)-CH2-COOH-CH2-COOH
05

Structure of 2-hydroxybenzoic acid (salicylic acid)

Salicylic acid is a benzoic acid derivative with an ortho hydroxyl (-OH) group on the benzene ring. The structure is: OH-C6H4-COOH
06

Structure of 2-oxopropanoic acid (pyruvic acid)

Pyruvic acid has a propanoic acid backbone with a keto group (=O) on the second carbon. The structure is: CH3-CO-COOH

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Key Concepts

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

2-aminopropanoic acid
2-aminopropanoic acid, also known as alanine, is an alpha-amino acid used in the biosynthesis of proteins. Its structure consists of three carbon atoms forming a propanoic acid backbone. This backbone has an amino group (-NH2) attached to the second carbon. The general formula for alanine is:
CH3-CH(NH2)-COOH.
To create this structure, follow these steps:
  • Begin with the propanoic acid backbone: CH3-CH2-COOH
  • Replace one hydrogen on the second carbon with an amino group (-NH2).
Now you have the complete structure for alanine!
2-hydroxypropanoic acid
2-hydroxypropanoic acid, commonly known as lactic acid, plays a significant role in various biochemical processes like metabolism. This molecule has a propanoic acid backbone, and a hydroxyl group (-OH) is attached to the second carbon. The structure is given by:
CH3-CH(OH)-COOH.
Here’s a step-by-step way to visualize and draw it:
  • Start with the propanoic acid backbone: CH3-CH2-COOH
  • Replace one hydrogen on the second carbon with a hydroxyl group (-OH).
Now you have lactic acid, a key player in muscle function and fermentation processes.
(Z)-2-methyl-2-butenoic acid
The molecule (Z)-2-methyl-2-butenoic acid features a butenoic acid backbone with a methyl group (-CH3) on the second carbon. The 'Z' configuration indicates a cis-geometry of the double bond, meaning hydrogen atoms are on the same side of the double bond. The structure is:
CH3-C(CH3)=CH-COOH.
To draw this molecule accurately, follow these steps:
  • Begin with the butenoic acid backbone: CH2=CH-COOH
  • Add a methyl group (-CH3) to the second carbon
  • Ensure the double bond uses the cis (Z) configuration.
Now, you have (Z)-2-methyl-2-butenoic acid.
2-hydroxy-1,2,3-propanetricarboxylic acid
2-hydroxy-1,2,3-propanetricarboxylic acid, known as citric acid, is an essential part of the citric acid cycle in metabolism. This molecule has a propane backbone with three carboxylic acid groups (-COOH) and one hydroxyl group (-OH). The structure is:
HO-C-(COOH)-CH2-COOH-CH2-COOH.
Steps to draw citric acid:
  • Start with a propane backbone: CH3-CH-CH3
  • Replace the hydrogen atoms with three carboxylic acid groups (-COOH), and one hydroxyl group (-OH).
Now, you have successfully drawn citric acid, an important metabolic intermediate.
2-hydroxybenzoic acid
2-hydroxybenzoic acid, also called salicylic acid, is a benzoic acid derivative regularly used for its anti-inflammatory properties. The molecule consists of a benzene ring with a carboxylic acid group (-COOH) and an ortho hydroxyl group (-OH) attached to it. The structure is:
OH-C6H4-COOH.
Here’s how to sketch it:
  • Begin with a benzene ring: C6H6
  • Add the carboxylic acid group (-COOH) to one of the carbons in the benzene ring.
  • Attach a hydroxyl group (-OH) to the ortho position, which is the adjacent carbon to the carboxylic group.
Salicylic acid thus plays a crucial role in drug synthesis and skin care treatments.
2-oxopropanoic acid
2-oxopropanoic acid, known as pyruvic acid, is pivotal in cellular respiration and fermentation. This molecule contains a propanoic acid backbone with a keto group (=O) on the second carbon. The structure is given by:
CH3-CO-COOH.
To draw pyruvic acid, follow these steps:
  • Start with the propanoic acid backbone: CH3-CH2-COOH
  • Replace one hydrogen on the second carbon with a keto group (=O).
Now you have pyruvic acid, a key intermediate in metabolic processes. Now you have 2-oxopropanoic acid, an integral part of various biochemical pathways.

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Most popular questions from this chapter

Write names for the following carboxylic acids: (a) CC(C)C=CCC(=O)O (b) CC(C)C(N)C(=O)O (c) O=C(O)C1CC(O)C1 (d) O=C(O)c1cccc(O)c1 (e) CC(CC(=O)O)c1ccccc1 (f) C=C(C)C(CC(=O)O)c1ccccc1 (g) CC(CC(=O)O)C(Br)(Br)Br (h) NCC(O)C(=O)O

Draw a structure for each of the following compounds: (a) lithium 2-hydroxy-3-methylpentanoate (b) sodium benzoate (c) sodium 2-amino-3-phenylpropanoate (d) potassium acetate (e) potassium 3-butenoate

CC(C)C(C… # Identify the functional groups in the following compounds: (a) COC(=O)C(C=O)C=CC(C)C(=O)O (b) O=C1CCC(O)C(=O)O1 (c) C#CC(C)C(CC(=O)CCl)c1ccccc1 (d) O=C(Cl)c1ccc(CBr)cc1

Explain why the \(\mathrm{p} K_{\mathrm{a}}\) of the protonated aminoacetic acid (2.3) is lower than the \(\mathrm{p} K_{\mathrm{a}}\) of acetic acid (4.76).

Estragole (1) is a major constituent of tarragon oil. Treatment of \(\mathbf{1}\) with hydrogen bromide in the presence of peroxides affords 2 . Compound 2 reacts with \(\mathrm{Mg}\) in ether, followed by addition of carbon dioxide and acidification, to give 3 . Spectral data for compound 3 are summarized below. Deduce the structure of compound 3 and propose structures for estragole and compound \(\mathbf{2}\). Compound 3 Mass spectrum: \(m / z=194(\mathrm{p})\) IR (Nujol): 3330-2500 (br, s), \(1696(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 1.95\) (quintet, \(J=7 \mathrm{~Hz}, 2 \mathrm{H}\) ) \(2.34(\mathrm{t}, J=7 \mathrm{~Hz}, 2 \mathrm{H})\) \(2.59(\mathrm{t}, J=7 \mathrm{~Hz}, 2 \mathrm{H})\) \(3.74(\mathrm{~s}, 3 \mathrm{H})\) \(6.75(\mathrm{~d}, J=8 \mathrm{~Hz}, 2 \mathrm{H})\) \(7.05(\mathrm{~d}, J=8 \mathrm{~Hz}, 2 \mathrm{H})\) \(11.6(\mathrm{~s}, 1 \mathrm{H})\)
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