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Chapter 17: Problem 37

Name the carboxylic acid and alcohol from which each ester is derived. (a) COC(=O)C1CCCCC1 (b) CC(=O)OC1CCC(OC(C)=O)CC1 (c) CCC=CC(=O)OC(C)C (d) CCOC(=O)CCC(=O)OCC

Short Answer

Expert verified
Question: Determine the carboxylic acid and alcohol for each ester provided. (a) Ester: COC(=O)C1CCCCC1 Carboxylic acid: Cyclohexane carboxylic acid Alcohol: Methanol (b) Ester: CC(=O)OC1CCC(OC(C)=O)CC1 Carboxylic acids: Ethanoic acid and Propanoic acid Alcohol: 4-oxopentyl alcohol (c) Ester: CCC=CC(=O)OC(C)C Carboxylic acid: Trans-but-2-enoic acid Alcohol: 2-butanol (d) Ester: CCOC(=O)CCC(=O)OCC Carboxylic acid: Butanoic acid Alcohol: Ethyl ethanoate

Step by step solution

01

(a) Identify RCOO and OR' groups

For the ester with the SMILES notation COC(=O)C1CCCCC1, first find the ester functional group, which is the C(=O)O part. The two groups attached to this functional group are: 1. RCOO group: C(=O)C1CCCCC1 2. OR' group: OC
02

(a) Name the carboxylic acid and alcohol

Based on the identified groups, the carboxylic acid is cyclohexane carboxylic acid, and the alcohol is methanol.
03

(b) Identify RCOO and OR' groups

For the ester with the SMILES notation CC(=O)OC1CCC(OC(C)=O)CC1, we identify the ester functional groups as C(=O)O and OC(=O)C. The two groups attached to these functional groups are: 1. RCOO group: C(=O)C and OC(=O)C 2. OR' group: OC1CCC(OR)CC1
04

(b) Name the carboxylic acid and alcohol

Based on the identified groups, the carboxylic acids are ethanoic acid and propanoic acid, and the alcohol is 4-oxopentyl alcohol.
05

(c) Identify RCOO and OR' groups

For the ester with the SMILES notation CCC=CC(=O)OC(C)C, the two groups attached to the ester functional group C(=O)O are: 1. RCOO group: CC(=O)C=CC 2. OR' group: OC(C)C
06

(c) Name the carboxylic acid and alcohol

Based on the identified groups, the carboxylic acid is trans-but-2-enoic acid, and the alcohol is 2-butanol.
07

(d) Identify RCOO and OR' groups

For the ester with the SMILES notation CCOC(=O)CCC(=O)OCC, we identify the R and R' groups attached to the ester functional group C(=O)O as: 1. RCOO group: CCC(=O)O 2. OR' group: OC(=O)CCC
08

(d) Name the carboxylic acid and alcohol

Based on the identified groups, the carboxylic acid is butanoic acid, and the alcohol is ethyl ethanoate.

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Key Concepts

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

Carboxylic Acids
Carboxylic acids are organic compounds that contain a carboxyl group (\(-COOH\)). This group is known for its acidity, making carboxylic acids an essential part of many chemical reactions including esterification.
In esterification, the carboxylic acid reacts with an alcohol to form an ester and water. It's worth noting that the carboxyl group's acidic hydrogen is replaced by an alkoxyl group (\(-OR\)) from the alcohol.
Carboxylic acids have diverse uses, ranging from culinary uses as vinegar (acetic acid) to industrial applications in the production of plastics and pharmaceuticals.
Alcohols
Alcohols are organic compounds characterized by the presence of a hydroxyl group (\(-OH\)) attached to a carbon atom. They are quite versatile in organic chemistry and have a wide range of applications.
In esterification, alcohols act as nucleophiles, reacting with carboxylic acids to form esters. During this process, the hydrogen from the hydroxyl group is replaced, helping to create the structural formula of an ester.
Different types of alcohols can lead to varying types of esters, each with its own unique scent and industrial application. Examples include methanol, used to make biodiesel, and ethanol, used in perfumery.
SMILES Notation
SMILES (Simplified Molecular Input Line Entry System) notation is a text-based format used to describe the structure of chemical molecules. It's widely used in cheminformatics.
In SMILES notation, atoms are represented by their chemical symbols and bond types are depicted by specific characters, allowing chemists to easily encode and decode molecular structures.
For instance, the ester SMILES notation "CCO" represents a simple structure with two carbons and an oxygen, interpreted according to the context provided by additional symbols such as \(=\). This system is particularly useful for sharing chemical information in databases and software systems.
Organic Chemistry Nomenclature
Organic chemistry nomenclature is the systematic method of naming organic chemical compounds as recommended by IUPAC. This ensures each compound has a unique and universally recognized name.
In naming esters, the name is derived from the alcohol and the acid that participate in the esterification process. The alcohol name gives the "alkyl" part and the carboxylic acid provides the "-oate" part of the ester name.
For example, the ester formed from methanol and acetic acid is named "methyl acetate." Understanding this naming system is crucial for accurately communicating chemical structures and reactions in the scientific community.

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Most popular questions from this chapter

Given here are \({ }^{1} \mathrm{H}-\mathrm{NMR}\) and \({ }^{19} \mathrm{C}-\mathrm{NMR}\) spectral data for nine compounds. Each compound shows strong absorption between 1720 and \(1700 \mathrm{~cm}^{-1}\), and strong, broad absorption over the region \(2500-3300 \mathrm{~cm}^{-1}\). Propose a structural formula for each compound. Refer to Appendices 4,5, and 6 for spectral correlation tables. $$ \begin{aligned} &\text { (a) } \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 0.94(\mathrm{t}, 3 \mathrm{H}) & 180.71 \\ 1.39(\mathrm{~m}, 2 \mathrm{H}) & 33.89 \\ 1.62(\mathrm{~m}, 2 \mathrm{H}) & 26.76 \\ 2.35(\mathrm{t}, 2 \mathrm{H}) & 22.21 \\ 12.0(\mathrm{~s}, 1 \mathrm{H}) & 13.69 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (b) } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{19} \text { C-NMR } \\ \hline 1.08(\mathrm{~s}, 9 \mathrm{H}) & 179.29 \\ 2.23(\mathrm{~s}, 2 \mathrm{H}) & 47.82 \\ 12.1(\mathrm{~s}, 1 \mathrm{H}) & 30.62 \\ & 29.57 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (c) } \mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{4}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 0.93(\mathrm{t}, 3 \mathrm{H}) & 170.94 \\ 1.80(\mathrm{~m}, 2 \mathrm{H}) & 53.28 \\ 3.10(\mathrm{t}, 1 \mathrm{H}) & 21.90 \\ 12.7(\mathrm{~s}, 2 \mathrm{H}) & 11.81 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (d) } \mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{4}\\\ &\begin{array}{cr} \hline{ }^{1} \text { H-NMR } & { }^{19} \text { C-NMR } \\ \hline 1.29(\mathrm{~s}, 6 \mathrm{H}) & 174.01 \\ 12.8(\mathrm{~s}, 2 \mathrm{H}) & 48.77 \\ & 22.56 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (e) } \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 1.91(\mathrm{~d}, 3 \mathrm{H}) & 172.26 \\ 5.86(\mathrm{~d}, 1 \mathrm{H}) & 147.53 \\ 7.10(\mathrm{~m}, 1 \mathrm{H}) & 122.24 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 18.11 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (f) } \mathrm{C}_{3} \mathrm{H}_{4} \mathrm{Cl}_{2} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{19} \text { C-NMR } \\ \hline 2.34(\mathrm{~s}, 3 \mathrm{H}) & 171.82 \\ 11.3(\mathrm{~s}, 1 \mathrm{H}) & 79.36 \\ & 34.02 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (g) } \mathrm{C}_{5} \mathrm{H}_{8} \mathrm{Cl}_{2} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 1.42(\mathrm{~s}, 6 \mathrm{H}) & 180.15 \\ 6.10(\mathrm{~s}, 1 \mathrm{H}) & 77.78 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 51.88 \\ & 20.71 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (h) } \mathrm{C}_{5} \mathrm{H}_{9} \mathrm{BrO}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 0.97(\mathrm{t}, 3 \mathrm{H}) & 176.36 \\ 1.50(\mathrm{~m}, 2 \mathrm{H}) & 45.08 \\ 2.05(\mathrm{~m}, 2 \mathrm{H}) & 36.49 \\ 4.25(\mathrm{t}, 1 \mathrm{H}) & 20.48 \\ 12.1(\mathrm{~s}, 1 \mathrm{H}) & 13.24 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (i) } \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{3}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.62(\mathrm{t}, 2 \mathrm{H}) & 177.33 \\ 3.38(\mathrm{~s}, 3 \mathrm{H}) & 67.55 \\ 3.68(\mathrm{~s}, 2 \mathrm{H}) & 58.72 \\ 11.5(\mathrm{~s}, 1 \mathrm{H}) & 34.75 \\ \hline \end{array} \end{aligned} $$

On a cyclohexane ring, an axial carboxyl group has a conformational energy of \(5.9 \mathrm{~kJ}\) (1.4 kcal)/mol relative to an equatorial carboxyl group. Consider the equilibrium for the alternative chair conformations of trans-1,4-cyclohexanedicarboxylic acid. Draw the less stable chair conformation on the left of the equilibrium arrows and the more stable chair on the right. Calculate \(\Delta G^{0}\) for the equilibrium as written, and calculate the ratio of the more stable chair to the less stable chair at \(25^{\circ} \mathrm{C}\).

The reaction of an \(\alpha\)-diketone with concentrated sodium or potassium hydroxide to give the salt of an \(\alpha\)-hydroxyacid is given the general name benzil-benzilic acid rearrangement. It is illustrated by the conversion of benzil to sodium benzilate and then to benzilic acid. Propose a mechanism for this rearrangement. O=C(O)C(=O)C(=O)c1ccccc1 O=C(O)C(O)(c1ccccc1)C(O)(c1ccccc1)c1ccccc1 Benzil Sodium benzilate Benzilic acid

Using your roadmaps as a guide, show how to convert propane into propyl propanoate. You must use propane as the source of all carbon atoms in the target molecule. Show all reagents needed and all molecules synthesized along the way. CCCOC(=O)CC Propane Propyl propanoate

We have studied Fischer esterification, in which a carboxylic acid is reacted with an alcohol in the presence of an acid catalyst to form an ester. Suppose that you start instead with a dicarboxylic acid such as terephthalic acid and a diol such as ethylene glycol. Show how Fischer esterification in this case can lead to a macromolecule with a molecular weight several thousands of times that of the starting materials. O=C(O)c1ccc(C(=O)O)cc1 1,4-Benzenedicarboxylic acid \(\quad 1,2\)-Ethanediol (Terephthalic acid) (Ethylene glycol) As we shall see in Section \(29.5 \mathrm{~B}\), the material produced in this reaction is a highmolecular-weight polymer, which can be fabricated into Mylar films, and into the textile fiber known as Dacron polyester.
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