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Ethyl caprate is an ester used in the manufacture of wine bouquets. It is sometimes called "cognac essence." Combustion analysis of a sample of ethyl caprate shows it to be 71.89\(\% \mathrm{C}\) , 12.13\(\%\) hydrogen, and 15.98\(\%\) O by mass. Hydrolysis (reaction with water) of the ester produces ethanol and a carboxylic acid. The molar mass of the carboxylic acid is 172 \(\mathrm{g} / \mathrm{mol}\) , and the \(\mathrm{R}\) group attached to the COOH group of the acid is a straight chain alkane. What is the structure of ethyl caprate?

Short Answer

Expert verified
The structure of ethyl caprate is C13H26O2.

Step by step solution

01

Convert mass percentages to molar percentages

In order to convert mass percentages to molar percentages, we need to use the molar masses of carbon, hydrogen, and oxygen. Divide the mass percentages by their respective molar masses: Carbon: \(\frac{71.89\%}{12.01 \text{ g/mol}} = 5.99\text{ moles}\) Hydrogen: \(\frac{12.13\%}{1.01 \text{ g/mol}} = 12.01\text{ moles}\) Oxygen: \(\frac{15.98\%}{16.00 \text{ g/mol}} = 1.00\text{ moles}\)
02

Determine the empirical formula of ethyl caprate

Now we can find the ratio of carbon, hydrogen, and oxygen in the compound. To do this, divide the number of moles of each element by the smallest number of moles: Carbon: \(\frac{5.99 \text{ moles}}{1.00 \text{ mole}} = 6\) Hydrogen: \(\frac{12.01 \text{ moles}}{1.00 \text{ mole}} = 12\) Oxygen: \(\frac{1.00 \text{ moles}}{1.00 \text{ mole}} = 1\) The empirical formula for ethyl caprate is C6H12O.
03

Find the molecular formula of the carboxylic acid

We know that the molar mass of the carboxylic acid is 172 g/mol. The empirical formula of the COOH group is CH2O2, which has a molar mass of 45 g/mol. Subtracting the molar mass of the COOH group from the total molar mass of the carboxylic acid will give us the molar mass of the R group: Molar mass of R group: 172 g/mol - 45 g/mol = 127 g/mol Since the R group is a straight chain alkane, which consists of carbon (C) and hydrogen (H2) atoms, we can determine the number of carbon atoms by dividing the molar mass of the R group by (12.01 g/mol + 2(1.01 g/mol)): Number of carbon atoms in the R group: \(\frac{127 \text{ g/mol}}{(12.01 \text{ g/mol} + 2(1.01 \text{ g/mol}))} = 10\) Therefore, the molecular formula of the carboxylic acid is C11H22O2.
04

Determine the structure of ethyl caprate

We know that ethyl caprate hydrolyzes to form ethanol (C2H5OH) and the carboxylic acid with a molecular formula of C11H22O2. The ethyl group in ethyl caprate comes from the ethanol, and the carboxylic acid group comes from the carboxylic acid molecule. Thus, the structure of ethyl caprate is given by combining the molecular formulas of ethanol and the carboxylic acid without the hydroxyl (OH) group and the hydrogen atom, respectively: Ethyl caprate structure: C2H5O - C11H21O The structure of ethyl caprate is C13H26O2.

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Key Concepts

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

Ester
Esters are fascinating compounds often encountered in chemistry and biochemistry due to their unique properties and reactions. These organic compounds are formed by the reaction of an alcohol and a carboxylic acid. During this process, an ester linkage is created, which is typically represented by the formula \(-COOR\), where \(-R\) stands for an alkyl or aryl group.

Esters are known for their distinct aromas and are commonly found in fragrances and flavorings. For instance, ethyl caprate is a well-known ester used to mimic fruity aromas and is sometimes even called "cognac essence." This property makes esters highly valuable in the beverage and cosmetics industries.

In chemical synthesis, esters can undergo various reactions, one of which is hydrolysis. This process breaks the ester back into its original alcohol and carboxylic acid, a concept crucial in understanding ester functionality and chemistry.
Empirical Formula
The empirical formula of a compound is the simplest ratio of the atoms of each element in that compound. It provides insight into the basic composition of the compound, though not its actual structure.

To determine the empirical formula, a combustion analysis is often employed, which involves the burning of a sample to find the percentages of its constituent elements. In the case of ethyl caprate, we discovered the molecular percentages of carbon, hydrogen, and oxygen by converting their mass percentages using their respective atomic masses.

Once the mole ratios are established, these can be simplified to express the smallest whole number ratio, thereby giving us the empirical formula. For ethyl caprate, based on the combustion analysis, the empirical formula was determined to be \( ext{C}_6 ext{H}_{12} ext{O}\). Understanding the empirical formula is essential in guiding us towards identifying the molecular formula and ultimately, the compound's structure.
Carboxylic Acid
Carboxylic acids are organic acids characterized by the presence of at least one carboxyl group (\(-COOH\)). These are one of the most important functional groups in organic chemistry due to their reactivity and prevalence.

The carboxyl group imparts acidic properties to the compound, allowing it to donate a proton (H鈦) in solution, making it a weak acid. Carboxylic acids are found in biological systems and are also key intermediate substances in chemical synthesis.

In chemistry, an understanding of carboxylic acids is important due to their role in forming esters. The carboxylic acid involved can react with an alcohol to form an ester and water, through a condensation reaction. In the problem of ethyl caprate, hydrolysis of the ester results in the production of ethanol and a carboxylic acid. The molecular formula of this carboxylic acid was determined to be \( ext{C}_{11} ext{H}_{22} ext{O}_{2}\), which provided essential information about the ester鈥檚 complete structure.
Hydrolysis
Hydrolysis is a fundamental chemical reaction in which a compound reacts with water, causing it to break down into two or more smaller molecules. This process is especially important in biochemistry and organic chemistry for the transformation and breakdown of various compounds.

For esters, hydrolysis involves breaking the ester linkage to form an alcohol and a carboxylic acid. This reaction is crucial in understanding the stability and reactivity of esters. In the context of ethyl caprate, hydrolysis facilitated the identification of the ester by revealing its components, ethanol and a carboxylic acid, thereby highlighting its chemical structure.

Hydrolysis can be catalyzed by acidic or basic conditions, which can influence the rate and product distribution. This versatility makes it an invaluable reaction in synthetic and analytical chemistry, helping chemists deduce and verify molecular structures.

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Most popular questions from this chapter

Explain why methyl alcohol is soluble in water in all proportions, while stearyl alcohol \(\left[\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{OH}\right]\) is a waxy solid that is not soluble in water.

Mycomycin, a naturally occurring antibiotic produced by the fungus Nocardia acidophilus, has the molecular formula \(\mathrm{C}_{13} \mathrm{H}_{10} \mathrm{O}_{2}\) and the systematic name \(3,5,7,8\) -tridecatetraene- \(10,12\) -diynoic acid. Draw the structure of mycomycin.

What types of interactions can occur between the side chains of the following amino acids that would help maintain the tertiary structure of a protein? $$ \begin{array}{ll}{\text { a. cysteine and cysteine }} & {\text { c. glutamic acid and lysine }} \\ {\text { b. glutamine and serine }} & {\text { d. proline and leucine }}\end{array} $$

For each of the following, fill in the blank with the correct response. All of these fill-in-the-blank problems pertain to material covered in the sections on alkanes, alkenes and alkynes, aromatic hydrocarbons, and hydrocarbon derivatives. a. The first 鈥渙rganic鈥 compound to be synthesized in the laboratory, rather than being isolated from nature, was , which was prepared from . b. An organic compound whose carbon鈥揷arbon bonds are all single bonds is said to be . c. The general orientation of the four pairs of electrons around the carbon atoms in alkanes is . d. Alkanes in which the carbon atoms form a single unbranched chain are said to be alkanes. e. Structural isomerism occurs when two molecules have the same number of each type of atom but exhibit different arrangements of the between those atoms. f. The systematic names of all saturated hydrocarbons have the ending added to a root name that indicates the number of carbon atoms in the molecule. g. For a branched hydrocarbon, the root name for the hydrocarbon comes from the number of carbon atoms in the continuous chain in the molecule. h. The positions of substituents along the hydrocarbon framework of a molecule are indicated by the of the carbon atom to which the substituents are attached. i. The major use of alkanes has been in reactions, as a source of heat and light. j. With very reactive agents, such as the halogen elements, alkanes undergo reactions, whereby a new atom replaces one or more hydrogen atoms of the alkane. k. Alkenes and alkynes are characterized by their ability to undergo rapid, complete reactions, by which other atoms attach themselves to the carbon atoms of the double or triple bond. l. Unsaturated fats may be converted to saturated fats by the process of . m. Benzene is the parent member of the group of hydrocarbons called hydrocarbons. n. An atom or group of atoms that imparts new and characteristic properties to an organic molecule is called a group. o. A alcohol is one in which there is only one hydrocarbon group attached to the carbon atom holding the hydroxyl group. p. The simplest alcohol, methanol, is prepared industrially by the hydrogenation of . q. Ethanol is commonly prepared by the of certain sugars by yeast. r. Both aldehydes and ketones contain the group, but they differ in where this group occurs along the hydrocarbon chain. s. Aldehydes and ketones can be prepared by of the corresponding alcohol. t. Organic acids, which contain the group, are typically weak acids. u. The typically sweet-smelling compounds called result from the condensation reaction of an organic acid with an .

Draw all the structural isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\) . Ignore any cyclic isomers.

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