Question

An unknown organic compound is found on elemental analysis to contain \(72.0 \%\) carbon, \(12.0 \%\) hydrogen, and \(16.0 \%\) oxygen by mass. It has a molecular weight of \(100 \mathrm{~g} / \mathrm{mol}\) and is slightly soluble in water. Upon careful oxidation it is converted into a ketone which contains \(73.4 \%\) carbon, \(10.3 \%\) hydrogen, and \(16.3 \%\) oxygen by mass. Indicate two or more reasonable structures for the unknown.

Short answer

The unknown compound has a molecular formula of \(C_6H_{12}O\) and is likely an alcohol, as it forms a ketone upon oxidation. Two reasonable structures for the compound are 2-hexanol (\(CH_3CH_2CH(OH)CH_2CH_2CH_3\)) and 3-hexanol (\(CH_3CH_2CH_2CH(OH)CH_2CH_3\)). Both of these secondary alcohols would form a ketone upon oxidation.

Step-by-step solution

Calculate the empirical formula of the unknown compound

Given the percentage composition of the compound: - Carbon: \(72.0\%\) by mass - Hydrogen: \(12.0\%\) by mass - Oxygen: \(16.0\%\) by mass Assume we have 100g of the compound: - 72g of carbon - 12g of hydrogen - 16g of oxygen First, we need to convert these amounts to moles: - \(n(C) = \frac{72\,\text{g}}{12.01\,\text{g/mol}} \approx 6\,\text{mol}\ C\) - \(n(H) = \frac{12\,\text{g}}{1.008\,\text{g/mol}} \approx 12\,\text{mol}\ H\) - \(n(O) = \frac{16\,\text{g}}{16\,\text{g/mol}} = 1\,\text{mol}\ O\) So, the empirical formula is \(C_6H_{12}O\).

Determine the molecular formula of the unknown compound

The molecular weight of the unknown compound is given as \(100 \, \text{g/mol}\). Divide the molecular weight by the empirical formula weight to find the ratio of the two: \(Ratio = \frac{100\,\mathrm{g/mol}}{(6 \times 12.01 + 12 \times 1.008 + 16)\,\mathrm{g/mol}} \approx 1\) Since the ratio is close to 1, the molecular formula is the same as the empirical formula, \(C_6H_{12}O\).

Identify the functional group before oxidation to form the ketone

The unknown compound is slightly soluble in water and forms a ketone upon oxidation. These facts suggest that the unknown compound is likely an alcohol. Oxidation of alcohols typically results in the formation of a ketone or an aldehyde. However, given the oxidized product has a higher carbon percentage, this reaction likely formed a ketone, consistent with the given information.

Propose two or more reasonable structures for the unknown compound

With the molecular formula \(C_6H_{12}O\) and considering the alcohol functional group, we can propose several reasonable structures: 1. 2-hexanol: This is a secondary alcohol (\(CH_3CH_2CH(OH)CH_2CH_2CH_3\)) that would form 2-hexanone upon oxidation. 2. 3-hexanol: This is another secondary alcohol (\(CH_3CH_2CH_2CH(OH)CH_2CH_3\)) that would form 3-hexanone upon oxidation. There are other isomers of hexanol that could be possible structures, but their oxidation wouldn't yield a ketone. Since the problem asks for two or more reasonable structures, the two listed above should suffice.

Key concepts

Empirical Formula

In organic chemistry, determining the empirical formula is the first step in understanding the composition of a compound. An empirical formula shows the simplest integer ratio of the elements in a compound. To calculate it, you need the mass percentage of each element, which allows conversion into moles. Assume 100g of the substance to directly translate percentage to mass.
For instance, with 72% carbon, 12% hydrogen, and 16% oxygen, consider:

• 72g of carbon, equivalent to \( \frac{72}{12.01} \approx 6 \) moles,
• 12g of hydrogen, equivalent to \( \frac{12}{1.008} \approx 12 \, \) moles,
• 16g of oxygen, equating exactly to 1 mole.

Thus, the empirical formula is \( C_6H_{12}O \). This information helps in determining the molecular structure of our unknown compound.

Molecular Weight

In chemistry, molecular weight (M.W.) is an essential property as it describes the mass of a single molecule of the substance. For the unknown compound, a given molecular weight of 100 g/mol helps to identify the actual formula from its empirical version. To determine if the empirical formula also represents the molecular formula, divide the molecular weight by the empirical formula weight:\[ Ratio = \frac{100 \, \text{g/mol}}{(6 \times 12.01 + 12 \times 1.008 + 16) \, \text{g/mol}} \approx 1 \]Here, the ratio is 1, indicating that the empirical formula \( C_6H_{12}O \) is also the molecular formula. Knowledge of molecular weight, therefore, can critically simplify molecular identification.

Functional Groups

Functional groups are specific groups of atoms within molecules that determine the chemical reactions of those molecules. In this exercise, the unknown compound's functional group is key to hypothesizing its structure and subsequent behavior during chemical reactions. Because it forms a ketone upon oxidation, it likely contains an alcohol group initially. This functional group is vital in predicting how the compound reacts.
For instance, alcohols are notable for their solubility properties and reactivity patterns:

• They can be oxidized under certain conditions, producing ketones or aldehydes.
• The oxidative formation of a ketone from the compound suggests a secondary alcohol functional group.

Considering functional groups enhances understanding, prediction, and identification of chemical reaction outcomes.

Alcohol Oxidation

The process of alcohol oxidation is a common investigation in organic chemistry. It involves converting an alcohol to a ketone or aldehyde, depending on the type of alcohol. The problem indicates that the unknown compound forms a ketone upon oxidation; hence, it contains a secondary alcohol group. Characteristics of alcohol oxidation include:

• Primary alcohols usually oxidize to aldehydes and then to acids.
• Secondary alcohols typically produce ketones, as is the case here with the resulting formation of a ketone with higher carbon content.
• Tertiary alcohols do not oxidize easily because they lack a hydrogen atom on the carbon atom with the -OH group.

Understanding the type of alcohol and its oxidation pathway is crucial to predicting the structure and reactions of organic compounds.