Problem 120 - Chemistry An Atoms First Approach

Question

A compound containing only carbon and hydrogen is \(85.63 \%\) C by mass. Reaction of this compound with \(\mathrm{H}_{2} \mathrm{O}\) produces a secondary alcohol as the major product and a primary alcohol as the minor product. (See Exercise 62.) If the molar mass of the hydrocarbon is between 50 and 60 g/mol, name the compound.

Step-by-Step Solution

Answer

The compound is 2-butene, with the molecular formula C4H8. This hydrocarbon contains 85.63% carbon and 14.37% hydrogen, has a molar mass between 50 and 60 g/mol, and reacts with water to produce a secondary alcohol as the major product and a primary alcohol as the minor product.

Step 1: Calculate the mass percentage of hydrogen

Since the compound contains only carbon and hydrogen, the mass percentage of hydrogen can be found by subtracting the mass percentage of carbon from 100%. Mass percentage of Hydrogen (H) = 100 - 85.63 = 14.37%

Step 2: Determine a reasonable ratio of C atoms to H atoms

Using the mass percentages we found in step 1 and the molar masses of carbon and hydrogen, we can find a reasonable ratio of C atoms to H atoms that would lie close to an integer. Molar mass of carbon (C) = 12 g/mol Molar mass of hydrogen (H) = 1 g/mol Moles of carbon = mass percentage of carbon (C) / atomic mass of carbon (C) = 85.63 / 12 Moles of hydrogen = mass percentage of hydrogen (H) / atomic mass of hydrogen (H) = 14.37 / 1 C:H mole ratio (rounded to nearest whole numbers) = (85.63 / 12):(14.37 / 1) = 7:14 which can be simplified to 1:2.

Step 3: Propose the molecular formula

Based on the mole ratio we found in step 2, the molecular formula can be assumed to be CnH2n. As we know that the molar mass is between 50 and 60 g/mol, substituting the various values of 'n' into the formula will help us find which number of carbon atoms match the given molar mass range. When n = 4, Molar mass of CnH2n = 4 × 12 + 2 × 4 × 1 = 48 + 8 = 56 g/mol. This falls into the given range.

Step 4: Consider potential carbon structures

Since we found that there are 4 carbon atoms and 8 hydrogen atoms in the hydrocarbon (C4H8), we have to think about potential carbon structures that would give a major product as a secondary alcohol and a minor product as a primary alcohol when reacted with water. The most reasonable structure that would produce this result is 2-butene. We can represent the compound as: \(CH_{3}-CH=CH-CH_{3}\) So, the name of the compound is 2-butene.

Key Concepts

Empirical Formula

An empirical formula is the simplest expression of the relative number of each type of atom in a compound. It provides a straightforward way to depict how the atoms are proportioned but not their actual arrangement.
To find the empirical formula, we start by identifying the percentage composition of each element within a compound. In our case, the hydrocarbon consists solely of carbon and hydrogen.

• Carbon is 85.63% of the total mass.
• Hydrogen is calculated as 100% - 85.63% = 14.37%.

The next step involves converting these percentages into a mole ratio by dividing each by their respective molar masses. This calculation gives a ratio of carbon to hydrogen, which simplifies to a 1:2 relationship. Therefore, the empirical formula is denoted as \(C_nH_{2n}\).

Hydrocarbon

Hydrocarbons are organic compounds made exclusively of carbon (C) and hydrogen (H) atoms. They serve as a fundamental category in organic chemistry, acting as the foundation for more complex chemical structures. Hydrocarbons can be simple, like methane, owing their relevance due to their role in fuel production and numerous industrial applications.
In this problem, we focus on the hydrocarbon type that forms the basis of the compound 2-butene. This particular hydrocarbon has a molecular formula of \(C_4H_8\), indicating four carbon atoms are bonded to eight hydrogen atoms. 2-butene is an example of an alkene, which is characterized by the presence of at least one carbon-carbon double bond.

Molecular Structure

The molecular structure illustrates how atoms are arranged in a compound, dictating its chemical properties and reactivity. Molecular structures can be illustrated via structural formulas, which show the connectivity between atoms.
For 2-butene, the structural formula can be simplified as \(CH_3-CH=CH-CH_3\). This molecular formula identifies it as a symmetric alkene, implying the existence of a carbon double bond featuring between its second and third carbon atoms. The presence of this double bond classifies it as a secondary alcohol precursor due to its reaction pattern with water to form alcohols upon hydration.

Chemical Reactions

Chemical reactions are processes where substances (reactants) are transformed into new substances (products), involving the rearrangement of atoms. In the realm of organic chemistry, understanding reactions is essential as they reveal insights into how compounds interact.
In the exercise, the hydrocarbon, 2-butene, undergoes a reaction with water to predominantly yield a secondary alcohol. This reaction aligns with the hydration of alkenes—an organic chemistry essential.

• The reaction predominantly leads to a secondary alcohol due to Markovnikov’s rule, which favors the addition of the water's OH group to the more substituted carbon atom.
• The minor product, a primary alcohol, results from the less common anti-Markovnikov addition pathway.

These chemical interactions depict the versatility of hydrocarbons like 2-butene and highlight the importance of structural forms in chemical reactions.