Question

For which compounds containing a heteroatom (an atom other than carbon or hydrogen) does the molecular ion have an even-numbered mass? For which does it have an odd-numbered mass? (a) A chloroalkane with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Cl}\) (b) A bromoalkane with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Br}\) (c) An alcohol with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{OH}\) (d) A primary amine with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n-1} \mathrm{NH}_{2}\) (e) A thiol with the molecular formula \(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{SH}\)

Short answer

Answer: Thiol has an odd-numbered molecular ion mass.

Step-by-step solution

Find the molecular ion's mass for each compound

For each given compound, we need to sum up the atomic masses of all its atoms. The atomic masses are: carbon (C) = 12, hydrogen (H) = 1, chlorine (Cl) = 35.5, bromine (Br) = 79.9, oxygen (O) = 16, nitrogen (N) = 14, and sulfur (S) = 32.1. (a) Chloroalkane (\(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Cl}\)): Mass = 12n + (2n+1) + 35.5 (b) Bromoalkane (\(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{Br}\)): Mass = 12n + (2n+1) + 79.9 (c) Alcohol (\(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{OH}\)): Mass = 12n + (2n+1) + 16 + 1 (d) Primary amine (\(\mathrm{C}_{n} \mathrm{H}_{2 n-1} \mathrm{NH}_{2}\)): Mass = 12n + (2n-1) + 14 + 2(1) (e) Thiol (\(\mathrm{C}_{n} \mathrm{H}_{2 n+1} \mathrm{SH}\)): Mass = 12n + (2n+1) + 32.1 + 1

Determine the parity of the molecular ion's mass

We need to find if the mass of each molecular ion is even or odd. (a) Chloroalkane: 12n is even, (2n+1) is odd, and 35.5 is odd (half-integer values are considered odd for this analysis). Even + odd + odd = even. Therefore, a chloroalkane has an even-numbered mass. (b) Bromoalkane: 12n is even, (2n+1) is odd, and 79.9 is odd. Even + odd + odd = even. Therefore, a bromoalkane has an even-numbered mass. (c) Alcohol: 12n is even, (2n+1) is odd, 16 is even, and 1 is odd. Even + odd + even + odd = even. Therefore, an alcohol has an even-numbered mass. (d) Primary amine: 12n is even, (2n-1) is odd, 14 is even, and 2(1) is even. Even + odd + even + even = even. Therefore, a primary amine has an even-numbered mass. (e) Thiol: 12n is even, (2n+1) is odd, 32.1 is odd, and 1 is odd. Even + odd + odd + odd = odd. Therefore, a thiol has an odd-numbered mass.

Conclude

In conclusion: - Chloroalkane, bromoalkane, alcohol, and primary amine have even-numbered molecular ion masses. - Thiol has an odd-numbered molecular ion mass.

Key concepts

Heteroatoms in Organic Chemistry

Organic chemistry is the branch of chemistry that deals with organic molecules, primarily consisting of carbon and hydrogen atoms. However, many organic compounds also contain other types of atoms, known as heteroatoms. Heteroatoms are any atoms in a molecule that are neither carbon nor hydrogen. Common examples include oxygen, nitrogen, sulfur, halogens (like chlorine, bromine), and more. These elements have distinct physical and chemical properties that they confer to the organic compounds they are part of.

For example, in the given exercise, compounds containing chlorine, bromine, oxygen, nitrogen, and sulfur are considered because they contain heteroatoms, which affect the molecular mass and reactivity of the compounds. Heteroatoms can also influence the electronegativity, polarity, and overall shape of the molecule, which are crucial for understanding the behavior of the compounds. With heteroatoms, molecules can obtain an even or odd-numbered mass, and this can be significant in determining compound types using techniques such as mass spectrometry.

Mass Spectrometry in Organic Chemistry

Mass spectrometry is an analytical technique used to measure the mass-to-charge ratio of ions. It is a powerful tool in organic chemistry for identifying the composition of a chemical sample. The method involves ionizing chemical compounds to generate charged molecules or molecule fragments and measuring their mass-to-charge ratios.

Using mass spectrometry, chemists can determine the molecular weight of a compound, and its chemical structure, and even identify unknown substances. The process often begins with the creation of a molecular ion, which is the ionized form of the molecule being analyzed, usually lacking one electron. After ionization, the molecular ion may break down into smaller fragments, and the mass of these ions or the intact molecular ion can provide valuable insights into the molecular formula and structure of the organic compound. As seen in the exercise, the technique is used to evaluate whether the compounds containing heteroatoms have even- or odd-numbered molecular ion masses, which is a fundamental aspect of interpreting mass spectrometry data.

Molecular Formulas of Organic Compounds

The molecular formula of an organic compound expresses the number of each type of atom present in a molecule. It is a foundational concept in organic chemistry that paves the way for understanding molecular structure, reactivity, and properties of the compound. The molecular formula is often written in a way that lists the number of carbon atoms first, followed by hydrogen, and then in alphabetical order for the heteroatoms present.

In the exercise, we have different types of organic molecules each with unique molecular formulas containing heteroatoms, such as chloroalkane (\(\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{Cl}\)), and bromoalkane (\(\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{Br}\)). These formulas help describe the exact composition of the compounds. It is crucial for students to understand how to calculate the molecular ion mass from these formulas, as they are the first step in predicting and interpreting results from mass spectrometry analyses. Understanding molecular formulas also aids in grasping isomerism and the diverse range of compounds that can exist even with the same molecular formula.