Compound \(\mathbf{A}\) has the formula \(\mathrm{C}_{8} \mathrm{H}_{7} \mathrm{BrO}_{2}\). The two most intense peaks in the mass spectrum are at \(m / z=216\) and 214 , and they appear in almost equal amounts. What ion gives rise to each of these peaks? Compound \(\mathbf{A}\) is an aldehyde. Explain why the next most intense peaks in the mass spectrum come at \(m / z\) values of 215 and 213 .

In mass spectrometry, molecular ion peaks are crucial for identifying the molecular weight of a compound. When a molecule is ionized, it often loses an electron without fragmenting, forming a molecular ion. This ion retains the molecular weight of the compound and appears as a distinct peak in the mass spectrum.
A molecular ion peak helps in determining the molecular formula by providing the exact mass of the molecule. For instance, in our problem, peaks at \(m/z = 216\) and \(m/z = 214\) are the molecular ion peaks. Their presence indicates the molecular ion of compound \(\text{C}_8\text{H}_7\text{BrO}_2\). Identifying these peaks is the first step in understanding the compound's structure.

Isotopic distribution refers to the presence of different isotopes of elements within a molecule, giving rise to multiple peaks in a mass spectrum. Elements like Bromine and Chlorine have significant isotopic patterns due to their naturally occurring isotopes. Bromine, for example, has two main isotopes: \(^{79}\text{Br}\) and \(^{81}\text{Br}\), which appear in nearly equal amounts.
In the case of compound \(\text{C}_8\text{H}_7\text{BrO}_2\), these isotopes create noticeable molecular ion peaks at \(m/z = 216\) and \(m/z = 214\). The near-equal intensity of these peaks is due to the roughly 50-50 natural abundance of the two isotopes. This isotopic distribution provides valuable clues in mass spectrometry analysis.

Fragment ions are pieces of the original molecule that result when it breaks apart during the ionization process in mass spectrometry. These fragments can also produce peaks in the mass spectrum and help in deducing the structure of the molecule. The formation of fragment ions often follows predictable patterns.
For compound \(\text{C}_8\text{H}_7\text{BrO}_2\), the peaks at \(m/z = 215\) and \(m/z = 213\) are due to the loss of a hydrogen atom, resulting in the ions \(\text{C}_8\text{H}_6\text{^{79}BrO}_2^+\) and \(\text{C}_8\text{H}_6\text{^{81}BrO}_2^+\). This is a common fragmentation pattern in aldehydes, where the molecule tends to lose a hydrogen proton readily.

Calculating the molecular mass of a compound is essential in mass spectrometry. This involves summing up the average atomic masses of all atoms in the molecule. For the compound \(\text{C}_8\text{H}_7\text{BrO}_2\), we calculate its mass as follows:

• Carbon (C): \(12\) units, 8 atoms: \(12 \times 8 = 96\)
• Hydrogen (H): \(1\) unit, 7 atoms: \(1 \times 7 = 7\)
• Bromine (Br): \(79/81\) units, 1 atom: average of \(80\)
• Oxygen (O): \(16\) units, 2 atoms: \(16 \times 2 = 32\)

Summing these gives the nominal molecular mass: \(96 + 7 + 80 + 32 = 215\). However, the molecular ion peaks observed are at \(m/z = 216\) and \(m/z = 214\), indicating the presence of isotopes and slightly differing masses. Thus, careful calculation and consideration of isotopic distribution are vital.