Question

The mass spectrum of compound A shows the molecular ion at \(m / z 85\), an \(\mathrm{M}+1\) peak at \(\mathrm{m} / z 86\) of approximately \(6 \%\) abundance relative to \(M\), and an \(M+2\) peak at \(m / z 87\) of less than \(0.1 \%\) abundance relative to \(\mathrm{M}\). (a) Propose a molecular formula for compound A. (b) Draw at least ten possible structural formulas for this molecular formula.

Short answer

a) Based on the mass spectrum data and our analysis, there are two possible molecular formulas for compound A: \(\textbf{C}_6\textbf{H}_{13}\) and \(\textbf{C}_6\textbf{H}_{11}\textbf{N}\). b) Here are ten possible structural formulas based on these molecular formulas: For \(\textbf{C}_6\textbf{H}_{13}\): 1. Hexane (straight-chain) 2. 2-Methylpentane 3. 3-Methylpentane 4. 2,2-Dimethylbutane 5. 2,3-Dimethylbutane For \(\textbf{C}_6\textbf{H}_{11}\textbf{N}\): 6. 1-Hexene-1-amine (primary amine) 7. 1-Hexene-2-amine (primary amine) 8. 1-Hexene-3-amine (primary amine) 9. 2-Methyl-1-pentene-1-amine (primary amine) 10. N,1-Dimethylpentene-1-amine (secondary amine)

Step-by-step solution

Analyze the Molecular Ion Peak

The molecular ion peak (\(m/z\)) reveals the molecular weight of compound A, which is 85 atomic mass units (amu). This gives us a starting point to predict the molecular formula.

Consider the M+1 Peak

An M+1 peak has a mass 1 amu greater than the molecular ion peak. The relative abundance of the M+1 peak is approximately 6% compared to the molecular ion peak. This suggests the presence of a significant amount of carbon, as the relative abundance of ^13C (the isotope responsible for the M+1 peak) in nature is about 1%. With 6% abundance, we can estimate the number of carbons present in compound A. For simplicity, we can approximate to 1% per carbon. Thus, there could be around 6 carbon atoms present in the molecular formula of compound A.

Determine the Molecular Formula

To get the molecular formula, we need to account for the mass of the remaining elements. The molecular ion peak for compound A is 85 amu, and the mass of 6 carbon atoms is 6(12) = 72 amu. So there is 85 - 72 = 13 amu left to assign to the remaining elements. We can assume that the compound contains hydrogen and possibly other elements like nitrogen, oxygen, or halogens. However, the M+2 peak (<0.1% abundance relative to M) implies a low probability of having elements like oxygen, sulfur, or any halogens with their isotope peaks at the same level. Thus, it's reasonable to consider only carbon, hydrogen, and possibly nitrogen for compound A. If the compound were to have nitrogen, it would have to contain one nitrogen atom. Thus, the possible molecular formulas are \(\textbf{C}_6\textbf{H}_{13}\) (no nitrogen) and \(\textbf{C}_6\textbf{H}_{11}\textbf{N}\) (with one nitrogen atom).

Structural Formulas for C6H13

Here are some possible structural formulas for \(\textbf{C}_6\textbf{H}_{13}\): 1. Hexane (straight-chain) 2. 2-Methylpentane 3. 3-Methylpentane 4. 2,2-Dimethylbutane 5. 2,3-Dimethylbutane

Structural Formulas for C6H11N

Here are some possible structural formulas for \(\textbf{C}_6\textbf{H}_{11}\textbf{N}\): 1. 1-Hexene-1-amine (primary amine) 2. 1-Hexene-2-amine (primary amine) 3. 1-Hexene-3-amine (primary amine) 4. 2-Methyl-1-pentene-1-amine (primary amine) 5. N,1-Dimethylpentene-1-amine (secondary amine) There are other possible structural formulas for this exercise.