Question

The mass spectrum of compound A shows the molecular ion at \(m / z 85\), an \(\mathrm{M}+1\) peak at \(m / z 86\) of approximately \(6 \%\) abundance relative to \(\mathrm{M}\), and an \(\mathrm{M}+2\) 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 10 possible structural formulas for this molecular formula.

Short answer

Based on the mass spectrum data and the given molecular ion peak, M+1 peak, and M+2 peak abundances, we determined that compound A has a molecular formula of C6H13. Afterward, we drew 10 possible structural formulas for the determined molecular formula. The actual structure can be determined using additional analytical techniques, such as NMR-spectroscopy or IR-spectroscopy.

Step-by-step solution

Find the molecular weight of compound A

Since the molecular ion peak has an m/z of 85, this means the molecular weight of compound A is 85.

Calculate the number of carbon atoms in the molecule

The M+1 peak has a relative abundance of 6% compared to the molecular ion peak. This M+1 peak arises primarily from the presence of \(^{13}\mathrm{C}\) isotope in the molecule. Since the natural abundance of \(^{13}\mathrm{C}\) is roughly 1%, we can assume that the molecule contains about six carbon atoms (\(6\% / 1\% = 6\) carbon atoms).

Calculate the number of hydrogen atoms in the molecule

The M+2 peak has a very low abundance, less than 0.1% relative to the molecular ion peak. Since masses of common elements in organic compounds are C (12), H (1), O (16), and N (14), and we've already accounted for carbon, we can deduce that there is only a small contribution of other elements with higher isotopes that lead to the M+2 peak. Based on this information, we can estimate that the molecule contains mostly hydrogen atoms and possibly oxygen and/or nitrogen atoms. Let's calculate the number of hydrogen atoms considering molecular weight and number of carbon atoms: \((\text{number of carbon atoms} \times \text{atomic mass of carbon}) + (\text{number of hydrogen atoms} \times \text{atomic mass of hydrogen}) = \text{molecular weight}\) \((6 \times 12) + (x \times 1) = 85\) \(72 + x = 85\) So, \(x=13\) hydrogen atoms.

Determine the molecular formula of the compound

Based on the obtained results, compound A has a molecular formula of \(\text{C}_6 \text{H}_{13}\).

Draw possible structural formulas

Now we can draw at least 10 possible structural formulas for compound A with the molecular formula \(\text{C}_6 \text{H}_{13}\): 1. CH\(_3\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-CH\(_3\) (hexane) 2. CH\(_3\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-(CH\(_3\))\(_2\)CH (3-Methylpentane) 3. CH\(_3\)-CH\(_2\)-CH(CH\(_3\))-CH\(_2\)-CH\(_3\) (2-Methylpentane) 4. (CH\(_3\))\(_2\)CH-CH\(_2\)-CH\(_2\)-CH\(_3\) (2,3-Dimethylbutane) 5. CH\(_3\)-CH\(_2\)-(CH\(_3\))\(_2\)CH-CH\(_3\) (3,3-Dimethylpentane) 6. CH\(_3\)-CH(CH\(_3\))-CH(CH\(_3\))-CH\(_3\) (2,2-Dimethylbutane) 7. (CH\(_3\))\(_2\)CH-CH(CH\(_3\))-CH\(_3\) (2,2,3-Trimethylpentane) 8. (CH\(_3\))\(_3\)C-CH\(_2\)-CH\(_3\) (2,3,3-Trimethylpentane) 9. (CH\(_3\))\(_3\)CH-(CH\(_3\))\(_2\)CH (4,4-Dimethyl-2-pentene) 10. CH\(_3\)-C(CH\(_3\))\(_2\)-C(CH\(_3\))\(_2\)-CH\(_3\) (3,3-Dimethyl-2-butene) These are 10 possible structural formulas for compound A. The actual structure would need to be determined using information from additional analytical techniques, such as NMR-spectroscopy or IR-spectroscopy.

Key concepts

Molecular Ion Peak

Mass spectrometry is a powerful analytical tool in organic chemistry, used to determine the molecular weight and structure of organic compounds. One of the key features in a mass spectrum is the molecular ion peak, often designated as M+.

The molecular ion peak corresponds to the unfragmented molecule of the compound being ionized. This peak provides the molecular weight of the compound. For instance, when mass spectrometry shows a molecular ion peak at an m/z (mass-to-charge ratio) of 85, it indicates that the molecular weight of the compound is 85.

In the given exercise, students were able to deduce the molecular weight from the highest m/z value, which corresponds to the molecular ion peak. In the context of an educational online platform, it's crucial to emphasize that the exact position of the molecular ion peak offers direct insight into the mass of the compound, which is the first step towards understanding the compound's possible structure.

M+1 Peak Analysis

Following the identification of the molecular ion peak, the M+1 peak analysis provides additional insights into the isotopic composition of the compound. The M+1 peak is primarily due to the naturally occurring isotope ¹³C, which has a natural abundance of approximately 1.1%.

By measuring the relative intensity of the M+1 peak to the molecular ion peak, chemists can estimate the number of carbon atoms present in the molecule. For example, if the M+1 peak is about 6% of the intensity of the M peak, we can hypothesize there are approximately six carbon atoms because each contributes roughly 1% to this M+1 peak intensity (6% divided by 1.1% per carbon atom).

This kind of analysis is particularly useful for providing content that helps students see beyond the numbers and understand the reason why certain peaks are present in the mass spectrum. It bridges the gap between observing a spectrum and interpreting the underlying molecular structure, enhancing comprehension.

Structural Formulas for Organic Compounds

Once the molecular formula of an organic compound is known, the next step is to explore the possible structural formulas. Structural formulas represent the unique arrangement of atoms within the molecule. Multiple structural formulas can correspond to a single molecular formula due to isomerism.

In our exercise scenario, when compound A is determined to have a molecular formula of C₆H₁₃, there are many structural isomers possible. These structures include straight-chain and branched alkanes, as well as cycloalkanes, if any ring structures are possible with that number of carbon atoms. The student's task is to draw at least 10 possible isomers to illustrate the concept of structural diversity.

By engaging with exercises that require the drawing of structural formulas, students learn how the same molecular formula can give rise to different molecules with unique chemical and physical properties. Providing numerous examples, as in the step-by-step solution, not only describes but also visually demonstrates the variety of organic structures – a revelation that often excites students about the richness of organic chemistry.