Question

(a) Compound \(\mathbf{Z}\) is a liquid at \(298 \mathrm{K}\) High-resolution mass spectrometry gives an exact mass of \(61.01638,\) and \(\mathbf{Z}\) contains \(19.7 \%\) C \(, 4.9 \%\) \(\mathrm{H}\) and \(22.9 \% \mathrm{N} .\) Suggest a possible formula for \(\mathrm{Z}\) (b) The low-resolution mass spectrum of Z shows intense peaks at 61,46,30 and \(15 .\) Deduce a possible structure of \(\mathbf{Z}\)

Short answer

The molecular formula is \( \text{C}_2\text{H}_5\text{N}_2 \), and a possible structure is ethyl-diazene.

Step-by-step solution

Determine the Molecular Formula from Composition

First, calculate the relative molar amounts of C, H, and N. Assume 100 g of Z for convenience. That provides 19.7 g C, 4.9 g H, and 22.9 g N. Converting to moles: C: \( \frac{19.7}{12.01} \approx 1.64 \text{ moles} \), H: \( \frac{4.9}{1.008} \approx 4.86 \text{ moles} \), N: \( \frac{22.9}{14.01} \approx 1.64 \text{ moles} \). The smallest mole ratio is \( 1.64 \), so divide all by this: C: \( 1 \), H: \( 2.96 \approx 3 \), N: \( 1 \). This suggests a formula of \( \text{C}_1\text{H}_3\text{N}_1 \).

Correct and Verify the Molecular Formula

The empirical mass of \( \text{C}_1\text{H}_3\text{N}_1 \) is approximately 29.038. Since the observed mass is 61.01638, a possible molecular formula is \( \text{C}_2\text{H}_5\text{N}_2 \) (actual mass 61.048). This fits the exact mass given.

Analyze Low-resolution Mass Spectrum

The intense peaks at 61, 46, 30, and 15 suggest fragment ions from the molecule. The base peak at 61 corresponds to the molecular ion \( \text{C}_2\text{H}_5\text{N}_2^+ \); 46 likely relates to the loss of \( \text{NH}_3 \) (mass 17), leaving \( \text{C}_2\text{H}_3 \); 30 may suggest \( \text{CH}_3\text{N} \); 15 might represent \( \text{CH}_3^+ \).

Determine a Possible Structure

Given the formula \( \text{C}_2\text{H}_5\text{N}_2 \), a possible structure is ethyl-diazene \( \text{CH}_3\text{CHRN}_2 \), where R indicates further bonding interactions like \( \text{CH}_3\text{CNH}_2 \), creating plausible fragmentation pathways matching the mass spectrum.

Key concepts

Molecular Formula Determination

Molecular formula determination is an essential concept in analytical chemistry. It involves assessing the composition of a compound to reveal its molecular makeup. In mass spectrometry, you start with percentage composition, which in this example was 19.7% carbon, 4.9% hydrogen, and 22.9% nitrogen.
To find out the molecular formula, you assume a 100-gram sample of the compound and convert these percentages into grams: it's like having 19.7 grams of carbon, 4.9 grams of hydrogen, and 22.9 grams of nitrogen. Then, these gram amounts are converted into moles by dividing by their atomic masses:

• Carbon: Relating 19.7g to its atomic weight (12.01 g/mol) gives approximately 1.64 moles.
• Hydrogen: 4.9g divided by 1.008 g/mol results in about 4.86 moles.
• Nitrogen: 22.9g divided by 14.01 g/mol equals about 1.64 moles.

To simplify, divide each by the smallest value, in this case, 1.64, to obtain a rudimentary formula, eliminating decimals wherever possible. This process resulted in an empirical formula of \(\text{C}_1\text{H}_3\text{N}_1\), which aligns with the given percentages.

Molecular Ion

A molecular ion is a key component in mass spectrometry, representing the ionized form of the entire molecule. When a molecule is bombarded with electrons in the mass spectrometer, it loses an electron, converting it into a positively charged molecule.
This gives you the molecular ion, denoted by a peak in the mass spectrum, representing the molecule's full mass with a single positive charge.
In the given problem, the base peak at 61 on the spectrum indicates the molecular ion, corresponding to \(\text{C}_2\text{H}_5\text{N}_2^+\). Understanding the molecular ion peak helps determine the full mass of the molecular formula and confirms the potential molecular structure.

Fragment Ions

Fragment ions are smaller pieces of a molecule that break off in a mass spectrometer, providing essential clues for understanding the structure of the original molecule. When the molecular ion is formed, it can fracture into fragments, each showing up as a peak in the mass spectrum. These peaks are critical for deducing molecular structure.
For compound \(Z\), peaks at 46, 30, and 15 indicate the presence of specific fragment ions that help identify simpler structures within the molecule.

• The peak at 46 corresponds to losing an \(\text{NH}_3\) group, producing \(\text{C}_2\text{H}_3\).
• The peak at 30 might relate to the fragment \(\text{CH}_3\text{N}\).
• Lastly, a peak at 15 suggests the presence of a \(\text{CH}_3^+\) ion.

Analyzing these fragment ions helps in piecing together the complete structure of the compound, offering insights into how different atoms join within the molecule.

Empirical Formula

An empirical formula represents the simplest ratio of elements in a compound. Whereas a molecular formula shows the actual number of each type of atom in a molecule, an empirical formula indicates the lowest whole number ratio.
For a compound like \(\text{Z}\), we first found an empirical formula from the experiment: \(\text{C}_1\text{H}_3\text{N}_1\). This formula suggests there is one carbon atom for every three hydrogen atoms and one nitrogen atom.
However, to match the actual mass given (61.01638), it was necessary to scale this empirical formula up. Doing so suggests the likely molecular formula \(\text{C}_2\text{H}_5\text{N}_2\), holding two of each ratio found in the empirical formula. This example illustrates the role of empirical formulas in establishing a foundation for determining a compound's full molecular structure, critical in various research and industrial applications.