When reactions are run in THF, it is often possible to isolate traces of an impurity, compound \(1 .\) Compound \(\mathbf{1}\) gives \(81.76 \% \mathrm{C}\) and \(10.98 \% \mathrm{H}\) upon elemental analysis. Spectral data for \(\mathbf{1}\) are summarized below. Deduce the structure of compound \(\mathbf{1}\) and explain your reasoning. Compound 1 Mass spectrum: \(m / z=220(\mathrm{M}, 23 \%), 205(100 \%), 57(27 \%)\) IR (Nujol): 3660 (m, sharp) cm \(^{-1}\) \({ }^{1} \mathrm{H}\) NMR \(\left(\mathrm{CDCl}_{3}\right): \delta 1.43(\mathrm{~s}, 18 \mathrm{H}), 2.27(\mathrm{~s}, 3 \mathrm{H}), 5.00(\mathrm{~s}, 1 \mathrm{H},\), shifts with \(\left.\mathrm{D}_{2} \mathrm{O}\right), 6.98(\mathrm{~s}, 2 \mathrm{H})\) \({ }^{13} \mathrm{C}\) NMR \(\left(\mathrm{CDCl}_{3}\right): \delta 21.2(\mathrm{q}), 30.4(\mathrm{q}), 34.2(\mathrm{~s}), 125.5(\mathrm{~d})\), \(128.2(\mathrm{~s}), 135.8(\mathrm{~s}), 151.5(\mathrm{~s})\) (hydrogen coupling indicated in parentheses)

Step by step solution

01

Calculate the empirical formula using elemental analysis

Begin by converting the percentage composition to grams starting with 100 grams of the compound. Convert the mass of each element to moles using their atomic masses:Carbon: \[ \frac{81.76 \, \text{g}}{12.01 \, \text{g/mol}} = 6.81 \, \text{mol} \]Hydrogen: \[ \frac{10.98 \, \text{g}}{1.008 \, \text{g/mol}} = 10.89 \, \text{mol} \]Find the ratio of moles by dividing by the smallest value:\[\frac{6.81}{6.81} = 1\quad\text{and}\quad \frac{10.89}{6.81} = 1.6 \]Approximate the ratio to the nearest whole number if possible. Here, the ratio is close to 1:1.6 (which is close to 1:2). Thus, the empirical formula is approximated as \( \text{C}_5 \text{H}_{10} \).

02

Determine the molecular formula

From the mass spectrum data, the molecular ion peak \( (M) \) is at \( m/z = 220 \). For the empirical formula \( \text{C}_5 \text{H}_{10} \), the molar mass is calculated as: \[ 5(12.01) + 10(1.008) = 70.12 \, \text{g/mol} \]Since \( 220 \, \text{g/mol} \) is the molecular weight and \( 70.12 \) is the empirical formula weight, find the ratio \( n \) by dividing the molecular weight by the empirical formula weight: \[ \frac{220}{70.12} \approx 3.14 \]This ratio suggests the molecular formula is roughly three times the empirical formula. Thus, the molecular formula is approximated as \( \text{C}_{15} \text{H}_{30} \).

03

Analyze the mass spectrum

The mass spectrum has peaks at \( m/z = 220 \) (molecular ion), \( 205 \) (base peak), and \( 57 \). The peaks suggest fragmentation patterns and possible structure. The peak at \( 205 \) likely indicates the loss of a methyl group (\( -15 \)).

04

Analyze the IR spectrum

The IR spectrum has a peak at \( 3660 \, \text{cm}^{-1} \) indicating an O-H stretch, suggesting the presence of hydroxyl or alcohol groups.

05

Analyze the NMR spectra

The \(^1 \text{H} \) NMR shows signals at \( \delta 1.43 \) (singlet, 18 H), \( \delta 2.27 \) (singlet, 3 H), \( \delta 5.00 \) (singlet, 1 H, shifts with \( \text{D}_2 \text{O} \)), and \( \delta 6.98 \) (singlet, 2 H). The singlet at \( 5.00 \) shifts with \( \text{D}_2 \text{O} \), indicating a hydroxyl proton. The \( ^{13} \text{C} \) NMR shows seven peaks which suggest aliphatic and aromatic carbons.

06

Propose the structure

Consider the molecular formula \( \text{C}_{15} \text{H}_{30} \), the presence of an O-H group, and the pattern of methyl groups. Combine the interpretation to suggest the structure of the impurity. Likely, it could be a compound bearing enough symmetrical methyl groups and a benzyl alcohol structure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

elemental analysis

Elemental analysis is a key technique in organic chemistry used to determine the percentage composition of a compound by weight. This technique helps us find out how much carbon, hydrogen, nitrogen, and other elements are present in a sample. For example, if a compound gives 81.76% C and 10.98% H, as in our problem, we can convert these percentages to grams assuming we have 100 grams of the compound. This gives us 81.76 grams of carbon and 10.98 grams of hydrogen.
Next, we convert these masses to moles using the atomic masses of carbon (12.01 g/mol) and hydrogen (1.008 g/mol). Dividing the masses by their respective atomic masses, we get the number of moles of each element. The mole ratio of the elements helps us determine the empirical formula. By dividing by the smallest number of moles, we find the simplest whole number ratio, which approximates the formula. Understanding elemental analysis is crucial for figuring out the empirical formula of unknown compounds.

mass spectrometry

Mass spectrometry (MS) is an analytical tool used for measuring the mass-to-charge ratio of ions. It provides important information such as the molecular weight of the compound and patterns of fragmentation. In our problem, the mass spectrum shows peaks at m/z = 220 (M+, the molecular ion peak), 205 (base peak), and 57.
The molecular ion peak at m/z 220 indicates the mass of the whole molecule, while the base peak (the most intense peak) often indicates a stable fragment. For instance, the peak at m/z 205 suggests a loss of a 15 mass unit fragment (which could be a methyl group). These fragmentation patterns help in deducing the structure of the compound by showing how the molecule breaks apart, revealing critical structural information.

IR spectroscopy

Infrared (IR) spectroscopy is a technique used to identify functional groups within a molecule by measuring the absorption of infrared light at various wavelengths. Different bonds absorb IR light at characteristic frequencies.
For example, in our problem, the IR spectrum shows a peak at 3660 cm^-1, which is indicative of an O-H bond, most commonly found in alcohols. The sharp nature of the peak provides further confirmation. Recognizing these peaks helps us understand the types of chemical bonds in the sample, giving clues about its functional groups and, hence, its structure.

NMR spectroscopy

Nuclear Magnetic Resonance (NMR) spectroscopy is used to determine the local environment of hydrogen and carbon atoms in a compound. There are two types of NMR used here:

• ^1H NMR (Proton NMR): This detects hydrogen atoms and how they are connected in the molecule. For instance, in our problem, the signals at δ 1.43 (s, 18H), δ 2.27 (s, 3H), δ 5.00 (s, 1H, shifts with D2O), and δ 6.98 (s, 2H) tell us about the different hydrogen environments.
• ^13C NMR (Carbon-13 NMR): This detects carbon atoms. Here, the δ values provide information about the number and types of carbon atoms in the compound. Signals at δ 21.2, 30.4, 34.2, 125.5, 128.2, 135.8, and 151.5 suggest a combination of aliphatic and aromatic carbons.

These insights help in piecing together the molecular structure by showing how different atoms are connected.

empirical formula calculation

Empirical formula calculation involves determining the simplest whole-number ratio of elements in a compound. This process starts from the results of the elemental analysis. For our problem, we convert the given percentages of carbon and hydrogen into grams, assuming a 100g sample. Then, we convert those grams into moles.
For instance, 81.76 grams of carbon becomes 6.81 mol, and 10.98 grams of hydrogen becomes 10.89 mol. We then find the simplest mole ratio by dividing by the smallest number of moles, which here approximates to a ratio close to 1:2. Hence, the empirical formula is C5H10.
This formula helps in further steps like determining the molecular formula, which can be found by comparing the empirical formula weight to the molecular ion peak in the mass spectrum.