Question

An alcohol \(\mathbf{X}\) has a composition of \(64.8 \% \mathrm{C}\) and \(13.6 \%\) H. The mass spectrum shows a parent ion at \(m / z=74 .\) The \(^{1} \mathrm{H}\) NMR spectrum of \(\mathbf{X}\) dissolved in \(\mathrm{CDCl}_{3}\) has signals at \(\delta 3.71\) (sextet, \(1 \mathrm{H}\) ), 2.37 (singlet, \(1 \mathrm{H}\) ), 1.46 (multiplet, \(2 \mathrm{H}\) ), 1.17 (doublet, 3H), 0.93 (triplet, \(3 \mathrm{H}\) ) ppm; in the \(^{13} \mathrm{C}\) NMR spectrum, four resonances are observed. Use these data to suggest a structure of \(\mathbf{X}\) and comment on isomer possibilities that retain the \(\mathrm{OH}\) functionality.

Short answer

The compound is 2-butanol, C4H10O. Possible isomers are 1-butanol, 2-methyl-1-propanol, tert-butanol; 2-butanol aligns with NMR data.

Step-by-step solution

Determine Empirical Formula

First, use the provided mass percentages to find the empirical formula. Assume you have 100 g of substance. Then, you have 64.8 g of carbon (C) and 13.6 g of hydrogen (H). Compute moles: \[\text{Moles of C} = \frac{64.8}{12.01} \approx 5.40\]\[\text{Moles of H} = \frac{13.6}{1.008} \approx 13.49\]Compute the simplest ratio by dividing each by the smallest number of moles:\[\frac{5.40}{5.40} = 1, \quad \frac{13.49}{5.40} \approx 2.50\]A simple multiple would be C2H5. So the empirical formula is approximately C2H5.

Determine Molecular Formula

Use the molecular ion (parent ion) peak from mass spectrometry, which gives the molecular weight as 74 g/mol. Assume the empirical formula's weight, C2H5, is approximately: \[\text{C}_2\text{H}_5 = 2(12.01) + 5(1.008) = 29.028 \text{ g/mol}\]Determine the multiple of the empirical formula weight in the molecular weight: \[\frac{74}{29.028} \approx 2.55 \Rightarrow 3.\]Hence, the molecular formula could be C4H11.

Identify Functional Group

The problem indicates that the compound is an alcohol, which contains an OH group. Based on C4H11, adjust for the oxygen as C4H10O for the final molecular formula, accounting for an OH group.

Analyze Proton NMR Data

Examine the proton NMR data:- \(\delta 3.71\) ppm (sextet, 1H) suggests a CH group next to other non-equivalent CH groups or CH2 near electronegative O (alcohol).- \(\delta 2.37\) ppm (singlet, 1H) implies a CH directly attached to electronegative atoms like O in OH.- \(\delta 1.46\) ppm (multiplet, 2H) typically indicates a CH2 group within a complex environment.- \(\delta 1.17\) ppm (doublet, 3H) and \(\delta 0.93\) ppm (triplet, 3H) likely point to a terminal CH3 adjacent to a CH2 (ethyl group).

Analyze Carbon-13 NMR Data

The Carbon-13 NMR shows four distinct signals, indicating four different carbon environments, consistent with a more symmetric alcohol like butanol. The resonances correspond to primary (CH3), secondary (CH2), tertiary (CH), and quaternary (C-OH) carbons.

Suggest Structure and Evaluate Isomers

Considering the molecular formula C4H10O and the NMR spectra, suggest a structure that fits: 2-butanol. It is an alcohol with a secondary carbon OH group near CH3 (signal at \(\delta 3.71\) ppm) and has four distinct carbon signals. Possible isomers retaining OH are: 1-butanol, 2-methyl-1-propanol, and tert-butanol. Only 2-butanol is fully consistent with the NMR data, given quaternary carbon among the isomers.

Key concepts

Molecular Formula Determination

Molecular formula determination is a fundamental step in understanding the composition of a chemical substance. It entails finding out how many atoms of each type are present in a compound. In organic chemistry, this often begins with calculating the empirical formula, which gives the simplest ratio of the atoms. For example, if a substance is composed of 64.8% carbon (C) and 13.6% hydrogen (H), you begin by converting these percentages to mass, assuming a total mass—typically 100 grams.

• Calculate moles of carbon: Divide the mass of carbon by its atomic mass (12.01 g/mol).
• Calculate moles of hydrogen: Divide the mass of hydrogen by its atomic mass (1.008 g/mol).

To get the simplest whole-number ratio, divide each mole number by the smallest value among them. This empirical formula must be scaled to reach the true molecular formula using the molar mass determined from mass spectrometry. For instance, if the empirical formula mass is 29.028 g/mol and the molecular weight from mass spectrometry is 74 g/mol, divide the two to find the multiple, thereby adjusting the empirical formula to the molecular formula.

Proton NMR Analysis

Proton NMR (\( ^1 H \) NMR) analysis is a powerful technique used to study hydrogen environments in organic molecules. It relies on the concept that hydrogen nuclei in different chemical environments absorb different radiofrequency energies.
In our example, different types of hydrogens exhibit peaks at:

• \( \delta 3.71 \) ppm – suggests a hydrogen (CH group) near an electronegative atom like oxygen (OH group in alcohols).
• \( \delta 2.37 \) ppm – usually corresponds to a hydrogen directly attached to a carbon near an oxygen, indicative of a hydroxyl (OH) group's hydrogen.
• \( \delta 1.46 \) ppm – represents complex binding environments,” suitable for CH2 groups.
• \( \delta 1.17 \) ppm (doublet) and \( \delta 0.93 \) ppm (triplet) – typically found in terminal CH3 groups adjacent to CH2.

The splitting patterns like singlets, doublets, and triplets arise from neighboring hydrogen atoms, providing information on how hydrogen atoms are situated in the molecular framework.

Carbon-13 NMR

Carbon-13 NMR (\( ^{13} C \) NMR) provides insight into the carbon skeleton of organic compounds by observing chemical shifts of carbon atoms in distinctive environments. Unlike hydrogen, \( ^{13} C \) nuclei give less intense signals due to their lower natural abundance.
In the provided analysis, four distinct resonances in the \( ^{13} C \) NMR spectrum suggest four unique carbon environments in the compound. This diversity in signals points to varied carbon attachments:

• Primary (CH3) carbons – often appear at lower ppm values.
• Secondary (CH2) carbons – show shifts dependent on surrounding groups.
• Tertiary (CH) carbons – are distinct due to fewer neighboring hydrogens.
• Quaternary (C-OH) carbons – are identifiable by their occurrence near oxygenated functional groups like OH.

By understanding these signals, structural subunits within the molecule, such as alkyl groups or functional components, become apparent.

Mass Spectrometry in Organic Chemistry

Mass spectrometry is a powerful analytical tool used in organic chemistry to determine molecular mass and molecular formula. It relies on ionizing chemical compounds to generate charged molecules or molecule fragments and measuring their mass-to-charge ratios (\( m/z \)).
For the compound given, the mass spectrometry data indicates a molecular ion peak at \( m/z = 74 \), corresponding to the molecular weight of the molecule. This value helps verify the molecular formula determined from the empirical formula by ensuring the calculated mass of the formula aligns with the mass spectrometry result.
The mass spectrum can also give insights into structural features via fragmentation patterns:

• Base peak – the tallest peak in the spectrum representing the most stable ion fragment.
• Parent ion (molecular ion peak) – corresponds to the entire molecule and aids in molecular formula determination.
• Fragmentation – provides clues on the structure by breaking the molecule at specific bonds yielding recognizable smaller ions.

By collating molecular ion data with fragmentation patterns, chemists can deduce the comprehensive structure of organic molecules.