Question

Biomolecule Researchers isolated an unknown substance, \(X\), from rabbit muscle. They determined its structure from the following observations and experiments. Qualitative analysis showed that \(X\) was composed entirely of \(C, H\), and \(O\). \(A\) weighed sample of \(X\) was completely oxidized, and the \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) produced were measured; this quantitative analysis revealed that \(\mathrm{X}\) contained \(40.00 \% \mathrm{C}, 6.71 \% \mathrm{H}\), and \(53.29 \% \mathrm{O}\) by weight. The molecular mass of \(\mathrm{X}\), determined by mass spectrometry, was \(90.00\) u (atomic mass units; see Box 1-1). Infrared spectroscopy showed that \(X\) contained one double bond. X dissolved readily in water to give an acidic solution that demonstrated optical activity when tested in a polarimeter. a. Determine the empirical and molecular formula of \(X\). b. Draw the possible structures of \(X\) that fit the molecular formula and contain one double bond. Consider only linear or branched structures and disregard cyclic structures. Note that oxygen makes very poor bonds to itself. c. What is the structural significance of the observed optical activity? Which structures in (b) are consistent with the observation?

Short answer

Empirical formula: CH\(_2\)O. Molecular formula: C\(_3\)H\(_6\)O\(_3\). Glyceraldehyde is optically active due to its chiral center.

Step-by-step solution

Calculate the Empirical Formula

To find the empirical formula, assume 100 g of the substance. Then, the masses of C, H, and O are 40.00 g, 6.71 g, and 53.29 g, respectively.**Convert masses to moles:**\[\text{Moles of } C = \frac{40.00}{12.01} = 3.33\]\[\text{Moles of } H = \frac{6.71}{1.008} = 6.66\]\[\text{Moles of } O = \frac{53.29}{16.00} = 3.33\]**Find the smallest whole number ratio:**\[\text{C:H:O} = 3.33:6.66:3.33 \approx 1:2:1\]The empirical formula is \(\text{CH}_2\text{O}\).

Determine the Molecular Formula

Find the ratio of the molecular mass to the empirical formula mass.**Calculate the empirical formula mass:**\[\text{Empirical formula mass of } \text{CH}_2\text{O} = 12.01 + (2 \times 1.008) + 16.00 = 30.03\]**Calculate the ratio:**\[\text{Ratio} = \frac{90.00}{30.03} \approx 3\]Multiply the empirical formula by this ratio:\[\text{Molecular formula} = (\text{CH}_2\text{O})_3 = \text{C}_3\text{H}_6\text{O}_3\]

Propose Possible Structures with One Double Bond

Since the molecular formula is \(\text{C}_3\text{H}_6\text{O}_3\) and there is one double bond, consider linear or branched structures.Possible structures:1. **Glyceraldehyde**: CHO-CH(OH)-CH2OH2. **Dihydroxyacetone**: CH2OH-C(=O)-CH2OHBoth of these have one C=O double bond and conform to the formula.

Evaluate Optical Activity

Optical activity indicates chirality. Only molecules with asymmetric (chiral) centers show optical activity. **Check chirality in proposed structures:** - **Glyceraldehyde** has a chiral center at the middle carbon, so it's optically active. - **Dihydroxyacetone** does not have a chiral center and is not optically active. Considering optical activity, glyceraldehyde is consistent with the observation.

Key concepts

Optical Activity and Chirality

Optical activity is a property of certain molecules to rotate the plane of polarized light. This property is observed in molecules that are chiral. Chirality refers to a molecular geometry where a molecule cannot be superimposed on its mirror image, similar to how our left and right hands are mirror images but not identical.
This chirality arises from the presence of an asymmetric carbon atom, which is a carbon atom bonded to four different atoms or groups. In the exercise, optical activity was used to identify if the molecule was chiral, which was necessary for determining the structure of molecule X.
In the proposed structures, glyceraldehyde contains a chiral center that contributes to its optical activity, whereas dihydroxyacetone does not have a chiral center. Therefore, its lack of optical activity is consistent with its structural arrangement.

Mass Spectrometry in Biochemistry

Mass spectrometry is a critical tool in biochemistry that allows for the determination of molecular mass by measuring the mass-to-charge ratio of ions. This is particularly useful for determining the molecular weight and structure of biomolecules.
In the exercise provided, mass spectrometry was utilized to find that the molecular mass of substance X was 90.00 u, which helped in confirming the molecular formula as \( \text{C}_3\text{H}_6\text{O}_3 \).
By comparing the experimental molecular mass to the calculated empirical formula mass, we can deduce possible molecular structures. This method is instrumental in narrowing down the possible configurations and compositions of unknown substances, thus aiding in structural analysis and characterization.

Infrared Spectroscopy Analysis

Infrared (IR) spectroscopy is a technique used to identify functional groups in a molecule based on the absorption of infrared light at specific frequencies. Each type of bond interacts with infrared light differently, leading to unique spectral patterns.
In this exercise, the infrared spectrum indicated the presence of a double bond within molecule X, as characteristic IR peaks for double bonds, like carbonyl groups, appear at distinct frequency ranges.
This information hinted that substance X might contain a carbonyl group. Consequently, the structures of glyceraldehyde and dihydroxyacetone both feature a C=O bond, matching the data obtained from IR spectroscopy. This ensures that molecular structures align with chemical observations.

Chemical Composition Analysis

Chemical composition analysis involves determining the elemental makeup and proportion of a compound. This process often includes both qualitative and quantitative analysis to reveal the elements present and their respective percentages.
For the compound X, chemical composition analysis revealed a percentage weight of 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen. This information was crucial for calculating the empirical formula, which is based on mole ratios derived from these percentages.
Once the empirical formula was calculated as \( \text{CH}_2\text{O} \), it was then multiplied by the integer ratio derived from comparing the empirical formula mass to the molecular mass to determine the actual molecular formula \( \text{C}_3\text{H}_6\text{O}_3 \). This step is vital in uncovering the fundamental structure of the molecule through analytical reasoning.

Double Bond Identification in Biomolecules

Identifying double bonds within biomolecules is key to elucidating their functional properties and reactivities. Double bonds, particularly carbon-carbon and carbon-oxygen, influence the shape and behavior of the molecule.
In the given problem, it was noted that the substance X must contain one double bond. Initially, through infrared spectroscopy, the presence of such a bond was confirmed.
Knowledge of potential double bonds guided the logical deduction of the structural candidates: glyceraldehyde and dihydroxyacetone. Each contains a carbonyl double bond \( \text{C}=\text{O} \), fulfilling both the molecular formula and spectroscopic findings. This understanding of bonding patterns is fundamental in organic chemistry, allowing scientists to predict molecular interactions and stability.