15.6d Structural Determination From All Spectra Example 4 | Organic Chemistry
TLDRThe video script discusses the process of identifying an unknown molecule using four different spectroscopic techniques: mass spectrometry, infrared (IR) spectroscopy, carbon-13 (C13) NMR, and proton (H1) NMR. The analysis begins with the mass spectrum, revealing a molecular weight of 170 and the presence of bromine. The IR spectrum suggests the presence of an aromatic ring, likely a benzene ring, due to specific peak patterns. The C13 NMR spectrum confirms the aromatic nature with four distinct signals, indicating symmetry. Finally, the H1 NMR spectrum provides the most detailed information, allowing the identification of a methyl group and the symmetrical substitution on the benzene ring. By piecing together the information from all spectra, the molecule is deduced to be a benzene ring with a bromine on one side and a methyl group on the other, totaling a molecular weight of 170.
Takeaways
- 🔍 **Mass Spectrum Analysis**: The molecular weight of the compound is 170, indicated by the molecular ion peak at m/z 170.
- 🌊 **Presence of Bromine**: An M+2 peak at m/z 172 suggests the presence of bromine in the molecule, as it is a characteristic isotope peak for bromine.
- 🚫 **Absence of Nitrogen**: The even molecular weight implies the absence of an odd number of nitrogen atoms in the compound.
- 🔬 **IR Spectrum Examination**: The IR spectrum reveals peaks outside the fingerprint region, indicating the presence of sp2 and sp3 hybridized carbon-hydrogen bonds.
- 💠 **Aromatic Ring Detection**: A signal in the IR spectrum suggests the presence of an aromatic ring, likely a benzene ring, which is further confirmed by the NMR spectra.
- 📊 **Carbon-13 NMR Spectrum**: The carbon-13 NMR spectrum shows six distinct signals, indicating symmetry within the aromatic region of the molecule.
- 🧬 **Symmetry in Aromatic Carbons**: The benzene ring's carbons are symmetrical, as evidenced by only four signals in the aromatic region of the carbon-13 NMR spectrum.
- 🧲 **Proton NMR Spectrum**: The proton NMR spectrum provides further confirmation of the benzene ring and identifies a singlet peak, indicating a methyl group with no neighboring protons.
- 🔢 **Molecular Formula Deduction**: The total molecular weight of 170 is consistent with the sum of the weights of a benzene ring, a bromine atom, and a methyl group.
- 🧩 **Molecule Structure Assembly**: The final structure has a benzene ring in the center with a bromine atom on one side and a methyl group on the other, maintaining the symmetry.
- 📐 **Bonding Considerations**: Bromine, with seven valence electrons, forms only one bond, while the benzene ring can form multiple bonds, positioning it centrally in the molecule.
Q & A
What is the recommended order to analyze the given spectra?
-The recommended order is to start with the mass spectrum, followed by the infrared (IR) spectrum, then the carbon-13 (C13) NMR spectrum, and finally the proton (H) NMR spectrum.
What does the presence of an M plus 2 peak in the mass spectrum indicate?
-The presence of an M plus 2 peak indicates that the compound contains bromine, as most compounds do not have a significant M plus 2 peak.
What does the absence of a significant signal around 1650 cm⁻¹ in the IR spectrum suggest?
-The absence of a significant signal around 1650 cm⁻¹ suggests that the compound does not contain an alkene, as this region is typically associated with C=C stretching in alkenes.
How can the presence of a benzene ring be inferred from the IR spectrum?
-The presence of a benzene ring can be inferred from the IR spectrum by the presence of one, two, or three peaks in the region slightly less than 1500 cm⁻¹ to slightly more than 1600 cm⁻¹.
What does the number of signals in the carbon-13 NMR spectrum indicate about the compound's structure?
-The number of signals in the carbon-13 NMR spectrum indicates the number of distinct carbon environments in the compound. In this case, six signals suggest a symmetrical benzene ring with substituents.
What does the singlet peak in the alkane region of the proton NMR spectrum represent?
-A singlet peak in the alkane region of the proton NMR spectrum represents a methyl group with no neighboring protons, indicating it is not adjacent to any other carbon atoms with hydrogen atoms.
How does the symmetry in the aromatic region of the proton NMR spectrum help in determining the structure?
-Symmetry in the aromatic region of the proton NMR spectrum suggests that there are two substituents on the benzene ring in a para-disubstituted pattern on opposite sides, maintaining the symmetry.
What is the significance of the multiplet pattern in the proton NMR spectrum?
-The multiplet pattern in the proton NMR spectrum represents two hydrogens on the benzene ring, indicating the presence of a symmetrical benzene ring with substituents.
How does the molecular weight of 170 and the presence of bromine help in determining the molecular formula?
-The molecular weight of 170 and the presence of bromine (which weighs 79) help in determining the molecular formula by allowing the calculation of the remaining mass that must be accounted for by other parts of the molecule, such as the benzene ring and the methyl group.
What is the role of the benzene ring in the final structure of the molecule?
-The benzene ring serves as the central structure to which the bromine and methyl group are attached. It is the only part of the molecule that can make multiple bonds due to its stability and the nature of its resonance.
What is the final structure of the molecule based on the analysis of the four spectra?
-The final structure of the molecule is a benzene ring with a bromine atom on one side and a methyl group on the other, with the benzene ring in the middle due to its ability to form multiple bonds and maintain the symmetrical structure.
Outlines
🔍 Analyzing Spectra to Determine Molecular Structure
The speaker begins by introducing the task of analyzing various spectra to determine the structure of a compound. They lay out a process for examining the mass spectrum, infrared (IR) spectrum, carbon-13 NMR, and proton NMR in sequence. The mass spectrum reveals a molecular weight of 170 with a noticeable M+2 peak, suggesting the presence of bromine. The IR spectrum indicates the presence of an aromatic ring, likely benzene, due to specific peak patterns. The carbon-13 NMR provides further evidence of aromaticity with signals in the 110 to 160 ppm range, and the proton NMR confirms the benzene ring and identifies a methyl group. The speaker emphasizes the importance of quickly analyzing the mass spec, IR, and carbon-13 NMR to spend more time on the proton NMR, which offers the most information for structural determination.
🧩 Piecing Together the Molecular Formula and Structure
The speaker then calculates the molecular formula based on the given mass and known presence of a bromine atom and a benzene ring. They deduce that the compound consists of a benzene ring with a mass of 72 (excluding hydrogen), four hydrogen atoms (adding up to 4), and a methyl group with a mass of 15, summing up to the molecular weight of 170. The speaker concludes by constructing the molecule, placing the bromine on one side and the methyl group on the other, with the benzene ring in the middle. The resulting structure is symmetrical, reflecting the information obtained from the spectra.
Mindmap
Keywords
💡Mass Spectrum
💡Infrared (IR) Spectrum
💡Carbon-13 NMR Spectrum
💡Proton NMR Spectrum
💡Degrees of Unsaturation
💡Molecular Weight
💡Brominated Compound
💡Benzene Ring
💡Symmetry
💡Methyl Group
💡Valence Electrons
Highlights
The process begins with analyzing the mass spectrum to identify the molecular ion peak, which reveals the molecular weight of the compound.
The presence of an M plus 2 peak at 172 indicates the compound contains bromine.
The absence of a significant peak around 1650 cm⁻¹ in the IR spectrum suggests the absence of an alkene.
A signal in the IR spectrum between 1500 and 1600 cm⁻¹ provides evidence for an aromatic compound.
The carbon-13 NMR spectrum shows four signals, indicating the presence of an alkane and aromatic carbons.
The carbon-13 NMR spectrum also suggests the presence of symmetry in the benzene ring.
The proton NMR spectrum confirms the presence of a benzene ring and an alkane region.
A singlet peak in the alkane region of the proton NMR spectrum indicates a methyl group with no neighboring protons.
The pattern of doublets in the proton NMR spectrum represents two hydrogens each, characteristic of a symmetrical benzene ring.
The molecular weight of 170 is consistent with the sum of the weights of bromine, a benzene ring, and a methyl group.
The structure of the molecule is deduced to have a benzene ring in the center with a bromine on one side and a methyl group on the other.
The importance of quickly analyzing mass spec, IR, and carbon-13 NMR before spending more time on the proton NMR is emphasized.
The process demonstrates how to construct a molecule's structure from spectral data without knowing its formula beforehand.
The use of symmetry in the benzene ring to simplify the interpretation of the carbon-13 NMR spectrum.
The identification of the molecular weight and the presence of bromine as key initial steps in spectral analysis.
The significance of the M plus 2 peak for identifying halogen atoms in the molecular structure.
The method for differentiating between sp2 and sp3 hybridized carbon-hydrogen bonds using the IR spectrum.
The strategy for confirming the presence of an aromatic ring using a combination of IR and carbon-13 NMR spectra.
The interpretation of the proton NMR spectrum to identify the functional groups and their arrangement in the molecule.
Transcripts
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