For each problem, you must: 1) Calculate the degree of unsaturation. 2) Assign the principal IR absorption bands above 1500 cm¹ 3) Draw the structure of the compound 4) Label the protons on your structure with letters and assign them to peaks on the NMR spectrum (see the example below). B OHD A C 2H .0 3.5 3.0 D 1H 2.5 2.0 B 2H A 3H 1.0 PPM

Organic Chemistry: A Guided Inquiry
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Author:Andrei Straumanis
Publisher:Andrei Straumanis
ChapterL3: Carbon (13c) Nmr Spectroscopy
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### Analysis of Compound with Molecular Formula C\(_6\)H\(_6\)O\(_3\)

#### Infrared (IR) Spectrum
The IR spectrum is shown above, where the y-axis represents the percentage transmittance, and the x-axis gives the wavenumber in cm\(^{-1}\), ranging from 4000 cm\(^{-1}\) to 650 cm\(^{-1}\). Key peaks are observed at:

- **3238 cm\(^{-1}\):** Typically associated with O-H stretching vibrations, indicating the presence of hydroxyl groups.
- **3015 cm\(^{-1}\):** Corresponding to C-H stretching in aromatic rings.
- **1659 cm\(^{-1}\):** Indicative of C=O stretching, possibly from carbonyl groups.
- **1512 cm\(^{-1}\):** Corresponding to aromatic C=C stretching vibrations.
- **1444 cm\(^{-1}\):** Aromatic C=C stretching.
- **1211 cm\(^{-1}\) and 1157 cm\(^{-1}\):** C-O stretching, indicative of alcohols, ethers, esters, or carboxylic acid functional groups.
- **780 cm\(^{-1}\):** Characteristic of aromatic C-H out-of-plane bending vibrations.
- **650 cm\(^{-1}\):** Another significant region often associated with substitution patterns on the aromatic ring.

This spectrum provides insights into the functional groups present in the compound.

#### Nuclear Magnetic Resonance (NMR) Spectrum
The NMR spectrum is depicted below the IR spectrum. The x-axis represents the chemical shift in parts per million (PPM), ranging from 12 PPM to 6 PPM.

- **Signal at ~10 PPM (1H):** Suggests the presence of a proton in an aldehyde group.
- **Signal at ~7-8 PPM (4H):** This multiplet corresponds to aromatic protons, typically occurring in this region.
- **Signal at ~12 PPM (1H):** A singlet characteristic of an OH group possibly in a carboxylic acid or phenol.

By analyzing both spectra, one can identify and confirm the functional groups and structure of the compound C\(_6\)H\(_6\)O\(_3\). These analytical techniques, IR and NMR spectroscopy, are crucial in structural determination and verification in organic chemistry.
Transcribed Image Text:### Analysis of Compound with Molecular Formula C\(_6\)H\(_6\)O\(_3\) #### Infrared (IR) Spectrum The IR spectrum is shown above, where the y-axis represents the percentage transmittance, and the x-axis gives the wavenumber in cm\(^{-1}\), ranging from 4000 cm\(^{-1}\) to 650 cm\(^{-1}\). Key peaks are observed at: - **3238 cm\(^{-1}\):** Typically associated with O-H stretching vibrations, indicating the presence of hydroxyl groups. - **3015 cm\(^{-1}\):** Corresponding to C-H stretching in aromatic rings. - **1659 cm\(^{-1}\):** Indicative of C=O stretching, possibly from carbonyl groups. - **1512 cm\(^{-1}\):** Corresponding to aromatic C=C stretching vibrations. - **1444 cm\(^{-1}\):** Aromatic C=C stretching. - **1211 cm\(^{-1}\) and 1157 cm\(^{-1}\):** C-O stretching, indicative of alcohols, ethers, esters, or carboxylic acid functional groups. - **780 cm\(^{-1}\):** Characteristic of aromatic C-H out-of-plane bending vibrations. - **650 cm\(^{-1}\):** Another significant region often associated with substitution patterns on the aromatic ring. This spectrum provides insights into the functional groups present in the compound. #### Nuclear Magnetic Resonance (NMR) Spectrum The NMR spectrum is depicted below the IR spectrum. The x-axis represents the chemical shift in parts per million (PPM), ranging from 12 PPM to 6 PPM. - **Signal at ~10 PPM (1H):** Suggests the presence of a proton in an aldehyde group. - **Signal at ~7-8 PPM (4H):** This multiplet corresponds to aromatic protons, typically occurring in this region. - **Signal at ~12 PPM (1H):** A singlet characteristic of an OH group possibly in a carboxylic acid or phenol. By analyzing both spectra, one can identify and confirm the functional groups and structure of the compound C\(_6\)H\(_6\)O\(_3\). These analytical techniques, IR and NMR spectroscopy, are crucial in structural determination and verification in organic chemistry.
### Problem Set Instructions

For each problem, you are required to:

1. **Calculate the degree of unsaturation.**
2. **Assign the principal IR absorption bands above 1500 cm⁻¹.**
3. **Draw the structure of the compound.**
4. **Label the protons on your structure with letters and assign them to peaks on the NMR spectrum (see the example below).**

### Example

#### Structure and Proton Labeling

The given structure is an alcohol, where the hydroxyl group (-OH) is attached to a carbon chain. The structure is labeled as follows:

- Proton A: Attached to the carbon next to the hydroxyl group, with 3 protons.
- Proton B: Attached to the carbon directly linked to the hydroxyl-group-bearing carbon, with 2 protons.
- Proton C: Attached to a carbon that is one carbon away from the hydroxyl group, with 2 protons.
- Proton D: The hydroxyl proton itself, with 1 proton.

#### NMR Spectrum Interpretation

Below the structure, there is an NMR spectrum ranging from 4.0 to 0.5 ppm (parts per million). The spectrum shows several peaks corresponding to different protons in the molecule:

- **Peak at approximately 3.6 - 3.4 ppm (C, 2H):** This corresponds to the protons labeled as C.
- **Peak at approximately 2.4 ppm (D, 1H):** This corresponds to the hydroxyl proton labeled as D.
- **Peak at approximately 1.6 ppm (B, 2H):** This corresponds to the protons labeled as B.
- **Peak at approximately 1.1 ppm (A, 3H):** This corresponds to the protons labeled as A.

Each peak is assigned based on the chemical environment of the protons in the molecule, illustrating how NMR spectroscopy can be used to determine the structure of a compound.

---

This example provides a framework for how to approach similar problems, including interpreting NMR spectra and understanding the relationship between proton environments and their respective NMR signals.
Transcribed Image Text:### Problem Set Instructions For each problem, you are required to: 1. **Calculate the degree of unsaturation.** 2. **Assign the principal IR absorption bands above 1500 cm⁻¹.** 3. **Draw the structure of the compound.** 4. **Label the protons on your structure with letters and assign them to peaks on the NMR spectrum (see the example below).** ### Example #### Structure and Proton Labeling The given structure is an alcohol, where the hydroxyl group (-OH) is attached to a carbon chain. The structure is labeled as follows: - Proton A: Attached to the carbon next to the hydroxyl group, with 3 protons. - Proton B: Attached to the carbon directly linked to the hydroxyl-group-bearing carbon, with 2 protons. - Proton C: Attached to a carbon that is one carbon away from the hydroxyl group, with 2 protons. - Proton D: The hydroxyl proton itself, with 1 proton. #### NMR Spectrum Interpretation Below the structure, there is an NMR spectrum ranging from 4.0 to 0.5 ppm (parts per million). The spectrum shows several peaks corresponding to different protons in the molecule: - **Peak at approximately 3.6 - 3.4 ppm (C, 2H):** This corresponds to the protons labeled as C. - **Peak at approximately 2.4 ppm (D, 1H):** This corresponds to the hydroxyl proton labeled as D. - **Peak at approximately 1.6 ppm (B, 2H):** This corresponds to the protons labeled as B. - **Peak at approximately 1.1 ppm (A, 3H):** This corresponds to the protons labeled as A. Each peak is assigned based on the chemical environment of the protons in the molecule, illustrating how NMR spectroscopy can be used to determine the structure of a compound. --- This example provides a framework for how to approach similar problems, including interpreting NMR spectra and understanding the relationship between proton environments and their respective NMR signals.
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