20) An unknown compound has the chemical formula CH₂O. It's IR spectrum and H NMR spectrum are shown below. Propose a structure for the unknown molecule. Calculate the Degrees of Unsaturation (DU), assign any important peaks in the IR spectrum to vibrational modes and briefly explain why the structure fits the 'H NMR spectrum that is provided Degrees of unsaturation (show work below): 642 Du = (2(3)+2)-6-8-6-2²/1/2 = 1 2 3437 1739 16-o mpy m Vibrational Assignments of IR Peaks Wavenumber (cm-¹) Vibrational Assignment 2 0-4 stretch C=0 strekn Peak 1: 9.79 ppm 1H Singlet 9 8 Brief rationalization for proposed structure: (explain why DU, IR, and NMR are consistent with proposed structure) ppm Peak 2 2.45 ppm 2H Quartet Peak 3 1.11 ppm 3H Triplet Proposed structure of unknown:

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**Analysis of an Unknown Compound \(C_4H_{10}O\) Using IR and \(^{1}H\) NMR Spectroscopy**

**Objective:**
Analyze an unknown compound with the chemical formula \(C_4H_{10}O\) using IR and \(^{1}H\) NMR spectra to propose a structure. Determine the Degrees of Unsaturation (DU) and make vibrational mode assignments in the IR spectrum.

**Degrees of Unsaturation Calculation:**

\[
DU = \frac{(2C + 2 + N - H - X)}{2} = \frac{(2(4) + 2 - 10)}{2} = \frac{8 - 10}{2} = 1
\]

**Spectral Analysis:**

**1. IR Spectrum:**
- **Significant Peaks Identified:**
  - **3487 cm\(^{-1}\):** Indicative of O-H stretch
  - **1739 cm\(^{-1}\):** Indicative of C=O stretch

**2. \(^{1}H\) NMR Spectrum:**
- **Peak 1:** 9.79 ppm, 1H, Singlet – consistent with an aldehyde proton
- **Peak 2:** 2.46 ppm, 2H, Quartet – likely corresponds to adjacent to an oxygen atom
- **Peak 3:** 1.11 ppm, 3H, Triplet – likely corresponds to a methyl group adjacent to a methylene group

**Vibrational Assignments of IR Peaks:**

| Wavenumber (cm\(^{-1}\)) | Vibrational Assignment |
|--------------------------|------------------------|
| 3487                     | O-H stretch            |
| 1739                     | C=O stretch            |

**Proposed Structure of the Unknown:**

- The compound is likely an alcohol or a carbonyl-containing compound, considering the chemical formula \(C_4H_{10}O\), the O-H stretch band, and the C=O stretch.

**Conclusion:**

- The analysis of IR and \(^{1}H\) NMR data suggests that the structure might be an aldehyde due to the aldehyde proton appearing downfield at 9.79 ppm, corroborated by the presence of the C=O stretch.
- The degrees of unsaturation suggest one ring or double bond, consistent
Transcribed Image Text:**Analysis of an Unknown Compound \(C_4H_{10}O\) Using IR and \(^{1}H\) NMR Spectroscopy** **Objective:** Analyze an unknown compound with the chemical formula \(C_4H_{10}O\) using IR and \(^{1}H\) NMR spectra to propose a structure. Determine the Degrees of Unsaturation (DU) and make vibrational mode assignments in the IR spectrum. **Degrees of Unsaturation Calculation:** \[ DU = \frac{(2C + 2 + N - H - X)}{2} = \frac{(2(4) + 2 - 10)}{2} = \frac{8 - 10}{2} = 1 \] **Spectral Analysis:** **1. IR Spectrum:** - **Significant Peaks Identified:** - **3487 cm\(^{-1}\):** Indicative of O-H stretch - **1739 cm\(^{-1}\):** Indicative of C=O stretch **2. \(^{1}H\) NMR Spectrum:** - **Peak 1:** 9.79 ppm, 1H, Singlet – consistent with an aldehyde proton - **Peak 2:** 2.46 ppm, 2H, Quartet – likely corresponds to adjacent to an oxygen atom - **Peak 3:** 1.11 ppm, 3H, Triplet – likely corresponds to a methyl group adjacent to a methylene group **Vibrational Assignments of IR Peaks:** | Wavenumber (cm\(^{-1}\)) | Vibrational Assignment | |--------------------------|------------------------| | 3487 | O-H stretch | | 1739 | C=O stretch | **Proposed Structure of the Unknown:** - The compound is likely an alcohol or a carbonyl-containing compound, considering the chemical formula \(C_4H_{10}O\), the O-H stretch band, and the C=O stretch. **Conclusion:** - The analysis of IR and \(^{1}H\) NMR data suggests that the structure might be an aldehyde due to the aldehyde proton appearing downfield at 9.79 ppm, corroborated by the presence of the C=O stretch. - The degrees of unsaturation suggest one ring or double bond, consistent
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