**QUESTION 21**Elemental analysis determined an unknown compound to contain 40.0 % carbon, 6.7 % hydrogen. The unknown compound had an M+ = 60 m/z; three key IR absorbances: strong at 1710, strong below 3000, and broad from 2500 to 3500. The proton NMR showed an absorbance at 1.7 ppm (lowest whole number ratio of H =3), and an absorbance at 11.5 ppm (lowest whole number ratio of H = 1). The compound was able to turn blue litmus paper red, and neutralize a base. What is this compound?- ☐ CH₃CO₂H- ☐ Formaldehyde- ☐ HO- ☐ CH₂(OH)CHO- ☐ CH(OH)=CH(OH)**Explanation**:- **Elemental Analysis**: The compound is composed of 40.0 % carbon and 6.7 % hydrogen.- **Mass Spectrometry**: Molecular ion peak (M+) is at 60 m/z.- **Infrared (IR) Spectroscopy**: Key absorbances include: - Strong absorbance at 1710 cm⁻¹ (indicating a carbonyl group, C=O) - Strong absorbance below 3000 cm⁻¹ (indicative of C-H stretches) - Broad absorbance from 2500 to 3500 cm⁻¹ (indicative of O-H stretches)- **Proton Nuclear Magnetic Resonance (NMR)**: - Absorbance at 1.7 ppm, suggesting alkyl protons (CH₃) - Absorbance at 11.5 ppm, likely due to acidic proton (carboxylic OH)- **Other Properties**: The compound turns blue litmus paper red, indicating that it is acidic, and it can neutralize a base.The correct answer is **CH₃CO₂H** (acetic acid), as it matches the molecular formula and the spectroscopic data provided.

Given: Percentage of carbon in the compound = 40.0 % Percentage of hydrogen in the compound = 6.67 %…

Elemental analysis determined an unknown compound to contain 40.0 % carbon, 6.7 % hydrogen. The unknown compound had an M+ = 60 m/z; three key IR absorbances: strong at 1710, strong below 3000, and broad from 2500 to 3500. The proton NMR showed an absorbance at 1.7 ppm (lowest whole number ratio of H =3), and an absorbance at 11.5 ppm (lowest whole number ratio of H = 1). The compound was able to turn blue litmus paper red, and neutralize a base. What is this compound? O CH3CO2H O Formaldehyde O HO C = CH2 но O CH2(OH)CHO O CH(OH)=CH(OH)

Elemental analysis determined an unknown compound to contain 40.0 % carbon, 6.7 % hydrogen. The unknown compound had an M+ = 60 m/z; three key IR absorbances: strong at 1710, strong below 3000, and broad from 2500 to 3500. The proton NMR showed an absorbance at 1.7 ppm (lowest whole number ratio of H =3), and an absorbance at 11.5 ppm (lowest whole number ratio of H = 1). The compound was able to turn blue litmus paper red, and neutralize a base. What is this compound? O CH3CO2H O Formaldehyde O HO C = CH2 но O CH2(OH)CHO O CH(OH)=CH(OH)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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