Elemental analysis determined unknown Compound G to contain 40.0 % carbon, 6.7 % hydrogen. Compound G had an M+ = 60 m/z on its MS. Its IR displayed three 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). Compound G was able to turn blue litmus paper red, and neutralize a base. What is Compound G? CH3CO2H Formaldehyde HO \ C = CH2 /HO CH2(OH)CHO CH(OH)=CH(OH)

Unknown compound can be identified from NMR, IR and mass spectroscopy.

Elemental analysis determined unknown Compound G to contain 40.0 % carbon, 6.7 % hydrogen. Compound G had an M+ = 60 m/z on its MS. Its IR displayed three absorbances: strong at 1710, strong below 3000, and broad from 2500 to 3500.

Elemental analysis determined unknown Compound G to contain 40.0 % carbon, 6.7 % hydrogen. Compound G had an M+ = 60 m/z on its MS. Its IR displayed three absorbances: strong at 1710, strong below 3000, and broad from 2500 to 3500.

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter17: Carboxylic Acids

Section: Chapter Questions

Problem 17.17P

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Question

Elemental analysis determined unknown Compound G to contain 40.0 % carbon, 6.7 % hydrogen. Compound G had an M+ = 60 m/z on its MS.

Its IR displayed three 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).

Compound G was able to turn blue litmus paper red, and neutralize a base. What is Compound G?

CH₃CO₂H

Formaldehyde

HO

\

C = CH₂

/

HO

CH₂(OH)CHO

CH(OH)=CH(OH)

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