Explain how you would be able to distinguish both acetophenone and 1-phenylethanol using Proton NMR spectroscopy.You must indicate the number of signals, their predicted chemical shift values as well as the splitting patterns.
Nuclear Magnetic Resonance (NMR) Spectroscopy has become one of the very important tools used to study the composition of organic molecules. This method is non destructive, meaning that the sample can be collected after the analysis and it is not destroyed like in the case of mass spectrometry.
The different functional groups in a molecule gives rise to signals in very specific regions. For instance protons on aromatic rings can always be seen down field, while protons on a methyl group will always be seen up field in the region of 1- 3 ppm
Answer and Explanation:
Below is the structures of the two products in question. The protons have been labelled on each.
- The most obvious difference between these two compounds on a H NMR spetra will be the presence of the methyl protons next to the ketone in acetophenone (labelled d). These 3 protons will show as a singlet in the region of 2-3 ppm. The methyl group in phenyethanol will be doublet (coupling with 1 proton) in the region of between 1 and 2 ppm.
- Also, since the phenyl ring is attached to an electron withdrawing group in acetophenone, it can be expected that protons a may be more down field than the other aromatic protons.
Below is a list of the expected peaks:
Protons a: doublet; 7.5 - 8 ppp
Protons b: multiplet; 7- 8 ppm
Proton c: multiplet; 7- 8 ppm
Proton d: singlet; 2-3 ppm
Protons a - c: multiplets; 7 - 8 ppm
Proton d: quartet; 4.5 and 5.5 ppm
Protons e: doublet; 1 - 2 ppm
Proton f; broad singlet; 2 - 5 ppm
Become a member and unlock all Study Answers
Try it risk-free for 30 daysTry it risk-free
Ask a question
Our experts can answer your tough homework and study questions.Ask a question Ask a question
Learn more about this topic:
from Organic Chemistry: Help & ReviewChapter 2 / Lesson 2