UV vis (1 Viewer)

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I am a bit confused for this cause as im not sure how the wavelength for the peak intensity is used as a distance for measuring

 

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Also shouldnt this be A instead of B as there is only 1 hydrogen environment?

 

[Luukas.2]

View attachment 40931
Also shouldnt this be A instead of B as there is only 1 hydrogen environment?

The question is asking about the NMR of the monomer, which is 2-hydroxy-2-methylpropanoic acid... i.e. (CH₃)₂C(OH)COOH, which has 3 x ¹H and 3 x ¹³C environments.

 

[Luukas.2]

View attachment 40930
I am a bit confused for this cause as im not sure how the wavelength for the peak intensity is used as a distance for measuring

You need to put in light of a wavelength that will be absorbed (so 625 nm rather than 510 nm) and then measure the intensity the remains after passing through the sample.

 

[panikinprogress]

The question is asking about the NMR of the monomer, which is 2-hydroxy-2-methylpropanoic acid... i.e. (CH₃)₂C(OH)COOH, which has 3 x ¹H and 3 x ¹³C environments.

wait sorry, but I thought the H atom of an OH group won't split into its own signal on a proton nmr?

 

[Luukas.2]

wait sorry, but I thought the H atom of an OH group won't split into its own signal on a proton nmr?

It's still an H environment that will give rise to a signal, whether it is involved in splitting or not.

In this monomer, none of the H environments are close enough for splitting to happen.

 

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tysm

 

[=)(=]

It's still an H environment that will give rise to a signal, whether it is involved in splitting or not.

In this monomer, none of the H environments are close enough for splitting to happen.

ohh is this why the hydrogen on an amine shows up but doesn't split?

 

[Luukas.2]

ohh is this why the hydrogen on an amine shows up but doesn't split?

Yes, every H environment should appear. Alcohol and amine signals can be broaden and look unlike a typical signal, but they should still appear. The reasons for these atypical features are beyond the syllabus.