Using a H NMR spectrum to determine the structure of a protein supplement

Question

This question is about the H NMR spectrum of creatine.

Creatine is a supplement taken by athletes to increase body mass. Recently, supplements containing derivatives of creatine have been marketed. These are usually more lipophilic in an attempt to improve uptake.

The 1 H NMR spectrum of one of these supplements in D2O is shown, with some parts magnified:

This particular supplement exists in an ionised form at pH1 but does not exist in an ionised form at pH 12 (I am unsure about this and will come back to it in my question below!)

Using this information we are asked to suggest a structure for the supplement. I was almost right. The answer is:

The mark scheme says that three things can be extracted from the NMR spectrum and the information given:

My question(s):

1. I don't understand the relevance of the pH statement. How does that relate to the structure? How does that relate to the ester group? (this is mentioned in the mark scheme which is attached just above this)

2. How can we use the integrals to calculate that there are 10 C-H protons? I measured the heights of the traces and got, 2.7cm, 4.75cm and 6.9cm but I'm not sure how this leads to '10 C-H protons'...

Would really appreciate help on this question and a breakdown of the key points discussed. Many thanks!

Answer 1

Let me try to explain the point about the 10 atoms and the integral.

The area under the peaks and the ratios of these areas correspond to the ratios of the hydrogen atoms. The blue line adds all these areas up as it goes along. That means at the first peak on the left you simply get the area of that peak. But at the second peak you still have the amount from the first peak plus the new area, which is why you have to look at the difference. If we denote the three peak areas as $A_1$, $A_2$, and $A_3$, the height of the blue line goes from $A_1$ to $A_1 + A_2$, then lastly $A_1 + A_2 + A_3$.

With your values we have, $$\begin{aligned} A_1 &= \pu{2.7 cm} \\ A_1+A_2 &= \pu{4.75 cm} \\ A_1+A_2+A_3 &= \pu{6.9 cm} \end{aligned}$$ Which allows us to easily determine $A_2 = \pu{2.05 cm}$ and $A_3 = \pu{2.15 cm}$. (Alternatively you could have simply read off the difference of the blue line between two peaks directly when you measured it).

This gives us the ratios of $\pu{2.7 cm} : \pu{2.05 cm} : \pu{2.15 cm}$. If we assume that the signal at $\pu{1.25 ppm}$ (i.e. $A_3$) is a methyl group with $3$ hydrogen atoms, we can divide all values by $\pu{2.15 cm}$ and multiply by 3 to normalize the ratio. We get approximately hydrogen atom ratios of $3.76 : 2.86 : 3.00$. Rounding the result gives a ratio of $4:3:3$ which agrees with $10$ atoms in total.

Answer 2

So I can answer the first question.

Esters hydrolyse readily in acidic conditions to the constituent acid and alcohols. As there is a signal in the 4.2 to 4.3 range, which falls in the 3.5 to 5.5 range shown by an alpha Hydrogen to an ether group or ester group.

To decide between an ether and ester, the question mentions the fact that the form is ionised at pH=1 and not at pH=12

Answer 3

About Q1:

The whole thing with information that creatine derivative in question exist in ionised form at pH = 12 and doesn't exist in ionised form at pH = 1 tells us that this compound tends to bind protons and doesn't show tendency to dissociate with proton release (as binding proton makes whole compound to adopt positive charge and dissotiation with proton release results in taking an anion form). Simple conclusion coming from this is that this compound has a basic group and no acidic group (that would be stronger acid than H2O).

Therefore logic conclusion from the information that creatine derivative in question doesn't have an acidic group would be that it has somehow modified carboxylic group of the parent compound. How it is modified? Well H NMR spectrum can be of some help. You can see a quartet from oxygen bonded methylene group (at about 4.24 ppm) that is coupled to methyl protons giving rise to triplet (at about 1.25 ppm) - that is some hint of ethyl ester being present in this creatine derivative.

More, this basic group that binds proton in acidic pH and gives rise to ionised form is that nitrogen-rich group, that sometimes is called in general guanidyl group (as it is related to the structure of guanidine). Nitrogen atoms there posses lone electron pairs and can react as base (similiary to amines).