Question

General Note: If a printer is not accessible, the problem set may be completed on regular paper, but please ensure all answers are clear. Notations for IR and NMR may be written out and listed by wavenumbers and ppm, respectively. if this is the case or provided space is insufficient 1. (3 points) For the structure and 'H NMR spectrum shown below, match the labeled hydrogen atoms on the molecule (A-D) with the corresponding signals in the spectrum using the boxes provided. Integration and splitting are noted. 3H, S 1H, m PPM Instructions for Problems 2-4: Provide the structure of the compound that corresponds to the set of spectra. Your notations and analysis must be shown for full credit. This includes diagnostic peaks in the IR, if any MS fragmentations are notable, and degrees of unsaturation (DU's). In regards to NMR: signals integration (for HNMR), chemical shift, and splitting (for HNMR) are relevant for the analysis and elucidation of fragments to guide toward full structure elucidation.

Answer 1

1. Looking at the given molecule, we can see that all the hydrogens are attached to saturated carbons,

Since all the hydrogens are attached to saturated carbons, this means that their 1H NMR peaks would lie in the more shielded region of the spectra which lies from 0 to 5 ppm.

Now, on examining the compound one can say, that there are 4 kinds of hydrogens in this molecule, which are marked by arrows. This fact is verified by the 1H NMR spectra of this molecule, which shows 4 peaks.

Now, let's come to the distribution of the peaks in the region of 0 to 5 ppm.

The most deshielded hydrogens are on the carbon which is marked as B by the arrows.

Firstly, it is better to know what deshielded means, deshielded means that the electron density is less surrounding the atom. The most common reason for this deshielding is because of the difference in electronegativity of the atoms, so the electrons are more drawn towards the more electronegative atom, and the less electronegative atom as a consequence of this would have reduced electron density around it. And this means that the less electronegative atom is deshielded.

So, the more deshielded an atom is, the more energy it would absorb and hence would resonate at higher frequency as compared to a shielded atom. In simple terms, this means that the signal for a deshielded atom would appear downfield or low field as compared to a shielded atom.

Now, let's get back to the hydrogens on the carbon which is marked as B by the arrows. Since this carbon is attached directly to an oxygen atom, the oxygen atom would draw the electron density towards itself due to the electronegativity difference, leaving the carbon and the hydrogens attached to it deshielded. This is the reason the signal for these two hydrogens would appear much downfield as compared to all the hydrogens in this molecule.

So, the signal near 4 ppm would be because of the 2 hydrogens on the carbon marked as B with an arrow.

Now, let's look at the carbon marked as A by an arrow.

Since this carbon is attached to the carbon of a C=O group which is sp2 hybridized, this is more deshielded as compared to the carbons marked as C and D by the arrows. But, since it is not directly attached to a more electronegative atom, the hydrogens on this carbon would appear upfield when compared to the hydrogens on the carbon marked as B.

Now, let's look at the carbon marked as C by an arrow.

This carbon is attached to an sp3 carbon (marked as B) which is attached to an oxygen atom, so this would experience a small amount of deshielding, but it would comparatively less than that of carbon marked as A. So, the signal for the hydrogen on the carbon marked as C would appear somewhere near the signal of hydrogens on carbon marked as A, but in the upfield region.

Now, let's look at the carbons marked as D.

There are two carbons marked as D because both of these carbons are identical to each other and so are the hydrogens attached to them. They are identical because of the bond rotation which happens in case of single bonds, and due to this bond rotation, both of these carbons and their hydrogens would experience the similar environment, for example, they would experience similar field effects, they would experience similar shielding. And since these carbons are not directly attached to any atom which has a different electronegativity or which would cause any kind of deshielding, the signals for the hydrogens attached to these carbons would appear in the upfield region or the more shielded region.

The marked spectra is in the image given below:

энэ 1Hm PPM

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