# Nmr  (Page 3/6)

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We can then use $\Delta$ E to give us some information about the nature of the atoms to which a particular hydrogen atom is attached.

The position of a signal in the NMR spectrum is known as its chemical shift. It has become conventional to reference the NMR signals for hydrogen atoms to the signal found for TMS (tetramethylsilane, $\text{Si}\left({\text{CH}}_{3}{\right)}_{4}$ ), which has been assigned a value of 0. This scale gives the chemical shifts for other atoms in terms of $\delta$ which has frequency units of parts per million (ppm). These values are constant regardless of the field strength of the magnet being used. Most hydrogen atoms attached to organic molecules will resonate in the region $\delta$ = 0 to 10 ppm. Often, the more acidic the proton is, the more positive this value becomes (acid protons in carboxylic acids will often be seen at>+10 ppm). This means that the proton is experiencing less electron density around it, a phenomenon known as deshielding. The less electron density there is about a given hydrogen atom, the more deshielded it is. On the opposite extreme, the more electron density a hydrogen atom experiences, the more shielded it is. While most hydrogen atoms are found in the NMR spectra at positive $\delta$ values, some may be found at negative $\delta$ values. This often occurs when protons are attached to metals (metal hydrides). In these cases, it is typical to find hydrogen signals with $\delta$ = 0 to -30 ppm. Chart 1 gives a general guideline for where certain types of protons are found in NMR spectra.

Chart 1. NMR Chemical Shift Assignments

## Integration

One of the most useful aspects of NMR spectroscopy is that the signal intensities are directly proportional to the amount of a particular type of H atom present. This can be used in two ways.

First, it can be tell us the number of hydrogen atoms of a given type in relation to other hydrogen atoms present in the molecule. In order to make use of this information, we need to realize that symmetrically equivalent H atoms will have identical chemical shifts.

Thus, if we have an ethyl group, there are three protons from the methyl $\left({\text{CH}}_{3}\right)$ group and two protons from the methylene $\left({\text{CH}}_{2}\right)$ group. We expect to see two signals - one for the methyl and one for the methylene - with relative peak areas of 3:2. The process of determining the peak areas is known as integration and the resultant peak areas are known as the integral of the spectrum.

In order to use this information, you need to determine what types of atoms are the same and what types are different within a molecule. Atoms may appear the same for two reasons.

• One is that they are related by symmetry. That means that the molecule has some symmetry element such as a mirror plane, rotation axis, inversion center, etc. that will make the atoms equivalent.
• The other possibility is that there is some molecular motion which causes the atoms to adopt an averaged environment. For example, in the ethyl group above, we can draw a static structure which makes the H atoms on the ${\text{CH}}_{3}$ group unequal because their proximity to the X group would be different. However, there is free rotation about the C-C bond, so all of the H atoms are whizzing around at very fast speeds, resulting in an averaged environment. There is only one signal observed for this methyl group.

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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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