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Butane

In the molecule CH 3 CH 2 CH 2 CH 3 size 12{ ital "CH" rSub { size 8{3} } ital "CH" rSub { size 8{2} } ital "CH" rSub { size 8{2} } ital "CH" rSub { size 8{3} } } {} , the methyl groups are equivalent and the methylene groups are equivalent. As described earlier, we would expect to see two signals in a 3:2 ratio. If we now consider coupling, we see that the methyl groups are attached to methylene groups. The CH 3 size 12{ ital "CH" rSub { size 8{3} } } {} protons will thus be split into triplets by the methylene protons. Likewise, the methylene protons will be split into quartets by the protons on the methyl groups. However, notice that because the methylene groups are equivalent, they do not couple to each other. Therefore, we expect to see the same pattern as that predicted for the ethyl group shown above.

More Complex Coupling Patterns

If a hydrogen atom is attached to a carbon atom to which nonequivalent carbon atoms with different numbers of protons are attached, then a more complex pattern is observed. One arrives at the total number of signals expected by multiplying the individual couplings. For example, consider 1,2-dichloropropane with three carbon atoms having different protons attached:

The CH 3 size 12{ ital "CH" rSub { size 8{3} } } {} protons will see one adjacent proton and the signal will be split into a doublet. The CH 2 size 12{ ital "CH" rSub { size 8{2} } } {} group will likewise be split into a doublet by the CH proton. Coupling between the CH 3 size 12{ ital "CH" rSub { size 8{3} } } {} group and the CH 2 size 12{ ital "CH" rSub { size 8{2} } } {} group will generally not be observed. However, the CH group in the middle is adjacent to two different types of carbon atoms with different numbers of protons attached. This CH group will be split into a quartet by the CH 3 size 12{ ital "CH" rSub { size 8{3} } } {} group and into a triplet by the CH 2 size 12{ ital "CH" rSub { size 8{2} } } {} group. The overall pattern will have 4 x 3 lines, or a total of 12, if our instrument is able to resolve all the lines. This pattern would be described as either a triplet of quartets or a quartet of a triplet, depending upon which coupling constant is larger. Often, these complex patterns are not cleanly resolved and a very complex spectrum results. The two limiting possibilities for the CH group would look something like the following. Notice that there are two different coupling constants in each case.

Quartet of triplets

Triplet of quartets

Experimental

You will be given a packet of several spectra. Using the chart and the information on chemical shifts, coupling patterns, and integrated intensities given above as a guide, identify the unknown organic compounds. Answer some review questions to refresh your mind. We will work on number 2 and 5 with you.

Lab Revision Questions: NM

(Total 20 Points)

On my honor, in preparing this report, I know that I am free to use references and consult with others. However, I cannot copy from other students’ work or misrepresent my own data.

…………………………………………………………….. (signature)

Print Name: _________________________________________

1. The atomic weight of Br is listed as 79.909 on the periodic table, but we use 79, not 80, when calculating the molecular weights of Br-containing compounds for mass spectrometry – why? (2 points)

2.

(a) Draw trans-3-bromo-1-butylcyclohexane in planar form (no chair). Be sure to indicate the stereochemistry in your drawing. (2 points)

(b) Now draw both chair conformations of trans-3-bromo-1-butylcyclohexane. (2 points)

(c) Circle the conformer above that is lower in energy. (1 point)

3.

(a) How many degrees of unsaturation does C 5 H 7 ClO size 12{C rSub { size 8{5} } H rSub { size 8{7} } ital "ClO"} {} have? (2 points)

(b) Draw an isomer of C 5 H 7 ClO size 12{C rSub { size 8{5} } H rSub { size 8{7} } ital "ClO"} {} that contains a ketone (There are many correct answers.). (2 points)

(c) Draw an isomer of C 5 H 7 ClO size 12{C rSub { size 8{5} } H rSub { size 8{7} } ital "ClO"} {} that does not contain a ketone. (Again, there are many correct answers.). (2 points)

4. Explain why 3-Chloropentane shows four different absorbencies in the 1 H size 12{ {} rSup { size 8{1} } H} {} NMR spectrum. (3 points)

5. Write the structure from the data given. (4 points)

Molecular formula C x H y NBr size 12{C rSub { size 8{x} } H rSub { size 8{y} } ital "NBr"} {}

IR:

1 H size 12{ {} rSup { size 8{1} } H} {} NMR:

13 C size 12{ {} rSup { size 8{"13"} } C} {} NMR:

Report QuestionsNMR Laboratory

(Total 70 Points)

On my honor, in preparing this report, I know that I am free to use references and consult with others. However, I cannot copy from other students’ work or misrepresent my own data.

…………………………………………………………….. (signature)

Print Name: _________________________________________

1. Explain each of the following observations. (3 X 2= 6 points)

(a) The compound CH 3 CH 2 CH = CH 2 size 12{ ital "CH" rSub { size 8{3} } ital "CH" rSub { size 8{2} } ital "CH"= ital "CH" rSub { size 8{2} } } {} shows an IR absorbance around 1550 cm–1, but its structural isomer CH 3 CH = CHCH 3 size 12{ ital "CH" rSub { size 8{3} } ital "CH"= ital "CHCH" rSub { size 8{3} } } {} does not-why?

(b) The 1 H size 12{ {} rSup { size 8{1} } H} {} -decoupled 13 C size 12{ {} rSup { size 8{"13"} } C} {} NMR spectrum of CHF 3 size 12{ ital "CHF" rSub { size 8{3} } } {} shows a single peak that is split into a quartet.

2.

(a) Draw trans-4-(1-ethylpropyl)cyclohexanol in planar form (no chair). Be sure to indicate the stereochemistry. (2 points)

(b) Now draw trans-4-(1-ethylpropyl)cyclohexanol in its lowest energy conformation. (2 points)

(c) Now draw cis-4-(1-ethylpropyl)cyclohexanol in its lowest energy conformation. (2 points)

(d) Considering your answers to (b) and (c), which is lower in energy, cis or trans-4-(1-ethylpropyl)cyclohexanol? Explain your answer briefly (one sentence). (2 points)

3. In the following compound, label symmetry-equivalent H atoms with the same letter and nonequivalent H atoms with different letters, as we did in class. Then predict the 1 H size 12{ {} rSup { size 8{1} } H} {} NMR spectrum of the compound, indicating the approximate chemical shift, integration, and multiplicity for each resonance that you expect to see. (20 points)

4. Below are the IR spectra of two isomers having the formula C 7 H 7 Cl size 12{C rSub { size 8{7} } H rSub { size 8{7} } ital "Cl"} {} . (Total 8 points)

1) What is the unsaturation number of these isomers? _____ (2 points)

2) Why is there such a difference in the intensity of the C-H stretching and C-H bending region between these two isomeric compounds? (2 points)

3) What structure would you assign to Compound A? Indicate on the spectrum which IR bands support your structure assignment. (4 points)

4. What structure would you assign to Compound B? Indicate which IR bands support your structure assignment. Note: NUJOL is a paraffin oil used to prepared the IR sample. (8 points)

5. Find the structure of the compound and explain each spectrum to support your answer.

(a) Compound contains C, H, and O. (10 points)

MS:

IR:

1 H size 12{ {} rSup { size 8{1} } H} {} NMR:

13 C size 12{ {} rSup { size 8{"13"} } C} {} NMR:

(b) Compound contains C, H, and O. (10 points)

MS:

IR:

1 H size 12{ {} rSup { size 8{1} } H} {} NMR:

13 C size 12{ {} rSup { size 8{"13"} } C} {} NMR:

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Source:  OpenStax, Chem 215 spring08. OpenStax CNX. Mar 21, 2008 Download for free at http://cnx.org/content/col10496/1.8
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