Practical - solutions to NMR practice problems PDF

Preview

Summary

Solutions to NMR Practice Problems...

Description

1 CHEM2001:

NMR Practice Problems

1. Indicate the number of different chemical environments for 1H and compounds. Compound

13C

nuclei in the following

Number of 1H Environments

Number of 13C Environments

2

3

Cl

1

1

NH2

3

2

2

3

3

4

O O Cl

Cl

Cl Cl

Cl

Cl

Cl

Cl

1

2

Cl

Br

2

4

4

6

2

2

4

3

6

6

3

4

Cl

Br OH

OH

OH O

NH2 I

2 2. Describe each pattern below in terms of multiplicity (singlet, doublet, doublet of doublets, etc).

(a)

(b) triplet

quartet

(e) doublet of quartets

(i)

(f) triplet

(j) quartet AND triplet

septet

(c) doublet

(d) doublet

(g) quintet

(h) doublet of triplets

(k) quartet AND triplet (again)

(l) doublet of doublets

3. For (a), (b), (e), (g), and (i) above, draw a structure that includes a proton (or protons) that will lead to such an NMR signal. For example, in the 1H NMR spectrum of 1-bromo-2,2-dichloroethane, the CH2 group would appear as a signal similar to (c). (b) (i)

O H

CH3CH2Br

(e) (another example) (a)

(e)

(g)

Br

3 4. The 60 MHz 1H NMR spectrum of 1-nitropropane (CH3CH2CH2NO2) is shown below.

(i)

Use chemical shift arguments to assign each signal to particular hydrogen nuclei in 1nitropropane. δ 4.49 ppm is most downfield, must be closest to electronegative atoms of NO2 so due to CH3CH2CH2NO2 δ 1.03 ppm is most upfield, must be farthest away from electronegative atoms of NO2 so due to CH3CH2CH2NO2 δ 2.09 ppm is in between the extremes, must be due to CH3CH2CH2NO2

(ii)

Estimate the chemical shift of each signal, to two decimal places. The chemical shift corresponds to the centre of the multiplet. see part (i)

(iii)

Indicate the relative areas for the signals. Normally you would obtain integral information from the spectrum, but for this question, use your answers in (i) to guess the relative areas of the signals. δ 4.49 ppm ==> 2H δ 2.09 ppm ==> 2H δ 1.03 ppm ==> 3H

(iv)

Give the multiplicity of each signal (singlet, doublet, triplet etc). δ 4.49 ppm ==> triplet δ 2.09 ppm ==> looks like a sextet, but really it is a triplet of quartets. You could call it an apparent sextet. In real life, most people would just be slack and call it a sextet. δ 1.03 ppm ==> triplet

(v)

Explain what causes the multiplicity of each signal. δ 4.49 ppm ==> split into a triplet by coupling to the two protons of the adjacent CH2 group. δ 2.09 ppm ==>shows a quartet splitting due to coupling to the protons of the adjacent CH3 group, and shows a triplet splitting due to coupling to the protons of the adjacent CH3 group. Because the type of bonding between the protons of the central CH2 group and the protons of the CH3 group is similar to the type of bonding between the protons of the central CH2

4 group and the protons of the adjacent CH2 group, the coupling constants for the two types of coupling are the same, so some lines withinthe triplet of quartets pattern overlap and the overall pattern looks like a sextet. In other words, the CH2 and CH3 protons that do the coupling are acting as "equivalent protons" so the signal due to the central CH2 group is split as if it was coupled to 5 equivalent protons (n+1 rule says 5 equivalent H will cause splitting to give a sextet). δ 1.03 ppm ==> split into a triplet by coupling to the two protons of the adjacent CH2 group.

(vi)

The middle signal is actually more complicated than it appears. Explain. see (v)

5. The aromatic region of the 300 MHz 1H spectrum of 2-nitrophenol is shown on the next page, along with a structure showing the assignments of the signals to specific protons. The spectrum shows four signals for the four hydrogen nuclei attached to the benzene ring. Each of the signals shows splitting. Signals (a) and (c) look roughly like triplets, but each of the three lines in these apparent triplets has additional fins splitting. Similarly, signals (b) and (d) look roughly like doublets, but each of the two main lines in these apparent doublets has additional fine splitting. The doublet and triplet splittings in these multiplets are about 7 Hz. By consideration of the typical sizes of coupling constants between aromatic protons (see the table on coupling constants in the lecture notes), explain: (i) how do the apparent doublets and triplets arise? Relatively large couplings between protons that are ortho to each other give rise to the apparent doublets and triplets. A proton that has one ortho neighbour gives an apparent doublet signal, a proton that has two ortho neighbours gives an apparent triplet signal. (ii) what causes the additional fine splitting? Smaller couplings between protons that are meta or para to each other give rise to the fine splitting

5

6.

The 1H NMR spectra of the aromatic-ring portions of the three isomers of nitroaniline are shown below. The labels s, d, t indicate apparent singlet, apparent doublet, and apparent triplet respectively. Assign each spectrum to the correct isomer and give reasons for your answers. d NH2

t

d

d NH2

d

NH2

t

s

d NO2

meta-nitroaniline

O2N

d d

para-nitroaniline

t d

NO2

ortho-nitroaniline

The red letters indicate the apparent splitting (singlet/doublet/triplet for s/d/t) that the signal for the proton attached to each carbon will experience due to the relatively large couplings to adjacent (ortho) protons. Looking at the splittings allows the spectra below to be asssigned as indicated.

ortho-nitroaniline

metanitroaniline

para-nitroaniline

6 Para-nitroaniline is easy...only two hydrogen environments therefore only two signals. Paradisubstituted benzenes are often easy to recognise by 1H NMR because they often show a pair of doublet-like signals in the aromatic region. Sometimes the outside lines of the pair of doublets are smaller than the inside ones so they look like AFL goal posts.

7. The 1H NMR spectrum of trans-4-hexen-3-one is provided below. The numbering scheme for the molecule is also provided.

O

H

H

H H

C

2

C

4

C

6

1

C

3

C

5

C

H H

H

H

H

H

(a)

Re-draw the molecule to show all H atoms clearly. see above

(b)

Using only chemical shifts to justify your answer, indicate which signals correspond to the hydrogen atoms attached to C4 and C5, and which signals correspond to hydrogen atoms at C1, C2, and C6. Signals e and d are in the region expected for protons attached to C=C bonds and so must be due to hydrogen nuclei on C4 and C5. It is not possible to assign specifically which H is on which C in the double bond using chemical shift arguments based on material covered in CHEM2001, but we can assign based on splitting patterns (see below). Signals a, b, and c are in the region expected for saturated alkyl groups (ie hydrogen nuclei bonded to carbon atoms involved only in single bonds), so these signals must be due to the hydrogen nuclei attached to C1, C2, and C6.

7

(c)

Determine the relative number of 1H nuclei contributing to signals a - e. a = 3H,

(d)

b = 3H,

c = 2H,

d = 1H,

e = 1H

Using the integral information, determine which signal corresponds to the hydrogen atoms attached to C2. Must be signal c, since that is the only one that integrates as 2H.

(e)

In the spectrum, the multiplicity of signals is: a = triplet, b= doublet. c = quartet, d= doublet, e = doublet of quartets. Use this information to fully assign all the signals in the spectrum, and explain the splitting patterns in each signal. a = triplet, corresponds to protons attached to C1, since they will be split into a triplet by coupling to the two protons of the adjacent CH2 group. b = doublet, corresponds to protons attached to C6, since they will be split into a doublet by coupling to the single proton of the adjacent CH group. c = quartet, corresponds to protons attached to C2, since they will be split into a quartet by coupling to the three protons of the adjacent CH3 group. d = doublet, corresponds to proton attached to C4, since it will be split into a doublet by coupling to the single proton of the adjacent CH group. e = doublet of quartets, corresponds to proton attached to C5, since it will be split into a doublet by coupling to the single proton of the adjacent CH group (C4) and will be split into a quartet by coupling to the three protons of the adjacent CH3 (C6).

7.

The 300 MHz 1H NMR spectrum and the IR spectrum of an unknown compound of molecular formula C8H7OBr are provided below.

8

(i) Determine the index of hydrogen deficiency for the compound. Grrr, had to look up the formula... IHD = (2 x 8 + 2 -7 - 1)/2 = 5 (ii) Determine the relative number of protons contributing to each signal. going from left to right: 1H 1H 1H 1H 3H

(iii) Determine the structure of the compound. see below (iv) Assign the signals in the 1H NMR spectrum as fully as possible. Hint: In the aromatic region of the 1H NMR spectrum, going from downfield to upfield the signals are an apparent singlet, an apparent doublet, another apparent doublet, and an apparent triplet. both of these are apparent doublets, one is at 7.87 ppm, the other is at 7.67 ppm, can't tell which one is which apparent triplet, 7.33 ppm

O

O 2.6 ppm

Br

apparent singlet, 8.1 ppm

(v) Assign any significant peaks in the IR spectrum. sharp strong signal at 1700 cm-1 is due to C=O stretch

Br

9 8. The 1H NMR spectrum of CH3OCH2CHCl2 will consist of: (a) three singlets (b) a singlet, a doublet, and a triplet (c) two triplets and a quartet (d) two triplets and a quintet

The 13C CPD NMR spectrum of CH3OCH2CHCl2 will consist of:

9.

(a) three singlets

(b) two doublets and one triplet

(c) a quartet and a triplet and a doublet

10. The 13C DEPT 135 NMR spectrum of CH3OCH2CHCl2 will consist of: (a) three +ve singlets

(b) two +ve singlets and one -ve singlet

(c) a quartet and a triplet and a doublet

11.

(d) one +ve singlet and one -ve singlet

This question refers to this structure: 4 1

CH3

O CH2 2

CH3 C

H3

CH3 5

Use the words UPFIELD, DOWNFIELD, SINGLET, DOUBLET, TRIPLET, QUARTET, QUINTET, SEXTET, SEPTET, EQUIVALENT, or INEQUIVALENT to fill in the blanks in the sentences below. * The protons labelled 2 are DOWNFIELD relative to the protons labelled 1. * The protons labelled 1 are INEQUIVALENT to the protons labelled 4. * The protons labelled 4 are EQUIVALENT to the protons labelled 5. * The NMR signal from the protons labelled 2 is split into a QUARTET. * The NMR signal from the protons labelled 3 is split into a SEPTET. * A less shielded proton appears DOWNFIELD in an NMR spectrum relative to a more shielded proton.

10 12. From the list PPM, MINUTES, WAVENUMBERS, AMU, HERTZ, select the usual units for the horizontal axis in:

(a) a 1H NMR spectrum

units = PPM

(b) a mass spectrum

units = AMU (= ATOMIC MASS UNITS, SOMETIMES CALLED DALTONS, Da)

(c) an infrared spectrum

units = WAVENUMBERS

(d) a 13C NMR spectrum

units = PPM

The next question summarises how I approach problems where I need to work out a structure using spectroscopic information. I recommend that students follow this approach too. In addition, students may also use the index of hydrogen deficiency, but only if it helps. I don't use it, but that's just me, some students find it very helpful...do what works for you. Use this approach when tackling the problems in Expt 1 and Expt 2 of the lab manual. Note that the problems in Expt 1 and 2 include mass spectra, so you would need to include an interpretation of mass spectra too when you are working out the structures for problems in those experiments. Also, note that this question was taken from the 2009 exam, and there will be a similar question in the 2011 exam. The question shows marks for each part. You will see that only a small proportion of the marks come from getting the right structure. Most of the marks come from your interpretation of data and your reasoning. 13.

NMR and infrared spectra for Compound E, C10H12O, are on the following page. Use the spectra to determine the structure of Compound E. Use points (a) - (d) as a guide to outlining your interpretation of the spectra.

(a) Infrared spectrum. What does the infrared spectrum indicate about the presence or absence of common oxygen-containing functional groups? Sharp strong signal near 1700 cm-1 indicates a C=O group is present. Absence of strong broad signals in the region 2500-3500 cm-1indicates there are no OH (alcohol or carboxylic acid) groups. (b) 13C NMR spectrum. (Note: the signal near 77 ppm is the solvent signal.) Briefly summarise each signal, in the format: δ (chemical shift), multiplicity (C, CH, CH2, or CH3), and, if possible, any other conclusions that you can make about the functional group giving rise to the signal. 14 ppm, CH3 (positive peak in DEPT, so must be CH3 or CH. Chemical shift suggests not due to CH) 18 ppm, CH2 (negative peak in DEPT, so must be CH2) 41 ppm, CH2 (negative peak in DEPT, so must be CH2) 128, 128.5 and 133, 3 x CH. (positive peak in DEPT, so must be CH3 or CH. Chemical shift range indicates these are not CH3 signals and suggests they may be due to carbons of an aromatic ring. 137, C (not seen in DEPT therefore no attached H). Chemical shift suggests may be due to carbon of an aromatic ring. 201 ppm, C (not seen in DEPT therefore no attached H). Chemical shift suggests may be due to carbon of C=O group

11 (c) 1H NMR spectrum. Briefly summarise each signal, in the format: δ, integral, multiplicity, explanation for multiplicity, and if possible, any other conclusions that you can make about the functional group giving rise to the signal. 1.0 ppm, 3H, t, suggests CH3 group split by coupling to an adjacent CH2 (fragment: CH3—CH2) 1.75 ppm, 2H, apparent sextet, suggests CH2 group split by coupling to 5 nearly equivalent protons of adjacent CH3 and CH2 groups. (fragment: CH3—CH2—CH2, which includes the fragment mentioned above) 2.9 ppm, 2H, t, suggests CH2 group split by coupling to an adjacent CH2 (fragment: CH2—CH2, which is part of the fragment mentioned above) 7.4-7.6 ppm, 3H, second order multiplet, chemical shift suggests protons attached to aromatic ring. 7.95-8.05 ppm, 2H, second order multiplet, chemical shift suggests protons attached to aromatic ring. Taken together with the previous signal, suggests presence of a phenyl group, C6H5.

(d) Draw the structure of Compound E and indicate the 1H and 13C NMR chemical shifts for the various carbon and hydrogen nuclei.

1.75 ppm 1H

NMR O 1.0 ppm

multiplets at 7.4-7.6 and 7.95-8.05 ppm 2.9 ppm

18 ppm 201 ppm 13C

NMR

O 14 ppm

128, 128.5 and 133 ppm 41 ppm

137 ppm

12...