CHEM 8M 1H NMR Problems (McMurry, Organic Chemistry, 8th Edition) #14 – 18 = 5 pts toward Exp 2 pre-lab, day 2

13.8 | 1H NMR Spectroscopy and Proton Equivalence 473

• The fourth possibility arises in chiral molecules, such as (R)-2-butanol. The two ! CH2 ! hydrogens at C3 are neither homotopic nor enantio-topic. Since substitution of a hydrogen at C3 would form a second chiral-ity center, different diastereomers (Section 5.6) would result depending on whether the pro-R or pro-S hydrogen were substituted for. Such hydro-gens, whose substitution by X leads to different diastereomers, are said to be diastereotopic. Diastereotopic hydrogens are neither chemically nor electronically equivalent. They are completely different and would likely show different NMR absorptions.

The two hydrogens on C3are diastereotopic and havedifferent NMR absorptions.

The two possible substitutionproducts are diastereomers.

Replace eitherH or H with X

or

pro-S pro-R

2 41 3

H OH

HH

C CH3H3C C

H OH

XH

C CH3H3C C

H OH

HX

C CH3H3C C

Problem 13.13Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:

(c)(b)(a)H H

(d)

H H

H OH

(e) (f)

Br

O

O

O

O H H

C

H

C

H3C

H3C CH3

CH3H

CH3

H

H

Problem 13.14How many kinds of electronically nonequivalent protons are present in each of the follow-ing compounds, and thus how many NMR absorptions might you expect in each?(a) CH3CH2Br (b) CH3OCH2CH(CH3)2 (c) CH3CH2CH2NO2(d) Methylbenzene (e) 2-Methyl-1-butene (f ) cis-3-Hexene

Problem 13.15How many absorptions would you expect (S)-malate, an intermediate in carbohydrate metabolism, to have in its 1H NMR spectrum? Explain.

(S)-Malate

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13.9 | Chemical Shifts in 1H NMR Spectroscopy 475

Problem 13.16Each of the following compounds has a single 1H NMR peak. Approximately where would you expect each compound to absorb?

(e)(d) CH2Cl2

O(b) (c)(a)

CH3C CH3

C

H

C

O

OH

(f)

N CH3

H3C

H3C

Problem 13.17Identify the different kinds of nonequivalent protons in the following molecule, and tell where you would expect each to absorb:

H

H

C

H

H

CH3O

H

CH2CH3C

H

Table 13.3 Correlation of 1H Chemical Shift with Environment

Type of hydrogen Chemical shift (!)

Reference Si(CH3)4 0

Alkyl (primary) O CH3 0.7–1.3

Alkyl (secondary) O CH2 O 1.2–1.6

Alkyl (tertiary)CH

1.4–1.8

Allylic

CC C

H 1.6–2.2

Methyl ketone

CH3

O

C

2.0–2.4

Aromatic methyl Ar O CH3 2.4–2.7

Alkynyl O C q C O H 2.5–3.0

Alkyl halide

C Hal

H 2.5–4.0

Type of hydrogen Chemical shift (!)

AlcoholC O H

2.5–5.0

Alcohol, ether

C O

H 3.3–4.5

Vinylic

C

H

C

4.5–6.5

Aryl Ar O H 6.5–8.0

Aldehyde

H

O

C

9.7–10.0

Carboxylic acid

O H

O

C

11.0–12.0

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476 CHAPTER 13 | Structure Determination: Nuclear Magnetic Resonance Spectroscopy

13.10 Integration of 1H NMR Absorptions: Proton Counting

Look at the 1H NMR spectrum of methyl 2,2-dimethylpropanoate in Figure 13.12. There are two peaks, corresponding to the two kinds of protons, but the peaks aren’t the same size. The peak at 1.2 !, due to the (CH3)3C ! protons, is larger than the peak at 3.7 !, due to the ! OCH3 protons.

012345678910 ppmChemical shift (!)

TMSH3C CH3

CH3

CH3

C C O

O

Inte

nsity

Chem.shift1.203.65

Rel.area3.001.00

Figure 13.12 The 1H NMR spectrum of methyl 2,2-dimethylpropanoate. Integrating the peaks in a stair-step manner shows that they have a 1!3 ratio, corresponding to the ratio of the numbers of protons (3!9) respon-sible for each peak. Modern instruments give a direct digital readout of relative peak areas.

The area under each peak is proportional to the number of protons causing that peak. By electronically measuring, or integrating, the area under each peak, it’s pos-sible to measure the relative numbers of the different kinds of protons in a molecule.

Modern NMR instruments provide a digital readout of relative peak areas, but an older, more visual method displays the integrated peak areas as a stair-step line, with the height of each step proportional to the area under the peak, and therefore pro-portional to the relative number of protons causing the peak. For example, the two steps for the peaks in methyl 2,2-dimethylpropanoate are found to have a 1!3 (or 3!9) height ratio when integrated—exactly what we expect since the three ! OCH3 protons are equivalent and the nine (CH3)3C ! protons are equivalent.

Problem 13.18How many peaks would you expect in the 1H NMR spectrum of 1,4-dimethylbenzene (para-xylene, or p-xylene)? What ratio of peak areas would you expect on integration of the spec-trum? Refer to Table 13.3 for approximate chemical shifts, and sketch what the spectrum would look like. (Remember from Section 2.4 that aromatic rings have two resonance forms.)

p-Xylene

CH3

H3C

13.11 Spin–Spin Splitting in 1H NMR SpectraIn the 1H NMR spectra we’ve seen thus far, each different kind of proton in a molecule has given rise to a single peak. It often happens, though, that the absorption of a proton splits into multiple peaks, called a multiplet. For exam-ple, in the 1H NMR spectrum of bromoethane shown in Figure 13.13,

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CHEM 8M: #19 – 21 and 21.70 (next page) = 10 pts toward Exp 5 report

13.11 | Spin–Spin Splitting in 1H NMR Spectra 481

To interpret the NMR information, let’s look at each absorption individually. The three- proton absorption at 0.92 ! is due to a methyl group in an alkane-like environment, and the triplet splitting pattern implies that the CH3 is next to a CH2. Thus, our molecule contains an ethyl group, CH3CH2 ! . The six-proton singlet at 1.20 ! is due to two equivalent alkane-like methyl groups attached to a carbon with no hydrogens, (CH3)2C, and the two-proton quartet at 1.50 ! is due to the CH2 of the ethyl group. All 5 carbons and 11 of the 12 hydrogens in the molecule are now accounted for. The remaining hydrogen, which appears as a broad one- proton singlet at 1.64 !, is probably due to an OH group, since there is no other way to account for it. Putting the pieces together gives the structure: 2-methyl-2-butanol.

Solution

0.92 ! 2-Methyl-2-butanol

CH3

CH2CH3

OH

CCH3

1.50 !

1.64 !

1.20 !

Problem 13.19Predict the splitting patterns you would expect for each proton in the following molecules:(a) CHBr2CH3

(d)

CH3CHCOCH2CH3

(b) CH3OCH2CH2Br (c) ClCH2CH2CH2Cl

CH3

O (e) (f)

CH3CH2COCHCH3

CH3

O

Problem 13.20Draw structures for compounds that meet the following descriptions:(a) C2H6O; one singlet (b) C3H7Cl; one doublet and one septet(c) C4H8Cl2O; two triplets (d) C4H8O2; one singlet, one triplet, and one quartet

Problem 13.21The integrated 1H NMR spectrum of a compound of formula C4H10O is shown in Figure 13.17. Propose a structure.

TMS

Inte

nsity

012345678910 ppmChemical shift (!)

Chem.shift1.223.49

Rel.area1.501.00

Figure 13.17 An integrated 1H NMR spectrum for Problem 13.21.

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** Pay attention to “Chem Shift” and “Rel. area” table to determine number of peaks and relative integration values (number of H’s per signal).

868 CHAPTER 21 | Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

21.70 Assign structures to compounds with the following 1H NMR spectra:(a) C5H10O2 IR: 1735 cm!1

Inte

nsity

012345678910 ppmChemical shift (!)

TMS

Chem.shift1.222.014.99

Rel.area6.003.001.00

(b) C11H12O2 IR: 1710 cm!1

TMS

Inte

nsity

012345678910 ppm012345678910 ppmChemical shift (!)

Chem.shift1.324.246.417.367.497.68

Rel.area3.002.001.003.002.001.00

21.71 Propose structures for compounds with the following 1H NMR spectra:(a) C5H9ClO2 IR: 1735 cm!1

TMS

Inte

nsity

012345678910 ppm012345678910 ppmChemical shift (!)

Chem.shift1.262.773.764.19

Rel.area1.501.001.001.00

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