Into how many peaks will the signal for each of the indicated protons be split? a

The 1 H NMR spectrum of CH3OH recorded on a 500 MHz NMR spectrometer consists of two signals, one due to the CH3 protons at 1715 Hz and one due to the OH proton at 1830 Hz, both measured downfi eld from TMS. (a) Calculate the chemical shift of each absorption. (b) Do the CH3 protons absorb upfi eld or downfi eld from the OH proton?

The 1 H NMR spectrum of 1,2-dimethoxyethane (CH3OCH2CH2OCH3) recorded on a 300 MHz NMR spectrometer consists of signals at 1017 Hz and 1065 Hz downfi eld from TMS. (a) Calculate the chemical shift of each absorption. (b) At what frequency would each absorption occur if the spectrum were recorded on a 500 MHz NMR spectrometer?

How many 1 H NMR signals does each compound show? a. CH3CH3 c. CH3CH2CH2CH3 e. CH3CH2CO2CH2CH3 g. CH3CH2OCH2CH3 b. CH3CH2CH3 d. (CH3)2CHCH(CH3)2 f. CH3OCH2CH(CH3)2 h. CH3CH2CH2OH

How many different types of protons does CH3CH2CH2CH2CH2CH2CH2CH2Cl contain?

How many 1 H NMR signals does each dimethylcyclopropane show? a. CH3 CH3 b. CH3 CH3 c. CH3 CH3

Label the protons in each indicated CH2 group as enantiotopic, diastereotopic, or neither. a. CH3CH2CH2CH2CH2CH3 b. CH3CH2CH2CH2CH3 c. CH3CH(OH)CH2CH2CH3

How many 1 H NMR signals would you expect for each compound: (a) CH3CH(Cl)CH2CH3; (b) ClCH2CH(CH3)OCH3; (c) CH3CH(Br)CH2CH2CH3?

For each compound, which of the underlined protons absorbs farther downfi eld: (a) FCH 2CH2CH 2Cl; (b) CH3CH 2CH2CH 2OCH3; (c) CH 3OC(CH 3)3?

For each compound, fi rst label each different type of proton and then rank the protons in order of increasing chemical shift. a. ClCH2CH2CH2Br b. CH3OCH2OC(CH3)3 c. CH3 C CH2CH3 O

Which compounds give a 1 H NMR spectrum with two signals in a ratio of 2:3? a. CH3CH2Cl b. CH3CH2CH3 c. CH3CH2OCH2CH3 d. CH3OCH2CH2OCH3

A compound of molecular formula C8H14O2 gives three NMR signals having the indicated integration values: signal [A] 14 units, signal [B] 12 units, and signal [C] 44 units. How many protons give rise to each signal?

Compound A exhibits two signals in its 1 H NMR spectrum at 2.64 and 3.69 ppm and the ratio of the absorbing signals is 2:3. Compound B exhibits two signals in its 1 H NMR spectrum at 2.09 and 4.27 ppm and the ratio of the absorbing signals is 3:2. Which compound corresponds to CH3O2CCH2CH2CO2CH3 (dimethyl succinate) and which compound corresponds to CH3CO2CH2CH2O2CCH3 (ethylene diacetate)?

Into how many peaks will each indicated proton be split? a. C Cl O CH3CH2 c. C CH2CH2Br O CH3 e. CH3 CH3CH2 H C C H b. CH3 H Br C Br d. Br H Cl C C H f. CICH2CH(OCH3)2

Although Cl2CHCHCl2 and Br2CHCHCl2 each have only two hydrogens, these compounds have very different 1 H NMR spectra. For each compound, give the number of 1 H NMR signals and indicate into how many peaks each signal is split.

For each compound give the number of 1 H NMR signals, and then determine how many peaks are present for each NMR signal. a. O b. CH3 C OCH2CH2OCH3 O c. CH3 C H O d. CI2CHCH2CO2CH3

Sketch the NMR spectrum of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), giving the approximate location of each NMR signal.

How many peaks are present in the NMR signal of each indicated proton? a. (CH3)2CHCO2CH3 b. CH3CH2CH2CH2CH3 c. C Cl H H CH2Br C d. C Br H H H C (all H atoms)

Describe the 1 H NMR spectrum of each compound. State how many NMR signals are present, the splitting pattern for each signal, and the approximate chemical shift. a. CH3OCH2CH3 c. CH3OCH2CH2CH2OCH3 b. CH3CH2 OCH(CH3)2 C O d. H CH3CH2 CH2CH3 H C C

Identify A and B, isomers of molecular formula C3H4Cl2, from the given 1 H NMR data: Compound A exhibits signals at 1.75 (doublet, 3 H, J = 6.9 Hz) and 5.89 (quartet, 1 H, J = 6.9 Hz) ppm. Compound B exhibits signals at 4.16 (singlet, 2 H), 5.42 (doublet, 1 H, J = 1.9 Hz), and 5.59 (doublet, 1 H, J = 1.9 Hz) ppm.

How many signals are present in the 1 H NMR spectrum for each molecule? What splitting is observed in each signal: (a) (CH3)3CCH2OH; (b) CH3CH2CH2OH; (c) (CH3)2CHNH2?

What protons in alcohol A give rise to each signal in its 1 H NMR spectrum? Explain all splitting patterns observed for absorptions between 07 ppm. 87654 3 2 1 0 ppm C CH3 H OH A

Propose a structure for a compound of molecular formula C7H14O2 with an IR absorption at 1740 cm1 and the following 1 H NMR data: Absorption ppm Integration value singlet 1.2 26 triplet 1.3 10 quartet 4.1 6

Propose a structure for a compound of molecular formula C3H8O with an IR absorption at 36003200 cm1 and the following NMR spectrum: 87654 3 1 H 1 H 6 H 210 ppm

The 1 H NMR spectrum of melatonin, the chapter-opening molecule, is more complex than other examples we have encountered, but the chemical shift and splitting patterns observed for several peaks can be explained by what we have learned about 1 H NMR thus far. (a) Which protons in melatonin give rise to signals [A][D]? (b) Explain the splitting pattern observed in signal [C]. 9 87654 [A] melatonin [B] [C] [D] 3 210 ppm

Identify products A and B from the given 1 H NMR data. a. Treatment of CH2 CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1 H NMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2 H) ppm. What is the structure of A? b. Treatment of acetone [(CH3)2C O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1 H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?

How many lines are observed in the 13C NMR spectrum of each compound? a. CH3CH2CH2CH3 c. CH3CH2CH2 O CH2CH2CH3 b. CH3CH2 C OCH3 O d. H CH3CH2 H H C C

Draw all constitutional isomers of molecular formula C3H6Cl2. a. How many signals does each isomer exhibit in its 1 H NMR spectrum? b. How many lines does each isomer exhibit in its 13C NMR spectrum? c. When only the number of signals in both 1 H and 13C NMR spectroscopy is considered, is it possible to distinguish all of these constitutional isomers?

Identify the carbon atoms that give rise to each NMR signal. 200 180 160 140 120 100 80 60 40 20 0 ppm a. CH3CH(OH)CH2CH3 220 200 180 160 140 120 100 80 60 40 20 0 ppm

A compound of molecular formula C4H8O2 shows no IR peaks at 36003200 or 1700 cm1. It exhibits one singlet in its 1 H NMR spectrum at 3.69 ppm, and one line in its 13C NMR spectrum at 67 ppm. What is the structure of this unknown?

Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at > 160 ppm. Then draw the structure of an isomer of molecular formula C4H8O that has all of its 13C NMR signals at < 160 ppm.

How many different types of protons are present in each compound? a. (CH3)3CH e. O h. CH3 CH3 CH2CH3 Br C C f. H CH3CH2 H CH2CH3 C C i. CH3CH(OH)CH2CH3 g. CH3CH2CH2OCH2CH2CH3 j. O CH3 b. (CH3)3CC(CH3)3 c. CH3CH2OCH2CH2CH2CH2CH3 d. CH3CH CH2

How many 1 H NMR signals does each compound give? a. CH3 CH3 b. CH2CH3 c. CH3 CH3 d. CH3 CH3

How many 1 H NMR signals does each natural product exhibit? a. N N N O CH3 O CH3 CH3 N caffeine (from coffee beans and tea leaves) b. CHO OH OCH3 vanillin (from the vanilla bean) c. OH thymol (from thyme) d. O CH3O HO capsaicin (from hot peppers) N H

Using a 300 MHz NMR instrument: a. How many Hz downfi eld from TMS is a signal at 2.5 ppm? b. If a signal comes at 1200 Hz downfi eld from TMS, at what ppm does it occur? c. If two peaks are separated by 2 ppm, how many Hz does this correspond to?

Acetone exhibits a singlet in its 1 H NMR spectrum at 2.16 ppm. If CH2Cl2 exhibits a singlet 1570 Hz downfi eld from acetone on a 500 MHz NMR spectrometer, what is the chemical shift of the singlet due to CH2Cl2?

Which of the indicated protons in each pair absorbs farther downfi eld? a. CH3CH2CH2CH2CH3 or CH3CH2CH2OCH3 c. or CH3OCH2CH3 b. CH or 3CH2CH2I CH3CH2CH2F d. CH or 3CH2CHBr2 CH3CH2CH2Br

How could you use chemical shift and integration data in 1 H NMR spectroscopy to distinguish between each pair of compounds? The 1 H NMR spectrum of each compound contains only singlets. a. CH3CO2C(CH3)3 and CH3CO2CH3 c. CH3 CH3 and CH3 CH3 CH3 b. CH3OCH2CH2OCH3 and CH3OCH2OCH3

Which compounds give one singlet in the 1 H NMR spectrum? CH3CH3 CH3 C C CH3 O O Cl Cl Br Br Br Br Br (CH3)3C C(CH3)3 (CH3)3C C OC(CH3)3 O CH2 CHCH C O CH2 CH3 CH3 CH3 CH3 C C

For the five isomeric alkanes of molecular formula C6H14, label each type of proton and indicate how many peaks each will exhibit in its 1 H NMR signal.

Into how many peaks will the signal for each of the indicated protons be split? a. CH3CH(OCH3)2 d. CH3OCH2CHCl2 g. CH3CH2CH2CH2OH j. CH3CH2 CH3 H H C C b. CH3OCH2CH2 OCH3 C O e. (CH3)2CH OCH2CH3 C O h. CH3CH2CH2 OH C O k. Br CH3 H H C C c. CH2CH3 f. HOCH2CH2CH2OH i. CH3CH2 H C O l. CH3 H H H C C

How can you use 1 H NMR spectroscopy to distinguish between CH2 C(Br)CO2CH3 and methyl (2E)-3-bromo-2-propenoate, BrCH CHCO2CH3?

Label the signals due to Ha, Hb, and Hc in the 1 H NMR spectrum of acrylonitrile (CH2 CHCN). Draw a splitting diagram for the absorption due to the Ha proton. Jab = 11.8 Hz Jbc = 0.9 Hz Jac = 18 Hz C Hc Hb Ha CN C 6.6 chemical shift (ppm) 6.2 5.7

Draw the four constitutional isomers having molecular formula C4H9Br and indicate how many different kinds of carbon atoms each has.

Which compounds in Problem 14.42 give one signal in their 13C NMR spectra?

Explain why the carbonyl carbon of an aldehyde or ketone absorbs farther downfi eld than the carbonyl carbon of an ester in a 13C NMR spectrum.

Rank the indicated carbon atoms in each compound in order of increasing chemical shift. a. CH3CH2 OH C O b. CH3CH2CHCH2CH3 OH c. CH2CH3 C O d. CH2 CHCH2CH2CH2Br

Identify the carbon atoms that give rise to the signals in the 13C NMR spectrum of each compound. a. CH3CH2CH2CH2OH; 13C NMR: 14, 19, 35, and 62 ppm b. (CH3)2CHCHO; 13C NMR: 16, 41, and 205 ppm c. CH2 CHCH(OH)CH3; 13C NMR: 23, 69, 113, and 143 ppm

a. How many signals does dimethyl fumarate (CH3O2CCH CHCO2CH3, with a trans C C) exhibit in its 13C NMR spectrum? b. Draw the structure of an isomer of dimethyl fumarate that has each of the following number of signals in its 13C NMR spectrum: [1] three; [2] four; [5] fi ve.

Propose a structure consistent with each set of spectral data: a. C4H8Br2: IR peak at 30002850 cm1; NMR (ppm): d. C6H14O: IR peak at 36003200 cm1; NMR (ppm): 1.87 (singlet, 6 H) 0.8 (triplet, 6 H) 1.5 (quartet, 4 H) 3.86 (singlet, 2 H) 1.0 (singlet, 3 H) 1.6 (singlet, 1 H) b. C3H6Br2: IR peak at 30002850 cm1; NMR (ppm): e. C6H14O: IR peak at 30002850 cm1; NMR (ppm): 2.4 (quintet) 1.10 (doublet, 30 units) 3.5 (triplet) 3.60 (septet, 5 units) c. C5H10O2: IR peak at 1740 cm1; NMR (ppm): f. C3H6O: IR peak at 1730 cm1; NMR (ppm): 1.15 (triplet, 3 H) 2.30 (quartet, 2 H) 1.11 (triplet) 1.25 (triplet, 3 H) 4.72 (quartet, 2 H) 2.46 (multiplet) 9.79 (triplet)

Identify the structures of isomers A and B (molecular formula C9H10O). Compound A: IR peak at 1742 cm1; 1 H NMR data (ppm) at 2.15 (singlet, 3 H), 3.70 (singlet, 2 H), and 7.20 (broad singlet, 5 H). Compound B: IR peak at 1688 cm1; 1 H NMR data (ppm) at 1.22 (triplet, 3 H), 2.98 (quartet, 2 H), and 7.287.95 (multiplet, 5 H).

Compound C has a molecular ion in its mass spectrum at 146 and a prominent absorption in its IR spectrum at 1762 cm1. C shows the following 1 H NMR spectral data: 1.47 (doublet, 3 H), 2.07 (singlet, 6 H), and 6.84 (quartet, 1 H) ppm. What is the structure of C?

As we will learn in Chapter 20, reaction of (CH3)2CO with LiC CH followed by H2O affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 36003200, 3303, 2938, and 2120 cm1. D shows the following 1 H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?

Identify the structures of isomers E and F (molecular formula C4H8O2). Compound E: IR absorption at 1743 cm1 87654 3 23 210 ppm 1H NMR of E 29 30 Compound F: IR absorption at 1730 cm1 87654 3 18 210 ppm

Identify the structures of isomers H and I (molecular formula C8H11N). a. Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm1 ppm 8765 5 H 2 H 2 H 2 H 4 3 210 b. Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm1 ppm 8765 1 H 3 H 2 H 5 H 4 3 210

Propose a structure consistent with each set of data. a. Compound J: molecular ion at 72; IR peak at 1710 cm1; 1 H NMR data (ppm) at 1.0 (triplet, 3 H), 2.1 (singlet, 3 H), and 2.4 (quartet, 2 H) b. Compound K: molecular ion at 88; IR peak at 36003200 cm1; 1 H NMR data (ppm) at 0.9 (triplet, 3 H), 1.2 (singlet, 6 H), 1.5 (quartet, 2 H), and 1.6 (singlet, 1 H)

In the presence of a small amount of acid, a solution of acetaldehyde (CH3CHO) in methanol (CH3OH) was allowed to stand and a new compound L was formed. L has a molecular ion in its mass spectrum at 90 and IR absorptions at 2992 and 2941 cm1. L shows three signals in its 13C NMR at 19, 52, and 101 ppm. The 1 H NMR spectrum of L is given below. What is the structure of L? ppm 9 3 012 7 46 25 45678

Treatment of (CH3)2CHCH(OH)CH2CH3 with TsOH affords two products (M and N) with molecular formula C6H12. The 1 H NMR spectra of M and N are given below. Propose structures for M and N and draw a mechanism to explain their formation. 3 H 2 H 3 H 1 H ppm 1H NMR of M 9 0 8 1234567 3 H 3 H 3 H 2 H 2 H ppm 1H NMR of N 9 0

Compound O has molecular formula C10H12O and shows an IR absorption at 1687 cm1. The 1 H NMR spectrum of O is given below. What is the structure of O? ppm 9 0 1234

Compound P has molecular formula C5H9ClO2. Deduce the structure of P from its 1 H and 13C NMR spectra. ppm 1H NMR of P 2 H 2 H 2 H 3 H ppm 876543210 13C NMR of P 200 180 160 140 120 100 80 60 40 20 0 ppm

Treatment of 2-butanone (CH3COCH2CH3) with strong base followed by CH3I forms a compound Q, which gives a molecular ion in its mass spectrum at 86. The IR (> 1500 cm1 only) and 1 H NMR spectrum of Q are given below. What is the structure of Q? 8 1H NMR of Q 7654 3 H 1 H 6 H 3210 ppm % Transmittance 100 50 0 4000 2000 3500 1500 3000 2500 Wavenumber (cm1)

Low molecular weight esters (RCO2R) often have characteristic odors. Using its molecular formula and 1 H NMR spectral data, identify each ester. a. Compound R, the odor of banana: C7H14O2; 1 H NMR: 0.93 (doublet, 6 H), 1.52 (multiplet, 2 H), 1.69 (multiplet, 1 H), 2.04 (singlet, 3 H), and 4.10 (triplet, 2 H) ppm b. Compound S, the odor of rum: C7H14O2; 1 H NMR: 0.94 (doublet, 6 H), 1.15 (triplet, 3 H), 1.91 (multiplet, 1 H), 2.33 (quartet, 2 H), and 3.86 (doublet, 2 H) ppm

When 2-bromo-3,3-dimethylbutane is treated with K+ OC(CH3)3, a single product T having molecular formula C6H12 is formed. When 3,3-dimethyl-2-butanol is treated with H2SO4, the major product U has the same molecular formula. Given the following 1 H NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T. 1 H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) ppm 1 H NMR of U: 1.60 (singlet) ppm

In a BaeyerVilliger reaction, ketones (R2C O) are converted to esters (RCO2R) by using peroxy acids. With an unsymmetrical ketone, two possible esters can be formed, as shown for 3,3-dimethyl-2-butanone as starting material. How could you use spectroscopic techniques1 H NMR, IR, and MSto determine which ester (A or B) is formed? CH3 C(CH3)3 CH3O C(CH3)3 CH3 OC(CH3)3 R'CO3H C O C O C O by-product from the peroxy acid or + R'CO2H 3,3-dimethyl-2-butanone A B

The 1 H NMR spectrum of N,N-dimethylformamide shows three singlets at 2.9, 3.0, and 8.0 ppm. Explain why the two CH3 groups are not equivalent to each other, thus giving rise to two NMR signals. C O H N,N-dimethylformamide N(CH3)2

18-Annulene shows two signals in its 1 H NMR spectrum, one at 8.9 (12 H) and one at 1.8 (6 H) ppm. Using a similar argument to that offered for the chemical shift of benzene protons, explain why both shielded and deshielded values are observed for 18-annulene. 18-annulen

Explain why the 13C NMR spectrum of 3-methyl-2-butanol shows fi ve signals.

Since 31P has an odd mass number, 31P nuclei absorb in the NMR and, in many ways, these nuclei behave similarly to protons in NMR spectroscopy. With this in mind, explain why the 1 H NMR spectrum of methyl dimethylphosphonate, CH3PO(OCH3)2, consists of two doublets at 1.5 and 3.7 ppm.