Organic Chemistry -Study Guide / Solution Manual (Custom)
Organic Chemistry -Study Guide / Solution Manual (Custom)
4th Edition
ISBN: 9781259141072
Author: SMITH
Publisher: MCG
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Chapter 14, Problem 14.13P
Interpretation Introduction

(a)

Interpretation:

The number of peaks for each indicated proton is to be predicted.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. Protons which are present in the same chemical environment that is between the same group of atoms are known as chemically equivalent protons and in 1HNMR all chemically equivalent proton generates one signal or one peak, whereas non-equivalent proton generates different signals.

Expert Solution
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Answer to Problem 14.13P

The given compound shows two signals in 1H NMR spectrum. Ha gives triplet and Hb gives quartet peaks.

Explanation of Solution

The number of signals in each compound is equal to the number of hydrogen atoms present in a different chemical environment. The given compound is CH3HaCH2HbCOCl.

It has two different types of non-equivalent protons shown in red color. Therefore, it gives two signals in 1HNMR spectrum. The signal Ha gets split into a triplet by 2 Hb protons and the signal Hb gets split into a quartet by 3 Ha protons.

Conclusion

The given compound shows two signals in 1H NMR spectrum. Ha gives triplet and Hb gives quartet peaks.

Interpretation Introduction

(b)

Interpretation:

The number of peaks for each indicated proton is to be predicted.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. Protons which are present in the same chemical environment that is between the same group of atoms are known as chemically equivalent protons and in 1HNMR all chemically equivalent proton generates one signal or one peak, whereas non-equivalent proton generates different signals.

Expert Solution
Check Mark

Answer to Problem 14.13P

The given compound shows two signals in 1H NMR spectrum. Ha gives doublet and Hb gives quartet peaks.

Explanation of Solution

The given compound is shown below.

Organic Chemistry -Study Guide / Solution Manual (Custom), Chapter 14, Problem 14.13P , additional homework tip  1

Figure 1

It has two non-equivalent protons. Therefore, it gives two signals in 1H NMR spectrum.

The signal Ha gets split into a doublet by 1 Hb proton and the signal Hb gets split into a quartet by 3 Ha protons.

Conclusion

The given compound shows two signals in 1H NMR spectrum. Ha gives doublet and Hb gives quartet peaks.

Interpretation Introduction

(c)

Interpretation:

The number of peaks for each indicated proton is to be predicted.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. Protons which are present in the same chemical environment that is between the same group of atoms are known as chemically equivalent protons and in 1HNMR all chemically equivalent proton generates one signal or one peak, whereas non-equivalent proton generates different signals.

Expert Solution
Check Mark

Answer to Problem 14.13P

The given compound shows three signals in 1H NMR spectrum. Ha gives singlet, Hb and Hc give triplet peaks.

Explanation of Solution

The given compound is shown below.

Organic Chemistry -Study Guide / Solution Manual (Custom), Chapter 14, Problem 14.13P , additional homework tip  2

Figure 2

It has three non-equivalent protons. Therefore, it gives three signals in 1H NMR spectrum. The signal Ha has no adjacent non-equivalent protons, therefore; it gives only a single peak that is singlet.

The signal Hb gets split into a triplet by 2 Hc protons and also, the signal Hc gets split into a triplet by 2 Hb protons.

Conclusion

The given compound shows three signals in 1H NMR spectrum. Ha gives singlet, Hb and Hc give triplet peaks.

Interpretation Introduction

(d)

Interpretation:

The number of peaks for each indicated proton is to be predicted.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. Protons which are present in the same chemical environment that is between the same group of atoms are known as chemically equivalent protons and in 1HNMR all chemically equivalent proton generates one signal or one peak, whereas non-equivalent proton generates different signals.

Expert Solution
Check Mark

Answer to Problem 14.13P

The given compound shows two signals in 1H NMR spectrum. Both Ha and Hb give two peaks.

Explanation of Solution

The given compound is shown below.

Organic Chemistry -Study Guide / Solution Manual (Custom), Chapter 14, Problem 14.13P , additional homework tip  3

Figure 3

It has two non-equivalent protons. Therefore, it gives two signals in 1H NMR spectrum.

The signal Ha gets split into two peaks that is doublet by 1 Hb proton and also, the signal Hb gets split into a doublet by 1 Ha proton.

Conclusion

The given compound shows two signals in 1H NMR spectrum. Both Ha and Hb give two peaks.

Interpretation Introduction

(e)

Interpretation:

The number of peaks for each indicated proton is to be predicted.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. Protons which are present in the same chemical environment that is between the same group of atoms are known as chemically equivalent protons and in 1HNMR all chemically equivalent proton generates one signal or one peak, whereas non-equivalent proton generates different signals.

Expert Solution
Check Mark

Answer to Problem 14.13P

The given compound shows two signals in 1H NMR spectrum. Both Ha and Hb give two peaks.

Explanation of Solution

The given compound is shown below.

Organic Chemistry -Study Guide / Solution Manual (Custom), Chapter 14, Problem 14.13P , additional homework tip  4

Figure 4

It has two non-equivalent protons. Therefore, it gives two signals in 1H NMR spectrum. Both the protons are present on the same carbon atom.

The signal Ha gets split into two peaks that is doublet by 1 Hb proton and also, the signal Hb gets split into a doublet by 1 Ha proton.

Conclusion

The given compound shows two signals in 1H NMR spectrum. Both Ha and Hb give two peaks.

Interpretation Introduction

(f)

Interpretation:

The number of peaks for each indicated proton is to be predicted.

Concept introduction:

The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. Protons which are present in the same chemical environment that is between the same group of atoms are known as chemically equivalent protons and in 1HNMR all chemically equivalent proton generates one signal or one peak, whereas non-equivalent proton generates different signals.

Expert Solution
Check Mark

Answer to Problem 14.13P

The indicated protons Ha and Hb shows doublet and triplet peaks.

Explanation of Solution

The given compound is shown below.

Organic Chemistry -Study Guide / Solution Manual (Custom), Chapter 14, Problem 14.13P , additional homework tip  5

Figure 5

It has three non-equivalent protons. Therefore, it gives three signals in 1H NMR spectrum.

The signal Ha gets split into two peaks that is doublet by 1 Hb proton; the signal Hb gets split into three peaks that is triplet by 2 Ha protons. The signal Hc has no adjacent proton therefore; it gets split into a singlet.

Hence, the indicated protons which are shown in red color shows doublet and triplet peaks.

Conclusion

The indicated protons Ha and Hb shows doublet and triplet peaks.

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Chapter 14 Solutions

Organic Chemistry -Study Guide / Solution Manual (Custom)

Ch. 14 - Prob. 14.11PCh. 14 - Prob. 14.12PCh. 14 - Prob. 14.13PCh. 14 - For each compound give the number of 1H NMR...Ch. 14 - Prob. 14.15PCh. 14 - Prob. 14.16PCh. 14 - Describe the 1H NMR spectrum of each compound....Ch. 14 - Draw a splitting diagram for Hb in...Ch. 14 - Problem 14.20 Identify A and B, isomers of...Ch. 14 - Problem 14.21 How many signals are present in the ...Ch. 14 - What protons in alcohol A give rise to each signal...Ch. 14 - How many peaks are observed in the NMR signal for...Ch. 14 - Propose a structure for a compound of molecular...Ch. 14 - Problem 14.25 Propose a structure for a compound...Ch. 14 - Problem 14.26. Identify products A and B from the...Ch. 14 - How many lines are observed in the 13C NMR...Ch. 14 - Problem 14.28 Draw all constitutional isomers of...Ch. 14 - Esters of chrysanthemic acid are naturally...Ch. 14 - Prob. 14.29PCh. 14 - Problem 14.31 Identify the carbon atoms that give...Ch. 14 - Problem 14.32 A compound of molecular formula ...Ch. 14 - Problem 14.33 Draw the structure of a compound of...Ch. 14 - 14.34 (a) How many NMR signals does each of the...Ch. 14 - 14.35 (a) How many NMR signals does each compound...Ch. 14 - How many different types of protons are present in...Ch. 14 - How many 1H NMR signals does each compound give?...Ch. 14 - 14.37 How many NMR signals does each natural...Ch. 14 - Prob. 14.38PCh. 14 - Acetone exhibits a singlet in its 1H NMR spectrum...Ch. 14 - Prob. 14.40PCh. 14 - Prob. 14.41PCh. 14 - How could you use chemical shift and integration...Ch. 14 - Which compounds give one singlet in the 1H NMR...Ch. 14 - Prob. 14.44PCh. 14 - 14.43 How can you use NMR spectroscopy to...Ch. 14 - Prob. 14.46PCh. 14 - Prob. 14.47PCh. 14 - Which compounds in Problem 14.43 give one signal...Ch. 14 - Prob. 14.49PCh. 14 - How many 13C NMR signals does each compound...Ch. 14 - Rank the highlighted carbon atoms in each compound...Ch. 14 - 14.50 Identify the carbon atoms that give rise to...Ch. 14 - 14.51 a. How many signals does dimethyl...Ch. 14 - 14.53 Propose a structure consistent with each set...Ch. 14 - 14.54 Identify the structures of isomers A and B...Ch. 14 - 14.55 Reaction of with affords compound W,...Ch. 14 - 14.56 Treatment of with , followed by aqueous ...Ch. 14 - 14.57 Compound C has a molecular ion in its mass...Ch. 14 - 14.58 As we will learn in Chapter 20, reaction of ...Ch. 14 - 14.59 Identify the structures of isomers E and F...Ch. 14 - 14.59 Identify the structures of isomers H and I...Ch. 14 - 14.61 Propose a structure consistent with each set...Ch. 14 - 14.62 Reaction of with , followed by treatment...Ch. 14 - Reaction of aldehyde D with amino alcohol E in the...Ch. 14 - 14.64 Propose a structure consistent with each set...Ch. 14 - 14.65 In the presence of a small amount of acid, a...Ch. 14 - 14.66 Treatment of with affords two products (M...Ch. 14 - 14.67 Compound O has molecular formula and shows...Ch. 14 - 14.68 Compound P has molecular formula . Deduce...Ch. 14 - 14.69 Treatment of with strong base followed by ...Ch. 14 - Low molecular weight esters RCO2R often have...Ch. 14 - 14.70 When -bromo--dimethylbutane is treated with...Ch. 14 - 14.71 Propose a structure consistent with each set...Ch. 14 - 14.72 Reaction of unknown A with forms...Ch. 14 - Prob. 14.75PCh. 14 - 14.74 -Annulene shows two signals in its ...Ch. 14 - 14.75 Explain why the spectrum of-methylbutan--ol...Ch. 14 - 14.76 Because has an odd mass number, nuclei...Ch. 14 - 14.77 Cyclohex--enone has two protons on its...
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