ORGANIC CHEMISTRY SOLUTIONS MANUAL
ORGANIC CHEMISTRY SOLUTIONS MANUAL
5th Edition
ISBN: 9781260367546
Author: SMITH
Publisher: MCG CUSTOM
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Question
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Chapter 14, Problem 14.17P
Interpretation Introduction

(a)

Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.

Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In 1HNMR all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. The number of peaks is calculated by the formula,

P=n+1

Expert Solution
Check Mark

Answer to Problem 14.17P

The labeled proton splits into septet in NMR spectrum of the given compound.

Explanation of Solution

The given compound is (CH3)2CHCO2CH3. The labeled proton is bonded to a carbon atom that has two methyl groups that has six hydrogen atoms. The number of peaks is calculated by the formula,

P=n+1

Where,

➢   P is the number of peaks.

➢   n is the number of protons present on the adjacent carbon atoms.

Here, n=6.

Substitute the value of n=6 in the above formula to calculate the number of peaks.

P=6+1=7

Hence, the labeled proton gives seven peaks in NMR spectrum.

Conclusion

The labeled proton splits into septet in NMR spectrum of the given compound.

Interpretation Introduction

(b)

Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.

Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In 1HNMR all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. The number of peaks is calculated by the formula,

P=n+1

Expert Solution
Check Mark

Answer to Problem 14.17P

The labeled proton (a), (b), and (c) splits into triplet, multiplet and quintet in NMR spectrum of the given compound.

Explanation of Solution

The given compound is CHa3CHb2CHc2CH2CH3. There are three labeled protons, (a), (b), and (c). Proton (a) is bonded to a methylene group that has two hydrogen atoms. The number of peaks is calculated by the formula,

P=n+1

Where,

➢   P is the number of peaks.

➢   n is the number of protons present on the adjacent carbon atoms.

Here, n=2.

Substitute the value of n=2 in the above formula to calculate the number of peaks.

P=2+1=3

Hence, the labeled proton (a) gives three peaks (triplet) in NMR spectrum.

Proton (b) is bonded to one methylene group that has two hydrogen atoms and one methyl group that has three hydrogen atoms. The number of peaks is calculated by the formula,

P=(n+1)(m+1)

Where,

➢   P is the number of peaks.

➢   n and m are the number of protons present on the adjacent carbon atoms.

Here, n=2 and m=3.

Substitute the value of n=2 and m=3 in the above formula to calculate the number of peaks.

P=(2+1)(3+1)=3×4=12

Hence, the labeled proton (b) splits into a multiplet in NMR spectrum.

Proton (c) is bonded to two methylene groups that have four hydrogen atoms. The number of peaks is calculated by the formula,

P=n+1

Where,

➢   P is the number of peaks.

➢   n is the number of protons present on the adjacent carbon atoms.

Here, n=4.

Substitute the value of n=4 in the above formula to calculate the number of peaks.

P=4+1=5

Hence, the labeled proton (c) gives quintet in NMR spectrum.

Therefore, protons (a), (b), and (c) give 3, 12 and 5 peaks in NMR spectrum.

Conclusion

The labeled proton (a), (b), and (c) splits into triplet, multiplet and quintet in NMR spectrum of the given compound.

Interpretation Introduction

(c)

Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.

Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In 1HNMR all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. The number of peaks is calculated by the formula,

P=n+1

Expert Solution
Check Mark

Answer to Problem 14.17P

The labeled proton (a) and (b) splits into doublet and sextet in NMR spectrum of the given compound.

Explanation of Solution

The given compound is shown below.

ORGANIC CHEMISTRY SOLUTIONS MANUAL, Chapter 14, Problem 14.17P , additional homework tip  1

Figure 1

There are two labeled protons, (a) and (b). Proton (b) is bonded to a carbon atom that has one hydrogen and one methylene group. The number of peaks is calculated by the formula,

P=(n+1)(m+1)

Where,

➢   P is the number of peaks.

➢   n and m are the number of protons present on the adjacent carbon atoms.

Here, n=1 and m=2.

Substitute the value of n=1 and m=2 in the above formula to calculate the number of peaks.

P=(1+1)(2+1)=2×3=6

Hence, the labeled proton (b) splits into a sextet in NMR spectrum.

Proton (a) is bonded to a carbon atom that has only one hydrogen atom. The number of peaks is calculated by the formula,

P=n+1

Where,

➢   P is the number of peaks.

➢   n is the number of protons present on the adjacent carbon atoms.

Here, n=1.

Substitute the value of n=1 in the above formula to calculate the number of peaks.

P=1+1=2

Hence, the labeled proton (a) gives two peaks (doublet) in NMR spectrum.

Therefore, protons (a) and (b) give 2 and 6 peaks in NMR spectrum.

Conclusion

The labeled proton (a) and (b) splits into doublet and sextet in NMR spectrum of the given compound.

Interpretation Introduction

(d)

Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.

Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In 1HNMR all chemically equivalent protons generates one signal or one peak, whereas non-equivalent proton generates different signals. The number of peaks is calculated by the formula,

P=n+1

Expert Solution
Check Mark

Answer to Problem 14.17P

The labeled proton (a) and (b) splits into triplet, doublet and doublet in NMR spectrum of the given compound.

Explanation of Solution

The given compound is shown below.

ORGANIC CHEMISTRY SOLUTIONS MANUAL, Chapter 14, Problem 14.17P , additional homework tip  2

Figure 2

There are two labeled protons, (a) and (b). Proton (a) is bonded to a carbon atom that has two hydrogen atoms. The number of peaks is calculated by the formula,

P=n+1

Where,

➢   P is the number of peaks.

➢   n is the number of protons present on the adjacent carbon atoms.

Here, n=2.

Substitute the value of n=2 in the above formula to calculate the number of peaks.

P=2+1=3

Hence, the labeled proton (a) gives three peaks (triplet) in NMR spectrum.

Both protons (b) are bonded to a carbon atom that has only one hydrogen atom. The number of peaks is calculated by the formula,

P=n+1

Where,

➢   P is the number of peaks.

➢   n is the number of protons present on the adjacent carbon atoms.

Here, n=1.

Substitute the value of n=1 in the above formula to calculate the number of peaks.

P=1+1=2

Hence, the labeled protons (b) give two peaks (doublet) in NMR spectrum.

Therefore, protons (a) and (b) give 3 and 2 peaks in NMR spectrum.

Conclusion

The labeled proton (a) and (b) splits into triplet, doublet and doublet in NMR spectrum of the given compound.

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Chapter 14 Solutions

ORGANIC CHEMISTRY SOLUTIONS MANUAL

Ch. 14 - Prob. 14.11PCh. 14 - Problem 14.12 Which compound give a NMR spectrum...Ch. 14 - Prob. 14.13PCh. 14 - Prob. 14.14PCh. 14 - For each compound give the number of 1H NMR...Ch. 14 - Prob. 14.16PCh. 14 - Prob. 14.17PCh. 14 - Problem 14.18 Describe the NMR spectrum of each...Ch. 14 - Problem 14.19 Draw a splitting diagram for 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 - Problem 14.22 What protons in alcohol A give rise...Ch. 14 - How many peaks are observed in the NMR signal for...Ch. 14 - Problem 14.24 Propose a structure for a compound...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 - Problem 14.27 How many lines are observed in the ...Ch. 14 - Problem 14.28 Draw all constitutional isomers of...Ch. 14 - Problem 14.29 Esters of chrysanthemic acid are...Ch. 14 - Prob. 14.30PCh. 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 - Prob. 14.36PCh. 14 - 14.37 How many NMR signals does each natural...Ch. 14 - Prob. 14.38PCh. 14 - 14.39 What effect does increasing the operating...Ch. 14 - Prob. 14.40PCh. 14 - 14.41 How could you use chemical shift and...Ch. 14 - Prob. 14.42PCh. 14 - 14.43 How can you use NMR spectroscopy to...Ch. 14 - Prob. 14.44PCh. 14 - Prob. 14.45PCh. 14 - Prob. 14.46PCh. 14 - Prob. 14.47PCh. 14 - 14.48 How many NMR signals does each compound...Ch. 14 - 14.49 Rank the highlighted carbon atoms in each...Ch. 14 - 14.50 Identify the carbon atoms that give rise to...Ch. 14 - 14.51 a. How many signals does dimethyl...Ch. 14 - 14.52 Answer the following questions about each of...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 - 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. 14.73PCh. 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...