Organic Chemistry: Principles and Mechanisms (Second Edition)
Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
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Chapter 15, Problem 15.56P
Interpretation Introduction

(a)

Interpretation:

The IR spectrum of the given compound is to be sketched with the X-axis range from 400 - 4000 cm-1, paying attention to each absorption’s frequency, breadth, and intensity.

Concept introduction:

IR spectroscopy enables to identify different functional groups in the given molecule.

One of the greatest advantages of IR spectroscopy is that the frequency of a particular type of vibration is typically found within a characteristic range of frequencies, regardless of the functional group with which the vibration is associated. The fact that absorptions by certain vibrations appear with characteristic frequency ranges enables us to use IR spectroscopy to obtain structural information about a molecule. Stretching vibrations of Q-H bonds (where Q is a heavy atom such as C, N or O) occurs in the region between ~ 2500 - 4000 cm-1. Double bonds appear between ~ 1500 and 2000 cm-1. Single bonds between two heavy atoms appear between ~ 1000 - 1500 cm-1. The region below ~ 1400 cm-1 is called the fingerprint region.

Expert Solution
Check Mark

Answer to Problem 15.56P

The IR spectrum of the given molecule can be sketched as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  1

Explanation of Solution

The given molecule is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  2

The given molecule has a ketone and alcohol as functional groups. According to Table 15-2, the characteristic frequency of absorption of alcohols (OH bond) appears as a broad and strong peak between ~3200 - 3600 cm-1. The conjugated C=O group in ketones appears in at ~1690 cm-1. The aromatic C=C appears between ~1450 - 1550 cm-1, and it shows two-three variable peaks. The sp3 hybridized C-H is a variable peak at ~2800 - 3000 cm-1. The para substituent appears in the fingerprint region as a strong peak ~830-850 cm-1. Hence, the IR sketch of the given molecule could be drawn as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  3

Each significant peak is assigned in the above IR spectrum.

Conclusion

The absorptions by certain vibrations appear with characteristic frequency ranges, enabling us to use IR spectroscopy to obtain structural information about a molecule.

Interpretation Introduction

(b)

Interpretation:

The IR spectrum of the given compound is to be sketched with the X-axis range from 400 - 4000 cm-1, paying attention to each absorption’s frequency, breadth, and intensity.

Concept introduction:

IR spectroscopy enables to identify different functional groups in the given molecule.

One of the greatest advantages of IR spectroscopy is that the frequency of a particular type of vibration is typically found within a characteristic range of frequencies, regardless of the functional group with which the vibration is associated. The fact that absorptions by certain vibrations appear with characteristic frequency ranges enables us to use IR spectroscopy to obtain structural information about a molecule. Stretching vibrations of Q-H bonds (where Q is a heavy atom such as C, N or O) occurs in the region between ~ 2500 - 4000 cm-1. Double bonds appear between ~ 1500 and 2000 cm-1. Single bonds between two heavy atoms appear between ~ 1000 - 1500 cm-1. The region below ~ 1400 cm-1 is called the fingerprint region.

Expert Solution
Check Mark

Answer to Problem 15.56P

The IR spectrum of the given molecule can be sketched as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  4

Explanation of Solution

The given molecule is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  5

The given molecule has an aldehyde and a secondary amine as functional groups. According to Table 15-2, the characteristic frequency of absorption of amines (NH bond) appears as a medium single peak between ~3300 - 3500 cm-1. The aldehyde C=O group appears ~1680-1750 cm-1. The sp2 C-H in aldehydes appears as two peaks, 2750 and 2820 cm-1. The sp3 hybridized C-H appears as a variable peak ~2800 - 3000 cm-1. Hence, the IR sketch of the given molecule could be drawn as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  6

Each significant peak is assigned in the above IR spectrum.

Conclusion

The absorptions by certain vibrations appear with characteristic frequency ranges, enabling us to use IR spectroscopy to obtain structural information about a molecule.

Interpretation Introduction

(c)

Interpretation:

The IR spectrum of the given compound is to be sketched with the X-axis range from 400 - 4000 cm-1, paying attention to each absorption’s frequency, breadth, and intensity.

Concept introduction:

IR spectroscopy enables to identify different functional groups in the given molecule.

One of the greatest advantages of IR spectroscopy is that the frequency of a particular type of vibration is typically found within a characteristic range of frequencies, regardless of the functional group with which the vibration is associated. The fact that absorptions by certain vibrations appear with characteristic frequency ranges enables us to use IR spectroscopy to obtain structural information about a molecule. Stretching vibrations of Q-H bonds (where Q is a heavy atom such as C, N or O) occurs in the region between ~ 2500 - 4000 cm-1. Double bonds appear between ~ 1500 and 2000 cm-1. Single bonds between two heavy atoms appear between ~ 1000 - 1500 cm-1. The region below ~ 1400 cm-1 is called the fingerprint region.

Expert Solution
Check Mark

Answer to Problem 15.56P

The IR spectrum of the given molecule can be sketched as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  7

Explanation of Solution

The given molecule is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  8

The given molecule has an alkyne and a primary amide as functional groups. According to Table 15-2, the characteristic frequency of absorption of amides (O-C-NH bond) appears as medium two peaks between ~3350 - 3500 cm-1. The amide C=O group appears ~1630-1690 cm-1. The CC in alkynes appears as variable peak around ~2100 - 2260 cm-1. The spC-H in alkynes appears as a variable peak at ~ 2800 - 3000 cm-1. The sp3 hybridized C-H appears as a variable peak at ~2800 - 3000 cm-1. Hence, the IR sketch of the given molecule could be drawn as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  9

Each significant peak is assigned in the above IR spectrum.

Conclusion

The absorptions by certain vibrations appear with characteristic frequency ranges, enabling us to use IR spectroscopy to obtain structural information about a molecule.

Interpretation Introduction

(d)

Interpretation:

The IR spectrum of the given compound is to be sketched with the X-axis range from 400 - 4000 cm-1, paying attention to each absorption’s frequency, breadth, and intensity.

Concept introduction:

IR spectroscopy enables to identify different functional groups in the given molecule.

One of the greatest advantages of IR spectroscopy is that the frequency of a particular type of vibration is typically found within a characteristic range of frequencies, regardless of the functional group with which the vibration is associated. The fact that absorptions by certain vibrations appear with characteristic frequency ranges enables us to use IR spectroscopy to obtain structural information about a molecule. Stretching vibrations of Q-H bonds (where Q is a heavy atom such as C, N or O) occurs in the region between ~ 2500 - 4000 cm-1. Double bonds appear between ~ 1500 and 2000 cm-1. Single bonds between two heavy atoms appear between ~ 1000 - 1500 cm-1. The region below ~ 1400 cm-1 is called the fingerprint region.

Expert Solution
Check Mark

Answer to Problem 15.56P

The IR spectrum of the given molecule can be sketched as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  10

Explanation of Solution

The given molecule is

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  11

The given molecule has carboxylic acid, alkene, and cyanide as functional groups.

According to Table 15-2, the characteristic frequency of absorption of OH bond in carboxylic acids appears as a strong and broad peak around ~ 2500 - 3000 cm-1. The C=O group in carboxylic acids appears as a strong peak around ~ 1710 - 1780 cm-1. The CN in nitriles appears as a medium peak around ~2210 - 2260 cm-1. The C=C in non-conjugated alkenes appears as a variable peak around ~1620-1680 cm-1. The sp3 hybridized C-H in alkanes appears as a variable peak at ~2800 - 3000 cm-1. The sp2C-H in alkenes appears as a variable peak at ~ 3000 - 3100 cm-1. Apart from these peaks, two strong peaks appear in the fingerprint region at 910 and 990 cm-1, which are associated with a bending mode in alkenes. Hence, the IR sketch of the given molecule could be drawn as follows:

Organic Chemistry: Principles and Mechanisms (Second Edition), Chapter 15, Problem 15.56P , additional homework tip  12

Each significant peak is assigned in the above IR spectrum.

Conclusion

The absorptions by certain vibrations appear with characteristic frequency ranges, enabling us to use IR spectroscopy to obtain structural information about a molecule.

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Chapter 15 Solutions

Organic Chemistry: Principles and Mechanisms (Second Edition)

Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42PCh. 15 - Prob. 15.43PCh. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Prob. 15.46PCh. 15 - Prob. 15.47PCh. 15 - Prob. 15.48PCh. 15 - Prob. 15.49PCh. 15 - Prob. 15.50PCh. 15 - Prob. 15.51PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - Prob. 15.57PCh. 15 - Prob. 15.58PCh. 15 - Prob. 15.59PCh. 15 - Prob. 15.60PCh. 15 - Prob. 15.61PCh. 15 - Prob. 15.62PCh. 15 - Prob. 15.63PCh. 15 - Prob. 15.64PCh. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - Prob. 15.67PCh. 15 - Prob. 15.68PCh. 15 - Prob. 15.1YTCh. 15 - Prob. 15.2YTCh. 15 - Prob. 15.3YTCh. 15 - Prob. 15.4YTCh. 15 - Prob. 15.5YTCh. 15 - Prob. 15.6YTCh. 15 - Prob. 15.7YTCh. 15 - Prob. 15.8YTCh. 15 - Prob. 15.9YTCh. 15 - Prob. 15.10YTCh. 15 - Prob. 15.11YTCh. 15 - Prob. 15.12YTCh. 15 - Prob. 15.13YTCh. 15 - Prob. 15.14YTCh. 15 - Prob. 15.15YTCh. 15 - Prob. 15.16YTCh. 15 - Prob. 15.17YTCh. 15 - Prob. 15.18YTCh. 15 - Prob. 15.19YTCh. 15 - Prob. 15.20YTCh. 15 - Prob. 15.21YTCh. 15 - Prob. 15.22YTCh. 15 - Prob. 15.23YTCh. 15 - Prob. 15.24YTCh. 15 - Prob. 15.25YTCh. 15 - Prob. 15.26YTCh. 15 - Prob. 15.27YTCh. 15 - Prob. 15.28YTCh. 15 - Prob. 15.29YTCh. 15 - Prob. 15.30YT
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