Organic Chemistry
Organic Chemistry
4th Edition
ISBN: 9780073402772
Author: Janice G. Smith
Publisher: MCG
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Chapter 13, Problem 13.40P

How would each of the following pairs of compounds differ in their IR spectra?

a. Chapter 13, Problem 13.40P, How would each of the following pairs of compounds differ in their IR spectra? a. d. b.e. CH3CCCH3 , example  1 d.Chapter 13, Problem 13.40P, How would each of the following pairs of compounds differ in their IR spectra? a. d. b.e. CH3CCCH3 , example  2

b.Chapter 13, Problem 13.40P, How would each of the following pairs of compounds differ in their IR spectra? a. d. b.e. CH3CCCH3 , example  3 e. CH 3 C CCH 3 and CH 3 CH 2 C CH

c.Chapter 13, Problem 13.40P, How would each of the following pairs of compounds differ in their IR spectra? a. d. b.e. CH3CCCH3 , example  4 f. HC CCH 2 N(CH 2 CH 3 ) 2 and CH 3 (CH 2 ) 5 C N

Expert Solution
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Interpretation Introduction

(a)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR spectrum of cyclopentene is different from pentyne because the absorption of Csp2=Csp2 bond is observed at 1650 cm1 and absorption of CspCsp bond is observed at 2250 cm1.

Explanation of Solution

The given compounds are cyclopentene and pentyne as shown below.

Organic Chemistry, Chapter 13, Problem 13.40P , additional homework tip  1

Figure 1

Cyclopentene contains Csp2=Csp2 bond and shows absorption at 1650 cm1 whereas 1-pentyne contains CspCsp and shows absorption at 2250 cm1. Hence, the IR absorptions of cyclopentene is different from cyclopentane due to the bond strength.

Conclusion

The IR spectrum of cyclopentene is different from pentyne because the absorption of Csp2=Csp2 bond is observed at 1650 cm1 and absorption of CspCsp bond is observed at 2250 cm1.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR absorption of propanoic acid is different from methyl acetate due to broad peak of OH bond is observed.

Explanation of Solution

The given compounds are propanoic acid and methyl acetate as shown below.

Organic Chemistry, Chapter 13, Problem 13.40P , additional homework tip  2

Figure 2

Propanoic acid contains OH and shows very broad peak at 3400-2400 cm1 while there is no OH bond present in methyl acetate. Hence, the IR absorptions of propanoic acid is different from methyl acetate.

Conclusion

The IR absorptions of propanoic acid is different from methyl acetate due to broad peak of OH bond is observed.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR absorptions of butan-2-one is different from 2-buten-1-ol due to peak at approximately at 1700 cm1 because of C=O peak in butan-2-one and broad peak of OH bond in 2-buten-1-ol is observed.

Explanation of Solution

The given compounds are butan-2-one and 2-buten-1-ol as shown below.

Organic Chemistry, Chapter 13, Problem 13.40P , additional homework tip  3

Figure 3

Butan-2-one contains C=O and shows a peak approximately at 1700 cm1 while there is no C=O bond present in 2-buten-1-ol. In the IR spectra of 2-buten-1-ol, there is presence of very broad peak in the range of 3600-2600 cm1 due to OH bond which is absent in butan-2-one. Hence, the IR absorptions of butan-2-one is different from 2-buten-1-ol.

Conclusion

The IR absorptions of propanoic acid is different from methyl acetate due to broad peak of OH bond is observed.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR spectra of methyl hexanoate are different from 1,1-dimethyloxycyclohexane due to the presence of Csp2=O bond.

Explanation of Solution

The given compound are 1,1-dimethyloxycyclohexane and methyl hexanoate as shown below.

Organic Chemistry, Chapter 13, Problem 13.40P , additional homework tip  4

Figure 4

Methyl hexanoate contains Csp2=O bond and shows absorption around at 1730 cm1 there is no Csp2=O bond present in 1,1-dimethyloxycyclohexane. Hence, the IR spectra of methyl hexanoate is different from 1,1-dimethyloxycyclohexane due to the presence of Csp2=O bond.

Conclusion

The IR spectra of methylhexanoate is different from 1,1-dimethyloxycyclohexane due to the presence of Csp2=O bond.

Expert Solution
Check Mark
Interpretation Introduction

(e)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

But-1-yne contains CspH bond and shows absorption at 3300 cm1 whereas but-2-yne does not contain this bond.

Explanation of Solution

But-1-yne (CH3CH2CCH) contains CspH bond and shows absorption at 3300 cm1 whereas but-2-yne (CH3CCCH3)does not contain this bond. Hence, the IR absorptions of but-1-yne is different from but-2-yne.

Conclusion

But-1-yne contains CspH bond and shows absorption at 3300 cm1 whereas but-2-yne does not contain this bond.

Expert Solution
Check Mark
Interpretation Introduction

(f)

Interpretation: The given pair of compounds is to be differentiated on the basis of their IR spectra.

Concept introduction: IR spectroscopy is used to identify the functional group present in a compound. Each and every bond vibrates at a characteristic frequency.

Answer to Problem 13.40P

The IR spectra of N, N-diethylprop-2-yn-1-amine is different from heptanenitrile due to presence of Csp-H bond.

Explanation of Solution

The given compound are N, N-diethylprop-2-yn-1-amine ((HCCCH2N(CH2CH3)2)) and heptanenitrile (CH3(CH2)5CN). N, N-diethylprop-2-yn-1-amine contain Csp-H bond and shows absorption at 3300 cm1. There is no Csp-H bond present in heptanenitrile. Hence, the IR spectra of N, N-diethylprop-2-yn-1-amine is different from heptanenitrile due to the presence of Csp-H bond.

Conclusion

The IR spectra of N, N-diethylprop-2-yn-1-amine is different from heptanenitrile due to the presence of Csp-H bond.

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Chapter 13 Solutions

Organic Chemistry

Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - How do the IR spectra of the isomers cyclopentane...Ch. 13 - How do the three isomers of molecular formula...Ch. 13 - Problem 13.18 What functional groups are...Ch. 13 - Problem-13.19 What are the major IR absorptions in...Ch. 13 - Problem-13.20 What are the major IR absorptions in...Ch. 13 - Prob. 13.20PCh. 13 - Problem-13.22 Propose structures consistent with...Ch. 13 - 13.23 What major IR absorptions are present above ...Ch. 13 - Problem-13.24 The mass spectrum of the following...Ch. 13 - What molecular ion is expected for each compound?Ch. 13 - Which compound gives a molecular ion at m/z= 122,...Ch. 13 - Propose two molecular formulas for each molecular...Ch. 13 - Propose four possible structures for a hydrocarbon...Ch. 13 - Match each structure to its mass spectrum. a. b....Ch. 13 - 13.32 Propose two possible structures for a...Ch. 13 - 13.33 What cations are formed in the mass...Ch. 13 - 13.35 For each compound, assign likely...Ch. 13 - Prob. 13.32PCh. 13 - 13.37 Propose a structure consistent with each...Ch. 13 - 13.38 A low-resolution mass spectrum of the...Ch. 13 - Can the exact mass obtained in a high-resolution...Ch. 13 - 13.39 Primary alcohols often show a peak in their...Ch. 13 - 13.40 Like alcohols, ethers undergo α cleavage by...Ch. 13 - Which of the highlighted bonds absorbs at higher v...Ch. 13 - What major IR absorptions are present above...Ch. 13 - How would each of the following pairs of compounds...Ch. 13 - 13.44 Morphine, heroin, and oxycodone are three...Ch. 13 - Prob. 13.42PCh. 13 - 13.47 Match each compound to its IR spectrum Ch. 13 - 13.48 Propose possible structures consistent with...Ch. 13 - A chiral hydrocarbon X exhibits a molecular ion at...Ch. 13 - 13.50 A chiral compound has a strong absorption...Ch. 13 - 13.51 Treatment of benzoic acid with followed by...Ch. 13 - 13.52 Treatment of benzaldehyde with in aqueous ...Ch. 13 - Prob. 13.49PCh. 13 - 13.54 Reaction of 2-methylpropanoic acid with ...Ch. 13 - 13.55 Reaction of pentanoyl chloride with lithium...Ch. 13 - Prob. 13.52PCh. 13 - 13.57 Treatment of anisole with and forms P,...Ch. 13 - 13.58 Reaction of with forms compound ,...Ch. 13 - Problem-13.59 The carbonyl absorption of an amide...Ch. 13 - Prob. 13.56PCh. 13 - Problem-13.61 Explain why a ketone carbonyl...Ch. 13 - 13.62 Oxidation of citronellol, a constituent of...
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