Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
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Question
Chapter 15, Problem 15.45P
Interpretation Introduction
Interpretation:
The carbonyl stretch of
Concept introduction:
The lone pair on nitrogen atom can take part in resonance with
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Students have asked these similar questions
Why do aldehydes, esters, and amides all have a strong absorption in the
1630-1780 cm1 region of their IR spectra?
A) The bond between H and the sp³-hybridized C in these functional groups
vibrates in this energy range.
B) Each of these functional groups has at least two resonance structures, and
the different vibrations of the resonance structures give off energy in this
region.
C) The bond between O and the sp²-hybridized C in these functional groups
vibrates at a frequency in this energy range.
D) Light at this wavenumber causes the average C to O bond length to increase
which causes more of this light to be transmitted.
E) An electron in the bond of these functional groups gets excited to the *
orbital.
2
The IR and 1H-NMR spectra of a compound with molecular formula C4H7ClO2 are shown below. Your objective as a group is to propose a structure for this compound, explaining how you reach your decision. Using all the information you have been given, in a post with others in your group share your initial ideas about the possible structure of the compound. Then use comments to interact with the other students in the group and propose a final answer to the problem. In the comment phase, you should comment on the postings of at least two other students.
Hydrohalic acids (i.e., HCl, HBr, HI) add to the double bond of alkenes to yield alkyl halides. Shown below is the carbon NMR spectrum of an alkene of formula C5H10. What is the structure of this alkene?
Chapter 15 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
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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Similar questions
- In the 1H NMR spectra of 2-bromopropane (CH3)2CHBr and 1-bromopropane CH3CH2CH2Br, how many signals do you expect to see?arrow_forwardDescribe the differences between the infrared spectra of butanoic acid and 2-butanonearrow_forwardKeeping with the theme of autumn, one of Dr. Danahy’s favorite molecules is caffeic acid due to its presence in pumpkins. This structure serves as an antioxidant and is one of many found within pumpkins. Despite its name, it bears no resemblance to caffeine. Answer the following questions about caffeic acid. The carbonyl stretch for caffeic acid is unusually low for a carboxylic acid at 1646 cm-1. For reference, the carbonyl stretch for propanoic acid is 1716 cm-1. Explain why the carbonyl stretch occurs at a lower wavenumber for caffeic acid.arrow_forward
- The C=C bond in 2-cyclohexenone (shown below) produces an unusually strong signal. Explain using resonance structures. 14.06a1 Which of the following explains why the C=Cond in 2-cyclohexenone produces an unusually strong signal. O Conjugation with the C=O results in resonance, giving the C=C bond some single bond character (making it weaker). Conjugation with the C=O results in resonance, making the C=C bond more polar than usual. O Conjugation with the C=O results in resonance, making the C=C bond less polar than usual. O Conjugation with the C=O results in resonance, giving the C=C bond some single bond character (making it stronger).arrow_forwardIn Section 15.5c, we learned that the frequency of a C=0 stretch decreases when the c=0 bond is conjugated to a C=C bond. Draw the pertinent resonance contributor of a conjugated carbonyl (C=C-c=0) and, based on the resulting resonance hybrid, explain why the frequency decreases.arrow_forward(1) ¹H NMR spectrum of an alcohol (C5H12O) is shown with related integral values of the each proton peaks. Propose a structure for this alcohol and assign each protons in this structure. 1H N-> 2 2H elle 8μ (ppm) 6H 3H il 1 0arrow_forward
- 6. Given the ¹H NMR provided, what is the molecule? The molecular formula is C7H₁40. There is a strong peak in the IR spectrum at 1700 cm³¹. Note that the integration line is shown below on the spectrum to show you the ratios of protons in the molecule. Identify which peak corresponds to which protons on the molecule you drew. You can circle the protons on your molecule and use "2H," "3H," or "9H" to identify which peak belongs to which protons on your molecule. 6 5 4 Chemical shift (8) 3 2H 3H 2 r 9H 1 0 (ppm)arrow_forwardChemistry For the given 13C NMR chemical shifts of the carbonyl carbons in the order of ketones > aldehydes > carboxylic acids, how is the induction affects the chemical shifts? provide the explanation.arrow_forwardFollowing is a 1H-NMR spectrum of 2-butanol. Explain why the CH2 protons appear as a complex multiplet rather than as a simple quintet.arrow_forward
- Q4. The following is the 13C NMR spectrum of a highly concentrated solution of dicyclopentadiene obtained in chloroform-d. Despite the high purity of the compound based on ¹H NMR, there is a massive peak observed at 77.2 ppm in this 13C spectrum, which doesn't correspond to any resonances expected for dicyclopentadiene. What is this peak, and why do you not see a corresponding peak for this compound in your ¹H NMR spectrum? 79-751 <-136.22 -132.62 132.27 132.23 135 130 125 120 115 110 105 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30arrow_forwardUse the supplied NMR spectrum to gain insight into salicylic acid. Decipher the spectrum by assigning hydrogen peaks in the molecule and include it with your report. CO₂H OH 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0arrow_forwardFollowing are the 13C and 1H spectra for one of four isomeric bromoalkanes with formula C4H9Br. Assign a structure of the isomer.arrow_forward
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