Organic Chemistry
5th Edition
ISBN: 9780078021558
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
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Chapter 20, Problem 20.77P
Interpretation Introduction
Interpretation: An explanation corresponding to the fact that the
Concept introduction: On the horizontal axis, the position of absorption is generally referred to as chemical shift. Depending upon the electron density or the concentration of electron around carbon, the chemical shift values of carbon varies relative to the reference signal. The terms, upfield and downfield expresses the relative location of signals. The meaning of upfield is to the right and of downfield is to the left.
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Explain why the β carbon of an α,β-unsaturated carbonyl compound absorbs farther downfield in the 13C NMR spectrum than the α carbon, even though the α carbon is closer to the electron-withdrawing carbonyl group. For example, the β carbon of mesityl oxide absorbs at 150.5 ppm, whereas the α carbon absorbs at 122.5 ppm.
Explain why the β carbon of an α,β-unsaturated carbonyl compound absorbs farther downfield in the 13 C NMR spectrum than the α carbon, even though the α carbon is closer to the electron-withdrawing carbonyl group. For example, the β carbon of mesityl oxide absorbs at 150.5 ppm, while the α carbon absorbs at 122.5 ppm.
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Predict 1H NMR spectra of acetophenone and the protonated acetophenone.
Chapter 20 Solutions
Organic Chemistry
Ch. 20 - Prob. 20.1PCh. 20 - Which carbonyl groups in the anticancer drug taxol...Ch. 20 - Prob. 20.3PCh. 20 - Problem 20.4 What alcohol is formed when each...Ch. 20 - Problem 20.5 What aldehyde or ketone is needed to...Ch. 20 - Prob. 20.6PCh. 20 - Problem 20.7 Draw the products formed when is...Ch. 20 - Problem 20.8 Draw the products formed (including...Ch. 20 - Prob. 20.9PCh. 20 - Problem 20.10 Draw a stepwise mechanism for the...
Ch. 20 - Prob. 20.11PCh. 20 - Problem 20.12 Draw the products formed from ...Ch. 20 - Prob. 20.13PCh. 20 - Prob. 20.14PCh. 20 - Prob. 20.15PCh. 20 - Problem-20.16 Review the oxidation reactions using...Ch. 20 - Problem-20.17 Write the step(s) needed to convert ...Ch. 20 - Problem-20.18 Oct-1-yne reacts rapidly with ,...Ch. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Problem 20.21 Draw the product of each reaction.
...Ch. 20 - Problem 20.22 Draw the products (including...Ch. 20 - Problem 20.23 What Grignard reagent and carbonyl...Ch. 20 - Problem 20.24 Linalool (the Chapter 9 opening...Ch. 20 - Problem 20.25 What Grignard reagent and carbonyl...Ch. 20 - Prob. 20.26PCh. 20 - Draw the products formed when each compound is...Ch. 20 - Problem 20.28 What ester and Grignard reagent are...Ch. 20 - Prob. 20.29PCh. 20 - Problem 20.30 What reagent is needed to convert ...Ch. 20 - Prob. 20.31PCh. 20 - What carboxylic acid formed from each alkyl halide...Ch. 20 - Prob. 20.33PCh. 20 - Problem 20.34 Draw the product when each compound...Ch. 20 - Problem 20.35 Synthesize each compound from...Ch. 20 - Prob. 20.36PCh. 20 - 20.37 Devise a synthesis of each alcohol from...Ch. 20 - 20.38 Draw the products formed when pentanal is...Ch. 20 - 20.39 Draw the product formed when is treated...Ch. 20 - The stereochemistry of the products of reduction...Ch. 20 - Prob. 20.41PCh. 20 - 20.42 Draw the products or each reduction...Ch. 20 - Prob. 20.43PCh. 20 - 20.44 Draw all stereoisomers formed in each...Ch. 20 - Prob. 20.45PCh. 20 - 20.46 Treatment of ketone A with ethynylithium...Ch. 20 - 20.47 Explain why metal hydride reduction gives an...Ch. 20 - Prob. 20.48PCh. 20 - 20.49 Identify the lettered compounds in the...Ch. 20 - Prob. 20.50PCh. 20 - 20.51 Draw a stepwise mechanism for the following...Ch. 20 - 20.52 Draw a stepwise mechanism for the following...Ch. 20 - Prob. 20.53PCh. 20 - 20.54 Draw a stepwise mechanism for the following...Ch. 20 - Prob. 20.55PCh. 20 - Prob. 20.56PCh. 20 - 20.57 What ester and Grignard reagent are needed...Ch. 20 - 20.58 What organolithium reagent and carbonyl...Ch. 20 - 20.59 What epoxide and organometallic reagent are...Ch. 20 - Prob. 20.60PCh. 20 - 20.61 Propose two different methods to synthesize...Ch. 20 - 20.62 Synthesize each compound from cyclohexanol...Ch. 20 - 20.63 Convert propan-2-ol into each compound....Ch. 20 - 20.64 Convert benzene into each compound. You may...Ch. 20 - 20.65 Design a synthesis of each compound from...Ch. 20 - 20.66 Synthesize each compound from the given...Ch. 20 - Prob. 20.67PCh. 20 - Prob. 20.68PCh. 20 - 20.69 An unknown compound A (molecular formula )...Ch. 20 - 20.70 Treatment of compound C (molecular formula )...Ch. 20 - 20.71 Treatment of compound E (molecular formula )...Ch. 20 - 20.72 Reaction of butanenitrile () with methyl...Ch. 20 - 20.73 Treatment of isobutene with forms a...Ch. 20 - 20.74 Draw a stepwise mechanism for the following...Ch. 20 - Prob. 20.75PCh. 20 - 20.76 Lithium tri-sec-butylborohydride, also known...Ch. 20 - Prob. 20.77PCh. 20 - Prob. 20.78PCh. 20 - Prob. 20.79PCh. 20 - 20.80 Draw a stepwise mechanism for the following...Ch. 20 - Prob. 20.81P
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- Describe how you could use NMR spectroscopy to distinguish between benzoyl chloride and para-chlorobenzaldehyde. O The H NMR spectrum of para-chlorobenzaldehyde should have a signal at approximately 6.5-8 ppm corresponding to the aryle protone. The H NMR spectrum of benzoyl chloride should not have a signal near 6.5-8 ppm. O The H NMR spectrum of para-chlorobenzaldehyde should have a signal at approximately 12 ppm. The 'H NMR spectrum of benzoyl chloride should not have a signal near 12 ppm. O The 'H NMR spectrum of para-chlorobenzaldehyde should have a signal at approximately 10 ppm corresponding to the aldehydic protone. The H NMR spectrum of benzoyl chloride should not have a signal near 10 ppm. O The H NMR spectrum of benzoyl chloride should have a signal at approximately 12 ppm. The 'H NMR spectrum of para- chlorobenzaldehyde should not have a signal near 12 ppm.arrow_forwardWhen diphenyl ether is reacted under the same conditions as in Problem 22.35, a compound is produced whose 13C NMR spectrum shows six signals. Draw that product.arrow_forwardThe 1H NMR spectrum of N,N-dimethylformamide shows three singlets at 2.9, 3.0, and 8.0 ppm. Explain why the two CH3 groups are not equivalent to each other, thus giving rise to two NMR signals.arrow_forward
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