Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Question
Chapter 16, Problem 46P
Interpretation Introduction
Interpretation:
The reason for the involvement of only one nitrogen, out of the three available, in semicarbazone formation, is to be explained.
Concept introduction:
Semicarbazide is an ammonia derivative. It contains an amide functionality bonded by its carbonyl to a hydrazinyl moiety.
Semicarbazone is formed when the condensation reaction between the primary amino group of semicarbazide with the carbonyl group, results in a carbon–nitrogen double bond.
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Students have asked these similar questions
Both pyridine and pyrrole are nitrogen containing aromaticheterocyclic compounds. When treated with HCI, only pyridine forms a
hydrochloride salt, whereas pyrrole is unreactive Which of the following is not a valid explanation for this observed reactivity?
pyridine
ругrole
O The lone pair on pyridine is not part of the aromatic system.
O The lone pair on pyridine can be protonated without disrupting the aromatic stability.
O The lone pair on pyrrole is sp hybridized and is less prone to protonation.
O Protonation of pyrrole leads to a nonaromatic cation, which is less stable
O The lone pair on pyrrole is involved in making the compound aromatic and thus is less susceptible to protonation.
a) Put these three common types of carbonyl compound in order of decreasing reactivity
ester amide acid chloride
b) For the least reactive, show the interconversion to its other resonance form:
How does this electron delocalisation make it stable?
c) For the most reactive, draw the mechanism of its undergoing hydrolysis (reaction with H2O):
Why makes this type of carbonyl so reactive to nucleophiles?
Explain why the hydroxyl group present in phenol increases the rate of electrophilic aromatic substitution reactions compared with benzene. include a coordinate diagram in the explaination.
Chapter 16 Solutions
Organic Chemistry
Ch. 16 - PRACTICE PROBLEM 16.1 (a) Give IUPAC substitutive...Ch. 16 - Prob. 2PPCh. 16 - Prob. 3PPCh. 16 - Practice Problem 16.4
Provide the reagents and...Ch. 16 - Prob. 5PPCh. 16 - Prob. 6PPCh. 16 - Prob. 7PPCh. 16 - Prob. 8PPCh. 16 - Prob. 9PPCh. 16 - Practice Problem 16.10
Shown below is the...
Ch. 16 - Prob. 11PPCh. 16 - Practice Problem 16.12
What product would be...Ch. 16 - Prob. 13PPCh. 16 - Practice Problem 16.14
Dihydropyran reacts readily...Ch. 16 - Practice Problem 16.15 Show how you might use...Ch. 16 - Practice Problem 16.16 (a) Show how you might...Ch. 16 - Practice Problem 16.17
In addition to...Ch. 16 - Practice Problem 16.18
Triphenylphosphine can be...Ch. 16 - Prob. 19PPCh. 16 - PRACTICE PROBLEM 16.20
Give the structure of the...Ch. 16 - PRACTICE PROBLEM 16.21 What would be the major...Ch. 16 - Prob. 22PCh. 16 - 16.23 Write structural formulas for the products...Ch. 16 - Give structural formulas for the products formed...Ch. 16 - 16.25 What products would be obtained when...Ch. 16 - Predict the major organic product from each of the...Ch. 16 - 16.27 Predict the major product from each of the...Ch. 16 - 16.28 Predict the major product from each of the...Ch. 16 - Prob. 29PCh. 16 - 16.30 Write detailed mechanisms for each of the...Ch. 16 - Prob. 31PCh. 16 - Prob. 32PCh. 16 - Show how you would convert benzaldehyde into each...Ch. 16 - 16.34 Show how ethyl phenyl ketone could be...Ch. 16 - Show how benzaldehyde could be synthesized from...Ch. 16 - Give structures for compounds AE. Cyclohexanol...Ch. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - Prob. 42PCh. 16 - 16.43 The structure of the sex pheromone...Ch. 16 - Provide reagents that would accomplish each of the...Ch. 16 - Write a detailed mechanism for the following...Ch. 16 - Prob. 46PCh. 16 - Dutch elm disease is caused by a fungus...Ch. 16 - Prob. 48PCh. 16 - Compounds W and X are isomers; they have the...Ch. 16 - Compounds Y and Z are isomers with the molecular...Ch. 16 - Compound A (C9H18O) forms a phenylhydrazone, but...Ch. 16 - Compound B (C8H12O2) shows a strong carbonyl...Ch. 16 - Prob. 53PCh. 16 - Prob. 54PCh. 16 - Prob. 55PCh. 16 - (a) What would be the frequencies of the two...Ch. 16 - Prob. 57PCh. 16 - Prob. LGP
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