Organic Chemistry
Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 16, Problem 13PP
Interpretation Introduction

Interpretation:

The use of cyclic acetal in carrying out the given conversion is to be shown and the reason why the direct addition of methyl magnesium bromide fails to carry out the conversion is to be explained.

Concept introduction:

Electrophiles are electron deficient species, which has positive or partially positive charge. Lewis acids are electrophiles, which accept electron pair.

Nucleophiles are electron rich species, which has negative or partially negative charge. Lewis bases are nucleophiles, which donate electron pair.

Nucleophilic addition reaction is an addition reaction where electron deficient or electrophilic double or triple bond reacts with electron rich species (nucleophile), with disappearance of the double bond and creation of two single bonds.

When same compound contains two moieties with carbonyl groups, both are reactive toward nucleophilic addition. The more reactive one undergoes the reaction faster. The more reactive group has to be protected to make the less reactive one undergoes a reaction.

Ketone is more reactive as compared to the ester so it has to be protected for the reagent to react with ester group.

The protection can be done by converting ketone to cyclic acetals, because of which only ester group is available.

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Chapter 16 Solutions

Organic Chemistry

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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