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 57P
Interpretation Introduction

Interpretation:

The structure of the product on the basis of mass spectra, IR and NMR data is to be determined.

Concept introduction:

Hydrogenolysis reaction of benzyl alcohols results in the cleavage of the benzyl and oxygen bond. It may also involve group migration, which is usually carried out by the phenyl ring.

Infrared absorption bands indicate the functional groups present in a compound. For example, carbonyl group shows a band from 1600 to 1750 cm-1, hydroxyl group shows a band from 3400 to 3550 cm-1. So, if a compound shows a band in the carbonyl range only, it is an aldehyde or ketone, while if it shows one in the region of hydroxyl, it is an alcohol or a phenol. If two bands are observed with one in each region, it is a carboxylic acid.

NMR data indicates the number and types of protons or carbons present in a compound based on the number of signals obtained in 1H NMR or 13C NMR, respectively. 1H NMR signals are often with multiplicity such as doublets and triplets, which indicates the number of protons present on the carbons adjacent to that proton. The protons of an aromatic ring show a signal from 6.57.5 ppm, those of alkyl group from 0.5 to 1.5 ppm while alkyl groups adjacent to a carbonyl group or oxygen atom show signals from 2.0 to 3.0 ppm. Alcohol group shows a signal from 2.0 to 3.0 ppm but is always a singlet.

The mass spectra of a compound indicate the molecular ion peak, which give the molecular mass of the compound.

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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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