Hexamethylbenzene undergoes free-radical chlorination to give one monochlorinated product (C12H17Cl) and four dichlorinated products (C12H16Cl2). These products are easily separated by GC-MS, but the dichlorinated products are difficultto distinguish by their mass spectra. Draw the monochlorinated product and the four dichlorinated products, and explainhow 13C NMR would easily distinguish among these compounds.1

Hexamethylbenzene undergoes free-radical chlorination to give one monochlorinated product (C12H17Cl) and four dichlorinated products (C12H16Cl2). These products are easily separated by GC-MS, but the dichlorinated products are difficultto distinguish by their mass spectra. Draw the monochlorinated product and the four dichlorinated products, and explainhow 13C NMR would easily distinguish among these compounds.1

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter19: Aldehydes And Ketones: Nucleophilic Addition Reactions

Section19.SE: Something Extra

Problem 67AP: Acidcatalyzed dehydration of 3hydroxy3phenylcyclohexanone leads to an unsaturated ketone. What...

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Analyzing and Predicting Spectroscopic Data

The interaction of electromagnetic radiation with the structure of molecules in the sample and their study is referred spectroscopy. Several spectroscopic methods are used to acquire sample information, and these methods are proved to be accurate, fast, and non-destructive. The information obtained about the content and composition of molecules from these spectroscopic methods is called spectroscopic data.

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Hexamethylbenzene undergoes free-radical chlorination to give one monochlorinated product (C12H17Cl) and four dichlorinated products (C12H16Cl2). These products are easily separated by GC-MS, but the dichlorinated products are difficult
to distinguish by their mass spectra. Draw the monochlorinated product and the four dichlorinated products, and explain
how 13C NMR would easily distinguish among these compounds.
1

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Hexamethylbenzene undergoes free-radical chlorination to give one monochlorinated product (C12H17Cl) and four dichlorinated products (C12H16Cl2). These products are easily separated by GC-MS, but the dichlorinated products are difficultto distinguish by their mass spectra. Draw the monochlorinated product and the four dichlorinated products, and explainhow 13C NMR would easily distinguish among these compounds
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