Organic Chemistry
Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
Question
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Chapter 10, Problem 10.60AP
Interpretation Introduction

(a)

Interpretation:

The step which is rate-limiting due to which the primary isotope effect is observed while the oxidation of deuterated isopropyl alcohol is to be stated.

Concept introduction:

Primary and secondary alcohols can be oxidized into aldehydes and ketones using Cr(VI) compounds as they are very good oxidizing agents. Few forms of chromium used to oxidize are chromate CrO42- and dichromate Cr2O72- and chromium trioxide (CrO3). Primary alcohols are oxidized into aldehydes and secondary alcohols are oxidized into ketones.

Interpretation Introduction

(b)

Interpretation:

An explanation as to why the deuterated aldehyde is the major product when either (S) or (R)-1-deuterioethanol is oxidized with PCC in CH2Cl2 is to be stated.

Concept introduction:

Primary and secondary alcohols can be oxidized into aldehydes and ketones using Cr(VI) compounds as they are very good oxidizing agents. Few forms of chromium used to oxidize are chromate CrO42- and dichromate Cr2O72- and chromium trioxide (CrO3). Primary alcohols are oxidized into aldehydes and secondary alcohols are oxidized into ketones.

Interpretation Introduction

(c)

Interpretation:

An explanation as to why no deuterated aldehyde product is obtained on oxidation of Organic Chemistry, Chapter 10, Problem 10.60AP with alcohol dehydrogenase and NAD+ is to be stated.

Concept introduction:

The two hydrogens of ethanol are enantiotopic in nature. That is substitution of one hydrogen by another group will lead to R-isomer while for the other will lead to S-isomer. The alcohol dehydrogenase reaction of isotopically substituted ethanol can be carried out in order to distinguish them as the reaction is stereospecific in nature.

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Chapter 10 Solutions

Organic Chemistry

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