EBK ORGANIC CHEMISTRY AS A SECOND LANGU
EBK ORGANIC CHEMISTRY AS A SECOND LANGU
4th Edition
ISBN: 9781119234722
Author: Klein
Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT
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Chapter 5.3, Problem 5.14P
Interpretation Introduction

Interpretation:

Products for the given reaction has to be predicted.

EBK ORGANIC CHEMISTRY AS A SECOND LANGU, Chapter 5.3, Problem 5.14P

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Br CH3OH + Br-Br H3CO The mechanism proceeds through a first cationic intermediate, intermediate 1. Nucleophilic attack leads to intermediate 2, which goes on to form the final product. In cases that involve a negatively charged nucleophile, the attack of the nucleophile leads directly to the product. +Br + CH3OH Br Intermediate 1 Intermediate 2 (product) In a similar fashion, draw intermediate 1 and intermediate 2 (final product) for the following reaction. OH + Br2 + HBr Br racemic mixture • Pay attention to the reactants, they may differ from the examples. In some reactions, one part of the molecule acts as the nucleophile. • You do not have to consider stereochemistry. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate intermediate 1 and intermediate 2 using the → symbol from the dropdown menu.
Acyl transfer (nucleophilic substitution at carbonyl) reactions proceed in two stages via a "tetrahedral intermediate." Draw the tetrahedral intermediate as it is first formed in the following reaction. CH3OH H3C CI • • • • You do not have to consider stereochemistry. Include all valence lone pairs in your answer. Do not include counter-ions, e.g., Na+, I", in your answer. In cases where there is more than one answer, just draw one.
Alcohols are acidic in nature. Therefore, a strong base can abstract the acidic hydrogen atom of the alcohol in a process known as deprotonation. The alcohol forms an alkoxide ion by losing the proton attached to the oxygen atom of the hydroxyl ( -OH) group. The alkoxide formed can act as a base or a nucleophile depending on the substrate and reaction conditions. However, not all bases can abstract the acidic proton of alcohols and not all alcohols easily lose the proton. Deprotonation depends on the strength of the base and the acidity of the alcohol. Strong bases, such as NaNH2, can easily abstract a proton from almost all alcohols. Likewise, more acidic alcohols lose a proton more easily. Determine which of the following reactions would undergo deprotonation based on the strength of the base and the acidity of the alcohol. Check all that apply. ► View Available Hint(s) CH3CH,OH + NH3 →CH,CH,O-NH CH3 CH3 H3C-C-H+NH3 → H3 C-C-H OH O-NH CH3CH2OH + NaNH, → CH3CH,O-Na* + NH3 CHC12 Cl₂…
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