EBK GET READY FOR ORGANIC CHEMISTRY
EBK GET READY FOR ORGANIC CHEMISTRY
2nd Edition
ISBN: 9780321830555
Author: KARTY
Publisher: VST
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Chapter 17, Problem 17.36P
Interpretation Introduction

(a)

Interpretation:

The mechanism for the given reactions is to be proposed.

Concept introduction:

A nucleophilic addition reaction involves a polar π bond which has an atom that is relatively electron-poor (i.e electrophile), so it tends to undergo the addition of a nucleophile, which is electron-rich. The most common polar π bond that participates in nucleophilic addition reactions is the one in the carbonyl (C = O) group. To avoid exceeding the octet on the C atom attacked by the nucleophile, the π bond is broken and the pair of electrons from the π bond becomes a lone pair on the more electronegative atom. The immediate product of nucleophilic addition is often a strong base because it possesses a relatively unstable negative charge. Sometimes, species already present in the reaction mixture, such as the solvent, are acidic enough to protonate that product. Otherwise, we can carry out this protonation by adding an acid in a subsequent acid workup.

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Chapter 17 Solutions

EBK GET READY FOR ORGANIC CHEMISTRY

Ch. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22PCh. 17 - Prob. 17.23PCh. 17 - Prob. 17.24PCh. 17 - Prob. 17.25PCh. 17 - Prob. 17.26PCh. 17 - Prob. 17.27PCh. 17 - Prob. 17.28PCh. 17 - Prob. 17.29PCh. 17 - Prob. 17.30PCh. 17 - Prob. 17.31PCh. 17 - Prob. 17.32PCh. 17 - Prob. 17.33PCh. 17 - Prob. 17.34PCh. 17 - Prob. 17.35PCh. 17 - Prob. 17.36PCh. 17 - Prob. 17.37PCh. 17 - Prob. 17.38PCh. 17 - Prob. 17.39PCh. 17 - Prob. 17.40PCh. 17 - Prob. 17.41PCh. 17 - Prob. 17.42PCh. 17 - Prob. 17.43PCh. 17 - Prob. 17.44PCh. 17 - Prob. 17.45PCh. 17 - Prob. 17.46PCh. 17 - Prob. 17.47PCh. 17 - Prob. 17.48PCh. 17 - Prob. 17.49PCh. 17 - Prob. 17.50PCh. 17 - Prob. 17.51PCh. 17 - Prob. 17.52PCh. 17 - Prob. 17.53PCh. 17 - Prob. 17.54PCh. 17 - Prob. 17.55PCh. 17 - Prob. 17.56PCh. 17 - Prob. 17.57PCh. 17 - Prob. 17.58PCh. 17 - Prob. 17.59PCh. 17 - Prob. 17.60PCh. 17 - Prob. 17.61PCh. 17 - Prob. 17.62PCh. 17 - Prob. 17.63PCh. 17 - Prob. 17.64PCh. 17 - Prob. 17.65PCh. 17 - Prob. 17.66PCh. 17 - Prob. 17.67PCh. 17 - Prob. 17.68PCh. 17 - Prob. 17.69PCh. 17 - Prob. 17.70PCh. 17 - Prob. 17.71PCh. 17 - Prob. 17.72PCh. 17 - Prob. 17.73PCh. 17 - Prob. 17.74PCh. 17 - Prob. 17.75PCh. 17 - Prob. 17.76PCh. 17 - Prob. 17.77PCh. 17 - Prob. 17.78PCh. 17 - Prob. 17.79PCh. 17 - Prob. 17.80PCh. 17 - Prob. 17.81PCh. 17 - Prob. 17.82PCh. 17 - Prob. 17.83PCh. 17 - Prob. 17.84PCh. 17 - Prob. 17.1YTCh. 17 - Prob. 17.2YTCh. 17 - Prob. 17.3YTCh. 17 - Prob. 17.4YTCh. 17 - Prob. 17.5YTCh. 17 - Prob. 17.6YTCh. 17 - Prob. 17.7YTCh. 17 - Prob. 17.8YTCh. 17 - Prob. 17.9YTCh. 17 - Prob. 17.10YTCh. 17 - Prob. 17.11YTCh. 17 - Prob. 17.12YTCh. 17 - Prob. 17.13YTCh. 17 - Prob. 17.14YT
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