INTRO.TO GENERAL,ORGAN...-OWLV2 ACCESS
INTRO.TO GENERAL,ORGAN...-OWLV2 ACCESS
12th Edition
ISBN: 9781337915977
Author: Bettelheim
Publisher: CENGAGE L
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Chapter 19, Problem 10P
Interpretation Introduction

Interpretation:

The number of stereocenters present in D-glucose and D-ribose should be identified. The number of stereoisomers possible for each monosaccharide should be calculated.

Concept Introduction:

  1. If 'n' is even (here n is the number of chiral centres): Number of enantiomers = 2 n - 1. Number of meso compounds = 2 n / 2 - 1. Total number of optical isomers = 2 n - 1 + 2 n / 2 - 1.
  2. If 'n' is odd: Number of enantiomers = 2 n - 1 - 2 (n - 1) / 2. Number of meso compounds = 2 (n - 1) / 2.

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. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. < cleavage Bond A • CH3 + 26. t cleavage 2°C• +3°C• Bond C Cleavage CH3 ZC '2°C. 26. E Strongest 3°C. 2C. Gund Largest BDE weakest bond In that molecule a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest C bond Produces A Weakest Bond Most Strongest Bond Stable radical Strongest Gund produces least stable radicals b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 人 8°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. methyl radical •CH3 formed in bund A Cleavage
Which carbocation is more stable?
Are the products of the given reaction correct?  Why or why not?

Chapter 19 Solutions

INTRO.TO GENERAL,ORGAN...-OWLV2 ACCESS

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