EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
6th Edition
ISBN: 9781305687875
Author: Gilbert
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 10.6, Problem 23E

(a)

Interpretation Introduction

Interpretation: The structure of the stereoisomers expected from the reaction of bromine with 1-methylcyclohexene should be drawn.

Concept introduction:Alkenes are considered electron-rich and undergo bromination addition reaction with Br2 in presence of CCl4 by electrophilic addition reaction.

The mechanism of Br2 addition occurs in two steps. In the first step, nucleophilic olefin abstracts the electrophilic bromine atom from Br2 and this results in attack of Br+ on the electron-rich olefinic bond that forms a sigma bond to each end of the olefinic bond. Such interaction leads to a three-membered bromonium ion intermediate.

In the second step, the strained bromonium ion intermediate opens while Br attack from behind the electron-deficient carbon analogous to SN2 attack.

In accordance with Cahn-Ingold-Prelog rules; if two high priority substituents happen to appear on opposite sides of olefinic bond then the absolute configuration for the alkene will be would be “E”.

(b)

Interpretation Introduction

Interpretation: Whether the stereochemistry of products drawn like or unlike should be identified.

Concept introduction:In order to assign absolute configuration of R and S, Cahn − Ingold − Prelog rules are used and the first step is to assign the priority order with the atomic number as the fundamental property. The one with the highest atomic number gest highest priority and is designated as “1” and so on.

In order to distinguish the relative stereochemistry of two chiral centers situated adjacent to one another, like or unlike notation is used.

For enantiomeric pairs that have same stereochemistry at adjacent carbon atoms, the notation “l” is designate “like” and notation “u” is designate “un-like”.

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Students have asked these similar questions
a. Identify two alkenes that react with HBr to form 1-bromo-1-methylcyclohexane without undergoing a carbocation rearrangement. b. Would both alkenes form the same alkyl halide if DBr were used instead of HBr? (D is an isotope of H, so D+ reacts like H+.)
The reaction of HBr with 1-butene produces only one product, 2- bromobutane. Why does this product form instead of 1-bromobutane? Select one: O A. Br bonds to the C-2 alkene carbon in the first step, forming the more stable carbon anion with (-) charge on C-1, followed by - bonding of H* electrophile to (-) charged C-1. OB. There is less steric hindrance if Br binds to C-2 and H* bonds to C-1. O C. H bonds to the C-1 alkene carbon in the first step, forming the more stable carbocation, with (+) charge on C-2, followed by bonding of Br nucleophile to positively charged C-2. O D. H and Br know what the "EL-NM" rule is and dutifully comply with it.
4. Write in the product, including stereochemistry where relevant, for these reactions. a) b) cyclohexene 1-pentene 1. Os04 2. NaHSO3 1. 03 2. Zn, H* →→

Chapter 10 Solutions

EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M

Ch. 10.2 - Prob. 11ECh. 10.2 - Prob. 12ECh. 10.2 - Prob. 13ECh. 10.2 - Prob. 14ECh. 10.2 - Prob. 15ECh. 10.2 - Prob. 16ECh. 10.2 - Prob. 17ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - Prob. 6ECh. 10.3 - Prob. 7ECh. 10.3 - Prob. 8ECh. 10.3 - Prob. 9ECh. 10.3 - Prob. 10ECh. 10.3 - Prob. 11ECh. 10.3 - Prob. 12ECh. 10.3 - Prob. 13ECh. 10.3 - Prob. 14ECh. 10.3 - Prob. 15ECh. 10.3 - Prob. 16ECh. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Prob. 19ECh. 10.3 - Prob. 20ECh. 10.3 - Prob. 21ECh. 10.3 - Prob. 22ECh. 10.3 - Prob. 23ECh. 10.3 - Prob. 24ECh. 10.3 - Prob. 25ECh. 10.3 - Prob. 26ECh. 10.3 - Prob. 27ECh. 10.3 - Prob. 28ECh. 10.3 - Prob. 29ECh. 10.3 - Prob. 30ECh. 10.3 - Prob. 31ECh. 10.3 - Prob. 32ECh. 10.5 - Prob. 1ECh. 10.5 - Prob. 2ECh. 10.5 - Prob. 3ECh. 10.5 - Prob. 4ECh. 10.5 - Prob. 5ECh. 10.5 - Prob. 6ECh. 10.5 - Prob. 7ECh. 10.5 - Prob. 8ECh. 10.5 - Prob. 9ECh. 10.5 - Prob. 10ECh. 10.5 - Prob. 11ECh. 10.5 - Prob. 12ECh. 10.5 - Prob. 14ECh. 10.5 - Prob. 15ECh. 10.5 - Prob. 16ECh. 10.5 - Prob. 17ECh. 10.5 - Prob. 18ECh. 10.5 - Prob. 19ECh. 10.5 - Prob. 20ECh. 10.5 - Prob. 23ECh. 10.6 - Prob. 1ECh. 10.6 - Prob. 2ECh. 10.6 - Prob. 3ECh. 10.6 - Prob. 4ECh. 10.6 - Prob. 5ECh. 10.6 - Prob. 6ECh. 10.6 - Prob. 7ECh. 10.6 - Prob. 8ECh. 10.6 - Prob. 9ECh. 10.6 - Prob. 10ECh. 10.6 - Prob. 11ECh. 10.6 - Prob. 12ECh. 10.6 - Prob. 13ECh. 10.6 - Prob. 14ECh. 10.6 - Prob. 15ECh. 10.6 - Prob. 16ECh. 10.6 - Prob. 17ECh. 10.6 - Prob. 18ECh. 10.6 - Prob. 20ECh. 10.6 - Prob. 21ECh. 10.6 - Prob. 22ECh. 10.6 - Prob. 23ECh. 10.6 - Prob. 24ECh. 10.6 - Prob. 25ECh. 10.6 - Prob. 26ECh. 10.6 - Prob. 28ECh. 10.6 - Prob. 29ECh. 10.6 - Prob. 30ECh. 10.7 - Prob. 1ECh. 10.7 - Prob. 2ECh. 10.7 - Prob. 3ECh. 10.7 - Prob. 4ECh. 10.7 - Prob. 5ECh. 10.7 - Prob. 6ECh. 10.7 - Prob. 7ECh. 10.7 - Prob. 8ECh. 10.7 - Prob. 9ECh. 10.7 - Prob. 10ECh. 10.7 - Prob. 11ECh. 10.7 - Prob. 12ECh. 10.8 - Prob. 1ECh. 10.8 - Prob. 2ECh. 10.8 - Prob. 4ECh. 10.8 - Prob. 5ECh. 10.8 - Prob. 6ECh. 10.8 - Prob. 7ECh. 10.8 - Prob. 8ECh. 10.8 - Prob. 9ECh. 10.8 - Prob. 10ECh. 10.8 - Prob. 11ECh. 10.8 - Prob. 12ECh. 10.8 - Prob. 13ECh. 10.8 - Prob. 14ECh. 10.8 - Prob. 15E
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