Organic Chemistry As a Second Language: First Semester Topics
4th Edition
ISBN: 9781119110668
Author: David R. Klein
Publisher: WILEY
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Chapter 13.5, Problem 13.28P
Interpretation Introduction
Interpretation:
The oxidation state of the carbon atom that is highlighted in the given atom has to be calculated.
Concept Introduction:
Oxidation state of an element is the total number of valence electron that it loses or gains to attain the stable octet structure.
To calculate the oxidation state of carbon atom, we treat all the bonds to be ionic. This means that the more electronegative atom will have the two electrons that is been shared with itself. An example for finding the oxidation state of carbon atom is given below,
By treating all the bonds as ionic, we find that the two electrons are given to chlorine atom. Between carbon and hydrogen, the two electrons are given to the carbon atom as carbon is slightly more electronegative than hydrogen atom. The electrons present between the carbon and carbon is shared equally because they are same. This gives the carbon atom to have five electrons around it.
We know that the valence electrons of carbon is four. In this case it has got one extra electron. Therefore, the oxidation state of this carbon atom can be said as “-1”.
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Can the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating
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Chapter 13 Solutions
Organic Chemistry As a Second Language: First Semester Topics
Ch. 13.1 - Prob. 13.2PCh. 13.1 - Prob. 13.3PCh. 13.1 - Prob. 13.4PCh. 13.1 - Prob. 13.5PCh. 13.2 - Prob. 13.7PCh. 13.2 - Prob. 13.8PCh. 13.2 - Prob. 13.9PCh. 13.2 - Prob. 13.10PCh. 13.3 - PROBLEMS For each pair of compounds, identify the...Ch. 13.3 - Prob. 13.13P
Ch. 13.3 - Prob. 13.14PCh. 13.3 - Prob. 13.15PCh. 13.3 - Prob. 13.16PCh. 13.3 - Prob. 13.17PCh. 13.4 - Prob. 13.19PCh. 13.4 - Prob. 13.20PCh. 13.4 - Prob. 13.21PCh. 13.4 - Prob. 13.22PCh. 13.5 - Prob. 13.24PCh. 13.5 - Prob. 13.25PCh. 13.5 - Prob. 13.26PCh. 13.5 - Prob. 13.27PCh. 13.5 - Prob. 13.28PCh. 13.5 - Prob. 13.29PCh. 13.5 - Prob. 13.31PCh. 13.5 - Prob. 13.32PCh. 13.5 - Prob. 13.33PCh. 13.5 - Prob. 13.34PCh. 13.5 - Prob. 13.35PCh. 13.5 - Prob. 13.36PCh. 13.6 - Prob. 13.38PCh. 13.6 - Prob. 13.39PCh. 13.6 - Prob. 13.40PCh. 13.6 - Prob. 13.41PCh. 13.6 - Prob. 13.42PCh. 13.6 - Prob. 13.43PCh. 13.6 - Prob. 13.44PCh. 13.6 - Prob. 13.45PCh. 13.6 - Prob. 13.46PCh. 13.6 - Prob. 13.47PCh. 13.6 - Prob. 13.48PCh. 13.6 - Prob. 13.49PCh. 13.7 - Prob. 13.51PCh. 13.7 - Prob. 13.52PCh. 13.7 - Prob. 13.53PCh. 13.7 - Prob. 13.54PCh. 13.7 - Prob. 13.55PCh. 13.7 - Prob. 13.56PCh. 13.8 - Prob. 13.58PCh. 13.8 - Prob. 13.59PCh. 13.8 - Prob. 13.60PCh. 13.8 - Prob. 13.61PCh. 13.9 - Prob. 13.63PCh. 13.9 - Prob. 13.64PCh. 13.9 - Prob. 13.65PCh. 13.9 - Prob. 13.66PCh. 13.9 - Prob. 13.67PCh. 13.9 - Prob. 13.68PCh. 13.10 - Prob. 13.70PCh. 13.10 - Prob. 13.71P
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