# Identifying if reaction is reduction or oxidation. Is there an error in this question?

I am looking at this reaction: I am trying to determine if it is reduction or oxidation.

In the initial molecule I calculate the overall oxidation numbers as:

ox(Cl)*6 + ox(H)*6 + ox(C)*6 = 0
-6 + 6 +ox(c)*6 = 0
hence, ox(c) = 0


In the subsequent molecule, I calculated the oxidation numbers as:

ox(Cl)*4 + ox(H)*6 + ox(C)*6 = 0
-4 + 6 +ox(c)*6 = 0
hence, ox(c) = 6/2 = 1/3


However, my textbook says the oxidation number of ALL carbon in the second molecule is 2. I am very confused by this, since if all the carbons have an oxidation number of 2 then the molecule would be charged overall? Then I am unsure if my oxidation numbers can be fractional. Is my textbook wrong, and I wrong or are we both wrong?

Overall, what I am really trying to get at is if the molecule is being oxidized or reduced, and how many electrons are being transferred. Is there a better way of doing this?

Thank you kindly

• There is nothing to calculate here. Simply seeing double bond between carbon atoms instead of bonds with chlorine should be all you need. – Mithoron Jan 24 at 15:59
• Consider something similar in reverse. If I gave you the product and converted the alkene into a diol, what would that be? – Zhe Jan 24 at 20:38
• Look at the polarity of vicinal C-Cl bonds. Carbon is positive, chlorine is negative. In the polarized double bond, one carbon is positive the other negative. The net result is a two electron reduction of the carbon framework. – user55119 Feb 24 at 23:13

## 1 Answer

Before the reaction, every carbon is attached to to other carbon atoms, one hydrogen atom and one chlorine atom. So oxidation state of each carbon = 0+0+-1(due to C-H bond) +1(due to C-Cl bond)= 0; average oxidation state of carbon is also zero. In the product only two carbon atoms suffer a change in bondings; now each of these two carbon atoms is attached to one carbon through double bond, another carbon through single bond and a hydrogen atom; so O.S. of this carbon = 0+0+-1 = -1; so to conclude four carbon atoms remain in zero oxidation states, where as the doubly bonded carbon atoms are in -1 oxidation state. so average oxidation state of carbon = -2/6 = -1/3

P.S.:Fractional oxidation state results only when we take average, and that too in some cases