In each of the following reactions, state whether C is oxidized or reduced.

Question

1. CH₃CH₂CH₂CH₃ → CH₃CH=CHCH₃

 

2. CH₃CH=CH₂ → CH₃CH₂CH₃

Answer 1

To determine if C is oxidized or reduced, you must first determine the change, if any, of the oxidation state of the C in question from the reactant side to the product side. The oxidation number of C can vary from -4 to +4. To determine the oxidation state of carbon you look at the attached atoms. If C is attached to 4 other C, the oxidation number is zero.  If it is attached to a H, the oxidation number of that C decreases by 1. If it is attached to a more electronegative element, the oxidation number of that C increases by 1.

Examples:  CH4.  C is attached to 4 H atoms, so each H atom is 1+ so C is 4- in this case.

CH3CH3.  C is attached to 3 H atoms, so each H is 1+, so C is 3-

H2=CH2.  C is attached to 2 H so C is 2-

H3C-CH=O (an aldehyde), the C is attached to an O (-2) and an H(+1), so C is 1+

 

In your question #1. the carbon atoms in question are C2 and C3 since C1 and C4 are the same (CH3). What has happened to C2 and C3.  They both go from having 2 H atoms to having 1 H atom.  So, on the left side each C atom has oxidation number of 2- and on the right side, each C atom has an oxidation number of 1-.  They have lost an electron.  

You should be able to tell if these C atoms have been oxidized or reduced.

 

Try #2 by yourself.  If no luck, respond.