Q61E
Question
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.26.
Step-by-Step Solution
Verifieda) From \(s{p^2}\) to \(s{p^3}\) hybridization after the reaction.
b) From \(s{p^2}\)to \(sp\) hybridization after the reaction.
First is the reaction of 2-butene with a chlorine molecule
2-butene is an alkene that has a double bond with Carbon \(2.\)
In the reaction with 1 mole of chlorine, each chlorine atom is added from one side of the double bond, forming alkane with two chlorine atoms.
Reaction (a)
The two carbon atoms in the middle that share a double bond are \(s{p^2}\)hybridized because of the planar arrangement that the double bond causes.
On the other hand, as they react, they tend to have 4 single bonds around them, like the other two carbon atoms.
The tetrahedral arrangement means \(s{p^3}\)hybridization after the reaction.
The second is the reaction of burning benzene.
When hydrocarbon undergoes burning, it reacts with oxygen from the air, forming \(C{O_2}\) and \({H_2}O\) as products.
A number of moles of \(C{O_2}\)match the number of \(C\)atoms in an alkane.
Reaction (b)
In benzene, there is a planar arrangement, and the whole structure is planar, which means that \(s{p^2}\)hybridization is present.
\(C{O_2}\)is made as a product by burning benzene.
Carbon dioxide is a linear molecule, having two double bonds with two oxygen atoms, one with each.
This type of structure can only take linear configuration, which means hybridization is \(s p.\)