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Magazine Chapter 20: Q63E (page 1146)
Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 20.51.
Short Answer
a) No change in hybridization.
b) From \(s{p^2}\)to \(s p.\)
Step by step solution
Step 1:To find the reaction of 1-butanol with acetic acid:
a) Reaction of 1-butanol with acetic acid
When acid and alcohol react together, they give an ester.
That ester has an acyl chain of an acid and an ester group of an alcohol, meaning that the product is butyl acetate.
Below are both types of reactions, with Lewis structures and condensed formulas.
In these reactions, water is another product because both OH and \(H\)are removed from these groups.
To find the hybridization change
In this case, the hybridization change doesn't occur.
No group changes its molecular geometry, and more or fewer bonds are not made either.
To find the reaction of propionic acid
b) Reaction of propionic acid with \(CaC{O_3}\)-
In this reaction, benzoic acid is an acid while\(CaCO{O_3}\)acts as a base. Thus it is an acid-base reaction.
Propionic acid loses its acidic proton and complexes with calcium. Since it's a bivalent cation, 2 molecules of propionic acid and anion are needed.
Another product is carbonic acid which decomposes to \(C{O_2}\)and \({H_2}O\) since it's an unstable acid.
The carbonate
In carbonate, there is a planar arrangement, and the whole structure is planar, which means that\(s{p^2}\)hybridization is present
By conversion to\(C{O_2},\)it gets 2 double bonds around that carbon atom
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 \(s p.\)
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