According to Valence Shell Electron Pair Repulsion (VSEPR) theory, which statement is correct? If a molecule has four pairs of e
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Answer:
See explanation
Explanation:
Our answer options for this question are:
a. 2-chlorobutanoic acid:_______ 2-chlorobutanoic acid:_______ 3-chlorobutanoic acid:______.
b. 2,4-dinitrobenzoic acid:______ p-nitrobenzoic acid:______ p-bromobenzoic acid:_______.
c. p-cyanobenzoic acid:________ benzoic acid:_______ p-aminobenzoic acid:______
We have to check each set of molecules
<u>a. 2-chlorobutanoic acid,</u> <u>3-chlorobutanoic acid</u>
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In this case, the difference between these molecules is the position of "Cl". If the chlorine atom is closer to the acid group, we will have a higher inductive effect. So, the bond O-H would be weaker and we will have more acidity. So, the molecule with more acidity is <u>2-chlorobutanoic acid</u> and the less acidic would be <u>3-chlorobutanoic acid.</u>
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<u>b. 2,4-dinitrobenzoic acid,</u> <u>p-nitrobenzoic acid,</u> <u>p-bromobenzoic acid</u>
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In this case, we have several structural differences. In all the structure, we have deactivating groups ( and
). If we have a deactivating group the acidity will increase. In the case of "Br", we have a weak deactivating, so, this will be the less acidic one (<u>p-bromobenzoic acid)</u>
in <u>2,4-dinitrobenzoic acid</u> we have two deactivating groups, therefore, this would be the most acid compound.
<u>c. p-cyanobenzoic acid</u>, <u>benzoic acid</u>, <u>p-aminobenzoic acid</u>
On these molecules, we have several structural differences. In <u>p-cyanobenzoic acid</u> we have a deactivating group, therefore in this molecule we will have more acidity. In the <u>p-aminobenzoic acid,</u> we have an activating group, so, this would be the less acidic compound.
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See figure 1
I hope it helps!
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