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Hello all,
Going back to one of the basics of OChem, i am stuck on something i can't wrap my head around.
Its regarding protonation site for a Carboxylic acid. Carboxylic acid has 2 oxygen sites (lets say site (a) is C==O, site (b) is -O-H). In determining which of the two sites will actually encourage protonation, resonance comes into play for site (a).
The question i have is regarding site (a). When H+ tries to attach to (a), oxygen will donate one of its electron pair, thereby developing a positive charge.
TBR OChem lists the following 3 resonance rules:
1. best structure allows for filled octets for all atoms (except H)
2. best structure reduces overall formal charge
3. (-)ve charge gets assigned to most electronegative atom
4. (+)ve charge gets assigned to least electronegative atom
So, when (a) develops (+)ve charge, AND
because C is less electronegative than O,
the C=O(+) will be converted into (+)C-O. This I get.
However, the resonance stabilization apparently doesn't stop there. O from site (b) apparently donates an electron pair to form C=O(+)ve at site (b), giving rise to the resonance hybrid #2. Per rule#(4) above, shouldn't the resonance hybrid allow C instead of O to carry the positive charge?
Hope the question makes sense.
Going back to one of the basics of OChem, i am stuck on something i can't wrap my head around.
Its regarding protonation site for a Carboxylic acid. Carboxylic acid has 2 oxygen sites (lets say site (a) is C==O, site (b) is -O-H). In determining which of the two sites will actually encourage protonation, resonance comes into play for site (a).
The question i have is regarding site (a). When H+ tries to attach to (a), oxygen will donate one of its electron pair, thereby developing a positive charge.
TBR OChem lists the following 3 resonance rules:
1. best structure allows for filled octets for all atoms (except H)
2. best structure reduces overall formal charge
3. (-)ve charge gets assigned to most electronegative atom
4. (+)ve charge gets assigned to least electronegative atom
So, when (a) develops (+)ve charge, AND
because C is less electronegative than O,
the C=O(+) will be converted into (+)C-O. This I get.
However, the resonance stabilization apparently doesn't stop there. O from site (b) apparently donates an electron pair to form C=O(+)ve at site (b), giving rise to the resonance hybrid #2. Per rule#(4) above, shouldn't the resonance hybrid allow C instead of O to carry the positive charge?
Hope the question makes sense.