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Why does carboxylic acid have tailing and low Rf on TLC? And how exactly does acetic acid help in that process?
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Onorato Campopiano
Why does carboxylic acid have tailing and low Rf on TLC? And how exactly does acetic acid help in that process?
Deprotonation of both acids leave an extra nonbonding electron pair on former-OH oxygen atom, so the charge -1 is left there. In carboxylate anion, pi-bound orbitals of adjacent carbonyl carbon and another oxygen can overlap with nonbonding orbital of former-OH oxygen, allow electron to flow among all three orbitals. This stabilises carboxylate anion. However, such stabilisation does not exist in peroxy anion as there is one extra oxygen atom separating aforementioned orbitals. Therefore, they cannot overlap, and the charge will fully remain on the former-OH oxygen atom. This makes peroxy anion more basic than carboxylate anion. Thus, peroxycarboxylic acid is less acidic than carboxylic acid, given that all other parts of the molecules are same.
Deprotonation of both acids leave an extra nonbonding electron pair on former-OH oxygen atom, so the charge -1 is left there. In carboxylate anion, pi-bound orbitals of adjacent carbonyl carbon and another oxygen can overlap with nonbonding orbital of former-OH oxygen, allow electron to flow among all three orbitals. This stabilises carboxylate anion. However, such stabilisation does not exist in peroxy anion as there is one extra oxygen atom separating aforementioned orbitals. Therefore, they cannot overlap, and the charge will fully remain on the former-OH oxygen atom. This makes peroxy anion more basic than carboxylate anion. Thus, peroxycarboxylic acid is less acidic than carboxylic acid, given that all other parts of the molecules are same.
A simplistic answer would be based on the most polar bond dissociating in water or other polar solvent and it is of course the O-H bond. The better founded consideration would include relative stability of the products of two heterolytic cleavages.
A) RCOOH -> RCOO- + H+
B) RCOOH -> RCO+ + OH-
In the first case, a negative charge is spread over three atoms of carboxyl group as can be seen from the two resonance structures describing the delocalization of the negative charge. A positive charge lies on a hydrogen which is ok due to its low electronegativity.
In the case B, a negative charge is a burden of a single oxygen atom as there is no resonance in the hydroxide ion. The positive charge lies on sp2 hybridized carbon (attached to oxygen) that has electronegativity around 3.0 (vs hydrogen’s 2.2 cf. case A). So that the products in reaction B are less stable, than the products A.
Of course, this is also simplified explanation as it totally ignores the interactions with another reactant and the solvent.
A simplistic answer would be based on the most polar bond dissociating in water or other polar solvent and it is of course the O-H bond. The better founded consideration would include relative stability of the products of two heterolytic cleavages.
A) RCOOH -> RCOO- + H+
B) RCOOH -> RCO+ + OH-
In the first case, a negative charge is spread over three atoms of carboxyl group as can be seen from the two resonance structures describing the delocalization of the negative charge. A positive charge lies on a hydrogen which is ok due to its low electronegativity.
In the case B, a negative charge is a burden of a single oxygen atom as there is no resonance in the hydroxide ion. The positive charge lies on sp2 hybridized carbon (attached to oxygen) that has electronegativity around 3.0 (vs hydrogen’s 2.2 cf. case A). So that the products in reaction B are less stable, than the products A.
Of course, this is also simplified explanation as it totally ignores the interactions with another reactant and the solvent.
Deprotonation of both acids leave an extra nonbonding electron pair on former-OH oxygen atom, so the charge -1 is left there. In carboxylate anion, pi-bound orbitals of adjacent carbonyl carbon and another oxygen can overlap with nonbonding orbital of former-OH oxygen, allow electron to flow among all three orbitals. This stabilises carboxylate anion. However, such stabilisation does not exist in peroxy anion as there is one extra oxygen atom separating aforementioned orbitals. Therefore, they cannot overlap, and the charge will fully remain on the former-OH oxygen atom. This makes peroxy anion more basic than carboxylate anion. Thus, peroxycarboxylic acid is less acidic than carboxylic acid, given that all other parts of the molecules are same.
Deprotonation of both acids leave an extra nonbonding electron pair on former-OH oxygen atom, so the charge -1 is left there. In carboxylate anion, pi-bound orbitals of adjacent carbonyl carbon and another oxygen can overlap with nonbonding orbital of former-OH oxygen, allow electron to flow among all three orbitals. This stabilises carboxylate anion. However, such stabilisation does not exist in peroxy anion as there is one extra oxygen atom separating aforementioned orbitals. Therefore, they cannot overlap, and the charge will fully remain on the former-OH oxygen atom. This makes peroxy anion more basic than carboxylate anion. Thus, peroxycarboxylic acid is less acidic than carboxylic acid, given that all other parts of the molecules are same.
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A simplistic answer would be based on the most polar bond dissociating in water or other polar solvent and it is of course the O-H bond. The better founded consideration would include relative stability of the products of two heterolytic cleavages.
A) RCOOH -> RCOO- + H+
B) RCOOH -> RCO+ + OH-
In the first case, a negative charge is spread over three atoms of carboxyl group as can be seen from the two resonance structures describing the delocalization of the negative charge. A positive charge lies on a hydrogen which is ok due to its low electronegativity.
In the case B, a negative charge is a burden of a single oxygen atom as there is no resonance in the hydroxide ion. The positive charge lies on sp2 hybridized carbon (attached to oxygen) that has electronegativity around 3.0 (vs hydrogen’s 2.2 cf. case A). So that the products in reaction B are less stable, than the products A.
Of course, this is also simplified explanation as it totally ignores the interactions with another reactant and the solvent.
A simplistic answer would be based on the most polar bond dissociating in water or other polar solvent and it is of course the O-H bond. The better founded consideration would include relative stability of the products of two heterolytic cleavages.
A) RCOOH -> RCOO- + H+
B) RCOOH -> RCO+ + OH-
In the first case, a negative charge is spread over three atoms of carboxyl group as can be seen from the two resonance structures describing the delocalization of the negative charge. A positive charge lies on a hydrogen which is ok due to its low electronegativity.
In the case B, a negative charge is a burden of a single oxygen atom as there is no resonance in the hydroxide ion. The positive charge lies on sp2 hybridized carbon (attached to oxygen) that has electronegativity around 3.0 (vs hydrogen’s 2.2 cf. case A). So that the products in reaction B are less stable, than the products A.
Of course, this is also simplified explanation as it totally ignores the interactions with another reactant and the solvent.
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