Order the following considering their boiling point in an aqueous solution.
1) pentan-2-one
2) pentan-2-ol
3) 2-aminopropanoic acid
The answer key says the order (increasing) is,
1 < 2 < 3 and states 1 only have weak permanent dipole dipole forces and Van der Walls forces.
But isn't there any chance that the Carbonyl group of an aqueous solution of a ketone(or even an aldehyde) can make H bonds with the H^+ ions in water?
So can we simply order them considering the reason the answer key states?
I think 3 should have the highest boiling point as it can make more than 1 H bond and what about the other two?
*the Oxygen atom in the Carbonyl group
You are right about 3. It does have the highest boiling poin; the key agrees with you. But the ketone has no to little chance of H bonding with another ketone molecule while the alcohol does which is why it is ketone<alcohol<aminoacid
But isn't there any chance that the Oxygen atom in the ketone molecule can make H bonds with the H^+ ions in water,not with another ketone molecule?
You raise a valid point regarding the possibility of hydrogen bonding between the carbonyl group of a ketone (or aldehyde) and the H⁺ ions in water. Hydrogen bonding can indeed occur between the oxygen atom of the carbonyl group and the H⁺ ions in water, but this interaction is relatively weak compared to other intermolecular forces like dipole-dipole interactions and Van der Waals forces.
In the case of the compounds you mentioned, pentan-2-one, pentan-2-ol, and 2-aminopropanoic acid, let's consider their intermolecular interactions in an aqueous solution.
1) Pentan-2-one: Pentan-2-one is a ketone, and while the carbonyl group can participate in hydrogen bonding with water, the strength of this interaction is relatively weak. Additionally, pentan-2-one does not have any other hydrogen atoms that can form hydrogen bonds. Therefore, the dominant intermolecular forces in pentan-2-one are weak dipole-dipole interactions and Van der Waals forces.
2) Pentan-2-ol: Pentan-2-ol is an alcohol, and it has an -OH group that can form hydrogen bonds with both water molecules and other pentan-2-ol molecules. The presence of these hydrogen bonds increases the boiling point of pentan-2-ol compared to pentan-2-one. The intermolecular forces in pentan-2-ol are stronger than in pentan-2-one due to the additional hydrogen bonding.
3) 2-Aminopropanoic acid: 2-Aminopropanoic acid is an amino acid, and it contains a carboxyl group (-COOH) and an amine group (-NH₂), both of which are capable of hydrogen bonding. This compound can form intramolecular hydrogen bonds within its structure and can also interact with water molecules via hydrogen bonding. The presence of multiple hydrogen bonding sites in 2-aminopropanoic acid makes its intermolecular forces stronger than in pentan-2-ol and pentan-2-one.
Considering the intermolecular forces described above, the boiling point order (increasing) is as follows:
1) Pentan-2-one
2) Pentan-2-ol
3) 2-Aminopropanoic acid
So, the answer key you provided seems to be incorrect. The boiling point of 2-aminopropanoic acid would be higher than that of pentan-2-ol due to the additional hydrogen bonding sites in the amino acid.