Ketones Lab Report

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I believe my molecule could be a Ketone because of its very high boiling point and solubility. The boiling point is 145.5˚C. The molecule I have been given is very soluble in water. There are dispersion and Dipole-dipole forces between each Ketone molecule. Oxygen is more electronegative than carbon, so it tends to pull the electrons towards the oxygen. One of the two pairs of electrons from the carbon-oxygen bond is easily drawn towards the oxygen. This makes the bond highly polar. This is why the dipole-dipole forces in ketones are so strong.

Because the carbonyl group interacts with water by hydrogen bonding, ketones are typically more soluble in water than other molecules. I have predicted that my molecule is a ketone because of …show more content…

Immediately I know that this cannot be my molecule as the boiling point is almost a third of my given boiling point. This is because of the intermolecular forces in this molecule. There are no hydrogens directly attached to an oxygen so that cancels out the possibility of any hydrogen bonding. However, there is dispersion forces and dipole-dipole interactions. These are the two weakest intermolecular forces, causing the low boiling point. Esters are fairly soluble in water, but the solubility decreases with the chain length. For ethylmethanoate its solubility is 10.5g/100mL. The reason for the solubility is that esters cannot hydrogen bond with themselves, they can hydrogen bond with water however. The poor solubility of ethylmethanoate also clarifies that my molecule is not an …show more content…

The -diol end to the name means that there is two alcohol functional groups. An alcohol can hydrogen bond with other molecules. Being the strongest intermolecular forces, the energy needed to separate them will be a lot greater. Having two alcohol functional groups makes the energy needed greater again. The double hydrogen bonding makes the boiling point very high at 195˚C. There is no definite solubility for 1,1-cyclopropanediol so it cannot be commented on. According to the boiling point, 1,1-cyclopropanediol is not my molecule.

The final isomer that hasn’t been explored yet is 1,3-dioxolane. The cyclic shape of this isomer changes the intermolecular forces in the molecule. There are no functional groups that branch off of this isomer, so there cannot be any hydrogen bonds or dipole-dipole interactions. The only possible intermolecular force is dispersion forces. Being the weakest force, the molecules are don’t need much energy to separate each other. This is the reason for the low boiling point. 1,3-dioxolane has a boiling point of 73˚C. It is impossible for my molecule to be 1,3-dioxolane because of its boiling

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