Question

Which of these three compounds are the most soluble in water? cis-2-pentene, 1- butanol or butanone? Give a brief explanation and draw that molecule showing how it interacts with water molecules.

Answer 1

Water is a polar protic solvent. polar protic solvents are those who are capable of H-bonding because they contain H atoms which directly attached to a electronegative atom.

1-butanol is also polar protic compound since it contains OH group.

Butanone is a polar aprotic compound. Polar aprotic compounds are those which do not have any H atom attached to a electronegative group. Therefore they can not form hydrogen bond among themselves. But they can form H-bond with water molecules. They are polar due to presence of an electronefative atom.

Cis-2-pentene is a nonpolar compound. Nonpolar compounds are those which involve bonds with atoms of similar electronegativity. They do not interact with polar compounds.

Now among these three compounds, butanol is most soluble in water and cis-2-pentene is least soluble. As both water and butanol are polar protic in nature, butanol can be dissolved in water immediately through extensive H-bonding. Where as cis-2-pentene is a nonpolar compound which can not interact with water molecules and remains insoluble in water. 1-Butanone has carbonyl oxygen atom and can interact with water through H bonding but not as extensively as butanol due to absence of attached H atom to the carbonyl oxygen. Therefore it's solublity is in the intermediate range among three given compounds.

The order of solubility in water are as follows

Cis-2-pentene < 1-butanone < butanol.

The molecular interaction of water with these compounds are shown in the attached file.

at SH 8 CH₃ CH₂ CH2 CH20- ..!!.. S ollut =O-H2CH2CH2CH3C stol Hot It-bonding interaction befseen water and I butanol. ili jina -00 & 87 8 .. . - TIDA O- Hrinio * Het : t-bonding between water and butanone . - 320 1 lge ? → No interaction Hst SH do No interaction between water and cis-pentene.