Question

In an experiment, two unknown compounds (one an alcohol and the other an ether) of equal molecular mass were dissolved in water. The result of the experiment is shown in the table. Solubility Comparison Unknown Compound Solubility (g/100 ml water) Unknown Compound A--> Solubility is 7.7 Unknown Compound B--> Solubility is 6.8 Which of the following correctly explains the identity of Compound B and its solubility? A. It is an alcohol because the lack of hydrogen bonding between the OH^- make is less soluble B. It is an ether because it is unable to to form a hydrogen bond, so it is less soluble than water C. It is an ether because in the absence of a highly electronegative atom like O, F, or N, it lacks hydrogen bonding and thus makes it less soluble D. It is an alcohol because dispersion forces between the OH^- group and organic chain make it less soluble.

Answer 1

Answer: B. It is an ether because it is unable to to form a hydrogen bond, so it is less soluble than water

Explanation: Alcohols $(ROH)$ are more soluble in water as they can form hydrogen bonding with water whereas ethers $(ROR)$ are less soluble as they do not form hydrogen bonds with water.

For formation of hydrogen bond, the electronegative atom (F, Oand N) must be attached to hydrogen and in ethers (ROR), there is no hydrogen directly attached to electronegative oxygen atom, thus are less soluble in water.