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)
A 7.7
B 6.8
Which of the following correctly explains the identity of Compound B and its solubility?
It is an alcohol because the lack of hydrogen bonding between the OH− groups make it less soluble.
It is an ether because it is unable to form a hydrogen bond, so it is less soluble than the alcohol.
It is an ether because in the absence of a highly electronegative atom like O, F, or N, it lacks hydrogen bonding and is thus less soluble.
It is an alcohol because dispersion forces between the OH− group and organic chain makes it less soluble.
The solubility in water of alcohols depends on the same two factors as previously, but which are here antagonistic:
* The carbon chain, hydrophobic, tends to make the molecule insoluble.
* The hydrophilic hydroxyl group (thanks to its hydrogen bonds) tends to render the molecule soluble.
Thus, the alcohols are all the more soluble in water that the number of alcohol functions is high. For example, butanediols are soluble in all proportions while butan-1-ol has a solubility of 77 g L-1.
Answer: The correct answer is B (It is an ether because it is unable to form a hydrogen bond, so it is less soluble than alcohol)
Explanation:
From the solubility data it is quite clear that the compound B has less solubility in water than compound A. Compound A would be an alcohol (-OH) as alcohols being polar molecules can form hydrogen bonding with water and are quite soluble in water.So, it can be inferred that compound B is ether which is non polar and which cannot form hydrogen bonds with water molecule, making it less soluble in water than alcohols.
The substances present before the reaction are the reactants. (As the reaction goes through, the substances that are produced are called the products of the reaction).
