Question

Arrange the following substances in order of increasing solubility of water. C6H14, C6H13Br, C6H13OH, C6H12(OH)2.

Answer 1

Answer:

C6H14 < C6H13Br  < C6H13OH < C6H12(OH)2

Explanation:

Hello,

In this case, since the solubility in water is related with the presence of polar bonds in the given molecules we can see that C6H12(OH)2 has the presence two O-H bonds which promote the highest solubility via hydrogen bonds as well as the C6H13OH but in a lower degree as only on O-H bond is present. Next since the bond C-Br in is slightly close to the polar bond C6H13Br rather than the C-C bonds only had by C6H14 we can infer that C6H13Br is more soluble in water than C6H14, therefore the required order is:

C6H14 < C6H13Br  < C6H13OH < C6H12(OH)2

Whereas C6H12(OH)2 is the most soluble and C6H14 the least soluble in water.

Best regards.