116k views
3 votes
Arrange the following substances in order of increasing solubility of water. C6H14, C6H13Br, C6H13OH, C6H12(OH)2.

1 Answer

4 votes

Answer:

C6H14 < C6H13Br < C6H13OH < C6H12(OH)2

Step-by-step 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.

User Gfpacheco
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.