Which statement about 2‑methyl‑1‑propanol, (ch3)2chch2oh , and 1,3‑propanediol, hoch2ch2ch2oh is true? 2‑methyl‑1‑propanol is more soluble in water than 1,3‑propanediol because 2‑methyl‑1‑propanol has a smaller molecular mass. 2‑methyl‑1‑propanol is more soluble in water than 1,3‑propanediol because 2‑methyl‑1‑propanol forms fewer hydrogen bonds with water. 1,3‑propanediol is more soluble in water than 2‑methyl‑1‑propanol because 1,3‑propanediol has a smaller molecular mass. 1,3‑propanediol is more soluble in water than 2‑methyl‑1‑propanol because 1,3‑propanediol can form multiple hydrogen bonds with water.

Propane-1,3-diol is more soluble in water than 2-methylpropan-1-ol because propane-1,3-diol can form more hydrogen bonds with water.

Explanation:

The structures of the two compounds are shown below.

The two compounds have almost the same molecular mass (72 vs 74), so the important factor must be the number of OH groups.

Thus, propane-1,3-diol is more soluble in water than 2 methyl 1 propanol because propane-1,3-diol can form more hydrogen bonds with water.

A and C are wrong. The molecular masses of the two compounds are almost identical.

B is wrong. Fewer hydrogen bonds would make 2-methylpropan-1-ol less soluble.

it's all four options.

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.