Decanoic acid and 1-aminodecane were separated by HPLC on a bonded (stationary)

Question

Decanoic acid and 1-aminodecane were separated by HPLC on a bonded (stationary) phase containing octadecyl groups (C18). The eluent was 20 % vol methanol in water adjusted to pH 3.0 with HCl.

Which compound is expected to be eluted first and why?

Decanoic acid will elute first. The carboxylic acid group will be depronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

Decanoic acid will elute first. The amino group will be pronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

1-aminodecane will elute first. The amino group will be pronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

1-aminodecane will elute first. The amino group will be depronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

Decanoic acid will elute first. The carboxylic acid group will be depronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

Decanoic acid will elute first. The amino group will be pronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

1-aminodecane will elute first. The amino group will be pronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

1-aminodecane will elute first. The amino group will be depronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

Explanation / Answer

Correct answer : 1-aminodecane will elute first. The amino group will be pronated at pH 3.0, this will give a charge to the molecule and make it more soluble in the polar eluent.

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