Question

Suppose you have crude reaction mixture containing napthalene, benzoic acid, and aniline dissolved in an organic solvent, and you wish to extract the different molecules by altering the solubility of each component in solution. Which of the following statements would be true?

a. Adding 5% HCl solution to the crude reaction mixture will deprotonate benzoic acid increasing its solubility in the aqueous solution.
b. Adding 5% NaOH solution to the crude reaction mixture will protonate napthalene making it more soluble in the aqueous solution.
c. Adding 5% HCl solution to the crude reaction mixture will protonate napthalene making it more soluble in the aqueous solution.
d. Adding 5% HCl solution to the crude reaction mixture will protonate aniline increasing its solubility in the aqueous solution.

Answer 1

The correct statement for separating components by solubility alteration is that adding 5% HCl will protonate aniline, increasing its solubility in aqueous solution. Benzoic acid is not deprotonated by HCl, and NaOH does not protonate naphthalene.(Option d)

Step-by-step explanation:

If you have a crude reaction mixture containing naphthalene, benzoic acid, and aniline dissolved in an organic solvent and wish to alter the solubility of each component in solution, the correct statement among the ones provided would be:

d. Adding 5% HCl solution to the crude reaction mixture will protonate aniline increasing its solubility in the aqueous solution.

This is because aniline is a basic compound and will react with HCl, a strong acid, to form anilinium chloride, which is water-soluble. Contrary to option a, adding an acid like HCl to benzoic acid would not deprotonate it but would instead reduce its solubility in the aqueous layer because it would remain as a non-ionized acid. In option b, NaOH would not protonate naphthalene since naphthalene is not acidic and does not readily undergo such a reaction. Likewise, option c is incorrect because HCl does not protonate naphthalene, which is nonpolar and lacks acidic hydrogens.

Answer 2

d. Adding 5% HCl solution to the crude reaction mixture will protonate aniline increasing its solubility in the aqueous solution.

Step-by-step explanation:

On this case, we have to check the structures of each compound (figure 1). For naphthalene we dont have any functional groups therefore, the addition of HCl or NaOH it will not affect naphthalene so we can discard "B" and"C".

When we add HCl solution we will have the production
`H^+` the presence of this hydronium ion will protonate the acid, so we can discard a.

Finally, for d when we add
`H^+` the hydronium ion will react with aniline (a base) and will produce an ammonium ion. This ammonium ion have a positive charge, therefore the polarity will increase and the molecule would be more soluble on water (figure 2).

I hope it helps!