Chapter 2, Problem 36P

Propranolol is an antihypertensive agent—that is, it lowers blood pressure. (a)

Which proton in propranolol is most acidic? (b) What products are formed when propranolol is treated with $\text{NaH}$? (c) Which atom is most basic? (d) What products are formed when propranolol is treated with $\text{HCl}$?

Interpretation Introduction

(a)

Interpretation: The most acidic proton in propranolol is to be identified.

Concept introduction: An acid is a substance that is capable to donate ${\text{H}}^{+}$ ions or protons. The loss of a proton by an acid results in the formation of a substance known as conjugate base and the gain of a proton by a base results in the formation of a substance known as conjugate acid. The acidity of an acid is directly proportional to the resonance stabilization of its conjugate base.

Answer to Problem 36P

The most acidic proton is $\text{H}$ of $-\underset{··}{\overset{··}{\text{O}}}\text{H}$ group.

Explanation of Solution

The given ball-and-stick model of propranolol is,

Figure 1

In ball-and-stick model, each colored ball represents a specific atom and each stick represents a bond. In this model, each black ball represents $\text{C}$ atom, each gray ball represents $\text{H}$ atom, each blue ball represents $\text{N}$ atom and each red ball represents $\text{O}$ atom.

Thus, the given compound is drawn as,

Figure 2

The acidity of an acid is directly proportional to the resonance stabilization of its conjugate base. In propranolol, when the $-\underset{··}{\overset{··}{\text{O}}}\text{H}$ group losses a proton it forms a conjugate base that is much more resonance stabilized.

Figure 3

Thus, the most acidic proton is $\text{H}$ of $-\underset{··}{\overset{··}{\text{O}}}\text{H}$ group.

Conclusion

The most acidic proton is $\text{H}$ of $-\underset{··}{\overset{··}{\text{O}}}\text{H}$ group.

Interpretation Introduction

(b)

Interpretation: The products formed by the treatment of propranolol with $\text{NaH}$ are to be stated.

Concept introduction: Sodium hydride $\left(\text{NaH}\right)$ is a metal hydride. It is a strong base. It abstracts the most acidic proton and forms alkoxide ion. The resultant ion is a conjugate base of parent one (alcohol).

Answer to Problem 36P

The product formed by the treatment of propranolol with $\text{NaH}$ is,

Explanation of Solution

Sodium hydride $\left(\text{NaH}\right)$ is a metal hydride. It is a strong base. It abstracts the most acidic proton and forms alkoxide ion. The formed ion is a conjugate base of parent one (alcohol).

The products formed by the treatment of propranolol with $\text{NaH}$ are shown in Figure 4.

Figure 4

Conclusion

The product formed by the treatment of propranolol with $\text{NaH}$ is an alkoxide ion.

Interpretation Introduction

(c)

Interpretation: The most basic atom in propranolol is to be identified.

Concept introduction: A base is a substance that is capable to accept a ${\text{H}}^{+}$ ion or a proton. The nitrogen atom of $\text{N}-\text{H}$ bond is more basic than the oxygen atom of $\text{O}-\text{H}$ bond. It is due to less electronegativity of nitrogen than that of oxygen atom. Moreover, the conjugate base of $-\text{OH}$ group is much more resonance stabilized.

Answer to Problem 36P

The most basic atom is $\text{N}$ of $-\stackrel{··}{\text{N}}\text{H}-$ group.

Explanation of Solution

A base is a substance that is capable to accept ${\text{H}}^{+}$ ion or hydrogen ion. The nitrogen atom of $\text{N}-\text{H}$ bond is more basic than the oxygen atom of $\text{O}-\text{H}$ bond. It is due to less electronegativity of nitrogen than that of oxygen atom. Moreover, the conjugate base of $-\text{OH}$ group is much more resonance stabilized than the $-\stackrel{··}{\text{N}}\text{H}-$ group.

Thus, the most basic atom is $\text{N}$ of $-\stackrel{··}{\text{N}}\text{H}-$ group.

Conclusion

The most basic atom is N of $-\stackrel{··}{\text{N}}\text{H}-$ group.

Interpretation Introduction

(d)

Interpretation: The products formed by the treatment of propranolol with $\text{HCl}$ are to be stated.

Concept introduction: Hydrogen chloride $\left(\text{HCl}\right)$ is a halogen acid. It is a strong acid. It donates ${\text{H}}^{+}$ ion to the most basic atom and forms a salt (includes halogen atom).

Answer to Problem 36P

The product formed by the treatment of propranolol with $\text{HCl}$ is,

Explanation of Solution

Hydrogen chloride $\left(\text{HCl}\right)$ is a halogen acid. It is a strong acid. It donates ${\text{H}}^{+}$ ion to the most basic atom and forms a salt (includes halogen atom).

The product formed by the treatment of propranolol with $\text{HCl}$ is shown in Figure 5.

Figure 5

Conclusion

The product formed by the treatment of propranolol with $\text{HCl}$ is a salt.