Chapter 2, Problem 2.71P

Answer the following questions about esmolol, a drug used to treat high blood pressure sold under the trade name Brevibloc.

a. Label the most acidic hydrogen atom in esmolol.

b. What products are formed when esmolol is treated with $\text{NaH}$?

c. What products are formed when esmolol is treated with $\text{HCl}$?

d. Label all $s{p}^2$ hybridized C atoms.

e. Label the only triagonal pyramidal atom.

f. Label all C’s that bear $\text{δ}+$ charge.

Interpretation Introduction

(a)

Interpretation: The most acidic hydrogen in the compound esmolol is to be labeled.

Concept introduction: The most acidic hydrogen $\left(\text{H}\right)$ atom or proton is the one which can be easily eliminated from the molecule. This higher acidity of $\text{H-atom}$ is dependent upon the electronegativity of the functional groups that are bonded with carbon atom containing hydrogen. The elimination of a proton leads to the formation of a conjugate base. The higher stability of the conjugate base corresponds to the higher acidity of the proton.

Answer to Problem 2.71P

The labeling of the most acidic hydrogen present in the compound esmolol is shown below.

Explanation of Solution

The most acidic hydrogen in the compound esmolol is shown as,

Figure 1

According to the structure of esmolol, there are three elements present in the compound to which protons are bonded. The acidity increases while moving left to right in the period due to an increase in the electronegativity of the elements. Thus, oxygen atom is located at the extreme right position in the periodic table and is highly electronegative than carbon and nitrogen. Thus, the proton that is attached to oxygen atom is the most acidic one as shown above.

Conclusion

The labeling of the most acidic hydrogen present in the compound esmolol is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The products formed by the reaction of esmolol with $\text{NaH}$ are to be predicted.

Concept introduction: An atom or a group of atoms that shows characteristic physical and chemical properties are collectively known as functional groups. The functional group is the most reactive part present in the molecule. The main functional groups are $\text{OH}$ (alcohol group), $\text{COOH(carboxylic acid group)}$, $\text{CHO}\left(\text{aldehyde}\text{group)}\right)$ and $\text{C=O}\left(\text{ketone}\text{group}\right)$.

According to Bronsted-Lowry theory, the species that easily accept the proton is known as base and the species that easily donate the proton is known as acid. The reaction of an acid with a base always leads to the formation of conjugate acid and base.

Answer to Problem 2.71P

The products formed by the reaction of esmolol with $\text{NaH}$ are shown below.

Explanation of Solution

The reaction of esmolol with $\text{NaH}$ is shown as,

Figure 2

In the above reaction, first of all, $\text{NaH}$ splits into ${\text{Na}}^{+}$ and ${\text{H}}^{-}$. Then, this hydride ion, ${\text{H}}^{-}$ acts as a base and abstracts a proton from hydroxyl group of esmolol which results in the formation of conjugate base and conjugate acid. The conjugate base reacts with ester intramolecularly and leads to the formation of a cyclized product.

Conclusion

The products formed by the reaction of esmolol with $\text{NaH}$ are shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The products formed by the reaction of esmolol with $\text{HCl}$ are to be predicted.

Concept introduction: An atom or a group of atoms that shows characteristic physical and chemical properties are collectively known as functional groups. The functional group is the most reactive part present in the molecule. The main functional groups are $\text{OH}$ (alcohol group), $\text{COOH(carboxylic acid group)}$, $\text{CHO}\left(\text{aldehyde}\text{group)}\right)$ and $\text{C=O}\left(\text{ketone}\text{group}\right)$.

According to Bronsted-Lowry theory, the species that easily tends to accept the proton is known as base and the species that easily donate the proton is known as acid. The reaction of an acid with a base always leads to the formation of conjugate acid and base.

Answer to Problem 2.71P

The products formed by the reaction of esmolol with $\text{HCl}$ are shown below.

Explanation of Solution

The reaction of esmolol with $\text{HCl}$ is shown as,

Figure 3

In the above reaction, a proton of $\text{HCl}$ is abstracted by the $\text{N}$ atom of esmolol which results in the formation of conjugate acid and conjugate base as the desired products.

Conclusion

The products formed by the reaction of esmolol with $\text{HCl}$ are shown in Figure 3.

Interpretation Introduction

(d)

Interpretation: The $s{p}^2$ hybridized . $\text{C}-$.atoms present in the compound, esmolol is to be labeled.

Concept introduction: The intermixing of $2s$ orbital with only two $2p$ orbital results in the $s{p}^2$ hybridization. The carbon atoms that have double bond possess $s{p}^2$ hybridization. The $s{p}^2$ hybridized $\text{C}-$ atom possesses one sigma bond and one pi bond which is perpendicular to the molecular plane.

Answer to Problem 2.71P

The labeling of $s{p}^2$ hybridized $\text{C}-$ atoms present in the compound, esmolol is shown below.

Explanation of Solution

The labeling of $s{p}^2$ hybridized $\text{C}-$ atoms present in the compound, esmolol is shown as,

Figure 4

According to the structure of esmolol, there are total seven $s{p}^2$ hybridized $\text{C}-$ atoms present in the compound.

Conclusion

The labeling of $s{p}^2$ hybridized $\text{C}-$ atoms present in the compound, esmolol is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation: The triagonal pyramidal atom present in the compound, esmolol is to be labeled.

Concept introduction: The atoms that possess $s{p}^3$ hybridization leads to the formation of triagonal pyramidal shape. The atoms with $s{p}^3$ hybridization possess single bond which is a sigma bond.

Answer to Problem 2.71P

The labeling of triagonal pyramidal atom present in the compound, esmolol is shown below.

Explanation of Solution

The labeling of triagonal pyramidal atom present in the compound, esmolol is shown as,

Figure 5

According to the structure of esmolol, nitrogen atom is present which is bonded to two carbon atoms and one hydrogen atom along with lone pair of electrons. This bonding of atoms with nitrogen results in the formation of $s{p}^3$ hybridization and triagonal pyramidal shape.

Conclusion

The labeling of triagonal pyramidal atom present in the compound, esmolol is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation: All $\text{C}-$ atoms that possess $\text{δ}+$ charge present in the compound, esmolol are to be labeled.

Concept introduction: The $\text{C}-$ atom which is bonded to more electronegative atom than itself always bear $\text{δ}+$ charge. This happens because the more electronegative atom attracts the shared pair of electrons towards itself that results in lowering the electron density on carbon atom. The elements present in a period which are more electronegative than carbon are nitrogen, oxygen and fluorine.

Answer to Problem 2.71P

The labeling of all $\text{C}-$ atoms that possess $\text{δ}+$ charge present in the compound, esmolol is shown below.

Explanation of Solution

The labeling of all $\text{C}-$ atoms that possess $\text{δ}+$ charge present in the compound, esmolol is shown as,

Figure 6

According to the structure of esmolol, the carbon atoms that are attached to nitrogen and oxygen possess $\text{δ}+$ charge.

Conclusion

The labeling of all $\text{C}-$ atoms that possess $\text{δ}+$ charge present in the compound, esmolol is shown in Figure 6.