Question: Given the values in Appendix A, which of the following bases are strong enough to deprotonate ${\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}$: (a) F ; (b) ${\left({\mathrm{CH}}_3\right)}_{\mathbf{3}}\mathbf{CO}$; (c)${{\mathbf{CH}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{}\mathbf{}\mathbf{}\mathbf{;}\mathbf{}\left(d\right)\mathbf{}\mathbf{}{{\mathbf{NH}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{}\mathbf{}\mathbf{}\left(e\right)\mathbf{}{\mathbf{Cl}}^{\mathbf{-}}$ ?

Species that can take up protons from other species are referred to as bases. There are various sort of bases, and they can be strong or weak.

The proton elimination from an acid like a Brønsted Lowry acid is referred to as deprotonation. The product obtained in deprotonation is called a conjugate base.

${\left({\mathrm{CH}}_3\right)}_3\mathrm{COOH}$ has a ${\mathrm{pk}}_{\mathrm{a}}$ value of 4.8. Any base possessing a conjugate acid of value greater than 4.8 can deprotonate ${\left({\mathrm{CH}}_3\right)}_3\mathrm{COOH}$.

a. The conjugate acid of ${\mathrm{F}}^{-}$, HF, possesses a ${\mathrm{pk}}_{\mathrm{a}}$value of 3.2, which is lesser than 4.8. Hence, is not strong enough to deprotonate ${\left({\mathrm{CH}}_3\right)}_3\mathrm{COOH}$.

b. The conjugate acid of${\left({\mathrm{CH}}_3\right)}_3{\mathrm{CO}}^{-}$ ,role="math" localid="1648718091857" ${\left({\mathrm{CH}}_3\right)}_3\mathrm{COOH}$ , possesses a value of 18, which is greater than 4.8. Hence, ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{CO}}^{-}$is strong enough to deprotonate ${\left({\mathrm{CH}}_3\right)}_3\mathrm{COH}$.

c. The conjugate acid of ${{\mathrm{CH}}_3}^{-}$,${\mathrm{CH}}_4$ possesses a ${\mathrm{pk}}_{\mathrm{a}}$value of 50, which is greater than 4.8. Hence, ${{\mathrm{CH}}_3}^{-}$is strong enough to deprotonate ${\left({\mathrm{CH}}_3\right)}_3\mathrm{COOH}$.

d. The conjugate acid of ${{\mathrm{NH}}_2}^{-}$, ${\mathrm{NH}}_3$, possesses a${\mathrm{pk}}_{\mathrm{a}}$ value of 38, which is greater than 4.8. Hence, ${{\mathrm{NH}}_2}^{-}$ is strong enough to deprotonate${\left({\mathrm{CH}}_3\right)}_3\mathrm{COOH}$ .

e. The conjugate base of ${\mathrm{Cl}}^{-}$, HCl, possesses a ${\mathrm{pk}}_{\mathrm{a}}$value of –7, which is lesser than 4.8. Hence,${\mathrm{Cl}}^{-}$ is not strong enough to deprotonate${\left({\mathrm{CH}}_3\right)}_3\mathrm{COOH}$ .