Question

Write these reactions as proton-transfer reactions. Label which reactant is the acid and which is the base, which product is the conjugate base of the original acid, and which is the conjugate acid of the original base. In addition, write Lewis structures for each reactant and product and use curved arrows to show the flow of electrons in each reaction. (a) \(\mathrm{CH}_{3} \mathrm{SH}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{~S}^{-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CH}_{2}=\mathrm{O}+\mathrm{HCl} \rightleftharpoons \mathrm{CH}_{2}=\mathrm{OH}^{+}+\mathrm{Cl}^{-}\)

Short answer

Question: In each of the given reactions, identify the acid, base, conjugate acid, and conjugate base along with their Lewis structures and electron flow. Reaction A:\( \mathrm{CH}_{3} \mathrm{SH}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{S}^{-}+\mathrm{H}_{2} \mathrm{O}\) Reaction B: \(\mathrm{CH}_{2}=\mathrm{O}+\mathrm{HCl} \rightleftharpoons \mathrm{CH}_{2}=\mathrm{OH}^{+}+\mathrm{Cl}^{-}\) Answer: Reaction A: Acid: \(\mathrm{CH}_{3} \mathrm{SH}\), Base: \(\mathrm{OH}^-\), Conjugate Acid: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate Base: \(\mathrm{CH}_{3} \mathrm{S}^-\) [Include the Lewis structures and electron flow from the content above] Reaction B: Acid: \(\mathrm{HCl}\), Base: \(\mathrm{CH}_{2}=\mathrm{O}\), Conjugate Acid: \(\mathrm{CH}_{2}=\mathrm{OH}^{+}\), Conjugate Base: \(\mathrm{Cl}^-\) [Include the Lewis structures and electron flow from the content above]

Step-by-step solution

Reaction A: Identify Acid, Base, Conjugate Acid, and Conjugate Base

To solve this problem, we must first look at the reaction: (a) \(\mathrm{CH}_{3} \mathrm{SH}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{S}^{-}+\mathrm{H}_{2} \mathrm{O}\) In the given reaction, \(\mathrm{CH}_{3} \mathrm{SH}\) loses a proton (\(\mathrm{H}^+\)), thus it is acting as the acid. As a result, the conjugate base of the acid is the resulting \(\mathrm{CH}_{3} \mathrm{S}^-\) species. On the other hand, the \(\mathrm{OH}^-\) ion gains the proton, so it is the base, yielding \(\mathrm{H}_{2}\mathrm{O}\) as the conjugate acid of the original base. To summarize: Acid: \(\mathrm{CH}_{3} \mathrm{SH}\), Base: \(\mathrm{OH}^-\), Conjugate Acid: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate Base: \(\mathrm{CH}_{3} \mathrm{S}^-\)

Reaction A: Draw Lewis structures and show electron flow

Now, let's draw the Lewis structures for the reactants and products and use curved arrows to show the movement of electrons. \(\hspace{20pt}\chemfig{*6(=-=(-[,0.8]S-[,,1]H)(-[::30]H)-(-[,0.8]H)-=)} \ \ + \ \ \chemfig{O*4(--[::-60]H-)} \ \ \rightleftharpoons \ \ \chemfig{*6(=-=(-[,0.8]S^{-})(-[::30]H)-(-[,0.8]H)-=)} \ \ + \ \ \chemfig{O*4(=[:-60]H--)}\)

Reaction B: Identify Acid, Base, Conjugate Acid, and Conjugate Base

Next, let's analyze the second reaction: (b) \(\mathrm{CH}_{2}=\mathrm{O}+\mathrm{HCl} \rightleftharpoons \mathrm{CH}_{2}=\mathrm{OH}^{+}+\mathrm{Cl}^{-}\) In this reaction, \(\mathrm{HCl}\) loses a proton and thus acts as the acid. The conjugate base of the original acid is the resulting \(\mathrm{Cl}^-\) species. On the other hand, \(\mathrm{CH}_{2}=\mathrm{O}\) gains the proton, so it is the base, yielding \(\mathrm{CH}_{2}=\mathrm{OH}^{+}\) as the conjugate acid of the original base. To summarize: Acid: \(\mathrm{HCl}\), Base: \(\mathrm{CH}_{2}=\mathrm{O}\), Conjugate Acid: \(\mathrm{CH}_{2}=\mathrm{OH}^{+}\), Conjugate Base: \(\mathrm{Cl}^-\)

Reaction B: Draw Lewis structures and show electron flow

Finally, let's draw the Lewis structures for the reactants and products of Reaction B and use curved arrows to show the movement of electrons. \(\hspace{10pt}\chemfig{*6(=-=(-[,0.8]O)-(-[,0.8]H)-=)} \ \ + \ \ \chemfig{H-Cl} \ \ \rightleftharpoons \ \ \chemfig{*6(=-=(-[,0.8]O-[,2]H^+)-(-[,0.8]H)-=)} \ \ + \ \ \chemfig{Cl^{-}}\) This completes the analysis and provides step-by-step solutions for the given exercise.