Classify each reactant and product as an acid or base according to the Bronsted theory.

[tex]\[
CH_3CH_2O^- + HC \equiv CH \rightleftharpoons HC \equiv C^- + CH_3CH_2OH
\][/tex]

Acid:

[tex]\[
\begin{array}{l}
\boxed{HC \equiv CH} \\
\boxed{CH_3CH_2OH}
\end{array}
\][/tex]

Base:

[tex]\[
\begin{array}{l}
\boxed{CH_3CH_2O^-} \\
\boxed{HC \equiv C^-}
\end{array}
\][/tex]

Answer Bank:

[tex]\[
\begin{array}{l}
HC \equiv C^- \\
HC \equiv CH \\
CH_3CH_2OH \\
CH_3CH_2O^-
\end{array}
\][/tex]



Answer :

Alright, let's classify each reactant and product in the given reaction according to the Bronsted-Lowry theory. According to the Bronsted-Lowry theory, an acid is a proton (H⁺) donor and a base is a proton acceptor.

The reaction provided is:
[tex]\[ CH_3CH_2O^- + HC \equiv CH \rightleftharpoons HC \equiv C^- + CH_3CH_2OH \][/tex]

1. HC≡CH (acetylen):
- This molecule donates a proton (H⁺) in the reaction, so it acts as an acid.
- When it donates the proton, it turns into [tex]\(HC≡C^{-}\)[/tex].

2. HC≡C^{-}:
- This is what remains after HC≡CH donates its proton.
- It is the conjugate base of the acid HC≡CH.

3. CH₃CH₂O⁻ (ethoxide ion):
- This molecule accepts a proton (H⁺) in the reaction, so it acts as a base.
- When it accepts the proton, it turns into [tex]\(CH_3CH_2OH\)[/tex].

4. CH₃CH₂OH (ethanol):
- This is what CH₃CH₂O⁻ becomes after accepting a proton.
- It is the conjugate acid of the base CH₃CH₂O⁻.

Thus, by following the Bronsted-Lowry theory:

- [tex]\( HC≡CH \)[/tex] is the acid because it donates a proton.
- [tex]\( HC≡C^{-} \)[/tex] is the base because it is the conjugate base that results from HC≡CH donating a proton.
- [tex]\( CH₃CH₂O^{-} \)[/tex] is the base because it accepts a proton.
- [tex]\( CH₃CH₂OH \)[/tex] is the acid because it is the conjugate acid that results from CH₃CH₂O^{-} accepting a proton.

So, we can classify them as follows:

Acids:
- [tex]\( HC≡CH \)[/tex]
- [tex]\( CH₃CH₂OH \)[/tex]

Bases:
- [tex]\( HC≡C^{-} \)[/tex]
- [tex]\( CH₃CH₂O^{-} \)[/tex]

Matching them up with the [tex]\( \square \)[/tex] entry and the provided answer bank:

- Acid: [tex]\( HC≡CH \)[/tex], [tex]\( CH₃CH₂OH \)[/tex]
- Base: [tex]\( HC≡C^{-} \)[/tex], [tex]\( CH₃CH₂O^{-} \)[/tex]