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]
