Advertisements
Advertisements
Question
Why are carboxylic acids more acidic than alcohols or phenols although all of them have hydrogen atom attached to a oxygen atom \[\ce{(-O-H)}\]?
Advertisements
Solution
The aliphatic carboxylic acids are stronger acids than alcohols and phenols. The difference in the relative acidic strengths can be understood if we compare the resonance hybrids of carboxylate ion and ethoxide phenoxide ion.
\[\ce{RCOOH ⇌ RCOO- + H+}\]
\[\ce{ROH ⇌ RO- + H+}\]
The resonance hybrids may be represented as:
\[\begin{array}{cc}
\phantom{.............}\ce{O^{δ-}}\\
\phantom{.........}//\\
\ce{R - C}\\
\phantom{........}\backslash\backslash\\
\phantom{............}\ce{O^{δ-}}\\
\end{array}\]
Carboxylate ion
Phenoxide ion
\[\ce{RO-}\]
(No resonance)
The resonance hybrids may be represented as: Carboxylate ion Phenoxide ion (No resonance)
The electron charge on the carboxylate ion is more dispersed in comparison to the phenate ion since there are two electronegative oxygen atoms in carboxylate ion as compared to only one oxygen atom in phenoxide ion. In other words, the carboxylate ion is relatively more stable as compared to phenate ion. Thus, the release of H+ ion from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.
APPEARS IN
RELATED QUESTIONS
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
- Methanal
- 2-Methylpentanal
- Benzaldehyde
- Benzophenone
- Cyclohexanone
- 1-Phenylpropanone
- Phenylacetaldehyde
- Butan-1-ol
- 2, 2-Dimethylbutanal
How will you convert ethanal into the following compound?
But-2-enal
How will you bring about the following conversion in not more than two steps?
Ethanol to 3-Hydroxybutanal
How will you bring about the following conversion in not more than two steps?
Benzaldehyde to 3-Phenylpropan-1-ol
Why is alpha (α) hydrogen of carbonyl compounds acidic in nature?
Write chemical equations of the following reaction :
Propanone is treated with dilute Ba (OH)2-.
What is substituted imine called?
Explain aldol condensation reaction in detail.
Which product is formed when the compound 
is treated with concentrated aqueous \[\ce{KOH}\] solution?
Compounds A and C in the following reaction are:
\[\ce{CH3CHO ->[(i) CH3MgBr][(ii) H2O] (A) ->[H2SO4, Δ] (B) ->[Hydroboration oxidation] (C)}\]
Which of the following compounds do not undergo aldol condensation?
(i) \[\ce{CH3 - CHO}\]
(ii) 
(iii) \[\begin{array}{cc}
\phantom{}\ce{O}\\
\phantom{}||\\
\ce{CH3 - C - CH3}
\end{array}\]
(iv) \[\begin{array}{cc}
\phantom{}\ce{CH3}\\
|\phantom{...}\\
\ce{CH3 - C - CHO}\phantom{..}\\
|\phantom{...}\\
\phantom{}\ce{CH3}\\
\end{array}\]
Why is there a large difference in the boiling points of butanal and butan-1-ol?
What product will be formed on reaction of propanal with 2-methylpropanal in the presence of \[\ce{NaOH}\]? What products will be formed? Write the name of the reaction also.
Identify A and B from the following reaction:
\[\begin{array}{cc}
\ce{CH3}\phantom{.................}\\
|\phantom{....................}\\
\phantom{}\ce{2CH3 - C = O ->[Ba(OH)2] A ->[Δ] B + H2O}
\end{array}\]
Convert the following:
Acetaldehyde to But-2-enal
When acetaldehyde is treated with dilute NaOH, the following reaction is observed.
\[\begin{array}{cc}
\ce{2CH3 - CHO ->[dil.NaOH] CH3 - CH - CH2 - CHO}\\
\phantom{...............}|\\
\phantom{.................}\ce{OH}
\end{array}\]
- What are the functional groups in the product?
- Can another product be formed during the same reaction? (Deduce the answer by doing atomic audit of reactant and product).
- Is this an addition reaction or condensation reaction?
What is aldol condensation? Explain it with suitable examples.
