Introduction to Carbonyl Compounds

What are Carbonyl Compounds?

Carbonyl compounds are organic substances characterized by the presence of a carbonyl group (C=O). This functional group is highly reactive and serves as the main structural feature of many important organic molecules.

Key Characteristics:

  • The carbonyl carbon is sp² hybridized
  • The C=O bond is polar due to oxygen’s higher electronegativity
  • Carbonyl compounds undergo nucleophilic addition reactions
  • They can be classified based on groups attached to carbonyl carbon

Classification of Carbonyl Compounds

Type General Formula Examples
Aldehydes RCHO Formaldehyde (HCHO), Acetaldehyde (CH₃CHO)
Ketones RCOR’ Acetone (CH₃COCH₃), Butanone (CH₃COC₂H₅)
Carboxylic Acids RCOOH Acetic acid (CH₃COOH), Benzoic acid (C₆H₅COOH)
Esters RCOOR’ Ethyl acetate (CH₃COOC₂H₅)
Amides RCONH₂ Acetamide (CH₃CONH₂)
Acyl Chlorides RCOCl Acetyl chloride (CH₃COCl)

Acidity of Carbonyl Compounds

Relative Acidities

Carboxylic acids are the most acidic among simple organic compounds due to resonance stabilization of their conjugate base (carboxylate ion).

RCOOH ⇌ RCOO⁻ + H⁺

Acidity Order:

Carboxylic acids > Phenols > Alcohols

pKa values: Carboxylic acids (~5), Phenols (~10), Alcohols (~16)

Effect of Substituents on Acidity

Electron-Withdrawing Groups INCREASE Acidity:

  • -Cl, -F, -Br, -I (Halogens)
  • -NO₂ (Nitro group)
  • -CN (Cyano group)

Electron-Donating Groups DECREASE Acidity:

  • -CH₃, -C₂H₅ (Alkyl groups)
  • -OH (Hydroxyl group)
  • -NH₂ (Amino group)
Compound Formula pKa
Trichloroacetic acid CCl₃COOH 0.9
Chloroacetic acid CH₂ClCOOH 2.9
Acetic acid CH₃COOH 4.7
Phenol C₆H₅OH 10.0
Ethanol CH₃CH₂OH ~16

Preparation Methods

Preparation of Benzoic Acid

Benzoic acid can be prepared by oxidation of alkyl benzenes using hot alkaline KMnO₄ followed by acidification.

C₆H₅CH₃ + 2KMnO₄ → C₆H₅COOK + 2MnO₂ + KOH + H₂O
C₆H₅COOK + HCl → C₆H₅COOH + KCl

Key Points:

  • Alkyl side chain is completely oxidized to -COOH group
  • Benzene ring remains intact during oxidation
  • Reaction requires heating under reflux conditions

Preparation of Acyl Chlorides

Acyl chlorides can be prepared from carboxylic acids using various chlorinating agents:

CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl
CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl
3CH₃COOH + PCl₃ → 3CH₃COCl + H₃PO₃

Note:

SOCl₂ is preferred because the byproducts (SO₂ and HCl) are gases that easily escape, leaving pure acyl chloride.

Multiple Choice Questions

1. Which reagent would you use to convert toluene to benzoic acid?
a) KMnO₄ and dilute acid
b) Fehling’s solution
c) PCl₅
d) SOCl₂

Correct Answer: a) KMnO₄ and dilute acid

Toluene (methylbenzene) is oxidized to benzoic acid using hot alkaline KMnO₄ followed by acidification with dilute acid.

2. What is the main product of the reaction between acetic acid and PCl₅?
a) Ethyl acetate
b) Acetyl chloride
c) Acetic anhydride
d) Acetone

Correct Answer: b) Acetyl chloride

PCl₅ reacts with carboxylic acids to form acyl chlorides: CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl

Short Answer Questions

i. Write the equation for the reaction of acetic acid with SOCl₂.
CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl

Explanation: Thionyl chloride (SOCl₂) converts carboxylic acids to acyl chlorides. The gaseous byproducts make purification easy.

ii. Explain why methanoic acid can be oxidized while other carboxylic acids cannot.

Answer: Methanoic acid (HCOOH) has a hydrogen atom attached directly to the carbonyl carbon, making it structurally similar to aldehydes. This allows it to be oxidized to CO₂ and H₂O. Other carboxylic acids have alkyl groups that resist oxidation under normal conditions.

Long Answer Questions

i. Describe the process of converting toluene to benzoic acid using hot alkaline KMnO₄ and then dilute acid. Include the mechanism and discuss the reasons for the choice of reagent.

Complete Process:

  1. Toluene is heated with alkaline KMnO₄ under reflux
  2. The methyl group is oxidized to carboxyl group
  3. The mixture is acidified with dilute acid

Chemical Equations:

C₆H₅CH₃ + 2KMnO₄ → C₆H₅COOK + 2MnO₂ + KOH + H₂O
C₆H₅COOK + HCl → C₆H₅COOH + KCl

Reason for Reagent Choice:

  • KMnO₄ is a strong oxidizing agent
  • It selectively oxidizes side chains without affecting benzene ring
  • Alkaline medium prevents further oxidation