Carbonyl Compounds Five Marks

1. Explain / Give the mechanism for Claisen Or Claisen-Schmidt reaction.

Claisen or Claisen-Schmidt reaction: Benzaldehyde reacts with aliphatic aldehydes or ketones in presence of NaOH forming a, β - unsaturated aldehyde or ketone.

                                                          NaOH

C6H5 – CHO + CH3 – CHO    →    C6H5 – CH = CH – CHO + H2O

Similarly,                                                 NaOH

C6H5 – CHO + CH3 – CO – CH3    →    C6H5 – CH = CH – CO – CH3 + H2O

Mechanism:

The carbanion formed from acetaldehyde or acetone brings forth a nucleophilic attack on the carbonyl carbon of Benzaldehyde. The aldol type of product undergoes β-elimination (of water).

Step 1: The carbanion is formed as the a-hydrogen atom is removed as a proton by the base.

Step 2: The carbanion (Nucleophile) attacks the carbonyl carbon of Benzaldehyde.

Step 3: β-elimination of water.

 2. Explain / Write the mechanism of crossed aldol condensation (of acetone).

Crossed aldol condensation: When an aldehyde and a ketone with a-hydrogen atom react in presence of NaOH forming Aldol.

                                                                         CH3

                                                                         |

CH3 – CO – CH3 + CH3 – CHO → CH3 – C – CH2 – CHO

                                                                         |

                                                                        OH

Mechanism:

Step 1: The carbanion is formed as the a-hydrogen atom is removed as a proton by the base.

Step 2: The carbanion attacks the carbonyl carbon of unionised ketone molecule.

Step 3: The alkoxide ion formed is protonated by water to give ‘Aldol’.

3. Explain the mechanism of aldol condensation in acetone.

Aldol condensation: When ketones with a-hydrogen atom react in presence of NaOH forming Ketol.

                                                                                CH3           O

                                                                                 |                 ||

CH3 – CO – CH3 + CH3 – CO – CH3 → CH3 – C – CH2 – C – CH3

                                                                                 |

                                                                                OH

Mechanism:

Step 1: The carbanion is formed as the a-hydrogen atom is removed as a proton by the base.

Step 2: The carbanion attacks the carbonyl carbon of another unionised ketone molecule.

Step 3: The alkoxide ion formed is protonated by water to give ‘Ketol’.

 4. Explain the mechanism of aldol condensation of acetaldehyde. Or Discuss the mechanism involved in aldol condensation of acetaldehyde.

Aldol condensation: When aldehydes with a-hydrogen atom react in presence of NaOH forming Aldol.

                                                                        CH3           O

                                                                         |                 ||

CH3 – CO – CH3 + CH3 – CHO → CH3 – C – CH2 – C – H

                                                                         |

                                                                        OH

Mechanism:

Aldol condensation is catalysed by base. The carbanion generated is nucleophilic in nature. Hence it can bring about nucleophilic attack on carbonyl group.

Step 1: The carbanion is formed as the a-hydrogen atom is removed as a proton by the base.

Step 2: The carbanion attacks the carbonyl carbon of another unionised acetaldehyde molecule.

Step 3: The alkoxide ion formed is protonated by water to give ‘Aldol’.

 5. Discuss the mechanism of aldol condensation.

Answer: Q.No. 2 or 3 or 4

6. Explain the mechanism of Cannizzaro reaction.

Cannizzaro reaction: Aldehydes without a-hydrogen atom when heated with concentrated NaOH involves self oxidation of one aldehyde molecule to carboxylic acid and self reduction of the other aldehyde molecule to a primary alcohol.

                                                                NaOH

C6H5 – CHO + C6H5 – CHO  → C6H5 – COONa + C6H5 – CH2 – OH

The mechanism involves the transfer of hydride ion from one molecule of benzaldehyde to the other molecule.

Step 1: Nucleophilic attack by OH – ion on carbonyl carbon

Step 2: Transfer of hydride ion from the anion to carbonyl carbon of another molecule.

Step 3: The benzyloxide ion picks up the acidic proton from benzoic acid to give benzyl alcohol.

7.  Explain 'Popott's rule' with an example.

During oxidation of unsymmetric ketones with oxidising agent which brings about the cleavage of C–C bond, the smaller alkyl group goes preferentially with the carbonyl group resulting in the carboxylic acids.

8. How is acetone converted to - i) mesityl oxide ii) mesitylene?

i) Acetone reacts with dry hydrogen chloride forms mesityl oxide.

                                                               Dry. HCl

CH3 – CO – CH3+ H2CH – CO – CH3  →   (H3C)2 – C = CH – CO – CH3

                                                                                       –H2O       

                                                                            Mesityl oxide

                                                                       4-methyl pent-3-ene-2-one

ii) In presence of Conc. H2SO4 three molecules of acetone condense to give mesitylene (1, 3, 5 trimethyl benzene)

9. Write notes on i) Perkin’s reaction and ii) Knoevenagal reaction iii) Stephen’s reaction.

i) Perkin’s reaction:  When benzaldehyde is heated with sodium salt of acetic acid in presence of acetic anhydride, it forms Cinnamic acid.

                                                                CH3COONa

C6H5–CH = O + CH3–CO– O –CO–CH3 → C6H5–CH = CH–COOH + CH3–COOH

                                Acetic anhydride            Δ        Cinnamic acid

Mechanism:

Sodium acetate is the base that generates a carbanion at the a-carbon in the acetic anhydride. This brings forth nucleophilic attack on the carbonyl carbon forming β-hydroxy acid, water gets removed from this by β-elimination.

ii) Knoevenagal reaction: Benzaldehyde condenses with malonic acid in presence of pyridine forming cinnamic acid, pyridine is the basic catalyst here.

                                                   Pyridine                                                  CO2

C6H5 – CH = O + H2C(COOH)2 → C6H5 – CH = C(COOH)2 → C6H5 –  CH = CH – COOH

                                                               Malonic acid                      Δ      Cinnamic acid

iii) Stephen’s reaction: Aldehyde can be prepared by the reduction of alkyl cyanide dissolved in ether with Stannous chloride and hydrochloric acid.

            H – H        SnCl2

CH3 – C ≡ N        →       CH3CH = NH.HCl

Methyl cyanide       HCl        Iminimum hydro chloride         

           O – H2        Hydrolysis

CH3 CH = NH.HCl    →  CH3CHO + NH4Cl

                                            Acetaldehyde 

10. Give the following reactions (i) Benzoin Condensation (ii) Knoevenagal reaction 
(i) Benzoin Condensation:

When benzaldehyde is refluxed with aqueous alcoholic potassium cyanide – α - hydroxy ketone called benzoin is formed. Cyanide ion (CN–) is the specific catalyst in this reaction.

Benzoin can be considered as dimer of benzaldehyde.

                                                            O                                               O

                                                            ||               alc. KCN                   ||

C6H5CH = O + H – C – C6H5  →  C6H5CHOH – C – C6H5

                                               Benzoin

(ii) Knoevenagal reaction:

Benzaldehyde condenses with malonic acid in presence of pyridine forming cinnamic acid; pyridine is the basic catalyst here.

           Pyridine                                   CO2

C6H5CH = O + H2C(COOH)2 → C6H5CH = C(COOH)2 → C6H5CH = CHCOOH

                                                                      Malonic acid             Δ    Cinnamic acid

11. Write the differences between Acetaldehyde and Benzaldehyde Or

Compare Aliphatic aldehyde and Aromatic aldehyde

Reactions	Acetaldehyde  CH3CHO	Benzaldehyde  C6H5CHO
	Aliphatic aldehyde	Aromatic aldehyde
1.  With Fehling’s solution	Gives a Red precipitate.	No reaction
2. With Ammonia	Forms Simple   Addition product	Forms Complex   Condensation product
3.  With Caustic soda	Undergoes Aldol condensation	Undergoes Cannizzaro reaction
4.  With Primary amines	Does not form Schiff’s base	Forms Schiff’s base.
5. With Chlorine 6. Polymerisation 7. Electrophilic substitution 8. With Schiff’s reagent	Does not form Acetyl chloride Undergoes Polymerisation Does not  undergo Gives Pink colour in cold	Forms Benzoyl chloride Does not polymerise Undergoes at the meta position Gives Pink colour

12. Write a note on  i) Clemmenson reduction     and         ii) Knoevenagel reaction

i) Clemmenson reduction : Aldehydes and ketones can be reduced to Hydrocarbons by Zinc amalgam and Con. HCl.

This reaction proceeds by electron addition to carbonyl carbon followed by protonation. Zinc metal is the electron source

In the absence of mercury, hydrogen gas will be evolved and the reduction is incomplete. This reduction is called Clemmenson reduction.

Clemmenson reduction	HCHO	CH3CHO	C6H5CHO	CH3COCH3	C6H5COCH3	C6H5COC6H5
Zn / Hg + Conc. Hcl	←--------- > C = O Group reduced to – CH2 –  Group to give Hydrocarbons ---------→    CH4      CH3CH3          C6 H5CH3     CH3CH2CH3      C6H5CH2CH3    C6H5CH2C6H5

ii) Knoevenagel reaction : Benzaldehyde condenses with malonic acid in presence of pyridine forming cinnamic acid; pyridine is the basic catalyst here.

           Pyridine                                   CO2

C6H5CH = O + H2C(COOH)2 → C6H5CH = C(COOH)2           → C6H5CH = CHCOOH

                                                                      Malonic acid           Δ   Cinnamic acid

13. Illustrate the reducing property of acetaldehyde with examples

Because aldehydes are easily oxidised, they are reducing agents. The reduce Ammonical silver nitrate (Tollen’s reagent) to metallic Silver and Fehling’s solution (Copper sulphate, Sodium potassium tartrate) to red Cuprous oxide.

CH3CHO + 2Ag+ + 3OH–  →  CH3COO– + 2Ag + 2H2O

Acetaldehyde                             Acetate ion      (Silver mirror)

CH3CHO + 2Cu2+ + 5OH– → CH3COO– + Cu2O + 3H2O

                     (Blue)                                      (Red precipitate)

                     Cupric ion                                  Cuprous ion

14. Write the differences between acetaldehyde and acetone.

S.No	Reactions	Acetaldehyde CH3CHO	Acetone CH3COCH3
1	With Fehling’s solution.	Gives a red precipitate	Does not react.
2	With Tollen’s reagent.	Gives silver mirror.	No silver mirror.
3	Oxidation	Gives acetic acid	Gives acetic acid with loss of one Carbon atom.
4	Reduction with NaBH4	Gives ethanol (primary alcohol)	Gives isopropyl alcohol (secondary alcohol)
5	With NH3	Simple addition product is formed.	Forms complex ketonic amine.
6	Iodoform reaction	Forms iodoform and formic acid.	Forms iodoform and acetic acid.
7	Polymerisation	Forms paraldehyde	Forms condensation products.
8	With Schiff’s reagent.	Pink colour appears in cold.	No pink colour in cold.
9	Warming with NaOH	A brown resinous mass.	No resinous mass.