Chapter 21, Problem 74PP

(a)

Interpretation Introduction

Interpretation:

Structure of the products obtained needs to be drawn when acetophenone is treated with given set of reagents.

Concept introduction:

The starting material here is a methyl ketone which reacts with iodine in presence of base namely sodium hydroxide to form carboxylate and a iodoform.  This undergoes reaction with aqueous acid to form carboxylic acid.  This reaction is known as haloform reaction.  Please do remember that this is a qualitative analysis for checking the presence of methyl ketone in compounds.

(b)

Interpretation Introduction

Interpretation:

Structure of the products obtained needs to be drawn when acetophenone is treated with given set of reagents.

Concept introduction:

The starting material here is ketone which reacts with bromine in presence of acetic acid to form ketone with bromine substituted at $\alpha$ position.

(c)

Interpretation Introduction

Interpretation:

Structure of the products obtained needs to be drawn when acetophenone is treated with given set of reagents.

Concept introduction:

Aldol condensation reaction is where the aldol addition product undergoes dehydration to give a conjugated enone.

In organic chemistry condensation reaction means loss of small molecules.  For aldol condensation reaction to take place the starting material should have two $\alpha$ hydrogens in the same carbon atom.

The driving force for aldol condensation reaction is the formation of conjugated system.  The steps involved in aldol condensation reaction are,

• Deprotonation of $\alpha$ carbon
• Attach of enolate ion with the aldehyde (nucleophilic attack)
• Protonation of the alkoxide ion to give aldol product
• Dehydration of the aldol addition product to form conjugated enone (aldol condensation product)

Please note that aldol condensation product is formed at higher temperature.