JEE Challenger
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ChemHardAdv 2016 • Paper 2

Major Product of Aldol and Cannizzaro Reaction Sequence

Identify the primary organic compound obtained from the given multi-step reaction scheme starting with the substituted ketone:

Question Diagram 1

Options

A
Option A
Correct
B
Option B
C
Option C
D
Option D

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and KetonesOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the major product of the multi-step reaction sequence, we analyze each step sequentially:

Step 1: Aldol Condensation and Crossed Cannizzaro Reaction

Reagents: HCHO (excess)/NaOH, heat\text{Reagents: } \text{HCHO (excess)} / \text{NaOH, heat}

  1. Aldol Addition: Isobutyrophenone, Ph-CO-CH(CH3)2\text{Ph-CO-CH(CH}_3)_2, contains a single acidic α\alpha-hydrogen. In the presence of sodium hydroxide (NaOH\text{NaOH}), deprotonation occurs at the α\alpha-carbon to form an enolate ion: Ph-CO-CH(CH3)2+OH−⇌Ph-CO-C−(CH3)2+H2O\text{Ph-CO-CH(CH}_3)_2 + \text{OH}^- \rightleftharpoons \text{Ph-CO-C}^-(\text{CH}_3)_2 + \text{H}_2\text{O}

    The nucleophilic enolate attacks the carbonyl carbon of formaldehyde (HCHO\text{HCHO}) to yield a β\beta-hydroxy ketone: Ph-CO-C−(CH3)2+HCHO→H2OPh-CO-C(CH3)2CH2OH+OH−\text{Ph-CO-C}^-(\text{CH}_3)_2 + \text{HCHO} \xrightarrow{\text{H}_2\text{O}} \text{Ph-CO-C}(\text{CH}_3)_2\text{CH}_2\text{OH} + \text{OH}^-

  2. Crossed Cannizzaro Reaction: Since formaldehyde is in excess and the reaction mixture is heated with strong base (NaOH\text{NaOH}), the newly formed β\beta-hydroxy ketone undergoes a crossed Cannizzaro reaction.

    Formaldehyde acts as a stronger hydride donor. Attack of OH−\text{OH}^- on formaldehyde forms a monoanionic intermediate that transfers a hydride ion (H−\text{H}^-) to the carbonyl carbon of the β\beta-hydroxy ketone, reducing it to a secondary alcohol group: Ph-CO-C(CH3)2CH2OH+HCHO→NaOH, heatPh-CH(OH)-C(CH3)2CH2OH+HCOO−\text{Ph-CO-C}(\text{CH}_3)_2\text{CH}_2\text{OH} + \text{HCHO} \xrightarrow{\text{NaOH, heat}} \text{Ph-CH(OH)-C}(\text{CH}_3)_2\text{CH}_2\text{OH} + \text{HCOO}^-

    The resulting product after workup is the 1,3-diol: 2,2-dimethyl-1-phenylpropane-1,3-diol.

Step 2: Cyclic Acetal Formation (Acetalization)

Reagents: HCHO / H+ (catalytic amount)\text{Reagents: } \text{HCHO / H}^+ \text{ (catalytic amount)}

When the 1,3-diol, Ph-CH(OH)-C(CH3)2CH2OH\text{Ph-CH(OH)-C}(\text{CH}_3)_2\text{CH}_2\text{OH}, is treated with formaldehyde (HCHO\text{HCHO}) in the presence of an acid catalyst (H+\text{H}^+), an acetalization reaction takes place.

The two hydroxyl (−OH-\text{OH}) groups of the 1,3-diol condense with the carbonyl carbon of formaldehyde, eliminating one molecule of water (H2O\text{H}_2\text{O}) to form a stable 6-membered cyclic acetal (5,5-dimethyl-4-phenyl-1,3-dioxane):

Ph-CH(OH)-C(CH3)2CH2OH+HCHO→H+5,5-dimethyl-4-phenyl-1,3-dioxane+H2O\text{Ph-CH(OH)-C}(\text{CH}_3)_2\text{CH}_2\text{OH} + \text{HCHO} \xrightarrow{\text{H}^+} \text{5,5-dimethyl-4-phenyl-1,3-dioxane} + \text{H}_2\text{O}

This structure corresponds to Option (A).

Final Answer:

The correct option is (A).

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