Organic Compounds Remaining After Halogenation Reactions of Acetone

To determine the organic compounds present in the reaction mixtures after completion, we analyze the mechanisms and stoichiometry of both halogenation reactions of acetone ($\text{CH}_3\text{COCH}_3$).

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1. Analysis of Reaction I (Base-Catalyzed Halogenation):

$\text{CH}_3\text{COCH}_3\,(1.0\text{ mol}) + \text{Br}_2\,(1.0\text{ mol}) \xrightarrow{\text{aqueous NaOH}} \text{Products}$

• Mechanism: Under basic conditions, halogenation occurs via the enolate intermediate. After the substitution of one hydrogen atom by a halogen atom, the electron-withdrawing nature of the bromine atom increases the acidity of the remaining $\alpha$-hydrogens on the same carbon. Consequently, subsequent bromination steps occur much faster than the initial step, rapidly forming tribromoacetone ($\text{CH}_3\text{COCBr}_3$).

  The intermediate $\text{CH}_3\text{COCBr}_3$ then undergoes a haloform cleavage by hydroxide ion ($\text{OH}^-$) to yield sodium acetate ($\text{CH}_3\text{COONa}$, compound $T$) and bromoform ($\text{CHBr}_3$, compound $U$).

• Stoichiometry: The complete stoichiometry of the haloform reaction is: $\text{CH}_3\text{COCH}_3 + 3\text{Br}_2 + 4\text{NaOH} \longrightarrow \text{CH}_3\text{COONa}\,(T) + \text{CHBr}_3\,(U) + 3\text{NaBr} + 3\text{H}_2\text{O}$

  Here, $3.0\text{ moles}$ of $\text{Br}_2$ are required to completely react with $1.0\text{ mole}$ of acetone.

  Since only $1.0\text{ mole}$ of $\text{Br}_2$ is provided for $1.0\text{ mole}$ of acetone: $\text{Moles of acetone consumed} = \frac{1}{3}\text{ mol}$ $\text{Moles of unreacted acetone remaining} = 1.0 - \frac{1}{3} = \frac{2}{3}\text{ mol}$

• Organic products/compounds present in Reaction I: $T\,(\text{CH}_3\text{COONa}), \quad U\,(\text{CHBr}_3), \quad \text{and unreacted acetone.}$

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2. Analysis of Reaction II (Acid-Catalyzed Halogenation):

$\text{CH}_3\text{COCH}_3\,(1.0\text{ mol}) + \text{Br}_2\,(1.0\text{ mol}) \xrightarrow{\text{CH}_3\text{COOH}} \text{Products}$

• Mechanism: Under acidic conditions ($\text{CH}_3\text{COOH}$), bromination proceeds via the enol mechanism. Monobromination forms bromoacetone ($\text{CH}_3\text{COCH}_2\text{Br}$, compound $P$).

  The presence of the electronegative bromine atom in bromoacetone destabilizes the protonated carbonyl intermediate ($\text{CH}_3\text{C}(\text{OH}^+)\text{CH}_2\text{Br}$), suppressing further enolization. Thus, acid-catalyzed halogenation stops cleanly after monobromination.

• Stoichiometry: $\text{CH}_3\text{COCH}_3\,(1.0\text{ mol}) + \text{Br}_2\,(1.0\text{ mol}) \xrightarrow{\text{CH}_3\text{COOH}} \text{CH}_3\text{COCH}_2\text{Br}\,(P, 1.0\text{ mol}) + \text{HBr}$

  Both reactants are present in a $1:1$ molar ratio, leading to complete consumption of both acetone and $\text{Br}_2$.

• Organic products/compounds present in Reaction II: $P\,(\text{CH}_3\text{COCH}_2\text{Br}).$

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Conclusion:

• Reaction I: $T, U$, acetone
• Reaction II: $P$

This corresponds to Option C.