Aldehydes and Ketones - Result Question 34
####35. After completion of the reactions (I and II), the organic compound (s) in the reaction mixtures is/are
(2013)
Reaction I
Reaction II
(a) reaction I $: P$ and reaction II $: P$
(b) reaction $I: U$, acetone and reaction II $: Q$, acetone
(c) reaction $I: T, U$, acetone and reaction II $: P$
(d) reaction $I: R$, acetone and reaction II $: S$, acetone
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Answer:
Correct Answer: 35. (a)
Solution:
- Plan When acetone reacts with $B r _2$ in basic medium, bromoform is formed.
Reaction I $CH _3 COCH _3+3 Br _2+4 NaOH$
$$ \begin{aligned} & \begin{array}{ll} 1 mol & 3 mol \ \frac{1}{3} mol & 1 mol \end{array} \ & \longrightarrow CH _3 COONa+CHBr _3+3 NaBr+3 H _2 O \ & (T) \ & (U) \end{aligned} $$
When $CH _3 COCH _3$ and $Br _2$ are in equimolar quantity, all the $Br _2$ (limiting reactant) is converted into desired products and $2 / 3$ mole of $CH _3 COCH _3$ remains unreacted, being in excess. When acetone reacts with $Br _2$ in acidic medium, there is monobromination of acetone.
Reactions II
$$ \begin{aligned} & CH _3 COCH _3+Br _2 \xrightarrow{CH _3 COOH} CH _3 COCH _2 Br+HBr \ & 1 mol \quad 1 mol \ & (P) \end{aligned} $$
$CH _3 COCH _3$ and $Br _2$ react in $1: 1$ mole ratio and $(P)$ is formed. In reaction $I,(U)$ and $(T)$ are formed and acetone (reactant) remains unreacted. In reaction II, $(P)$ is formed.
