Animated Solution for Chemistry - Organic Chemistry: In the following sequence of reactions, $C_3{H}_6\stackrel{H^{+}/H_{2}O}{\to }A\underset{\text{dil. KOH}}{\overset{KIO}{\to }}B+C$ The compounds $B$ and $C$ respectively are

Organic chemistry is often like solving a fascinating puzzle where each reagent acts as a specific clue. In this problem, we are given a two-step reaction sequence starting with propene ($C_3{H}_6$). Our ultimate mission is to identify the final products, $B$ and $C$.

Let's break down the journey of our starting molecule step by step.

The Acidic Hydration of Propene

The first transformation involves treating propene with water in the presence of an acid catalyst ($H^{+}/H_{2}O$). This is a classic electrophilic addition reaction known as acidic hydration.

When an unsymmetrical alkene like propene reacts with water, the addition follows Markovnikov's rule. The rule states that the negative part of the addendum (in this case, the $O{H}^{-}$ group from water) attaches to the carbon atom of the double bond that has fewer hydrogen atoms.

Mechanistically, the $H^{+}$ ion attacks the double bond to form the most stable carbocation. For propene, a secondary ($2^{\circ }$) carbocation is formed at the central carbon. The water molecule then attacks this carbocation, leading to the formation of a secondary alcohol.

Thus, our intermediate compound A is Propan-2-ol.

The Famous Iodoform Test

Now, we take Propan-2-ol and treat it with $KIO$ and dilute $KOH$. This specific combination of reagents is the hallmark of the Iodoform test. The reagent $KIO$ (potassium hypoiodite) is typically generated in situ by reacting iodine ($I_2$) with an alkali like $KOH$.

The Iodoform test is a highly specific analytical tool used to detect the presence of a methyl ketone group ($-CO-C{H}_3$) or a methyl carbinol group ($-CH(OH)-C{H}_3$) in a molecule.

Does Propan-2-ol fit the bill? Absolutely! It contains the exact $C{H}_3-CH(OH)-$ structural motif required for a positive test.

During the reaction, the hypoiodite first oxidizes the secondary alcohol to a ketone (acetone). Then, the methyl group adjacent to the carbonyl carbon undergoes exhaustive iodination, replacing all three hydrogen atoms with iodine atoms. Finally, the basic medium cleaves the molecule.

The cleavage results in two distinct products: 1. A pale yellow precipitate of Iodoform ($CH{I}_3$), which corresponds to compound B. 2. The potassium salt of the remaining two-carbon carboxylic acid, which is Potassium acetate ($C{H}_{3}COOK$), corresponding to compound C.

Matching our findings with the given options, we can confidently conclude that the correct answer is (d). Mastering these named reactions and their specific structural requirements is a surefire way to boost your score in organic chemistry!