Animated Solution for Chemistry - Organic Chemistry: In the reaction, $C{H}_{3}COOH\stackrel{LiAl{H}_4}{\to }A\stackrel{PC{l}_5}{\to }B\stackrel{alc.KOH}{\to }C$ the product $C$ is

Welcome to a classic organic chemistry transformation! In this problem, we are taking a simple molecule, acetic acid, and putting it through a three-step chemical journey. Each step introduces a specific reagent that alters the functional group while keeping the carbon skeleton intact. Let's break down the thought process behind each transformation.

The Power of Reduction

Our starting material is acetic acid, $C{H}_{3}COOH$. The first reagent we encounter is lithium aluminium hydride, $LiAl{H}_4$.

$LiAl{H}_4$ is known as a very strong reducing agent. When it reacts with a carboxylic acid, it doesn't just stop at an aldehyde; it drives the reduction all the way down to a primary alcohol.

So, the acetic acid gets reduced, and we obtain ethanol. This is our intermediate Product A. The two-carbon chain remains perfectly intact, but the highly oxidized acid group has been transformed into a hydroxyl group.

The Classic Chlorination

Next, we treat our newly formed ethanol with phosphorus pentachloride, $PC{l}_5$.

$PC{l}_5$ is a classic chlorinating agent used to convert alcohols into alkyl chlorides. It performs a nucleophilic substitution where the hydroxyl ($-OH$) group leaves, and a chlorine atom takes its place.

We now have ethyl chloride. This is our intermediate Product B. It's a straightforward substitution reaction that prepares the molecule for the final, crucial step.

The Solvent Trap

Finally, we bring in alcoholic $KOH$. This is where many students fall into a trap!

If we had used aqueous $KOH$, the hydroxide ion would act as a nucleophile, substituting the chlorine and taking us right back to ethanol. However, alcoholic $KOH$ creates a strongly basic environment (forming ethoxide ions). In this medium, the reagent acts as a strong base rather than a nucleophile, heavily favoring $\beta$-elimination.

The base removes a proton from the $\beta$-carbon, and the chlorine leaves from the $\alpha$-carbon. This concerted elimination forms a double bond, yielding ethylene as our final Product C.

By carefully tracking the function of each reagent, we successfully navigated from a carboxylic acid to an alkene. The final product is ethylene, which corresponds to option (c). Always remember to pay close attention to the solvent conditions, as seen in the final step, because they can completely dictate the reaction pathway!