Animated Solution for Chemistry - Organic Compounds Containing Halogens: In the following sequence of reactions the $P$ is

We are given a two-step reaction sequence starting with cyclopentyl chloride. The first step involves reacting this alkyl halide with magnesium metal in the presence of dry ether.

This is a classic setup that should immediately ring a bell. Whenever you see an alkyl or aryl halide paired with magnesium and dry ether, you are looking at the birth of a Grignard reagent.

Forming the Grignard Reagent

In the first step, the magnesium atom inserts itself directly into the carbon-chlorine bond.

This transforms our starting material into cyclopentylmagnesium chloride, which is our intermediate $A$.

Why is this step so important? By inserting magnesium, we completely flip the polarity of the carbon atom. Originally, the carbon attached to chlorine was electrophilic (electron-poor) because chlorine is highly electronegative.

However, magnesium is electropositive. In the new carbon-magnesium bond, the carbon atom pulls the electron density towards itself, gaining a significant partial negative charge ($\delta -$).

This makes the carbon atom highly nucleophilic and, crucially, a very strong base.

Now, we move to the second step. We introduce ethanol ($C_2{H}_{5}OH$) to our newly formed Grignard reagent.

Ethanol contains a hydrogen atom attached to an electronegative oxygen atom. This makes the proton slightly acidic.

Here is where many students fall into a trap. They see a nucleophile (the Grignard reagent) and an organic molecule, and they immediately try to perform a nucleophilic attack.

But remember this golden rule of organic chemistry: Acid-base reactions are always faster than nucleophilic substitutions or additions.

Because the Grignard reagent is an exceptionally strong base, it will not act as a nucleophile here. Instead, it will instantly abstract the acidic proton from ethanol.

The Quenching

The cyclopentyl carbanion grabs the $H^{+}$ from ethanol, neutralizing itself to form a stable alkane.

The remaining ethoxide ion ($C_2{H}_5{O}^{-}$) pairs up with the magnesium complex to form the byproduct $Mg(O{C}_2{H}_5)Cl$.

Thus, our final major product $P$ is simply cyclopentane.

This reaction perfectly illustrates why Grignard reactions must be carried out in strictly anhydrous (moisture-free) conditions. Even a tiny amount of water or any protic solvent will act as an acid and destroy the Grignard reagent, turning it into a useless alkane!