A crucial rule for Grignard reactions: they are incompatible with protic functional groups (like -OH, -NH, -SH, -COOH). If a protic reagent is added, a simple acid-base reaction occurs, protonating the Grignard reagent to form the corresponding alkane. This is often a trap in exam questions.
Step 1: Understanding the Question:
We are given a two-step reaction sequence starting with cyclopentyl chloride and need to identify the major organic product P. Step 2: Key Formula or Approach:
The reaction sequence involves:
Formation of a Grignard reagent: An alkyl halide reacts with magnesium metal in dry ether. R-Cl + Mg \(\rightarrow\) R-MgCl.
Reaction of Grignard reagent with a protic solvent: Grignard reagents are very strong bases and will react with any compound that has an acidic proton (like water, alcohols, carboxylic acids) in an acid-base reaction.
Step 3: Detailed Explanation: Step 1: Formation of A
Cyclopentyl chloride reacts with magnesium (Mg) in dry ether to form the Grignard reagent, cyclopentylmagnesium chloride.
\[ \text{C}_5\text{H}_9\text{Cl} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_5\text{H}_9\text{MgCl} \text{ [A]} \]
The structure [A] contains a highly basic cyclopentyl carbanion. Step 2: Formation of P
The Grignard reagent [A] is then reacted with ethanol (CH\(_3\)CH\(_2\)OH). Ethanol has an acidic proton on the oxygen atom. The Grignard reagent acts as a strong base and abstracts this proton.
\[ \underbrace{\text{C}_5\text{H}_9\text{MgCl}}_{\text{Strong Base}} + \underbrace{\text{CH}_3\text{CH}_2\text{OH}}_{\text{Acid}} \rightarrow \underbrace{\text{C}_5\text{H}_{10}}_{\text{Product P}} + \text{Mg(OCH}_2\text{CH}_3)\text{Cl} \]
The cyclopentyl anion (C\(_5\)H\(_9^-\)) takes the proton (H\(^+\)) from ethanol to form the alkane, cyclopentane (C\(_5\)H\(_{10}\)). Step 4: Final Answer:
The major organic product P is cyclopentane. Option (B) shows the structure of cyclopentane.
