Decoding Multi-Step Syntheses: Acetic Acid to Lactic Acid
In JEE Main and NEET Advanced Organic Chemistry, examiners rarely ask single-step questions. Instead, they give you a starting material and a string of 4 or 5 reagents, asking you to identify the final "Major Product P".
To solve these, you cannot just memorize reactions; you must understand what each specific reagent does to the functional group in front of it. Let's decode a fantastic PYQ that converts a simple carboxylic acid into a highly substituted biological acid.
Video Tutorial: The 4-Step Reaction Roadmap
Watch Abhishek Sengar sir from CHEMCA expertly map out the reaction intermediates, step-by-step, until the final target molecule is formed.
Step-by-Step Reaction Breakdown
Reactant: Acetic Acid (CH3COOH)
Reagents: 1) LiAlH4 → 2) PCC → 3) HCN → 4) H2O / H+ / Δ
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Step 1: Strong Reduction (LiAlH4)
Lithium Aluminum Hydride is an extremely powerful reducing agent. When it reacts with Acetic Acid, it reduces the Carboxylic Acid group all the way down to a Primary Alcohol.
Intermediate A: Ethanol (CH3CH2OH). -
Step 2: Controlled Oxidation (PCC)
Pyridinium Chlorochromate (PCC) is a mild, anhydrous oxidizing agent. It oxidizes the primary alcohol (Ethanol) but strictly stops at the Aldehyde stage, preventing over-oxidation back to a carboxylic acid.
Intermediate B: Acetaldehyde or Ethanal (CH3CHO). -
Step 3: Nucleophilic Addition (HCN)
The Cyanide ion (CN-) acts as a nucleophile, attacking the electrophilic carbonyl carbon of the aldehyde. The oxygen grabs the H+. This "steps up" the carbon chain and forms a Cyanohydrin.
Intermediate C: 2-Hydroxypropanenitrile (CH3-CH(OH)-CN). -
Step 4: Complete Hydrolysis (H3O+ / Δ)
When a Nitrile (-CN) group is heated with acidic water, it undergoes complete hydrolysis, converting into a Carboxylic Acid (-COOH) group.
Final Product (P): 2-Hydroxypropanoic Acid (commonly known as Lactic Acid!).
Fig: A classic 4-step organic roadmap combining reduction, oxidation, nucleophilic addition, and hydrolysis.
Practice Questions for JEE & NEET
The secret to organic chemistry is knowing why a specific reagent was chosen. Test your understanding below!
Question 1: In Step 2, what would have happened if we used Acidified Potassium Permanganate (KMnO4 / H+) instead of PCC?
Answer: The reaction sequence would be ruined by over-oxidation.
Reasoning:
KMnO4 is a very strong oxidizing agent. If you react Ethanol (CH3CH2OH) with it, the ethanol will oxidize to Acetaldehyde, but the reaction won't stop there! The strong oxidizer will immediately oxidize the aldehyde back into Acetic Acid (your starting material).
By using PCC (a mild, anhydrous oxidizer), we successfully trap and isolate the Acetaldehyde so it can undergo the next reaction with HCN.
Question 2: In Step 4, we perform complete hydrolysis of the Nitrile (-CN) group using water and heat. What would the product be if we only performed partial hydrolysis (using alkaline H2O2 or concentrated HCl at room temperature)?
Answer: An Amide instead of a Carboxylic Acid.
Reasoning:
Nitrile hydrolysis occurs in two stages.
Partial Hydrolysis: The -C≡N group adds one molecule of water to become an Amide (-CONH2). In our reaction, this would yield 2-Hydroxypropanamide.
Complete Hydrolysis (with heat): The Amide reacts with a second molecule of water, releasing Ammonia gas (NH3) and leaving behind the Carboxylic Acid (-COOH). Since the question specified heating with H3O+, complete hydrolysis occurred, giving us the acid.
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