Decoding Organic Reaction Charts: Propene Conversions
One of the most intimidating parts of JEE and NEET Organic Chemistry is the "Reaction Roadmap" or "Reaction Chart." These questions test your ability to link multiple chapters together—from Hydrocarbons to Amines—in a single flow.
Let's dissect a highly conceptual reaction chart starting with Propene. We'll identify 6 unknown products (A through F) by tracing the mechanism of each step.
Video Tutorial: Solving the Reaction Chart
Watch Abhishek Sengar sir from CHEMCA rapidly decode the entire reaction map, identifying each major product along the way.
Step-by-Step Reaction Breakdown
Fig: Complete Reaction Chart Flow
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Formation of A (Anti-Markovnikov Addition):
Propene reacts with HBr in the presence of peroxide. The Peroxide Effect (Kharasch effect) directs the Bromine to the terminal carbon, forming a primary alkyl halide.
CH3-CH=CH2 → (HBr / Peroxide) CH3-CH2-CH2-Br
[A] 1-Bromopropane -
Formation of B (Step-Up via Nucleophilic Substitution):
1-Bromopropane reacts with KCN. The cyanide ion (CN-) replaces Bromine. This is a crucial "step-up" reaction, increasing the carbon chain from 3 to 4 carbons.
CH3-CH2-CH2-Br → (KCN) CH3-CH2-CH2-CN
[B] Butanenitrile (Propyl Cyanide) -
Formation of C (Partial Hydrolysis):
When a nitrile reacts with concentrated HCl, it undergoes partial hydrolysis to form an amide.
CH3-(CH2)2-CN → (conc. HCl) CH3-(CH2)2-C(=O)NH2
[C] Butanamide -
Formation of F (Complete Hydrolysis):
If the nitrile instead reacts with dilute acid (H+ / H2O) and heat, it undergoes complete hydrolysis, converting all the way to a Carboxylic Acid.
CH3-(CH2)2-CN → (H+ / H2O, Δ) CH3-(CH2)2-COOH
[F] Butanoic Acid -
Formation of D (Reduction of Amide):
Treating Butanamide with a strong reducing agent like Lithium Aluminium Hydride (LiAlH4) reduces the carbonyl group (C=O) to a methylene group (CH2), preserving all 4 carbons.
CH3-(CH2)2-CONH2 → (LiAlH4) CH3-(CH2)2-CH2NH2
[D] Butan-1-amine -
Formation of E (Hoffmann Bromamide Degradation):
Treating the amide with Br2 / KOH triggers a "step-down" reaction. The carbonyl carbon is completely removed as a carbonate, leaving a primary amine with one less carbon atom.
CH3-CH2-CH2-CONH2 → (Br2 / KOH) CH3-CH2-CH2-NH2
[E] Propan-1-amine
Practice Questions for JEE & NEET
Make sure you've memorized the reagents driving these reactions with these concept checks.
Question 1: In Step 2, we used KCN to convert the alkyl halide into a Nitrile. What would have been the major product if we had used AgCN instead?
Answer: An Isocyanide (Isocyanopropane)
Reasoning:
- KCN is predominantly ionic. It provides free cyanide ions (CN-) in solution. Since Carbon is a better nucleophilic center than Nitrogen, attack occurs through Carbon, forming a Nitrile (Cyanide).
- AgCN is predominantly covalent. The carbon atom is tightly bound to Silver, leaving the Nitrogen atom free to donate its lone pair. Thus, the attack occurs through Nitrogen, forming an Isocyanide (Isocyanopropane: CH3CH2CH2NC).
Question 2: What are the byproducts formed alongside the primary amine during the Hoffmann Bromamide Degradation (Step 6)?
Answer: Potassium Carbonate, Potassium Bromide, and Water.
The complete balanced chemical equation for the Hoffmann Bromamide Degradation is a frequent target in JEE MCQs. For Butanamide:
R-CONH2 + Br2 + 4KOH → R-NH2 + K2CO3 + 2KBr + 2H2O
Notice that exactly 4 moles of KOH and 1 mole of Br2 are consumed per mole of amide. The "lost" carbonyl carbon ends up in the K2CO3 molecule.
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