Mastering the Propene Reaction Chart: Haloalkanes
In Organic Chemistry, one starting material can yield drastically different products depending entirely on the reagent used. One of the most frequently tested molecules in JEE and NEET is Propene (CH3-CH=CH2).
The key to mastering these reactions is knowing when the reagent attacks the double bond (Addition) versus when it attacks the carbon next to the double bond (Allylic Substitution).
Video Tutorial: The 5 Major Propene Reactions
Watch Abhishek Sengar sir from CHEMCA expertly navigate through 5 different reagents applied to Propene, predicting the exact major product and IUPAC name for each.
Breaking Down the Roadmap
Reagents like HBr or Br2/CCl4 break the π bond and add atoms across it. However, reagents like NBS and SO2Cl2 leave the double bond untouched! Instead, they substitute a hydrogen atom on the Allylic Carbon (the sp3 carbon directly attached to the double bond).
Fig: The 5 key reactions of Propene. Notice how Allylic Substitution preserves the double bond!
Naming the Allylic Products Correctly
When naming the products formed from NBS or SO2Cl2, students often make numbering errors. The IUPAC rules state that the double bond has higher priority than the halogen substituent. Therefore, numbering must start from the double bond end:
3CH2(Br) - 2CH = 1CH2
This is why the product is 3-Bromopropene, not 1-Bromopropene!
Practice Questions for JEE & NEET
Examiners will try to trick you using specific reagent conditions. Test your alertness with these two questions!
Question 1: You react Propene with HCl in the presence of a Peroxide. What is the major product formed?
Answer: 2-Chloropropane (Markovnikov Product).
Reasoning (The Trap):
This is a classic examiner trap! The Anti-Markovnikov "Peroxide Effect" (or Kharasch Effect) is strictly limited to HBr. It does not work with HCl or HI because the propagation steps for HCl/HI are endothermic and thermodynamically unfavorable for free radicals. Therefore, even if Peroxide is present, HCl will still follow standard Markovnikov addition, yielding 2-Chloropropane.
Question 2: What is the specific reaction mechanism taking place when Propene is treated with NBS (N-Bromosuccinimide)?
Answer: Free Radical Substitution.
Reasoning:
NBS provides a very low, steady concentration of Bromine radicals (Br•) in the presence of light (hν) or heat. The reaction abstracts a hydrogen atom from the allylic position, forming a highly resonance-stabilized allylic free radical intermediate. The radical then reacts with Br2 to form the substituted product, leaving the double bond intact.
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