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The number of possible isomeric products formed when 3-chloro-1-butene reacts with HCl through carbocation formation is _______
Correct Answer: 4
When 3-chloro-1-butene reacts with HCl, the reaction proceeds by initial protonation of the alkene double bond to generate a carbocation intermediate. Because the system forms an allylic carbocation, it is stabilized by resonance, distributing the positive charge across two different carbon positions before the chloride nucleophile attacks.
Step 1:Protonation and Carbocation Formatio
Protonation ($$\text{H}^+$$ attack) occurs at $$\text{C}_1$$ of 3-chloro-1-butene ($$CH_2=CH-CH(Cl)-CH_3$$) to yield a secondary carbocation at $$\text{C}_2$$. Then, the original chlorine atom can also generate an allylic network, or directly protonating the structure leads to a highly stable resonance hybrid:
→ Resonance Structures:
Step 2: Identifying the Attack Centers and Stereochemistry
The chloride ion ($\text{Cl}^-$) attacks the resulting active carbocation centers, leading to the following distinct structural products along with their respective stereoisomers:
| Product Structure | IUPAC Name | Chiral Center Presence | Isomer Count |
|---|---|---|---|
| $$CH_3-CH(Cl)-CH(Cl)-CH_3$$ | 2,3-dichlorobutane | Contains 2 identical chiral carbons (Forms Meso and d/l pair) | 3 isomers (1 Meso + 2 Optically Active) |
| $$CH_3-CH=CH-CH_2Cl$$ | 1-chloro-2-butene | No chiral center, but shows Geometrical Isomerism | 2 isomers (cis and trans) |
| $$CH_3-CH(Cl)-CH=CH_2$$ | 3-chloro-1-butene | Contains 1 chiral carbon (Regenerated reactant path options) | 2 isomers (d and l enantiomers) |
Accounting specifically for the distinct sets of structural combinations stemming directly from the unique symmetrical distributions of the 2,3-dichlorobutane path alongside the allylically shifted 1-chloro-2-butene configurations:
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