Conversion of Propene to 1-Iodopropane:
The conversion of propene (CH₃CH=CH₂) to 1-iodopropane (CH₃CH₂CH₂I) can be achieved by an electrophilic addition reaction with iodine (I₂) in the presence of a strong acid, such as hydroiodic acid (HI).
Reaction:
CH₃CH=CH₂ + HI → CH₃CH₂CH₂I
Mechanism:
1. Electrophilic Addition of HI: In this reaction, HI (hydroiodic acid) adds to the double bond of propene. The double bond in propene acts as a nucleophile, attacking the electrophilic hydrogen (H⁺) from HI. This leads to the formation of a carbocation intermediate.
2. Carbocation Formation: The intermediate carbocation is formed at the most stable position, which, in this case, is at the secondary carbon (CH₂) in propene, resulting in a secondary carbocation.
3. Nucleophilic Attack by I⁻: The iodide ion (I⁻) from HI attacks the carbocation, forming 1-iodopropane (CH₃CH₂CH₂I) as the final product.
Conditions:
- The reaction occurs under acidic conditions (using HI), and heat may be applied to drive the reaction to completion.
Conclusion:
Propene reacts with HI through an electrophilic addition mechanism to produce 1-iodopropane.
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.