Sodium methoxide is a strong base and nucleophile. What would be the product of an S,2 reaction of sodium methoxide and ethyl iodide? Write a chemical reaction with organic structures.
The Correct Answer and Explanation is :
In an (S_N2) reaction between sodium methoxide (NaOCH₃) and ethyl iodide (C₂H₅I), the product will be methanol (CH₃OH) and ethyl methoxide (C₂H₅OCH₃).
Chemical Reaction:
[
\text{C}_2\text{H}_5\text{I} + \text{CH}_3\text{O}^-\text{Na} \rightarrow \text{C}_2\text{H}_5\text{OCH}_3 + \text{I}^-
]
Explanation:
An (S_N2) (bimolecular nucleophilic substitution) reaction involves the following steps:
- Nucleophilic Attack: Sodium methoxide (CH₃O⁻), which is a strong nucleophile, attacks the carbon attached to the iodine in ethyl iodide (C₂H₅I). The carbon in C₂H₅I is electrophilic due to the polar nature of the C-I bond, with the iodine being a good leaving group. The nucleophilic methoxide ion (CH₃O⁻) uses its lone pair of electrons to form a new bond with the carbon atom.
- Transition State: The reaction proceeds through a transition state in which the bond between carbon and iodine is partially broken, and the bond between carbon and methoxide is partially formed. This is characteristic of the (S_N2) mechanism, which involves a backside attack by the nucleophile (the methoxide ion).
- Inversion of Configuration: Since the attack occurs from the opposite side of the leaving group, the stereochemistry of the carbon center in ethyl iodide undergoes inversion. However, in this case, the carbon is not chiral, so this inversion is not important for the overall outcome.
- Leaving Group Departure: The iodine ion (I⁻), a good leaving group, departs once the new bond is fully formed, leaving behind ethyl methoxide (C₂H₅OCH₃) as the main product.
Thus, the reaction produces ethyl methoxide (C₂H₅OCH₃) and iodide ion (I⁻), with methanol (CH₃OH) being released due to the involvement of sodium methoxide.