Identify the major product(s) in the reaction of 2 -bromo-3ethyipentane with sodium hydroxide.
I
II
III
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***B*** ∻Key Concept∻ ⚹Nucleophilic Substitution Reactions (SN2)⚹ ∻Explanation∻ ⚹The reaction of 2-bromo-3-ethylpentane with sodium hydroxide (NaOH) is a nucleophilic substitution reaction (SN2). In this reaction, the hydroxide ion (OH⁻) from NaOH acts as a nucleophile and attacks the carbon atom that is bonded to the bromine atom, resulting in the displacement of the bromine and formation of a new carbon-oxygen bond. The major product will be the one where the hydroxide ion has attacked the carbon atom at the position where the bromine was originally attached, leading to the formation of an alcohol. Since the reaction proceeds via an SN2 mechanism, the hydroxide ion attacks from the opposite side of the leaving group, resulting in an inversion of configuration at the carbon atom. Product II corresponds to this description, where the OH group is attached to the second carbon atom, which is the position previously occupied by the bromine atom◊What is the mechanism of the reaction of 2-bromo-3-ethylpentane with sodium hydroxide leading to the formation of the major product(s)?⍭ Generate me a similar question◊⚹

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