Answer :
Answer:
Strong nucleophile in an aprotic solvent favors [tex]S_{N}2[/tex] reaction on 2-bromo-3-methylbutane.
Explanation:
- [tex]S_{N}2[/tex] reaction proceeds through a tetragonal bipyramidal transition state where incoming nucleophile and leaving group simultaneously enter and leave the nucleophilic center.
- If weak nucleophile is used for nucleophilic substitution then detachment of leaving group can take place earlier to attack by nucleophile favoring [tex]S_{N}1[/tex] mechanism.
- If protic solvent is used then, it will make H-bonding with nucleophile. Therefore nucleophilicity is somewhat reduced and the substrate gets enough time to produce carbocation before it gets attacked by nucleophile.
So, strong nucleophile in an aprotic solvent favors [tex]S_{N}2[/tex] reaction on 2-bromo-3-methylbutane.
Strong nucleophile in an aprotic solvent favors reaction on 2-Bromo-3-methyl butane.
What are SN1 and SN2?
The SN1 reaction should proceed via the tetragonal bipyramidal transition state in which the incoming nucleophile and exit the group should enter and leave the center of nucleophilic. In the case when the weak nucleophile should be applied for nucleophilic substitution so detachment of exiting group could be taken place and favored SN2
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