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Chloride ion is a strong nucleophile and bromide is a good leaving group. the major product of treating (s)-2-bromobutane with nacl in ch3c(o)ch3 (acetone) is _________.

User Patan
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Final answer:

The major product of treating (S)-2-bromobutane with NaCl in acetone is (S)-2-chlorobutane, resulting from an SN2 reaction mechanism that proceeds with inversion of configuration due to the strong nucleophilicity of the chloride ion in polar aprotic solvent conditions.

Step-by-step explanation:

The question asks about the chemical reaction of (S)-2-bromobutane with NaCl in an acetone solvent, which involves a nucleophilic substitution. Chloride ion is a strong nucleophile, which means it is likely to attack the carbon atom that is bonded to a good leaving group, in this case, the bromide ion. Since acetone is a polar aprotic solvent, it enhances the reactivity of the nucleophile and does not stabilize the leaving group. The reaction mechanism is likely SN2, characterized by a bimolecular, one-step process where the nucleophile attacks the electrophilic carbon and displaces the leaving group, resulting in the inversion of configuration at that carbon.

The major product of treating (S)-2-bromobutane with NaCl in acetone is therefore (S)-2-chlorobutane. The reaction proceeds with inversion of stereochemistry, and because the starting material is in the (S)-configuration, the product will maintain this configuration due to the backside attack by the chloride ion.

User Alexlod
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