Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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 _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane

Sagot :

Answer:

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 _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane

Explanation:

The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:

The answer is (R)-2-chlorobutane.

The reaction take splace through [tex]S_{N} _2[/tex] mechansim and inversion in configuration happens.

View image almatheia