382,305 views
23 votes
23 votes
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

User Jpenzer
by
3.4k points

1 Answer

12 votes
12 votes

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

Step-by-step 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
S_(N) _2 mechansim and inversion in configuration happens.

Chloride ion is a strong nucleophile and bromide is a good leaving group. The major-example-1
User HydRAnger
by
3.0k points