SMC2UCLA2_

Senior Member
10+ Year Member
5+ Year Member
May 7, 2004
173
0
Status
The reaction of 2-chloro-2-methylhexane with sodium cyanide in ethanol results in
a. the retention of absolute configuration
b. the formation of a racemic mixture
c. the elimination of HCL
d. the inversion of absolute configuration
e. the formation of optically inactive products


The answer says E but im not sure why B wouldn't be right as well. Looks like this is an SN1 reaction with cyanide as the nucleophile and since this reaction proceeds with a carbocation intermediate, the nucleophile attacks from both sides resulting in a racemix mixture and thus an optically inactive product.

Am i missing something?
 

IndDent

|| Bharat Maata Ki Jai ||
10+ Year Member
5+ Year Member
Aug 29, 2005
174
0
Lost in India
Status
Yeah, I agree with you on that one. It is an SN1 reaction, thus will result in racemic mixture. So that makes both B and E true.
 

geckel

Senior Member
10+ Year Member
5+ Year Member
Nov 8, 2005
277
0
Status
Pre-Dental
SMC2UCLA2_ said:
The reaction of 2-chloro-2-methylhexane with sodium cyanide in ethanol results in
a. the retention of absolute configuration
b. the formation of a racemic mixture
c. the elimination of HCL
d. the inversion of absolute configuration
e. the formation of optically inactive products


The answer says E but im not sure why B wouldn't be right as well. Looks like this is an SN1 reaction with cyanide as the nucleophile and since this reaction proceeds with a carbocation intermediate, the nucleophile attacks from both sides resulting in a racemix mixture and thus an optically inactive product.

Am i missing something?

I see what you are saying.....but if you draw that bad boy out then you will see that there is no chirality(2nd carbon has a methyl and then there is a first carbon also.......basically to methyls attached to the second carbon) and thus E is as correct as you can get. A sample of racemic products is optically inactive but it is because they rotate polarized light in opposite directions and thus cancel each other, so to speak.

Long story short.......draw the molecule and make sure that all four bonds to your carbon are to different groups b/c you can't have enantiomers w/out a chiral center.

Also....you are right about it being an Sn1.
 
OP
SMC2UCLA2_

SMC2UCLA2_

Senior Member
10+ Year Member
5+ Year Member
May 7, 2004
173
0
Status
geckel said:
I see what you are saying.....but if you draw that bad boy out then you will see that there is no chirality(2nd carbon has a methyl and then there is a first carbon also.......basically to methyls attached to the second carbon) and thus E is as correct as you can get. A sample of racemic products is optically inactive but it is because they rotate polarized light in opposite directions and thus cancel each other, so to speak.

Long story short.......draw the molecule and make sure that all four bonds to your carbon are to different groups b/c you can't have enantiomers w/out a chiral center.

Also....you are right about it being an Sn1.
Touche!