Tricky Ochem Question

[sfoksn]

The reaction of 2-chloro-2-methylhexane with sodium cyanide in ethanol results in?

a. retention of absolute configuration (what does this mean?)
b. formation of racemic mixture
c. elimination of HCl
d. inversion of absolute configuration
e. formation of optically inactive products.


What do you guys think?
Thanks for the input!

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[299678]

I guess it is d) ?

 

[herkulease]

I'm in my review of Ochem but this looks like an SN1 reaction and inversion doesn't occur for sn1 reactions.

I like B. But I can totally be wrong.

So here's my sure fire help. absolutely configuration is the whole R/S designation.

So A says the product configuration for instance was R or S it would still be R or S.

answer D would be if it was R then it would be S and vice versa.

 

[sfoksn]

The answer that they give is E... Isn't it weird?

I agree with you herkulease

I mean, it is tertiary carbon, so practically Sn2 cannot happen even though the nucleophile is strong, due to steric hinderance right?

So I thought it would undergo Sn1 instead, for carbocation, and there will be no shift cuz its already at tertiary carbon, then CN- would attack from either side forming racemic mixture.

Maybe the solution is wrong?

 

[UndergradGuy7]

The reaction of 2-chloro-2-methylhexane with sodium cyanide in ethanol results in?

a. retention of absolute configuration (what does this mean?)
b. formation of racemic mixture
c. elimination of HCl
d. inversion of absolute configuration
e. formation of optically inactive products.


What do you guys think?
Thanks for the input!

Well after you gave the key...

since it is tertiary its sn1.
Inversion of the configuration is R to S etc.
Sn1 does not do inversion only sn2 does.
So its not D.

Not C since you are getting rid of Cl and it will combine with Na.
Not A since you form a carbocation and then it loses its configuration.

B would be possible if the R and S were both possible products.
E would be possible too because a racemic mixture is optically inactive.

So B or E?

 

[herkulease]

ah the wonders of google.

apparently someone posted this exact question. I'm wondering where it comes out(destroyer?) of but yeah I"m gonna quote Geckel

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.

you can read the original post here
http://forums.studentdoctor.net/showthread.php?t=239154

 

[sfoksn]

Haha, thank you.

How stupid, I didn't realize that.

Thank you!

 

[Maygyver]

I guess it is d) ?

D is not possible because there is no chirality in the first place. I believe this will be SN2 because it is a negatively charged nucleophile, but since there is no chirality there will be no inversion.

 

[evanyou]

ah the wonders of google.

apparently someone posted this exact question. I'm wondering where it comes out(destroyer?) of but yeah I"m gonna quote Geckel

you can read the original post here
http://forums.studentdoctor.net/showthread.php?t=239154

yeah I think this one is a good question!
first, CN- strong neucleophile (usually sn2), but undergoes sn1 b/c tertiary.
second, sn1 gives racemic mixture only if it was chiral.

 

[dtran95]

I think B and E are related. Racemic mixture are optically inactive. CN is a strong Nu so it will support SN instead of elimination, plus ethanol is a protic solvent and the reactant is tertiary substituted so this is SN1.
This is such a bad written question

 

[krnkimsung]

Is this from the Kaplan Subject Test?

I think the explanation states that loss of chirality due to SN1 reaction since the nucleophile can attack from either side of SP2 plane of tertiary carbocation.

I've got confused too, so I clarified this with my o-chem professor, and she said something like ... "It might be true that SN1 reaction produce racemic mixtures, however, the general ratio favors inverted configuration."

I am still confused haha

 

[luv8724]

-CN is a strong nucleophile under certain circumstances isnt it?

I don't think it's always the case there..

by the way... right off the bet we know this is Sn1 because of the

tertiary and the fact that Cl is a good leaving group..

and MUST KNOW! Sn1 is optically inactive...

so the answer is E

 

[herkulease]

-CN is a strong nucleophile under certain circumstances isnt it?

I don't think it's always the case there..

by the way... right off the bet we know this is Sn1 because of the

tertiary and the fact that Cl is a good leaving group..

and MUST KNOW! Sn1 is optically inactive...

so the answer is E

racemic mixtures are be definition optically inactive.

But I talked my ochem teacher this and she replied that the reactant was optically in active to begin with. You're not gonna make a optically active product.

But yeah she too said draw it out. you'll see that there's no chiral center so we wouldn't have an R/S designation anyhow. If you can't assign and R/S we can't produce products that have an R/S designation. and that's what racemic mixtures are. equal parts r/s.

So yeah what i quoted from a few years ago a couple of post before is the perfect plan. Draw it out. sometimes it consuming and we can use every second.

More I think about the more I see this question is merely trying to trick us by giving us enough reason to say its an SN1 and it a rush to get easy questions out of the way we pick B.