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Post by Mr Kwok on Mar 5, 2009 10:21:57 GMT -5
Hi everyone,
Let's now consider what is happening. When CH3CH2CH(Cl)CH3 is heated with KCN (alc), the product is able to rotate polarised light.
However when C6H5CH(Cl)CH3 is heated with KCN (alc), it is observed that a racemic mixture is obtained.
Suggest observations for this reaction.
Regards
Mr Kwok
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Post by seahyimin on Mar 6, 2009 21:27:07 GMT -5
When CH3CH2CH(Cl)CH3 is heated with KCN (alc), Sn2 took place. There is the inversion of the molecule and all the products rotate polarised light in the same direction.
When C6H5CH(Cl)CH3 is heated with KCN (alc), only Sn1 took place. Sn2 cannot take place because the nucleophile is repelled by the electron-rich benzene ring, and hence cannot attack the C-Cl bond from the back. The intermediate is formed when chlorine takes the 2 bonding electrons away to form a chlorine anion. Sn1 can take place because the intemediate is stable. Electrons can flow from the benzene ring to the carbocation to stablise it. After the carbocation is formed, the nucleophile can attack from both sides of the trigonal planar carbocation and so both optical isomers are formed. Hence a racemic mixture is formed.
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Post by joanna on Mar 7, 2009 0:31:37 GMT -5
CH3CH2CH(Cl)CH3 undergoes Sn2 reaction whilst C6H5CH(Cl)CH3 undergoes Sn1 reaction. The Sn2 reaction would form inversions of CH3CH2CH(CN)CH3 that would be able to polarise light. On the other hand, The Sn1 reaction would form a racemic mixture of C6H5CH(CN)CH3 as there is a chiral center(underlined) which would result in a pair of optical isomer.
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des
New Member
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Post by des on Mar 7, 2009 3:28:06 GMT -5
When CH3CH2CH(Cl)CH3 is heated with KCN (alc), Sn2 occurs and CN- can only attack from the back, resulting in an inversion of all the products. Hence, the product, being chiral and thus, optically active, will be able to rotate polarised light in the same direction.
In the case of C6H5CH(Cl)CH3, Sn1 has taken place. Sn2 cant take place because the nucleophile, CN-, is unable to approach the partial positive C due to electron repulsion from the electron rich benzene ring. As Sn1 has occurred, the nucleophile, CN-, will be able to attack the carbocation intermediate from both above and below the plane as the intermediate is planar. As such, the products are a pair of chiral enatiomer produced in 1:1 ratio (probability of above and below attack is 50-50). Hence, a racemic mixture is obtained.
des
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Post by Mr Kwok on Mar 7, 2009 8:33:47 GMT -5
hi, you have read the views from David, Joanna and Desmond. All three of them raised similar points. Are there any further points to add? Whose answer would you think is the most comprehensive and why?
Mr Kwok
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Post by belinda on Mar 9, 2009 6:21:28 GMT -5
hello
I think all of them are correct about the SN2 mechanism for CH3CH2CH(Cl)CH3 producing a product that has a mirror-image of the reactant.
I think yi-min's point about the stability of the carbocation is good and desmond's point on the ratio of chiral products form is also an important point.
just a side qn.. not inrelation to SN1 nucleophilic substitution..
can a racemic mixture be a mixture of equal amounts of more than 2 chiral enantiomers?
(i.e. can a racemic mixture be a mixture of 25:25:25:25 of 2 pairs of enantiomers?)
belinda
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kay
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Post by kay on Mar 9, 2009 9:50:40 GMT -5
hello:)
i think yimin's answer for the reaction of C6H5CH(Cl)CH3 is the most comprehensive cos he raised the point that the intermediate is stable, so a carbo cation can be formed, and that's why sn1 can take place, and therefore resulting in a racemic mixture to be formed. since that is an important characteristic of sn1, that's why i think it's pretty good justification for the explanation of a racemic mixture.
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Post by seahyimin on Mar 13, 2009 8:36:54 GMT -5
belinda: I suspect that either the mixture of 2 pairs of enantiomers of similar concentration is racemic. Either that or you get an opaque mixture. If you have 4 unrelated enantiomers that rotate light in the same direction, you'll probably get optical rotation in 1 direction(the magnetude of rotation may be the average of the 4, probably not the scalar sum).
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Post by Mr Kwok on Mar 13, 2009 9:43:00 GMT -5
belinda: I suspect that either the mixture of 2 pairs of enantiomers of similar concentration is racemic. Either that or you get an opaque mixture. If you have 4 unrelated enantiomers that rotate light in the same direction, you'll probably get optical rotation in 1 direction(the magnetude of rotation may be the average of the 4, probably not the scalar sum). Hi Yi-Min, could you clarify on the opaque? |
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Post by belinda on Mar 15, 2009 1:22:28 GMT -5
sorry yi-min, i don't quite get what u said..
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Post by seahyimin on Mar 29, 2009 7:19:42 GMT -5
The opaque thing is just a guess, I didn't go research on it whatsoever. I better not comment on how I arrived at it because it will sound very weird lol.
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grace
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Post by grace on Aug 16, 2013 15:32:13 GMT -5
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grace
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Post by grace on Sept 12, 2013 13:09:33 GMT -5
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grace
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Post by grace on Dec 4, 2013 11:47:59 GMT -5
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grace
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Post by grace on Jan 8, 2014 12:13:40 GMT -5
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