November 29, 2020, 08:21:24 AM
Forum Rules: Read This Before Posting


Topic: Neighbouring group participation (anchimeric): acetolysis  (Read 843 times)

0 Members and 1 Guest are viewing this topic.

Offline xshadow

  • Full Member
  • ****
  • Posts: 300
  • Mole Snacks: +1/-0
Neighbouring group participation (anchimeric): acetolysis
« on: February 13, 2020, 08:08:12 AM »
Hi
I have a doubt   this example :


The  neighbouring group participation helps the reaction
But this effect is stronger in the "Trans" isomer than tje cis one

Now I  have some doubts:

1)is thecetolysis  reaction a sn2 or sn1  for cis and trans isomer?

2) the  neighbouring group participation is stronger  in the trans isomer with a positive charged- cyclic intermediate

Now (3)  CAN this positive charged intermediate    be formed also in the intermediate (if it is a sn1?)of cis
reagent ?

Thanks!

Offline pgk

  • Chemist
  • Full Member
  • *
  • Posts: 892
  • Mole Snacks: +97/-24
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #1 on: February 14, 2020, 10:58:48 AM »
1). - Nucleophilic substitution of primary leaving groups, goes by SN2 mechanism.
- Nucleophilic substitution of tertiary leaving groups, goes by SN1 mechanism.
- But nucleophilic substitution of secondary leaving groups, may go either by SN2 or SN1 mechanism, depending on the reaction governing factors.
In this case:
- Acetolysis of the cis- isomer goes by SN2 mechanism, as usual.
- But acetolysis of the trans- isomer by SN2 mechanism, is not favored due to steric hindrance and thus, it finally goes by SN1 mechanism.
2). There is no neighbouring group participation in the case of cis- isomer.
But in the case of trans- isomer, neighbouring group participation stabilizes the so formed carbocation via the positive charged-cyclic intermediate, making acetolysis 670 folds faster.
3). No! A positive charged intermediate cannot be formed in the case of cis- isomer because carbocations are not formed during SN2 nucleophilic substitution.

Offline pgk

  • Chemist
  • Full Member
  • *
  • Posts: 892
  • Mole Snacks: +97/-24
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #2 on: February 14, 2020, 11:48:49 AM »
Question: SN1 nucleophilic substitution leads to a mixture of enantiomers. But acetolysis of this trans- isomer, is stereospecific. Why?

Offline xshadow

  • Full Member
  • ****
  • Posts: 300
  • Mole Snacks: +1/-0
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #3 on: February 14, 2020, 06:35:10 PM »
1). - Nucleophilic substitution of primary leaving groups, goes by SN2 mechanism.
- Nucleophilic substitution of tertiary leaving groups, goes by SN1 mechanism.
- But nucleophilic substitution of secondary leaving groups, may go either by SN2 or SN1 mechanism, depending on the reaction governing factors.
In this case:
- Acetolysis of the cis- isomer goes by SN2 mechanism, as usual.
- But acetolysis of the trans- isomer by SN2 mechanism, is not favored due to steric hindrance and thus, it finally goes by SN1 mechanism.
2). There is no neighbouring group participation in the case of cis- isomer.
But in the case of trans- isomer, neighbouring group participation stabilizes the so formed carbocation via the positive charged-cyclic intermediate, making acetolysis 670 folds faster.
3). No! A positive charged intermediate cannot be formed in the case of cis- isomer because carbocations are not formed during SN2 nucleophilic substitution.
Very helpful
Only one doubt:

In acetolysis  is  the nucleophile CH3COO- or CH3COOH ?

Because the tosylate the leaving group in tue cis-isomer is bounded to a secondary carbon

So I could have Sn1  or Sn2.
Sn1 if the nucleophile is acetic acid ("""weak""" nucleophile)
Sn2 if the nucleophile is acetate (charged ,good nucleophile)

Here in the cis isomer  is CH3COO- the nucleophile rather than acetic acid?


Thanks

Offline xshadow

  • Full Member
  • ****
  • Posts: 300
  • Mole Snacks: +1/-0
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #4 on: February 14, 2020, 06:38:07 PM »
Question: SN1 nucleophilic substitution leads to a mixture of enantiomers. But acetolysis of this trans- isomer, is stereospecific. Why?

Because I have the anchimeric assistance??...and the formation of a ciclyc  intermediate...." attack on the opposite site of the cycle?

Offline pgk

  • Chemist
  • Full Member
  • *
  • Posts: 892
  • Mole Snacks: +97/-24
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #5 on: February 17, 2020, 01:45:40 PM »
1). In both cases, CH3COO- is the nucleophile and CH3CO2H is the solvent. The concentration of the nucleophile remains constant during the reaction due to the equilibrium between the acid solvent (CH3CO2H) and the conjugated base (nucleophile, CH3COO-).
2). The tosylate, leaving group is bounded to a secondary carbon in both cases.
3). SN2 nucleophilic attack is always anti- and accompanied with inversion of the chirality in a chiral molecule (Walden inversion); which is not possible in the trans- isomer due to the steric hindrance of the existing acetate substituent and which forces to SN1 nucleophilic substitution.
Walden inversion - Wikipedia
https://en.wikipedia.org/wiki/Walden_inversion
4). Contrary, the formed carbocation during SN1 nucleophilic substitution leads to a mixture of enantiomers because the nucleophile can equally attack from both sides of the carbocation; which is not possible in the trans- isomer due to the steric hindrance of the existing acetate substituent that leads to stereospecific formation of the one enantiomer, only (and which is the correct answer to the posted question).


Offline xshadow

  • Full Member
  • ****
  • Posts: 300
  • Mole Snacks: +1/-0
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #6 on: February 17, 2020, 05:00:47 PM »
1). In both cases, CH3COO- is the nucleophile and CH3CO2H is the solvent. The concentration of the nucleophile remains constant during the reaction due to the equilibrium between the acid solvent (CH3CO2H) and the conjugated base (nucleophile, CH3COO-).
2). The tosylate, leaving group is bounded to a secondary carbon in both cases.
3). SN2 nucleophilic attack is always anti- and accompanied with inversion of the chirality in a chiral molecule (Walden inversion); which is not possible in the trans- isomer due to the steric hindrance of the existing acetate substituent and which forces to SN1 nucleophilic substitution.
Walden inversion - Wikipedia
https://en.wikipedia.org/wiki/Walden_inversion
4). Contrary, the formed carbocation during SN1 nucleophilic substitution leads to a mixture of enantiomers because the nucleophile can equally attack from both sides of the carbocation; which is not possible in the trans- isomer due to the steric hindrance of the existing acetate substituent that leads to stereospecific formation of the one enantiomer, only (and which is the correct answer to the posted question).
Okk for 1),2),3) answer...clear!

For 4) I think I've understood :
The acetate substituent of trans isomer forms a "bridge structure" (the neighbouring group assistance)  so the nucleophile can only attack from the opposite side


Is it correct?
Thanks!!!

Offline pgk

  • Chemist
  • Full Member
  • *
  • Posts: 892
  • Mole Snacks: +97/-24
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #7 on: February 18, 2020, 10:07:57 AM »
Yes and mainly as positive charged, cyclic intermediate.

Offline xshadow

  • Full Member
  • ****
  • Posts: 300
  • Mole Snacks: +1/-0
Re: Neighbouring group participation (anchimeric): acetolysis
« Reply #8 on: February 18, 2020, 12:41:46 PM »
Yes and mainly as positive charged, cyclic intermediate.
Thanks a lot!!

Sponsored Links