Understanding the Saytzeff and Hoffman Rule in Organic Chemistry: An Explanation

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Discussion Overview

The discussion revolves around the Saytzeff and Hoffmann rules in organic chemistry, particularly focusing on the mechanisms and outcomes of reactions involving fluoroalkanes treated with bases like KOH. Participants explore the conditions under which the Hoffmann product is favored over the Saytzeff product, and the implications of fluorine as a leaving group.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants request clarification on the mechanisms of the Saytzeff and Hoffmann rules, particularly in the context of reactions involving fluorine.
  • There is a suggestion that the major product from the reaction of fluoroalkanes with KOH is the Hoffmann product, which contradicts the Saytzeff rule.
  • One participant notes that Hoffmann elimination occurs due to the poor leaving group nature of fluorine, leading to double bond formation on the less substituted side.
  • Another participant mentions that Hoffmann elimination may also depend on the bulky nature of the catalyst used, suggesting that different bases might yield different results.
  • Concerns are raised about the stability of carbanions versus carbocations in these reactions, with discussions on how this affects the formation of double bonds.
  • Participants discuss the role of solvent polarity and the stability of intermediates in determining the outcome of the reactions.
  • There is a distinction made between the kinetic favorability of Hoffmann reactions versus the thermodynamic favorability of Saytzeff reactions.

Areas of Agreement / Disagreement

Participants express differing views on the mechanisms and conditions affecting the outcomes of the reactions, particularly regarding the role of fluorine as a leaving group and the nature of the base used. The discussion remains unresolved with multiple competing perspectives presented.

Contextual Notes

Participants note the importance of reaction conditions, such as the type of base and solvent used, which may influence the products formed. There are also references to the stability of intermediates, but these points are not fully resolved.

Who May Find This Useful

Students and enthusiasts of organic chemistry, particularly those interested in reaction mechanisms and the effects of different leaving groups in elimination reactions.

Suraj M
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Could you give me a link or just reply, which shows the mechanism of the saytzeff and hoffman's rule.
##Also I read somewhere that, when a fluoroalkane is treated with a base like KOH, then the major product is the hoffman product.(against saytzeff rule), how does that happen??
 
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still doesn't explain the hoffmann's reactions when fluorine is used.. only in case of fluorine and not for other halogens.
 
Suraj M said:
##Also I read somewhere that, when a fluoroalkane is treated with a base like KOH, then the major product is the hoffman product.(against saytzeff rule), how does that happen??
I think the KOH should be alcoholic, otherwise if aqueous nucleophilic reaction will take place.
 
Raghav Gupta said:
I think the KOH should be alcoholic, otherwise if aqueous nucleophilic reaction will take place.
sorry, yeah i forgot to mention that, anyway it is alcoholic, but my question is something else..the fluorine part!
 
Suraj M said:
sorry, yeah i forgot to mention that, anyway it is alcoholic, but my question is something else..the fluorine part!
Yeah, Now I will answer.
Hoffmann elimination takes place when in reactant there is a poor leaving group. Fluorine is very reactive and a poor leaving group.
So in fluoroalkanes, when elimination takes place a double bond comes on the less substituted side(against Zaitsev's rule or Saytzeff .)
 
Ohh okay, but my teacher actually told us that Hoffmann elimination takes place because of the bulky nature of the catalyst..it doesn't use KOH ##(alcoholic)##, but some other heavy alkali in alcohol! Is that right?
 
I think this video will explain it all. The 7-8 minutes of this video are worthy for your question.

Please reply or comment afterwards.
 
Explains the mechanism of hoffmann elimination, completely, thank you
I had to watch this to understand bettter:

what i still don't understand is: the guy in the video talks of only sp³-sp³ bonds but what about the intermediate carbanion
WIN_20150217_143700.JPG

Does the sp3sp3 bond weakness overule the carbanion stability??
And also when he mentions that F is a poor leaving group, is the difference between the ability of ##Cl ## and ##F## to leave that different?
 
  • #10
F is a poor leaving group than Cl because of it's small size.
You may know that HI is more acidic than HCl which is more acidic than HF, because I is having more size than H , it is kind of easy for H to leave I because the nucleus of both of them are very far accounting for less electrostatic attraction.
Similarly if we compare C and F both have same size and both have more electrostatic attraction making F poor leaving group.

Now in elimination reactions we should have polar protic solvent, as F is poor leaving group, H of the compound is attracted to base, making a carbanion. Now as discussed in video 1° carbanion is more stable because it is having less steric hindrance and H+ of water can easily come to it.

Not understanding your sp3-sp3 overruling. Can you elaborate?
 
  • #11
I suggest you to look from another way apart from video.
In Zaitsev's elimination carbocation is formed, and 3° carbocation are more stable because of hyper conjugation. So here obviously more substituted side will the double bond formed.

In case of Hoffmann elimination carbanion is formed and 1° carbanion is more stable because of less steric hindrance. So here obviously less substituted side will the double bond formed.
Main deciding factor for Zaitsev and Hoffmann-
Leaving group.
 
  • #12
Raghav, that thing about sp3sp3 bond is in the video i posted ##Post~9##
yes i understand that 1°carbanion is more stable that's why i was asking, in Zaitsev's elimination the carbanion formed is tertiary, which is very unstable, watch the second part of the video i posted.
He never mentions anything about 1° or 2° carbanions in the mechanism of zaitsev's elimination.
Raghav Gupta said:
I suggest you to look from another way apart from video.
In Zaitsev's elimination carbocation is formed, and 3° carbocation are more stable because of hyper conjugation. So here obviously more substituted side will the double bond formed.
Yes, first carbocation is formed but then there is a carbanion formed before the double bond is formed!
I'm talking about the stability of that carbanion because it decides the position of the new double bond!
 
  • #13
Now I understand your problem.
Well it took me much time to figure out the answer to your problem, as it is a good one.
In Zaitsev process the charge is not residing for a long time, and so the carbocation and carbanion quickly makes double bond accordingly in which orbitals are stable, so here not to look for the carbanion stability case.
Now you would have the question in your mind that then why the guy in video is talking of carbanion stability in Hoffmann video or the video that I have posted. Well the reaction is slow there because of poor leaving group fluorine.
Remember Hoffmann reaction is kinematically favored while Zaitsev thermodynamically.
If you carefully see the video of mine in the last seconds the guy is saying that the major product is less stable and why it is formed is because of kinematics.(This time allows to choose the side where carbanion is more stable.) As if we look at orbitals in Hoffmann reaction then major product is less stable.

I would also answer this
Suraj M said:
Ohh okay, but my teacher actually told us that Hoffmann elimination takes place because of the bulky nature of the catalyst..it doesn't use KOH ##(alcoholic)##, but some other heavy alkali in alcohol! Is that right?
Your teacher is correct.
Hoffmann elimination depends on 2 things.
More the bulkier the alkali is , more the steric hindrance for reaching H in compound which is analogue to F a poor leaving group both taking enough time .
 
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  • #14
:smile: wow, okay, that makes a lot of sense, thanks a lot.
 
  • #15
You are welcome.:smile:
Actually this thread helped me also in understanding some of the concepts as I am a student preparing for the exams.
Really like your signature or quote by Neil's Bohr.
The mistakes I have done has also helped me gain some more concepts and experience.
Can you do me a favour of clearing the question in thread?
You can consider it as Benjamin effect( See in wikipedia) also and I desperately need the answer of question in thread.
https://www.physicsforums.com/threads/why-reducing-nature-of-hydrides-increases-down-a-group.798142/
 
Last edited:
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  • #16
Raghav Gupta said:
I am a student preparing for the exams.
you and me both!
12th? CBSE?
 
  • #17
Yeah.
 

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