# Saytzeff and Hoffman's rule

1. Feb 16, 2015

### Suraj M

Could you give me a link or just reply, which shows the mechanism of the saytzeff and hoffman's rule.
$Also I read somewhere that, when a fluoroalkane is treated with a base like KOH, then the major product is the hoffman product.(against saytzeff rule), how does that happen?? 2. Feb 16, 2015 ### Greg Bernhardt ### Staff: Admin 3. Feb 17, 2015 ### Suraj M still doesn't explain the hoffmann's reactions when fluorine is used.. only in case of fluorine and not for other halogens. 4. Feb 17, 2015 ### Raghav Gupta I think the KOH should be alcoholic, otherwise if aqueous nucleophilic reaction will take place. 5. Feb 17, 2015 ### Suraj M sorry, yeah i forgot to mention that, anyway it is alcoholic, but my question is something else..the fluorine part!! 6. Feb 17, 2015 ### Raghav Gupta Yeah, Now I will answer. Hoffmann elimination takes place when in reactant there is a poor leaving group. Fluorine is very reactive and a poor leaving group. So in fluoroalkanes, when elimination takes place a double bond comes on the less substituted side(against Zaitsev's rule or Saytzeff .) 7. Feb 17, 2015 ### Suraj M Ohh okay, but my teacher actually told us that Hoffmann elimination takes place because of the bulky nature of the catalyst..it doesn't use KOH$(alcoholic)$, but some other heavy alkali in alcohol!! Is that right? 8. Feb 17, 2015 ### Raghav Gupta I think this video will explain it all. The 7-8 minutes of this video are worthy for your question. Please reply or comment afterwards. 9. Feb 17, 2015 ### Suraj M Explains the mechanism of hoffmann elimination, completely, thank you I had to watch this to understand bettter: what i still don't understand is: the guy in the video talks of only sp³-sp³ bonds but what about the intermediate carbanion Does the sp3sp3 bond weakness overule the carbanion stability?? And also when he mentions that F is a poor leaving group, is the difference between the ability of$Cl $and$F$to leave that different? 10. Feb 17, 2015 ### Raghav Gupta F is a poor leaving group than Cl because of it's small size. You may know that HI is more acidic than HCl which is more acidic than HF, because I is having more size than H , it is kind of easy for H to leave I because the nucleus of both of them are very far accounting for less electrostatic attraction. Similarly if we compare C and F both have same size and both have more electrostatic attraction making F poor leaving group. Now in elimination reactions we should have polar protic solvent, as F is poor leaving group, H of the compound is attracted to base, making a carbanion. Now as discussed in video 1° carbanion is more stable because it is having less steric hindrance and H+ of water can easily come to it. Not understanding your sp3-sp3 overruling. Can you elaborate? 11. Feb 17, 2015 ### Raghav Gupta I suggest you to look from another way apart from video. In Zaitsev's elimination carbocation is formed, and 3° carbocation are more stable because of hyper conjugation. So here obviously more substituted side will the double bond formed. In case of Hoffmann elimination carbanion is formed and 1° carbanion is more stable because of less steric hindrance. So here obviously less substituted side will the double bond formed. Main deciding factor for Zaitsev and Hoffmann- Leaving group. 12. Feb 17, 2015 ### Suraj M Raghav, that thing about sp3sp3 bond is in the video i posted$Post~9##
yes i understand that 1°carbanion is more stable thats 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.
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. Feb 17, 2015

### Raghav Gupta

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.

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 .

14. Feb 18, 2015

### Suraj M

wow, okay, that makes a lot of sense, thanks a lot.

15. Feb 18, 2015

### Raghav Gupta

You are welcome.
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.

Last edited: Feb 18, 2015
16. Feb 19, 2015

### Suraj M

you and me both!
12th? CBSE?

17. Feb 19, 2015

Yeah.