Cannizzaro? More like bizarro, because I can't remember the last time I saw a hydride kicked out like that! Thanks again for sharing all of this knowledge.
@@abdullabohra914 I'm always happy to oblige! By the way, if you ever want to hear a pun on a specific topic just let me know. Q: What do chemists do with boomboxes? A: Stereochemistry! 📾
@@godson200 he is awesome. but unfortunately its highly time consuming to watch his videos. one entire chapter is 15 hours long. and that is without pausing for taking notes and other stuff.
@@nrnjn8547 watch at 2x dude.. Obviously he doesnt teach chemistry anymore but he really teaches in detail every topic... And personally it never took me any time to pause the videos..
@@godson200 well I dont have a good memory, so I need to pause to take all the info I just gathered, that's why when I'm on a time constraint, I dont watch him. If i start watching him a year before my finals then maybe i can watch all of them. But unfortunately he never finished the last 2 chapters before my physics boards. So I went to other sources
I remember from orgo 1 that after a carbonyl group is attacked, it gets reformed by expelling an atom but I am pretty sure that a hydrogen atom (hydride ion) or a carbon atom shouldn't be expelled. I think I read this rule from "Organic chemistry as a second language" so my question is why in this reaction can we expell a hydrogen? Great content as always.
It appears the expulsion of the H is coupled to the reduction of the benzaldehyde. If reaction 1 + reaction 2 create a net - enthalpy, it’s still favored even if the one step is not. Very odd and rare indeed.
@@Hematite-Heart just a guess here from my experience...I think that most of the time the lone pair kicks out the hydroxide group that just attacked and this is an equilibrium reaction. However, once in a while it will kick out the hydride which is irreversible because the redox is favorable. Thus, over time more and more product 'leaks' past the unfavorable hydride loss
The mechanism proposed in this video is correct except for one step. When a hydroxide anion attacks a benzaldehyde the tetrahedral intermediate is formed. Then it can expel either hydroxide (that leads to the starting benzaldehyde and hydroxide) or hydride. Of course, expelling a hydride is incredibly improbable process, it is almost never a nucleophile but a base. And this is where a mistake is hidden. The real thing is that the tetrahedral intermediate can be deprotonated again forming a dianion. So you have now two oxygens with a minus charge attached to the same carbon atom. This process is much more likely to happen than losing a hydride. That’s why this reaction requires a very strong and concentrated base (like 50% KOH). After that it seems pretty obvious why this double charged intermediate expels exactly H(-) and not O(2-) or Ph(-). You can read more about the mechanism of Cannizzaro reaction in Organic Chemistry by J. Clayden et al.
@@tolikb8701 that makes a lot of sense-I understand how the hydride expulsion is irreversible and will disrupt prior equilibria But what prevents the hydride anion from immediately deprotonating the benzoic acid, creating benzoate and H2 gas irreversibly leaving the system? Does this happen to, but occasionally the hydride anion will react with the other benzaldehyde? Is benzaldehyde’s carbonyl electropositive enough to drive the hydride to attack it nucleophilicly? This really is a curious mechanism
you didn't include this in your aldehydes and ketones playlist that i got worried that nobody will properly explain this to me. thank god that's not the case.
Iis there any possible side product of this reaction? since my benzaldehyde has no color before I mix them, but the mixtures become yellow after rest for 24 hours.
Can you please tell me why the Hydrogen proton that was dispelled by the reaction intermediate attacks the carbonyl group of the other aldehyde. The carbon of the carbonyl group is electrophilic in nature so why is it accepting another electrophile (H+ proton)?
What is the difference that occurs when sodium hydroxide base and potassium hydroxide are used in the cannizzal reaction and how can the products of the reaction be separated
What is the driving force behind this reaction? I’m curious what drives the hydride anion to be ejected as a leaving group-it seems like a very unfavorable LG so I’m just curious why this reaction proceeds how it does
It's cuz the other molecule is an aldehyde too, also the alkoxide ion is then stabilised by resonance, and the formation of alkoxide outweighs the energy required for departure of hydride there can be more reasons like nucleophilic assistance of other aldehyde in breaking the CH bond @vcube1234 @@bhumikagupta5071
aldehyde is susceptible to nucliphielic addition. Because carbonyl carbon has a partial positive charge and the nucliphielic molecule (which is -OH) has a negative charge. So it attacks the carbon
I THINK KHOSI SAID SHE HAS BOYFRIEND OUTSIDE, WHAT HAPPENS TO HIS BOYRIEND OUTSIDE, KHOSI CAME TO DIS SHOW TO PLAY BOTH THE HOUSEMATES AND WE THE VIEWERS
Organic Chemistry - Video Lessons: www.video-tutor.net/organic-chemistry.html
Cannizzaro? More like bizarro, because I can't remember the last time I saw a hydride kicked out like that! Thanks again for sharing all of this knowledge.
bizarro reaction😂😂🤣
@@harishchad. I do what I can.
@boiledpotatoe I have so many more bad “puns” if you’d ever like to hear them.
@@PunmasterSTPlife's boring right now I would like to hear them😅
@@abdullabohra914 I'm always happy to oblige! By the way, if you ever want to hear a pun on a specific topic just let me know.
Q: What do chemists do with boomboxes?
A: Stereochemistry! 📾
Any jee aspirant?
🤚
👍
Yea !!
W
Neet aspirant here 🙋🏽♀️
I wish India had such teachers. Huge respect to u sir..😊
Physics wallah
@@godson200 he is awesome. but unfortunately its highly time consuming to watch his videos. one entire chapter is 15 hours long. and that is without pausing for taking notes and other stuff.
@@nrnjn8547 watch at 2x dude.. Obviously he doesnt teach chemistry anymore but he really teaches in detail every topic... And personally it never took me any time to pause the videos..
@@godson200 well I dont have a good memory, so I need to pause to take all the info I just gathered, that's why when I'm on a time constraint, I dont watch him. If i start watching him a year before my finals then maybe i can watch all of them. But unfortunately he never finished the last 2 chapters before my physics boards. So I went to other sources
#agerastudyclub videos hv too good quality and explained very politely. They do their best and will continue.
Thanks to your video I'm doing a great review for my exams next week. Meanwhile Im improving my English skills 😅💪 thanks you!
Always grateful for your great work sir! You’re such a complete and great scientist!
The way u teach was awesome sir . thank you so much sir .🤩🙏
Thanks mate for uploading this.
I remember from orgo 1 that after a carbonyl group is attacked, it gets reformed by expelling an atom but I am pretty sure that a hydrogen atom (hydride ion) or a carbon atom shouldn't be expelled. I think I read this rule from "Organic chemistry as a second language" so my question is why in this reaction can we expell a hydrogen? Great content as always.
It appears the expulsion of the H is coupled to the reduction of the benzaldehyde. If reaction 1 + reaction 2 create a net - enthalpy, it’s still favored even if the one step is not. Very odd and rare indeed.
@@Hematite-Heart just a guess here from my experience...I think that most of the time the lone pair kicks out the hydroxide group that just attacked and this is an equilibrium reaction. However, once in a while it will kick out the hydride which is irreversible because the redox is favorable. Thus, over time more and more product 'leaks' past the unfavorable hydride loss
The mechanism proposed in this video is correct except for one step. When a hydroxide anion attacks a benzaldehyde the tetrahedral intermediate is formed. Then it can expel either hydroxide (that leads to the starting benzaldehyde and hydroxide) or hydride. Of course, expelling a hydride is incredibly improbable process, it is almost never a nucleophile but a base. And this is where a mistake is hidden. The real thing is that the tetrahedral intermediate can be deprotonated again forming a dianion. So you have now two oxygens with a minus charge attached to the same carbon atom. This process is much more likely to happen than losing a hydride. That’s why this reaction requires a very strong and concentrated base (like 50% KOH). After that it seems pretty obvious why this double charged intermediate expels exactly H(-) and not O(2-) or Ph(-). You can read more about the mechanism of Cannizzaro reaction in Organic Chemistry by J. Clayden et al.
@@tolikb8701 that makes a lot of sense-I understand how the hydride expulsion is irreversible and will disrupt prior equilibria
But what prevents the hydride anion from immediately deprotonating the benzoic acid, creating benzoate and H2 gas irreversibly leaving the system?
Does this happen to, but occasionally the hydride anion will react with the other benzaldehyde? Is benzaldehyde’s carbonyl electropositive enough to drive the hydride to attack it nucleophilicly?
This really is a curious mechanism
Awesome explaination
Thanks sir , for your good teaching now I'm understood well
Thankyou for the free content keep up
Thank you so much ❤️ love from india 👍❤️❤️
you didn't include this in your aldehydes and ketones playlist that i got worried that nobody will properly explain this to me. thank god that's not the case.
Hello Sir, Your videos are awesome. Can you make videos on organic chemistry experiments? I'll really appreciate.
Helped a lot bro.. thnx😘
Iis there any possible side product of this reaction? since my benzaldehyde has no color before I mix them, but the mixtures become yellow after rest for 24 hours.
Can you please tell me why the Hydrogen proton that was dispelled by the reaction intermediate attacks the carbonyl group of the other aldehyde. The carbon of the carbonyl group is electrophilic in nature so why is it accepting another electrophile (H+ proton)?
It is not a hydrogen proton, but a hydride anion. The carbon of carbonyl group is electrophilic thus it gets attacked by the hydride anion.
It's hydride that's the speciality of this rxn hydride gets the kick this time and not H+ funny right😂😂
What is the difference that occurs when sodium hydroxide base and potassium hydroxide are used in the cannizzal reaction and how can the products of the reaction be separated
What is the driving force behind this reaction? I’m curious what drives the hydride anion to be ejected as a leaving group-it seems like a very unfavorable LG so I’m just curious why this reaction proceeds how it does
same question!!if you have the answer,please let me know if you know!!!
It's cuz the other molecule is an aldehyde too, also the alkoxide ion is then stabilised by resonance, and the formation of alkoxide outweighs the energy required for departure of hydride there can be more reasons like nucleophilic assistance of other aldehyde in breaking the CH bond @vcube1234 @@bhumikagupta5071
thanks so much
Thanks
𝙃𝙤𝙬 𝙢𝙖𝙣𝙮 𝙬𝙖𝙩𝙘𝙝𝙞𝙣𝙜 𝙛𝙧𝙤𝙢 𝙄𝙣𝙙𝙞𝙖? 𝙇𝙞𝙠𝙚
So further co2 and benzene ring will also be formed?
Thank you sir
🤔 so it wont ended with two alcohol ? why the O- wont ended up same way as the other alcohol?
Equilibrium dood
THANK YOUU
why would the oh- attack the carbon and not the hydrogen
Due to inductive effect of oxygen carbon will gain partial positive charge and oH is negatively charged therefore the are more reactive
aldehyde is susceptible to nucliphielic addition.
Because carbonyl carbon has a partial positive charge and the nucliphielic molecule (which is -OH) has a negative charge. So it attacks the carbon
Wow sir
اعمل ترجمه عربي وسوف ندعمك با المشا هدات في كل الفيديوهات تبعك
Just learn english
تم تحديث اليوتيوب وتوفرت الترجمة
Good 🙂
Good
I THINK KHOSI SAID SHE HAS BOYFRIEND OUTSIDE, WHAT HAPPENS TO HIS BOYRIEND OUTSIDE, KHOSI CAME TO DIS SHOW TO PLAY BOTH THE HOUSEMATES AND WE THE VIEWERS
Thanks