Something about Sal’s voice is just so soothing, hypnotic almost. Anytime I watch a Khan Academy video I always hope that it’s Sal narrating and not somebody else. Anybody else feel the same?
yeah, same. BUTTTT, it hurts even morrrre when the other narrator tries to imitate Sal's style. for such cases, i am always like stop, dont try to copy. dont fake that joy/tone/accent. like that forced friction accent - soooo irritating.
I guess im randomly asking but does any of you know of a trick to get back into an instagram account? I stupidly forgot my login password. I would appreciate any tips you can give me!
@acmilanshevachels the carbonyl carbon is the one that loses the electron. It is giving up its end of the bond with the alpha carbon to the other carbonyl carbon. That is why it is the alpha carbon that forms the bond.
You are wrong. The way you presented it looks like the carbonyl atom (masked within an enolate anion resonance structure) is performing a nucleophilic attack to the carbonyl carbon of the other molecule. If it would behave like this you'd get an alpha-hydroxycarbonyl compund as a product, wouldn't you? Arrow at 6:25 is WRONG. BTW, in nucleophilic reaction there is no "loosing of electrons", nucleophile SHARES electron with electrophilic partner by creating new COVALENT bond Loosing electron is called "oxidation", aldol reaction is not a red-ox. Peace.
dude, seriously. It blows my mind that you know all this material in all of these subjects!!(math, biology, chemistry, economics, etc). You are probably one of the most knowledgeable persons in the world
Aldol's reaction or Aldol's condensation is one the best known or all organic reactions. I've seen it on every professional test I've ever taken. It's routinely on grad and medical school entrance exams. If you're going to remember one, consider Aldol. Great post.
Well you just managed to explain something in 16 minutes,I didn´t understand in weeks!!! Great Video and also quite useful that you use so many different colours. Greetings from Germany
@khanacademy I hope this clarifies for those thinking as acmilanshevachels may have been (and as I was): I am being taught the convention of showing nucleophilic attacks from electron-pairs, as contrasted to giving/taking single electrons--as an aside, I believe that the electron-pairs convention is taught due to advanced molecular orbital theory. Because of the difference between Sal's way and the electron-pairs way, I was confused as to which carbon would bond with the orange carbon.
@ZacharySmith89 So if you are also having trouble seeing Sal's way, as I was, note that when an arrow is drawn from an atom to another--as Sal is doing between these carbonyl carbons--in the electron-pair convention that arrow is starting at and showing the action of an electron-PAIR, NOT a single electron, as is the case in Sal's way of drawing. Just be aware of this difference, and you should be able to understand this. I hope that helps more random people than it confuses! :)
hey thanx a lot ur helping so many ppl and on that u should get a Nobel prize can u pliz make more videos about organic chemistry? more about isomers... (like determing the number of isomers) more common naming like styrene, biologically importent groups and more thanx anyway :)
Dear Sir, alpha carbon in enolate anion is the nucleophilic one, (we can observe the negative charge there in the first resonance structure of the enolate anion) therefore arrow that you drawn at 6:25 is WRONG, even though the product is CORRECT. Purple bond is forming between alpha carbon of one molecule (enolate anion) and carbonyl carbon of the other, while your arrow connects two carbonyls (one of them "masked" as an resonance structure - enolate anion). Please correct this one because its confusing! Best regards!
Great video. During the nucleophillic attack of the enolate, I think it would be easier to see if you used the other electron to attach to the aldehyde. It makes it seem like that carbon is being attached to the carbonyl carbon, but when you draw it is actually the alpha carbon that attaches to caronyl carbon. Hope you get what i mean.
The pink arrow combining the resonated-enolate ion to the other aldehyde is misleading. That negative is coming from the alpha carbon, not the carbon directly connectex to the O- of that enolate ion.
@cmonbugmenot "R" used to stand for "radical", which used to mean a set of covalently linked atoms that acted as a single group in chemical reactions. This use of the term "radical" is falling in disfavor due to the ambivalence with the other meaning of "radical".
Isn't the reaction slightly different if ketones are used? The oxygen on the carbonyl group that is attacked by the enolate dissociates and turns into a pi bond. I was looking for a video on that mechanism.
Ran into an MCAT question on B-hydroxyaldehyde formation, and kept thinking, man I hope he mentions this. Saved right at the nick of time, glad I watched all the way through!
There are a few issues here - firstly there are a variety of different aldol reactions: Mr. Khan mentioned that aldehydes or ketones can undergo aldol reactions, and this is true, but he omitted to mention that there are two different pathways for the C-C bond formation. The mechanism is the enolate mechanism, but aldol reactions can also proceed by an acid-catalysed enol mechanism. Also troubling is the non-standard arrow-pushing, which at one point seems to make a bond 'from' the carbonyl carbon on the enolate, when the bond clearly forms at the α-carbon. I don't want to say that anyone's method for drawing mechanisms is 'wrong', but I think that the regular method of pushing full arrows from electron pairs is valuable for the sake of clarity - I've never seen an actual chemist draw the aldol mechanism like this. Finally, although this channel seems devoted to very simple organic chemistry, it's good to at least mention stereochemistry, if not show the Zimmerman-Traxler model.
There is an obvious mistake in arrow drawn at 6:25, should've started from nucleophilic alpha-carbon (the one possesing negative charge on the other resonance structure)
@@krzysztofgrudzien9313 I believe that the Enolate what gives this carbon such a character, cause in fact using such an OH will yield minimum amounts of the C-, it is true that the later is more true in the case of Ketones due to the Conj-base strength difference but I believe that the enolate will push the C- to do what it should do
Hey, found this very helpful, but I was confused on one point. Shouldn't the nucleophilic attack be off of the alpha carbon and not off of the carbonyl carbon? Because in the next step, you depict the new bond between the alpha carbon of one aldehyde to the carbonyl carbon of the other.
Your notation when drawing mechanisms is worrying. Full arrow heads and bonds usually correlate to pairs of electrons, not just one. This reaction is not a radical reaction but you're characterizing it as one. Just something to keep in mind.
why does he call the "R" groups "radical groups"? did he mean to say the "R" groups are alkyl groups? anyway, thanks for the video. to summarize: -bonds carbon to carbon, can form aldehyde-alcohol or ketone-alcohol -enolate ion acts as nucleophile -alpha-carbon hydrogens are more acidic because when one is removed, the electron pair it leaves behind can resonate with the carbonyl C=O bond
Thanks for uploading this! I have a quick question regarding the choice of where the enolate will form in a molecule with two carbonyl groups: one aldehydic, and the other a ketone. If we imagine 7C chain, 1st C = aldehyde, and 6th C = ketone, where will the enolate form? With the aldehyde or with the ketone? I feel like kinetically they're both the same thing (2 H's each), but thermodynamically... would the ketone carbonyl produce a more stable enolate? Thanks for any help!!
what's preventing it to go to a aldol-condensation? the OH group would be expelled and a double bond between ß and alpha carbon would form. since both just require NaOH i dont understand whats preventing the last step.
Hi Sal, some of the videos in this organic chem playlist are showing as private videos. please help :(:(. I've been following the playlist from the beginning video but at the middle of the playlist it shows private video. plz help
Is he saying that the two electrons are given or actually one electron (as in a radical?)? Because he's using double arrows and it doesn't make sense....
Aida K. Two electrons are in a bond, but H is only contributing one electron and carbon is contributing one electron. He is showing that the hydrogen leaves behind it's one electron, but it's not a radical reaction because there's another electron from carbon--it's an anion. In radical mechanisms, the hydrogen would take it's one electron rather than leave it behind (and the carbon would be left with a single electron--a radical)
at 4:15 why does Sal sir says that only a single electron goes to the O , giving it a negative charge, it should be both the pi electrons of the pi bond that should go to O , giving it the negative charge ?
Charge is always attained due to transfer of electrons not the bonds themselves. - (or -1) charge means that the atom has gained a *single* electron. In 4:15, the upper electron in the pi bond was already owned by the oxygen atom, it was the carbon's electron that got transferred from pi bond to the oxygen atom. That's why the (-) charge. If you watch Khan Academy's earliest Organic chemistry videos, then you'll understand this easily because Sir explained it very well in the beginning videos.
Something about Sal’s voice is just so soothing, hypnotic almost. Anytime I watch a Khan Academy video I always hope that it’s Sal narrating and not somebody else. Anybody else feel the same?
Yup. Totally agree. Love his voice
yeah, same.
BUTTTT, it hurts even morrrre when the other narrator tries to imitate Sal's style. for such cases, i am always like stop, dont try to copy. dont fake that joy/tone/accent. like that forced friction accent - soooo irritating.
I guess im randomly asking but does any of you know of a trick to get back into an instagram account?
I stupidly forgot my login password. I would appreciate any tips you can give me!
i like grant sanderson's better
simp
Finally, I finished the playlist...
thank you it really helped me for my exams. loved the way it was explained
please keep uploading organic chemistry mechanisms videos PLEASE!!!!!!!
No more organic Chemistry videos? Why? 😭
Watching all the videos in this playlist was such a good experience ; I want more of it! 😫
We neeed moooree organic chemistry!!!!!
@acmilanshevachels the carbonyl carbon is the one that loses the electron. It is giving up its end of the bond with the alpha carbon to the other carbonyl carbon. That is why it is the alpha carbon that forms the bond.
You are wrong. The way you presented it looks like the carbonyl atom (masked within an enolate anion resonance structure) is performing a nucleophilic attack to the carbonyl carbon of the other molecule. If it would behave like this you'd get an alpha-hydroxycarbonyl compund as a product, wouldn't you?
Arrow at 6:25 is WRONG.
BTW, in nucleophilic reaction there is no "loosing of electrons", nucleophile SHARES electron with electrophilic partner by creating new COVALENT bond Loosing electron is called "oxidation", aldol reaction is not a red-ox. Peace.
MORE ORGANIC CHEM VIDS PLEASE!!
I noticed your other videos in the playlist are set to private. The question is why? It would be nice to keep learning.
dude, seriously. It blows my mind that you know all this material in all of these subjects!!(math, biology, chemistry, economics, etc).
You are probably one of the most knowledgeable persons in the world
ur comment = 10 yrs ago. how has ur experience with youtube changed all these years?
Thanks a lot Sir! This video cleared my doubt completely!
you are better than my teacher! thanks
Oh man you have helped me so much!
Aldol's reaction or Aldol's condensation is one the best known or all organic reactions. I've seen it on every professional test I've ever taken. It's routinely on grad and medical school entrance exams. If you're going to remember one, consider Aldol. Great post.
thanks
Well you just managed to explain something in 16 minutes,I didn´t understand in weeks!!! Great Video and also quite useful that you use so many different colours. Greetings from Germany
HES BACKKK
@khanacademy
I hope this clarifies for those thinking as acmilanshevachels may have been (and as I was):
I am being taught the convention of showing nucleophilic attacks from electron-pairs, as contrasted to giving/taking single electrons--as an aside, I believe that the electron-pairs convention is taught due to advanced molecular orbital theory. Because of the difference between Sal's way and the electron-pairs way, I was confused as to which carbon would bond with the orange carbon.
This video cleared my doubt completely!
@ZacharySmith89 So if you are also having trouble seeing Sal's way, as I was, note that when an arrow is drawn from an atom to another--as Sal is doing between these carbonyl carbons--in the electron-pair convention that arrow is starting at and showing the action of an electron-PAIR, NOT a single electron, as is the case in Sal's way of drawing. Just be aware of this difference, and you should be able to understand this.
I hope that helps more random people than it confuses! :)
hey thanx a lot
ur helping so many ppl and on that u should get a Nobel prize
can u pliz make more videos about organic chemistry?
more about isomers... (like determing the number of isomers)
more common naming like styrene, biologically importent groups and more
thanx anyway :)
Dear Sir, alpha carbon in enolate anion is the nucleophilic one, (we can observe the negative charge there in the first resonance structure of the enolate anion) therefore arrow that you drawn at 6:25 is WRONG, even though the product is CORRECT. Purple bond is forming between alpha carbon of one molecule (enolate anion) and carbonyl carbon of the other, while your arrow connects two carbonyls (one of them "masked" as an resonance structure - enolate anion).
Please correct this one because its confusing!
Best regards!
Great video. During the nucleophillic attack of the enolate, I think it would be easier to see if you used the other electron to attach to the aldehyde. It makes it seem like that carbon is being attached to the carbonyl carbon, but when you draw it is actually the alpha carbon that attaches to caronyl carbon. Hope you get what i mean.
i am quite happy , i got what i need
Thanks
The pink arrow combining the resonated-enolate ion to the other aldehyde is misleading. That negative is coming from the alpha carbon, not the carbon directly connectex to the O- of that enolate ion.
we need more people like you and less people like HM
u r method of teaching is really excellent
How about an acid-catalyzed aldol reaction? Can you please show the mechanism for this as well? I will really appreciate it!
wow, comment from 7 yrs ago. how's current life goin'?
thanks for the video. i find videos bettr than books
plz upload all video
Really enjoy learning from you!!
@cmonbugmenot "R" used to stand for "radical", which used to mean a set of covalently linked atoms that acted as a single group in chemical reactions. This use of the term "radical" is falling in disfavor due to the ambivalence with the other meaning of "radical".
Can you do a video on a ring closure?
Isn't the reaction slightly different if ketones are used? The oxygen on the carbonyl group that is attacked by the enolate dissociates and turns into a pi bond. I was looking for a video on that mechanism.
Thank you for your lecture. It's so detailed and so easy to understand
My like was the 1k th like.. AMAZING TEACHING🙇👍
this topic is probably the hardest for me in ochem
Ran into an MCAT question on B-hydroxyaldehyde formation, and kept thinking, man I hope he mentions this. Saved right at the nick of time, glad I watched all the way through!
Yes it is helpful qnd entertaining
6:47 when I got a tic
There are a few issues here - firstly there are a variety of different aldol reactions: Mr. Khan mentioned that aldehydes or ketones can undergo aldol reactions, and this is true, but he omitted to mention that there are two different pathways for the C-C bond formation. The mechanism is the enolate mechanism, but aldol reactions can also proceed by an acid-catalysed enol mechanism.
Also troubling is the non-standard arrow-pushing, which at one point seems to make a bond 'from' the carbonyl carbon on the enolate, when the bond clearly forms at the α-carbon. I don't want to say that anyone's method for drawing mechanisms is 'wrong', but I think that the regular method of pushing full arrows from electron pairs is valuable for the sake of clarity - I've never seen an actual chemist draw the aldol mechanism like this.
Finally, although this channel seems devoted to very simple organic chemistry, it's good to at least mention stereochemistry, if not show the Zimmerman-Traxler model.
Sorry, *this mechanism in the 1st paragraph.
ur my true sunshine~~! u have helped me to get out of the darkness!! Thx.
this one is greatly helpful, but i want the mechanism of acid catalysed aldol condensation.......
can you please provide me the link...??
Very good explaining👍
Always wondering, why my teacher doesn't explain like you !
Again I got 100% on my Test, thanks!
Septimius Severus I'm sorry this is off topic, but are you a Potterhead too?
Nidhi Gupta lol we HP fans never miss a single chance huh! I'm a potterhead as well! 😌
The way you draw the mechanism makes it harder to see what's going on
There is an obvious mistake in arrow drawn at 6:25, should've started from nucleophilic alpha-carbon (the one possesing negative charge on the other resonance structure)
@@krzysztofgrudzien9313 I believe that the Enolate what gives this carbon such a character, cause in fact using such an OH will yield minimum amounts of the C-, it is true that the later is more true in the case of Ketones due to the Conj-base strength difference but I believe that the enolate will push the C- to do what it should do
Hey, found this very helpful, but I was confused on one point. Shouldn't the nucleophilic attack be off of the alpha carbon and not off of the carbonyl carbon? Because in the next step, you depict the new bond between the alpha carbon of one aldehyde to the carbonyl carbon of the other.
It's very entertaining, thank you very much :)
it's not only entertaining but very valuable. Thank you for the effort. I guess I kinda get what this is about now
Your notation when drawing mechanisms is worrying. Full arrow heads and bonds usually correlate to pairs of electrons, not just one. This reaction is not a radical reaction but you're characterizing it as one. Just something to keep in mind.
why does he call the "R" groups "radical groups"? did he mean to say the "R" groups are alkyl groups?
anyway, thanks for the video. to summarize:
-bonds carbon to carbon, can form aldehyde-alcohol or ketone-alcohol
-enolate ion acts as nucleophile
-alpha-carbon hydrogens are more acidic because when one is removed, the electron pair it leaves behind can resonate with the carbonyl C=O bond
@echelecopao no, he's like one of THE most knowledgeable persons on earth , hehe. :) Sal deserves a big THE.
thank you very much !
Anurag Pradhan u look so Smart
why are all the rest privated?
Why are the next vids in the playlist privated?
organic is going to be the death of me😡
I hate chemistry but I'm majboor😣😣
Thanks for uploading this! I have a quick question regarding the choice of where the enolate will form in a molecule with two carbonyl groups: one aldehydic, and the other a ketone. If we imagine 7C chain, 1st C = aldehyde, and 6th C = ketone, where will the enolate form? With the aldehyde or with the ketone? I feel like kinetically they're both the same thing (2 H's each), but thermodynamically... would the ketone carbonyl produce a more stable enolate? Thanks for any help!!
hope you got ur tests done.....sry for asking after 9 years
You are better than my " professor "
May I synthesise 2,6-dimethylhex-2-enone from 5-oxo-4-methylheptanal via intramolecular aldol condensation?
what's preventing it to go to a aldol-condensation?
the OH group would be expelled and a double bond between ß and alpha carbon would form. since both just require NaOH i dont understand whats preventing the last step.
you need to add heat to dehydrate the water to form the double bond
respect!!
How to watch remaining videos
good
Hi Sal, some of the videos in this organic chem playlist are showing as private videos. please help :(:(.
I've been following the playlist from the beginning video but at the middle of the playlist it shows private video. plz help
can u xpln crossed aldol condesation
Is he saying that the two electrons are given or actually one electron (as in a radical?)? Because he's using double arrows and it doesn't make sense....
Aida K. Two electrons are in a bond, but H is only contributing one electron and carbon is contributing one electron. He is showing that the hydrogen leaves behind it's one electron, but it's not a radical reaction because there's another electron from carbon--it's an anion. In radical mechanisms, the hydrogen would take it's one electron rather than leave it behind (and the carbon would be left with a single electron--a radical)
but then how can we call it " beta- hydroxycarbonyl compound" ??
at 4:15 why does Sal sir says that only a single electron goes to the O , giving it a negative charge, it should be both the pi electrons of the pi bond that should go to O , giving it the negative charge ?
Charge is always attained due to transfer of electrons not the bonds themselves.
- (or -1) charge means that the atom has gained a *single* electron.
In 4:15, the upper electron in the pi bond was already owned by the oxygen atom, it was the carbon's electron that got transferred from pi bond to the oxygen atom. That's why the (-) charge.
If you watch Khan Academy's earliest Organic chemistry videos, then you'll understand this easily because Sir explained it very well in the beginning videos.
anyone know any chemical reactions that use the aldol reaction?
Samridhi Kumaresan Robinson annulation
@intellegence63smart holy shit...I HOPE i can get at least a 90 on my first exam.
is this the same guy lolololol
I want your babies