Hi Everyone! I just want to note that there is an error in this video in regards to the Carboxylic Acid portion of the the mech. The first step I showed is not quite correct, and I have made a note in the video description with the correction as well with Annotations. Please try to NOT view this on a mobile device but rather a laptop so you can see the annotation corrections. Once again I apologize for any confusion, let me know if you have questions!
I love your videos! So funny that all my orgo 2 classmates also watch your videos so we all talk about how great of a teacher you are. Seriously grateful and keep up the great work.
Great refresher course, thank you. I loved Organic Chemistry in college and miss studying it. One question: in the carboxylic acid reduction, wouldn't you need two molar equivalents of LiAlH4? In the link provided in the description with the corrected reaction mechanism, it seems to imply this, but in the general reaction summary it only shows one molar equivalent, so I might be misunderstanding it.
One should be a little bit careful to be specific with the reagents and conditions being used when stating "X will not reduce Y". By *itself* and with non-forcing conditions NaBH4 cannot reduce carboxylic acids (or esters). However, depending upon the use of additional reagents or forcing conditions, the reducing power can be accentuated. For example, using MeOH with NaBH4 and THF with refluxing increases the reducing power and allows it to reduce esters. Using iodine and THF with NaBH4 allows it to reduce carboxylic acids via hydroboration. But a very nice video - well done.
tluedeke reactor7 Thanks! And ah I actually didn't know that. In regular undergrad level of Orgo classes most students don't learn it that in-depth so I didn't include any of that and just kept to what I was taught. Did you take Orgo honors?
Elvira Mendoza you’re totes right! I’ve been meaning to make one for a while now but it probably won’t come out until maybe Spring 2020 when the BU students get to that topic (I tend to make vids on topics right after I tutor it. 😶) there are some good ones in my Orgo 2 Made Easy Playlist tho! There’s a link to the 2nd semester playlist at this link 😅 (I forgot the link to the 2nd one haha) ruclips.net/p/PLP0TLbeMObSxSKD5QfePIfTNm3-XwXAhq
+Maria Alvarado Hahaha I'm happy to hear that! I loved making this video for you and the other peeps. :D BTW! Have you heard of these other Orgo peeps? ruclips.net/video/1QkGxQCI1es/видео.html
Thanks so much frank it be great if you could stank off to the side instead of the Center of the board at the end so we can see all the products clearly. Thanks again for your service... I love your videos and how thorough you are.
We had the reaction in EtOH (dried), so without any h2o. So now I'm curious if you showed just the simplified reaction or if this mechanism is prefered when working with h2o in acidic coniditions. (afaik 4 R-CO-R' + NaBH4 -> B-(O-C(-R)-R')4 + 4 EtOH -> 4 B(OEt)4 + 4 R-COH-R'). I'm missing the vocabular for all the reactants to describe the problem, but I hope the sum formula makes it clear enough, if not I will try to translate it
What happens if the aldehyde is formaldehyde ? Do you still get the secondary alcohol ? Also, does NaBH4 react with alcohol ? It seems to have a reaction with methanol . What is the mechanism of that reaction ?
I always see that teachers don't pay attention to by-products. What is the other product of the reaction with NaBH4, B2H6? B2H6 is a pretty explosive solid, however BH3 does not exist...
Your welcome Chris! Yeah sorry I wish I could cover it but I've been pretty caught up with end of semester affairs lately and haven't been able to get a chance to do it yet. If you get confused by it, there are some good mechanism walkthroughs online.
sir this video is very useful.....but my curiosity is to know the lewis dot structure of LiAlH4 and NaBH4 ...as sodium and aluminum have 3valence electrons so how can. they form 4 covalent bonds...????
Hi Frank! Great video! If you don't have 2eq of LiAlH4 with carboxylic acid then will there be NR or will the reducing agent reduce some molecules to a primary alcohol and leave the others alone? Thanks again for making these! 🖖
Hey Nicole! Great question! So one equivalent of LiAlH4 should technically be able to reduce a whole equivalent of Carboxylic Acid. 1 H is used to Deprotonate the COOH (not shown in my vid, see the correction link in the description) 1H is used to attack the Carbonyl C and is the H of our aldehyde intermediate, and then 1 more H ends up being the other H in our primary alcohol product in addition to the aldehyde H. So anyways unless your Prof teaches that you need more than 1 equiv of LiAlH4 I don't think you need to worry about it. ;)
Great question there! Bonga Msomi It cannot reduce an ester, as esters are almost as difficult to reduce as carboxylic acids. I'll be posting a video soon that explains why carboxylic acids are hard to reduce so keep an eye out! :]
It's a little complicated, so I would recommend checking it with your Professor. But usually it's taught that both NaBH4 and LiAlH4 cannot reduce alkenes. If the course is advanced enough, they might teach that conjugated alkenes can be reduced. (alkenes next to a C=O)
;) did that so you guys won't forget haha, What'd you think of my longer video? I personally felt it was a tad too long but there was just so much I wanted to fit in.
+Tracy Win Haha Yeaaaah! They're one of the peeps who inspired this channel! Btws there are bunch of perks/discounts you get access to as a subscriber to this channel make sure you've watched this vid! ruclips.net/video/1QkGxQCI1es/видео.html
Yup! You are correct, I made a pinned comment about it and mentioned it in the video description I think. Most schools don’t really teach the mech/students aren’t required to draw it so this shortcut mechanism usually does the job for solving problems. 😊
Nitha Chalil The aldehyde should get reduced down to a primary alcohol. But 2-butanal shouldn't be a real molecule by the way. Aldehydes are always at the end of a chain, so I would double check that.
Nitha Chalil Hmm I still don't think that's a possible compound. 1-methyl propanal wouldn't be that structure because the first carbon (carbon of the aldehyde) can't have that many bonds. It's already double bonded to a O, single bonded to an H, so it can only be bound to one more thing. Either the 2 other carbons to make the propanal or a methyl to make it ethanal. And I don't believe you can get a diol from an aldehyde treated with NaBH4. I would double check with your Prof. on that one.
Haha thanks! Unfortunately I do have some bad news. I accidentally mixed up the LiAlH4 Reduction Mechanism for Carboxylic Acids with Esters so my first two steps of the mechanism of the Carboxylic Acid reduction needs to be tweaked. Here's the correct mechanism: chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Conversion_of_carboxylic_acids_to_alcohols_using_LiAlH4. My bad! Sorry for any confusion.
Hi Everyone! We just released our new Orgo Made Easy Podcast series! Be sure to check it out, we had so much fun making this one for you and this is what we're envisioning the next phase of Orgo Made Easy to be like! ruclips.net/video/6yFsJLcvSzY/видео.html
Hi Everyone! I just want to note that there is an error in this video in regards to the Carboxylic Acid portion of the the mech. The first step I showed is not quite correct, and I have made a note in the video description with the correction as well with Annotations. Please try to NOT view this on a mobile device but rather a laptop so you can see the annotation corrections. Once again I apologize for any confusion, let me know if you have questions!
Sir at which part does correction occurs
Why you ad na+/li+ with oxygen in first step second line
Oh is bad leaving group but you did something different
WOW this was such a clear well-made video!! I also love your energy and enthusiasm, you have a natural gift for teaching 💯
Thanks for watching!!! 🥳 Glad you liked it! 😄
I love your videos! So funny that all my orgo 2 classmates also watch your videos so we all talk about how great of a teacher you are. Seriously grateful and keep up the great work.
I really appreciate your videos! Could you possibly do videos on practice problems for these reactions?
Wow, it actually makes a lot of sense. Thank you so much!!
yes i m here to figure out Lithiumborohydride reducing mechanism! bloops made me craked up. thanks for the video!
Yo man you are doing God's work keep it up 🤞🤞
Thanks! 😄 Haven’t been making videos lately but doing a ton of tutoring! Where are you taking Organic Chem at?
Thanks Frank! This video helps a lot, save my time to read through the notes tho.
Great refresher course, thank you. I loved Organic Chemistry in college and miss studying it. One question: in the carboxylic acid reduction, wouldn't you need two molar equivalents of LiAlH4? In the link provided in the description with the corrected reaction mechanism, it seems to imply this, but in the general reaction summary it only shows one molar equivalent, so I might be misunderstanding it.
Thanks Brian! I believe you are correct but I’ve never seen Profs or schools write 2 Equivalents 🤷🏻♂️ 😂 so good Q!
I highly recommend checking out our new Podcast series it’s super fun and I got a co-host who has a PhD ruclips.net/video/nvqP8Z45rGY/видео.html
One should be a little bit careful to be specific with the reagents and conditions being used when stating "X will not reduce Y". By *itself* and with non-forcing conditions NaBH4 cannot reduce carboxylic acids (or esters). However, depending upon the use of additional reagents or forcing conditions, the reducing power can be accentuated.
For example, using MeOH with NaBH4 and THF with refluxing increases the reducing power and allows it to reduce esters. Using iodine and THF with NaBH4 allows it to reduce carboxylic acids via hydroboration.
But a very nice video - well done.
tluedeke reactor7 Thanks! And ah I actually didn't know that. In regular undergrad level of Orgo classes most students don't learn it that in-depth so I didn't include any of that and just kept to what I was taught. Did you take Orgo honors?
Clear and precise. Outstanding job
Thanks Sharif! 😁
Happy studying! 💪🏼
FRANK YOURE A GOD!
Elvira Mendoza Gonzalez God says hi! 😄 jk (just a former struggling Orgo student 👨🎓 hehe 😀)
Elvira Mendoza you’re totes right! I’ve been meaning to make one for a while now but it probably won’t come out until maybe Spring 2020 when the BU students get to that topic (I tend to make vids on topics right after I tutor it. 😶) there are some good ones in my Orgo 2 Made Easy Playlist tho! There’s a link to the 2nd semester playlist at this link 😅 (I forgot the link to the 2nd one haha) ruclips.net/p/PLP0TLbeMObSxSKD5QfePIfTNm3-XwXAhq
I must say this, I might have fallen in love. Oh, and the video was nice too. Hahaha :) Thanks for the video, I just loved it.
+Maria Alvarado Hahaha I'm happy to hear that! I loved making this video for you and the other peeps. :D BTW! Have you heard of these other Orgo peeps? ruclips.net/video/1QkGxQCI1es/видео.html
hahahaha. Amazing. . That's exactly what i remembered from class at 2:43
Love your vids Frank, so dynamic !
Subbed !
Thanks CartoBalow ! I really appreciate the support. :]
Thanks so much frank it be great if you could stank off to the side instead of the Center of the board at the end so we can see all the products clearly. Thanks again for your service... I love your videos and how thorough you are.
Stand off*
We had the reaction in EtOH (dried), so without any h2o. So now I'm curious if you showed just the simplified reaction or if this mechanism is prefered when working with h2o in acidic coniditions. (afaik 4 R-CO-R' + NaBH4 -> B-(O-C(-R)-R')4 + 4 EtOH -> 4 B(OEt)4 + 4 R-COH-R'). I'm missing the vocabular for all the reactants to describe the problem, but I hope the sum formula makes it clear enough, if not I will try to translate it
This was great. Thanks Frank!
I can’t thank you enough!!! THANK YOU 🙏🏾
Very informative
YO Frank YOU THA NIGGAA!!!!!!!!!!!!!!!!!!!!!!!!!!
Mesa grant yousa N word pass. Enjoy.
Thank you! 😊
Would be cool to see the nitrostyrene reduction to amine by the same reducing agents.
What happens if the aldehyde is formaldehyde ? Do you still get the secondary alcohol ?
Also, does NaBH4 react with alcohol ? It seems to have a reaction with methanol . What is the mechanism of that reaction ?
Thanks.
Chem goat fr
I always see that teachers don't pay attention to by-products. What is the other product of the reaction with NaBH4, B2H6? B2H6 is a pretty explosive solid, however BH3 does not exist...
Thanks frank sir for this video its really very helpful
Thank you for this. Really wish you had the Wolf-Kishner :/
Your welcome Chris! Yeah sorry I wish I could cover it but I've been pretty caught up with end of semester affairs lately and haven't been able to get a chance to do it yet. If you get confused by it, there are some good mechanism walkthroughs online.
sir this video is very useful.....but my curiosity is to know the lewis dot structure of LiAlH4 and NaBH4 ...as sodium and aluminum have 3valence electrons so how can. they form 4 covalent bonds...????
are the hydrogens actually bound to the aluminum? because i thought aluminum only really bonds to three things since it has 3 valence electrons?
Hi Frank! Great video! If you don't have 2eq of LiAlH4 with carboxylic acid then will there be NR or will the reducing agent reduce some molecules to a primary alcohol and leave the others alone? Thanks again for making these! 🖖
Hey Nicole! Great question! So one equivalent of LiAlH4 should technically be able to reduce a whole equivalent of Carboxylic Acid. 1 H is used to Deprotonate the COOH (not shown in my vid, see the correction link in the description) 1H is used to attack the Carbonyl C and is the H of our aldehyde intermediate, and then 1 more H ends up being the other H in our primary alcohol product in addition to the aldehyde H. So anyways unless your Prof teaches that you need more than 1 equiv of LiAlH4 I don't think you need to worry about it. ;)
What happens when you add NaBH4 and or LiAlH4 to H2O... I heard it explodes. Is that true? Nice videos.
it reacts with water in the air.
Thaaaank youu !
Yeah it wasn't very clear that the animation at 4:15 was regarding the resonnance of the carbonyl. These arrows were too thick (although correct).
It helped me alot
thank you very much :)
In my text EtOH or MeOH is used with NaBH4 and EtO2 or THF is used with LAH. Is there a reason these would be more effective/ preferred over HCL?
Great video thanks man
2:45 ....based!
Nice video
so helpful :) thanku frank
Thank youuuuuuu
Are enamiomers of the alcohols also made?
are you one of the wong fu's production brother? hahahaha
Non with acid .first we have acidobasic réaction because hydrure ion is strong base.
With 2 eq of DIBAL H can we reduce carboxylic acids to alcohol ?
I don’t believe so! I think only LiAlH4 can do it.
Thank you sir
nice explanation but i need advantages of nabh4
Since NaBH4 is a mild reducing agent,can it reduce an ester?
Great question there! Bonga Msomi It cannot reduce an ester, as esters are almost as difficult to reduce as carboxylic acids. I'll be posting a video soon that explains why carboxylic acids are hard to reduce so keep an eye out! :]
Frank Wong Can LiAlH4 reduce alkenes ?
It's a little complicated, so I would recommend checking it with your Professor. But usually it's taught that both NaBH4 and LiAlH4 cannot reduce alkenes. If the course is advanced enough, they might teach that conjugated alkenes can be reduced. (alkenes next to a C=O)
Can NaBH4 reduce carbon carbon double bond to single bond?
Nope, NaBH4 and LiAlH4 but target polar double bonds and there is no dipole between carbon carbon double bonds. =]
+Frank Wong Thank you sir
Arnav Pareek If you want to reduce C=C you have to use H2/Ni
What is the stereochemistry? Is it retention or inversion.
You should get a racemic mixture of both enantiomers because the H have an equal chance of attacking from top/bottom.
Why NaBH4 is less reducing powr dan LiAlH4 or vice versa..
Sanjoy Saha Great question! I'm already one step ahead of ya ;) Here's that video: ruclips.net/video/oJyVWsygzEc/видео.html
Lol. Attacked twice. xD
;) did that so you guys won't forget haha, What'd you think of my longer video? I personally felt it was a tad too long but there was just so much I wanted to fit in.
I didn't find anything wrong with it.
Oh okay awesome! haha
who can give me an experimantal section of redection of acetophenone or benzophenon with lialh4
There's no explanaiton why NaBH4 can't reduce carb.acid to alchohol
nice guy shirt! wong fuuu
+Tracy Win Haha Yeaaaah! They're one of the peeps who inspired this channel! Btws there are bunch of perks/discounts you get access to as a subscriber to this channel make sure you've watched this vid! ruclips.net/video/1QkGxQCI1es/видео.html
Mechanism for carboxylic acid is wrong because there is acidic hydrogen so it will abstract Hydrogen but product will be the same.
Yup! You are correct, I made a pinned comment about it and mentioned it in the video description I think. Most schools don’t really teach the mech/students aren’t required to draw it so this shortcut mechanism usually does the job for solving problems. 😊
Loved Sleeping Beauty.
look at aluminium..... how it is written😊😊
what wud u get if u react 2-butanal with NaBH4?
Nitha Chalil The aldehyde should get reduced down to a primary alcohol. But 2-butanal shouldn't be a real molecule by the way. Aldehydes are always at the end of a chain, so I would double check that.
sorry i got the compound name wrong...its...
CH3-CH2-CH2(CH3)-CHO...1-methyl propanal
according to the answer given in the reference book that i have it says it will become a diol...if thats right can you tell me how?
Nitha Chalil Hmm I still don't think that's a possible compound. 1-methyl propanal wouldn't be that structure because the first carbon (carbon of the aldehyde) can't have that many bonds. It's already double bonded to a O, single bonded to an H, so it can only be bound to one more thing. Either the 2 other carbons to make the propanal or a methyl to make it ethanal. And I don't believe you can get a diol from an aldehyde treated with NaBH4. I would double check with your Prof. on that one.
k thanks..:)
Would prefer if you use "official terms" and not like "shoot in"... Thanks..
u r amazing. Here's a no homo heart
Haha thanks! Unfortunately I do have some bad news. I accidentally mixed up the LiAlH4 Reduction Mechanism for Carboxylic Acids with Esters so my first two steps of the mechanism of the Carboxylic Acid reduction needs to be tweaked. Here's the correct mechanism: chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Conversion_of_carboxylic_acids_to_alcohols_using_LiAlH4. My bad! Sorry for any confusion.
Isnt the end product the same though?
Yup! Exact same end product, just slightly different mechanism.
why is there no indian guy in this vediio
Attacked twice
Hi Everyone! We just released our new Orgo Made Easy Podcast series! Be sure to check it out, we had so much fun making this one for you and this is what we're envisioning the next phase of Orgo Made Easy to be like! ruclips.net/video/6yFsJLcvSzY/видео.html
Hindi me babu ji
stop wasting time and explain direct please🙂