NaBH4 and LiAlH4 Reduction Mechanism Made Easy! | Organic Chemistry

[@OrgoMadeEasy]

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!

[@sanjeetpurang]

Sir at which part does correction occurs

[@md.nournoby4523]

Why you ad na+/li+ with oxygen in first step second line

[@manikantaprasad721]

Oh is bad leaving group but you did something different

[@OrgoMadeEasy]

Going to follow up here: I didn’t realize when I was making this video that this was actually the mechanism for Ester reduction not Carboxylic Acid reduction. But most schools don’t ask for carb acid full mechanism so you can essentially think of it this way for the purposes of product prediction. You are correct though in the full correct mechanism the OH actually gets deprotonated and leaves in a different method. You will notice though in Organic Chem sometimes OH can leave despite being a bad leaving group (Wolff Kishner Reduction + Aldol Condensation). In both the product you’re making is so much more stable that the OH can just leave.

[@SyenPie]

WOW this was such a clear well-made video!! I also love your energy and enthusiasm, you have a natural gift for teaching 💯

[@OrgoMadeEasy]

Thanks for watching!!! 🥳 Glad you liked it! 😄

[@alip9683]

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.

[@nerverevermind]

yes i m here to figure out Lithiumborohydride reducing mechanism! bloops made me craked up. thanks for the video!

[@cabbagelettuce3340]

I really appreciate your videos! Could you possibly do videos on practice problems for these reactions?

[@selina2789]

Wow, it actually makes a lot of sense. Thank you so much!!

[@soumo2725]

Yo man you are doing God's work keep it up 🤞🤞

[@OrgoMadeEasy]

Thanks! 😄 Haven’t been making videos lately but doing a ton of tutoring! Where are you taking Organic Chem at?

[@BrianOSheaPlus]

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.

[@OrgoMadeEasy]

Thanks Brian! I believe you are correct but I’ve never seen Profs or schools write 2 Equivalents 🤷🏻‍♂️ 😂 so good Q!

[@OrgoMadeEasy]

I highly recommend checking out our new Podcast series it’s super fun and I got a co-host who has a PhD kzbin.info/www/bejne/pKfUgWuQaZqlfbs

[@omgsadaiqa]

Thanks Frank! This video helps a lot, save my time to read through the notes tho.

[@sharifturnquest50]

Clear and precise. Outstanding job

[@OrgoMadeEasy]

Thanks Sharif! 😁

[@OrgoMadeEasy]

Happy studying! 💪🏼

[@tluedeke]

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.

[@OrgoMadeEasy]

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?

[@Music.n.medicine]

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.

[@Music.n.medicine]

Stand off*

[@OrgoMadeEasy]

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! kzbin.info/www/bejne/bKqppH2CmNuGsLs

[@Slovas008]

hahahaha. Amazing. . That's exactly what i remembered from class at 2:43

[@Eneres33]

I can’t thank you enough!!! THANK YOU 🙏🏾

[@elviramendozagonzalez3281]

FRANK YOURE A GOD!

[@OrgoMadeEasy]

Elvira Mendoza Gonzalez God says hi! 😄 jk (just a former struggling Orgo student 👨‍🎓 hehe 😀)

[@OrgoMadeEasy]

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) kzbin.info/aero/PLP0TLbeMObSxSKD5QfePIfTNm3-XwXAhq

[@MariaAlvarado-sd6nf]

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.

[@OrgoMadeEasy]

+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? kzbin.info/www/bejne/Z4LOeKuHeK5km9U

[@MetalKabu]

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

[@mmoney1712]

This was great. Thanks Frank!

[@latamalviya1044]

Thanks frank sir for this video its really very helpful

[@CartoBalow]

Love your vids Frank, so dynamic ! Subbed !

[@OrgoMadeEasy]

Thanks CartoBalow ! I really appreciate the support. :]

[@zainabusman3391]

Thanks.

[@BM-bs1jj]

Chem goat fr

[@todoconciencia6010]

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...

[@simplifiedchemistrywithanita]

Very informative

[@nh9662]

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! 🖖

[@OrgoMadeEasy]

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. ;)

[@leenalsader8476]

Thank you! 😊

[@Mr_c-tm3hu]

What happens when you add NaBH4 and or LiAlH4 to H2O... I heard it explodes. Is that true? Nice videos.

[@dorthysavage6660]

it reacts with water in the air.

[@thrdel]

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 ?

[@suos3738]

It helped me alot thank you very much :)

[@birdog23]

Thank you for this. Really wish you had the Wolf-Kishner :/

[@OrgoMadeEasy]

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.

[@abdulshehata3213]

YO Frank YOU THA NIGGAA!!!!!!!!!!!!!!!!!!!!!!!!!!

[@ColmK83]

Mesa grant yousa N word pass. Enjoy.

[@mattnstewart262]

are the hydrogens actually bound to the aluminum? because i thought aluminum only really bonds to three things since it has 3 valence electrons?

[@kq6up]

Would be cool to see the nitrostyrene reduction to amine by the same reducing agents.

[@gasperkosmac7672]

Great video thanks man

[@mariatahir5718]

so helpful :) thanku frank

[@VivekYadav-bd1bt]

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...????

[@YouMockMe]

2:45 ....based!

[@shotmeindaface]

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).

[@noranfelemban3067]

Thaaaank youu !

[@RechargingBatteries]

Nice video

[@melissahoaglund5196]

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?

[@bentameurmiloud685]

Non with acid .first we have acidobasic réaction because hydrure ion is strong base.

[@nikitasingh4297]

Thank you sir

[@laibarana2674]

Thank youuuuuuu

[@bhaibhai.........50yearsag94]

Oh is poor leaving group so why you kicked it out Btw this video is helpful

[@OrgoMadeEasy]

@@bhaibhai.........50yearsag94 good Q! I think I made a comment on this + in the video description. I didn’t realize when I was making this video that this was actually the mechanism for Ester reduction not Carboxylic Acid reduction. But most schools don’t ask for carb acid full mechanism so you can essentially think of it this way for the purposes of product prediction. You are correct though in the full correct mechanism the OH actually gets deprotonated and leaves in a different method. You will notice though in Organic Chem sometimes OH can leave despite being a bad leaving group (Wolff Kishner Reduction + Aldol Condensation). In both the product you’re making is so much more stable that the OH can just leave.

[@bhaibhai.........50yearsag94]

@@OrgoMadeEasy oo I don't know about this thanks sir you really helped me Keep making this type of video I will support you😊

[@juststudy9989]

With 2 eq of DIBAL H can we reduce carboxylic acids to alcohol ?

[@OrgoMadeEasy]

I don’t believe so! I think only LiAlH4 can do it.

[@probill4703]

Are enamiomers of the alcohols also made?

[@mathumithamurugan9491]

nice explanation but i need advantages of nabh4

[@bongamsomi6778]

Since NaBH4 is a mild reducing agent,can it reduce an ester?

[@OrgoMadeEasy]

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! :]

[@wadjo1]

Frank Wong Can LiAlH4 reduce alkenes ?

[@OrgoMadeEasy]

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)

[@superbvstudio]

Mechanism for carboxylic acid is wrong because there is acidic hydrogen so it will abstract Hydrogen but product will be the same.

[@OrgoMadeEasy]

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. 😊

[@Msnanabubu]

are you one of the wong fu's production brother? hahahaha

[@sanjoysaha7530]

Why NaBH4 is less reducing powr dan LiAlH4 or vice versa..

[@OrgoMadeEasy]

Sanjoy Saha Great question! I'm already one step ahead of ya ;) Here's that video: kzbin.info/www/bejne/pXvch4qprsyte8U

[@arnavpareek7854]

Can NaBH4 reduce carbon carbon double bond to single bond?

[@OrgoMadeEasy]

Nope, NaBH4 and LiAlH4 but target polar double bonds and there is no dipole between carbon carbon double bonds. =]

[@arnavpareek7854]

+Frank Wong Thank you sir

[@ivanbombana7282]

Arnav Pareek If you want to reduce C=C you have to use H2/Ni

[@SoundsOfTheWild3]

What is the stereochemistry? Is it retention or inversion.

[@OrgoMadeEasy]

You should get a racemic mixture of both enantiomers because the H have an equal chance of attacking from top/bottom.

[@shoutitallloud]

There's no explanaiton why NaBH4 can't reduce carb.acid to alchohol

[@EvenelShadowPierre]

Lol. Attacked twice. xD

[@OrgoMadeEasy]

;) 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.

[@EvenelShadowPierre]

I didn't find anything wrong with it.

[@OrgoMadeEasy]

Oh okay awesome! haha

[@tracywin9617]

nice guy shirt! wong fuuu

[@OrgoMadeEasy]

+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! kzbin.info/www/bejne/Z4LOeKuHeK5km9U

[@charleschidsey6192]

Loved Sleeping Beauty.

[@dada_giri]

look at aluminium..... how it is written😊😊

[@studiespotatoe5757]

Would prefer if you use "official terms" and not like "shoot in"... Thanks..

[@besmasemoud113]

who can give me an experimantal section of redection of acetophenone or benzophenon with lialh4

[@duff2233]

what wud u get if u react 2-butanal with NaBH4?

[@OrgoMadeEasy]

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.

[@duff2233]

sorry i got the compound name wrong...its... CH3-CH2-CH2(CH3)-CHO...1-methyl propanal

[@duff2233]

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?

[@OrgoMadeEasy]

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.

[@duff2233]

k thanks..:)

[@RamiNajjar]

u r amazing. Here's a no homo heart

[@OrgoMadeEasy]

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.

[@RamiNajjar]

Isnt the end product the same though?

[@OrgoMadeEasy]

Yup! Exact same end product, just slightly different mechanism.

[@arshsingh1205]

why is there no indian guy in this vediio

[@nat1XP]

Attacked twice

[@umakantyadav1182]

Hindi me babu ji

[@ismasiddiqui2986]

stop wasting time and explain direct please🙂