Orgo 1 final exam tomorrow, and I know I'm not alone. For those of you cramming like I am, I wish you the best of luck. One final push, comrades. Let's do this.
@@gigachad5029 its awesome its kinda like math word problem its like solving a puzzle just look at it as if its a puzzle and make up million possibilities in your head then analyze them closely then take the one which makes the most sense play the video and oh boy when u are right that feeling u get is so lit
@1:46 you say "the SN1 reaction occurs in a single step" For anyone else thatt might be reading this, it's supposed to say "the SN2 reaction" not "SN1"
@@kevinhan2581dude the amount of times my proff said “keep this in mind it doesn’t make sense now, but it’ll make sense later” And half the time never comes around to what he meant. randomly he says “see remember what I told you this would make sense in a bit which is now” and elaborates nothing on the topic leaving it a mystery beginning to end. Recently it was nucleophiles before it’s chapter and bro swear it was some mind blowing thing we wouldn’t get yet. 2 weeks later it’s just e rich species. Niceee set upp.
My Doctorate lecturer, I just take this opportunity to express my sincere gratitude for enabling me to pass my exams and cats. Actually, you are a hero to be remembered. I revised all the problems and guess what. Exam was just a celebration!!. Long live my master.
I have to criticize a bit. Sn2 has 2 in its name not because it is second order overall but because it is a bimolecular nucleophilic substitution, emphasis on bi. Sn2 can occur as a pseudo-first order reaction in case of solvolysis(when the solvent is the nucleophile and is in excess)! But otherwise, thank you for this video.
preci preci i now have a girlfriend and a 102 in organic chemistry thanks to ur videos. also own a lamborghini and live in a gated community in the hills thanks to your hard work
First look at the Alpha C. Is is a primary, secondary, or tertiary C? If it is Primary, it will always be Sn2 (minus a couple instances), if it is secondary you need to look at what you are reacting with and the protic/aprotic conditions along with the leaving group, if it is tertiary is always favors Sn1.
It’s implied when we discuss sn1/sn2. The biggest thing to remember (what I learned) is that sn1 is usually a 2 step process while sn2 are 1 step:) hope it helps. There’s basic rules to help you immediately see the rxn that will happen when u see the reagent with starting material. So primary, sec, and tertiary substrates helps me know whether the rxn will proceed in either sn1 or sn2. Check the Carbon:)
@kuewayneennis9521 Yes! I was also confused because at 1:47 he says "The SN1 reaction." while he meant SN2! I hope there aren't more confusions? Also how ironic rn, your comment is 2 yo for me 😂😂
Why does the intermediate molecule at 14:21 let go of the hydrogen instead of the methyl group? What makes it more favorable? Why would the methanol want the hydrogen? Wouldn't it have to dissociate for the OH- to grab the H+ and become H2O, leaving a methyl ion CH3+, which is kinda unfavorable?
In your first example, isn't iodine a better leaving group, and thus a weaker base, than bromine? So the reaction will not proceed, or will at least favor the reactants if the reactions is reversible
hi not sure if you can see this comment since this is a pretty old video,, but im open to anyone who can answer my question: for 17:43 product, is it not going to be a racemic mixture just like the previous example since H2O is a neutral nucleophile?
01:41 Dear Princess Celestia, today I learned that Iodide is a naughty backstabber :J (it's also a good nucleophile, they say) 06:02 What do you mean "leaves"? It can't just break that bond, withdraw its electrons and leave all by itself, can it? It it's better off alone, why would it form a bond with that carbon to begin with? :q 08:18 Is there any way to separate those two products from the mixture?
@7:48- how do we know if the Nu: attacks from the front or back?? I know that you said it should attack from the back and why, but is that always the case? Can you explain this further? Should it always attack from the back WBR the Nu: is a Br-?
at 4:05 you were drawing 2-bromo-2-methylpropane but as you can see that you made mistake for that methyl (to the left of this molecule.) C and H suppose to be switched or else I will think that the carbon in the center of the molecule is connected to the hydrogen and that hydrogen is connected to the other carbon which that is crazy.
The simplest answer is: we've just done a nucleophile-electrophile reaction when that oxygen attacked the carbocation, so the next step shouldn't be another nucleophile-electrophile reaction, but an acid-base reaction instead. And acid-base reactions are all about giving/taking protons (hydrogen nuclei). Also hydrogens are relatively easy to detach, because they're very small when compared with any other atom. They can just lose one electron to the oxygen and leave when someone else offers them another electron. Bonds between carbon and oxygen are harder to break, because carbon is not very happy either when losing or gaining an electron.
For the last molecule, would it actually have 4 steroisomers? The methyl-substituted carbon is an asymmetric center, and I assume the methyl shift could add to either side.
Well the 3° carbon is chiral and no because only carbon-carbon bonds count, so the 2° carbon is bonded with 2 other carbons thats why its 2° and the 3° is bonded with 3 carbons and so its 3°
Formal charge of the electron is +1 formal charge = valence electrons of the free atom - non bonding electrons - 1/2 of the bonding electrons oxygen has 6 valence electrons and here 2 are not bonded, and it has 3 bonds (each bond consists of 2 electrons) 6 minus 2 minus 3 +1 on the oxygen :)
Now, I could be wrong here, but maybe, just maybe, he meant to say SN2....Ever consider that, genius? Also, backside attacks can occur in SN1 reactions...Take a look at his example at 7:37
U said that "since bromide repels oxygen ....inverted product will be formed"....but this is Sn1 reaction so the leaving group have already expelled ...then there will be no repulsion for oxygen right ??? Kindly answer
I don't get how the bromide affects the water if it attacks from the front of the molecule, but not the back? It's all in the same solution, right? So, why can't the bromide affect the water if it's attacking from the back?
It's not really about "front" and "back", but about "one side of the plane" and "another side of the plane". That carbocation is sp²-hybridized, so its geometry is trigonal planar, and normally the water molecule could attack it from both sides. But the negative bromide ion is still hanging around nearby, attracted by that positive carbocation, and partially blocking the way from one side for that water molecule. Therefore, that water molecule has still a better chance of attacking the carbocation from one side (the one without the bromide ion) than the other (where the bromide is still hanging around and blocking its way).
He determines the rotation of the molecule, assigning power to the atoms of greatest atomic mass attached to the stereogenic center, Hydrogen is always 4 if present on a stereogenic center as it i's AMU is 1, Bromine is heavier than the other molecules so it is assigned as 1, then the Ethyl is 2 and the Methyl is 3. You don't do this if any of the substituents are the same, Ethyl and Methyl while both being carbon attached to the sterogenic center can be exempt as you just assign power to the next atom it is connected to for example Methyl vs Ethyl carbon vs carbon = no winner so we determine power with the next atoms hydrogen vs carbon = carbon wins as there is a second carbon in the ethyl group and thus assigned higher power. Feel free to correct me, but this is my understanding.
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thank you goat
l want work sheets
"let's not worry about stereochemistry in this example"
I think that is the sexiest thing a chemistry teacher could ever say
Orgo 1 final exam tomorrow, and I know I'm not alone. For those of you cramming like I am, I wish you the best of luck. One final push, comrades. Let's do this.
Its friking hard
@@gigachad5029 its awesome its kinda like math word problem its like solving a puzzle just look at it as if its a puzzle and make up million possibilities in your head then analyze them closely then take the one which makes the most sense play the video and oh boy when u are right that feeling u get is so lit
@@akashsunil7464 still annoying thi
@Orion Jaiden Bot says what?
i hate organic cant wait ti get this over with
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@@Jdkrjj ?
@1:46 you say "the SN1 reaction occurs in a single step" For anyone else thatt might be reading this, it's supposed to say "the SN2 reaction" not "SN1"
yes
Yes
I was wondering about that. Thank you!
Yes, I just caught that as well!
I wish I would've read this comment 30 mins ago!!! I had to pop out my textbook cause I was so confused!!!! Thank you! two years later lol
I hope this man gets everything he wants in life
💯 always saving lives!
You teach better in 15 minutes then my "professors" who leave me so confused after 2 hours.
Sure. Same for me. Professor said: you will learn it later on lessons. For me, It feels like pre-view and "waste" of time.
@@kevinhan2581dude the amount of times my proff said “keep this in mind it doesn’t make sense now, but it’ll make sense later”
And half the time never comes around to what he meant. randomly he says “see remember what I told you this would make sense in a bit which is now” and elaborates nothing on the topic leaving it a mystery beginning to end.
Recently it was nucleophiles before it’s chapter and bro swear it was some mind blowing thing we wouldn’t get yet. 2 weeks later it’s just e rich species. Niceee set upp.
yea... im fucked
brandon mora I’m with you brother
i have my chem test in 5 hours
AAARHHH how’d it go?
@@aaarhhh4341 12 hours remain for me
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You are god!! Nobody explained me this clearly!! Thank you! You are peak
My Doctorate lecturer, I just take this opportunity to express my sincere gratitude for enabling me to pass my exams and cats. Actually, you are a hero to be remembered. I revised all the problems and guess what. Exam was just a celebration!!. Long live my master.
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You guys have no idea how I've been battling to get the Sn1 and Sn2 reactions. Thanks
Biology is my subject, i was never really good at chem, but watching your videos clarifies alot.
This channel is doing the Lord's work. Thank you.
I have to criticize a bit. Sn2 has 2 in its name not because it is second order overall but because it is a bimolecular nucleophilic substitution, emphasis on bi. Sn2 can occur as a pseudo-first order reaction in case of solvolysis(when the solvent is the nucleophile and is in excess)! But otherwise, thank you for this video.
bimolecular literally means it is a second order reaction
@@okayhaffy no rate = k [a]^2 you would say this is still second order ,I thought. (I could be wrong)
The amount of times youve saved my ass is out of this world. Youve basically been teaching me my whole college career. Thankyou ❣
Im here starting college yes he is saving us FR
What year are you in? Personally in my school (Belgium) I’m learning this during my final year of highschool so I’m wondering
@@nauwwww bro same but not from Belgium
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You save my life! I have to give in a test on this in like 2 days.
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Giving u my first born child
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this is very useful. Thank you for the wonderful explanations as always!
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Sir you so amazing my teacher makes it more complicated n I came to a conclusion that chemistry is hard but u took me to another level
Understood more in five minutes here than I did from 2 weeks (about 4-5 hours) of lectures from my ochem 1 professor. .
preci preci i now have a girlfriend and a 102 in organic chemistry thanks to ur videos. also own a lamborghini and live in a gated community in the hills thanks to your hard work
I hope this isn't too forward but... I love you.
Good work, but it was very confusing when you switch from sn1 to Sn2 without saying anything so i cant tell which is which
First look at the Alpha C. Is is a primary, secondary, or tertiary C? If it is Primary, it will always be Sn2 (minus a couple instances), if it is secondary you need to look at what you are reacting with and the protic/aprotic conditions along with the leaving group, if it is tertiary is always favors Sn1.
It’s implied when we discuss sn1/sn2. The biggest thing to remember (what I learned) is that sn1 is usually a 2 step process while sn2 are 1 step:) hope it helps. There’s basic rules to help you immediately see the rxn that will happen when u see the reagent with starting material. So primary, sec, and tertiary substrates helps me know whether the rxn will proceed in either sn1 or sn2. Check the Carbon:)
@@yasssgawwwd5643 bruh that message Is2 years ago, and he just made some mistakes in the video tbh said sn1 when he meant sn2
@kuewayneennis9521 Yes! I was also confused because at 1:47 he says "The SN1 reaction." while he meant SN2! I hope there aren't more confusions?
Also how ironic rn, your comment is 2 yo for me 😂😂
I love this channel so much
Why does the intermediate molecule at 14:21 let go of the hydrogen instead of the methyl group? What makes it more favorable? Why would the methanol want the hydrogen? Wouldn't it have to dissociate for the OH- to grab the H+ and become H2O, leaving a methyl ion CH3+, which is kinda unfavorable?
In your first example, isn't iodine a better leaving group, and thus a weaker base, than bromine? So the reaction will not proceed, or will at least favor the reactants if the reactions is reversible
I hope you know how important your videos to us are. Thanks
I love your videos, they're informative and kinda asmr lol thank you so much.
1:46 don't you mean Sn2?
Yes SN2 is Concerted single step, while SN1 is multi step.
That's what I was thinking too
Does Carbonation formation occur in SN1?
hi not sure if you can see this comment since this is a pretty old video,, but im open to anyone who can answer my question: for 17:43 product, is it not going to be a racemic mixture just like the previous example since H2O is a neutral nucleophile?
Its me from Nepal
Your videos are extraordinary sir 🥰
Hello! This was awesome, thanks alot for your tutoring!! It helps alot😁
thanks for your videos dude, very helpful!
01:41 Dear Princess Celestia, today I learned that Iodide is a naughty backstabber :J (it's also a good nucleophile, they say)
06:02 What do you mean "leaves"? It can't just break that bond, withdraw its electrons and leave all by itself, can it? It it's better off alone, why would it form a bond with that carbon to begin with? :q
08:18 Is there any way to separate those two products from the mixture?
@7:48- how do we know if the Nu: attacks from the front or back?? I know that you said it should attack from the back and why, but is that always the case? Can you explain this further? Should it always attack from the back WBR the Nu: is a Br-?
magnificent, all 360p's of it
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correct me if I'm wrong, but I'm pretty sure that the second example can only work with E2. SN1 does not work under strong basic conditions.
Substitution reactions prefer formation of the more stable base. Reacting iodine with an alkyl-bromide would produce no reaction. Please clarify this
at 4:05 you were drawing 2-bromo-2-methylpropane but as you can see that you made mistake for that methyl (to the left of this molecule.) C and H suppose to be switched or else I will think that the carbon in the center of the molecule is connected to the hydrogen and that hydrogen is connected to the other carbon which that is crazy.
I need to know this for a test in less than 1.5 hour. Thanks.
Very good effort👍
this video helped me out a lot
My guy, you are a B L E S S I N G
Thank u so much for this video
does SN2 and SN1 work for alkane, alkene and alkynes or only for alkyl halides?
Why will methanol remove the hydrogen and not the methyl Group on the oxygen? At 14:18
Is there any law on why it pick that?
The simplest answer is: we've just done a nucleophile-electrophile reaction when that oxygen attacked the carbocation, so the next step shouldn't be another nucleophile-electrophile reaction, but an acid-base reaction instead. And acid-base reactions are all about giving/taking protons (hydrogen nuclei).
Also hydrogens are relatively easy to detach, because they're very small when compared with any other atom. They can just lose one electron to the oxygen and leave when someone else offers them another electron. Bonds between carbon and oxygen are harder to break, because carbon is not very happy either when losing or gaining an electron.
Is this for the AS level?
Thank you so much sir !
Not me checking how to do this 5 minutes before the exam 🎉
the fact that i can understand him and not my professor!!!!
For the last molecule, would it actually have 4 steroisomers? The methyl-substituted carbon is an asymmetric center, and I assume the methyl shift could add to either side.
Very helpful
so helpful!
At 17:23, Why isnt the other carbon chiral? Also aren't the two carbons 4° and 3° respectively? Someone help!
Well the 3° carbon is chiral and no because only carbon-carbon bonds count, so the 2° carbon is bonded with 2 other carbons thats why its 2° and the 3° is bonded with 3 carbons and so its 3°
how do you know if it attacks from behind or in front
THANKYOU LIFESAVER!!!!
What is the link to video that has 75 practice problems?
on the last problem, how do you which is r and s? there's no methyl on a dashed or wedge bond.
Did you mean to say Sn2 at 1:46?
I think so
have my natural products test on Friday haha
at 1:46 you say "the SN1 reaction is a concerted reaction" when it should be SN2!
There are a few strange things:
1. I is a better LG than Br so it shouldn’t react
2. Methanol is acidic and would not accept protons
i tought I- was a better abandonat group than Br- . but in the video you use Br- as the abandonat group. i am kinda cofused. can somebody explain?
14:10 wait why does this intermediate form? Oxygen seems to have 8 electrons so why does it bare a positive charge?
Formal charge of the electron is +1
formal charge = valence electrons of the free atom - non bonding electrons - 1/2 of the bonding electrons
oxygen has 6 valence electrons and here 2 are not bonded, and it has 3 bonds (each bond consists of 2 electrons)
6 minus 2 minus 3
+1 on the oxygen
:)
do your videos have any order and organization? I don't want to jump from one video to the next.
1:46 is Sn2 (not Sn1)
yup, you are right.
My exam on this is today 😂😂
Any practice question for reaction with formic acid?
So, what type of nucleophile will determine if it will be a SN1 reaction or a SN2 reaction?
Why does the oxygen develop a partial positive charge instead of negative when bonding to the carbocation?
Guys, there is never a backside attack on an Sn1 reaction... Don't know what he was talking about there around 1:46
He said it is SN2 not SN1
@@StfuSiriusly nah he said sn1, he just misspoke there, he is talking about sn2
Now, I could be wrong here, but maybe, just maybe, he meant to say SN2....Ever consider that, genius?
Also, backside attacks can occur in SN1 reactions...Take a look at his example at 7:37
@@xantherxavier5429 obviously, doesn't mean it's not gonna confuse the crap out of some students
Ever consider that genius?
@@xantherxavier5429 Why such a dick? A glaring mistake like that is the video makers fault, people are allowed to call him out on it
U said that "since bromide repels oxygen ....inverted product will be formed"....but this is Sn1 reaction so the leaving group have already expelled ...then there will be no repulsion for oxygen right ??? Kindly answer
Hello
How do you know what's front and what's back? Isn't that relative?
Thanks
Thank you so much
What else could replace this method ? -In case it is needed.
What type of stereoisomers are the molecules in the last example?
Overall, this video is very helpful. However, I'd like to point out that at many points in the video, some terms are not very clearly pronounced.
I don't get how the bromide affects the water if it attacks from the front of the molecule, but not the back? It's all in the same solution, right? So, why can't the bromide affect the water if it's attacking from the back?
It's not really about "front" and "back", but about "one side of the plane" and "another side of the plane". That carbocation is sp²-hybridized, so its geometry is trigonal planar, and normally the water molecule could attack it from both sides. But the negative bromide ion is still hanging around nearby, attracted by that positive carbocation, and partially blocking the way from one side for that water molecule. Therefore, that water molecule has still a better chance of attacking the carbocation from one side (the one without the bromide ion) than the other (where the bromide is still hanging around and blocking its way).
Can you please reply asap. Is 1:46 right or not because I dont understand
he made a mistake. SN2 reactions are always one step while SN1 are multi-step reactions. Hope it clears it up for you
Is there any influence in percent of racemic mixtures if the bromine is put in front or at the back of carbon atom?
No because all reasoning is reversed
hmm why these guys are explaining things to us on youtube?!! this should be my professor
thanks!!!
You're welcome Ana.
Can someone explain what he does at 2:18?
He determines the rotation of the molecule, assigning power to the atoms of greatest atomic mass attached to the stereogenic center, Hydrogen is always 4 if present on a stereogenic center as it i's AMU is 1, Bromine is heavier than the other molecules so it is assigned as 1, then the Ethyl is 2 and the Methyl is 3.
You don't do this if any of the substituents are the same, Ethyl and Methyl while both being carbon attached to the sterogenic center can be exempt as you just assign power to the next atom it is connected to for example
Methyl vs Ethyl
carbon vs carbon = no winner
so we determine power with the next atoms
hydrogen vs carbon = carbon wins as there is a second carbon in the ethyl group and thus assigned higher power.
Feel free to correct me, but this is my understanding.
Isn’t iodide weaker than bromide making it unable to remove it?
Thank you! But why does the carbocation forms? I mean, how come the carbon-bromide bond just splits?
you dont really need to know those but it is because of the dipole interaction between carbon and bromine.