E2 reactions | Substitution and elimination reactions | Organic chemistry | Khan Academy

Courses on Khan Academy are always 100% free. Start practicing-and saving your progress-now: www.khanacadem...
E2 Elimination Reactions. Created by Sal Khan.
Watch the next lesson: www.khanacadem...
Missed the previous lesson? www.khanacadem...
Organic Chemistry on Khan Academy: Carbon can form covalent bonds with itself and other elements to create a mind-boggling array of structures. In organic chemistry, we will learn about the reactions chemists use to synthesize crazy carbon based structures, as well as the analytical methods to characterize them. We will also think about how those reactions are occurring on a molecular level with reaction mechanisms. Simply put, organic chemistry is like building with molecular Legos. Let's make some beautiful organic molecules!
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to Khan Academy’s Organic Chemistry channel: / channel
Subscribe to Khan Academy: www.youtube.co...

[@pontos78]

2 important things Sal forgot to mention. E2 reactions occur at very high temperatures, and they need a bulky base to prevent an SN2 reaction from occurring.

[@ish.theticaly8175]

11 years ago??? Omgggggg!!!

[@nataliabrand5182]

Thanks! I learned more with you in 4 videos then in 1 year at Rutgers lol

[@benwilliams1889]

Sal, you're still the undisputed master at explaining Organic Chemistry. Great video. Really wish you taught Orgo in every Khan Academy video.

[@Thinobard]

This is wrong. For an E2 reaction to take place the geometry must be anti periplanar. The alkyl halide must be in perspective formula to be a correct mechanism.

[@Smick10]

who cares

[@ShotClanTxter]

Smick10 i care, and hes 100% correct the molecule needs to be in anti comformation in order for an E2 reaction to occur

[@Smick10]

It's not like this happens in real life anyways, this is all just theories

[@ShotClanTxter]

Smick10 yes it is a theory, not a hypothesis. A theory is heavily backed by experimintation and years of research. It is very likely that it is real that is why you can recreate an e2 reaction in the lab. If you understand. Plus you should care if youre learning off this video you dont wanna get a question wrong on a test

[@Smick10]

I graduated in biomedical physiology years ago and now in medical school. I'm using this video to tutor my younger brother

[@curtpiazza1688]

You explain this so clearly! 😊

[@ilickcatnip]

His voice is so good, you just can't do anything but pay attention.

[@Oodstech]

The repetition might not be pleasant, however it IS useful.

[@bemnetgebru1101]

men...i love this guy....

[@lilangel0072]

i thought for E2 reactions the leaving H and Cl had to be antiperiplanar???

[@preethivenkatesan6940]

Yeah it has to be in anti periplanar in E2 reaction

[@606cole]

Will you be my Organic Professor! You are so much better than the one I have currently!

[@aimezmoi13]

What program do you use for these animations?

[@DopeFreshFitness]

No,Na+ and OCH3- where both free ions in the solution, "replace" usually suggests a causal relationship, and that H certain didn't cause the reaction to happen, rather OCH3- did.

[@TheMusicKing0906]

u are just a life saviour...........

[@dishonesttAbe]

You should cover quantum mechanics after the chemistry is finished. You make difficult math concepts easy to understand so I imagine you can do QM very well.

[@ankithsmenon1935]

You are one of the oldest comments let's see if you reply after 9 years

[@dishonesttAbe]

@@ankithsmenon1935 hi

[@ankithsmenon1935]

@@dishonesttAbe wow if you don't mind me asking what do you do now

[@dovelygodess4661]

Yay Khan Academy! To the rescue again!

[@DicedPeas]

@antonyfl Since the base is so strong, E2 will still occur here. Polar protic solvents CAN be used for E2/SN2 under the right conditions. Since polar protic solvents are much cheaper, this is actually how SN2/E2 is carried out industrially.

[@Dccole3214]

Strength of base and nucleophile take priority over solvent when determining which type of reaction it is.

[@danielafernandez6169]

The ch3o attacks that specific hydrogen because, according to zaitsev's rule, the hydrogen tnat should leave is the one bonded to the less hydrogenated carbon

[@ShimmyMD]

@dalcde Yup it does. We just leave it out bc they're not organic products.

[@TheMiiirna]

God bless you man.

[@s0m0c]

Gracias.

[@MsHellogoodbyee]

I also believe this would be best in an E1 reaction. The carbocation would be stable as tertiary and it's a polar protic solvent!

[@FindingCreatures]

The hydrogen that is taken in an elimination reaction is anticoplanar

[@BlueDeath88]

These videos are great supplemental learning tools!

[@zoudek123]

Too helpful

[@soulmatesr64]

thanx for help

[@antonyfl]

Quick note Sal: This reaction isn't the best to show an E2 mechanism because the solvent is CH3OH, a polar protic solvent, which actually would favour SN1/E1 . I think a better solvent for this example would be DMSO

[@sparky1807]

DMSO is polar aprotic and would favour SN2. E2 mechanisms favour polar protic.

[@starlight200816]

Seriously amazing

[@soulmatesr64]

thanx

[@shivanshsinghslanki5945]

That's too good.

[@miso_honey]

Thank you so much for doing this.

[@chijade2813]

shouldn't the Cl- bond with the Na+ and from NaCl since they are both ions?

[@josam4545]

that's what I was thinking. Did you ever figure out the answer?

[@ktkc1o7]

@@josam4545 If they are close enough to each other then they will bond. Unless they are dissolved in a solution. Which they are in this video. Just think back to hydration shells from gen chem.

[@bonbonpony]

Yes, that's what they will do as soon as you start evaporating the mixture. Normally these ions just float around dissolved in the solvent (surrounding by its molecules), but as soon as you start decreasing the concentration of the solvent (by evaporation), cations and ions start meeting each other and attract to form ionic compounds (salts). They will crystalize and precipitate out of the solution as NaCl (table salt).

[@soulmatesr64]

how can we know is the given base is strong or weak? is there any priority order for it? plz clear my confusion....

[@bonbonpony]

It's all about how much it wants that proton. The more it wants it, the stronger the base is. And how much does it want that proton depends on how unhappy it is without it. Negatively-charged atoms are very unhappy, they will grab the first proton that they can detach from another molecule (if it is detachable enough, i.e. acidic) to become neutral again. If an ion is big enough, it can bear the negative charge better because it can distribute it over a larger volume, which decreases charge density, and this makes it less likely to want that proton (weaker base). Resonance can stabilize that negative charge too, by distributing it over multiple atoms, and in that case it's hard to decide which of those atoms would get that proton, so it doesn't want it that much (even weaker base). Having no negative charge watsoever makes it even weaker base. Atoms with lone pairs of electrons can sometimes act like a base, using this lone pair to grab a proton, but only if someone pushes it at them, or from someone who is even weaker than them. But if an atom doesn't have any free electrons to share, or - worse still - it's got a positive charge, it can't act as a base at all.

[@starlight200816]

I love this

[@alexhahahahaha]

Man I love you!

[@elvenator]

@egyzzzz70 & martin8768 Boo! The repetition is one of the best parts. It helps with the memory.

[@proMediic]

E1 elimination only works with 3 degree halides

[@JEHill86]

Find the pKa in a literature source.

[@tasnimmohamed9566]

The differance between E2 and E1 is the transition state true????????

[@arshpatel5306]

I thought the leaving group and the H its was attacking were supposed to be trans to each other, 180 degree?

[@ktkc1o7]

Why did the methoxide take an H from the 3rd C instead of the 1st C? Presumably because the 3rd C is a secondary C and the 1st C is only a primary C and that affects the stability somehow. Do you have a video on this?

[@bonbonpony]

Stability of carbocations or carbanions is more important for SN1 and E1, but it's rather irrelevant in SN2 or E2 because in type 2 reactions carbocation intermediates don't form. Instead, base takes the most acidic proton in the molecule, which in elimination reactions is the β carbon (the one right next to the α carbon that the leaving group - in this case the chlorine - is attached to). So it can be either carbon 1 or carbon 3. And in the beaker, you will actually get both of these products. But one of them will dominate: the one predicted by the Zaitsev's rule, that is, the one in which the C=C is more branched.

[@Bryan-jv1vj]

I thought E1 reactions only happen for anti elimination?

[@cvjkctr]

Couldn't the OCH3 anion directly attack the Carbon with the Chlorine on it?

[@willis12491]

I was taught that och3- will act as a nucleophile and push off the cl

[@bonbonpony]

@@willis12491 Normally it could, but note how the solvent is polar protic, which weakens the nucleophile by solvation. Not only that, but the solvent is also the conjugate acid of the methoxide, which is a strong hint that an acid-base reaction will be favored here. Acid-base reactions are fast. The methoxide would rather just grab the most acidic proton in the molecule to become neutral methyl alcohol than go through the burden of nucleophilic attacks with bigger molecules. It can literally just grab the proton on its way to the electrophilic site.

[@TheBigEz]

@lilangel0072 only if its cyclohexane or something where you can draw chair conformations.

[@urdouchbag]

@antonyfl exactly what was going through my mind...

[@nihayaturrohmah4230]

#34

[@johnmandrake8829]

its okay the solvent is protic?

[@dpavankalyan660]

No for e2 we have to take aprotic solvent.. In case on e1 protuc will involve.

[@meetadityanihlani]

He first said the first carbon doesn't needs the Oxygen's electron so it passed to another carbon and then finally gave it to chlorine because it was very electronegative. But then again he said the hydrogen gets bonded with oxygen. this is sooo confusing >.

[@meetadityanihlani]

why Chlorine becomes an anion after gaining the electron? it already completes its configuration and hence number of protons=number of electrons so it should be neutral ?

[@eragon2121]

Chlorine has only 7 protons. If it has 8 electrons, it should be negative. ^^

[@DannyLrides]

the higher the pKa value...the stronger the base

[@bonbonpony]

It's like saying that the higher the temperature, the hotter it is. It's true, but it doesn't _explain_ anything.

[@rubezmess1]

or DMF :)

[@eerriiccccc]

correction *sodium methoxide not nitrogen methoxide

[@andrewlin7247]

Why did you need sodium methoxide? Would this reaction happen with sodium hydroxide?

[@bloggervista]

yes . OH- will be good electrophile and attack hydrogen and form double bond between alpha and beta carbon

[@agent475816]

+Andrew Lin Sure. any strong base will go through e2 or sn2.

[@karangujar317]

+Shirshak Bajgain lol,oh- is nucleophile.

[@psmanici4]

you saved me from my tutor

[@risingsun_gt]

@egyzzzz70 Thanks Einstein...

[@camilomunozschuler531]

Why alcoxyde didnt attack the left side Hβ to form 1-butene?

[@agent475816]

+Camilo Muñoz Watch the Zaitsev's rule video. Since a secondary alkene is more stable than a primary, the Zaitsev product is more thermodynamically favorable. The product you said is a Hoffman product, it is kinetically favorable. If you have a base that has a lot of steric hindrance, it will not be able to attack the hydrogen in a secondary or tertiary carbon so it will go with the carbon with least substituents. Since Methoxide is not sterically hindered, but-2-ene (the Zaitsev product) will be the dominant product. You will get some but-1-ene but very little because reactions go the path of least energy.

[@willis12491]

I was taught that Meo- acts as a nucleophile....

[@FourTwentyMagic]

Elimination and substitution reactions often compete

[@eerriiccccc]

why does the electron on the oxygen of the nitrogen methoxide compound attack the specific hydrogen on the third carbon? why not a hydrogen from the 1 carbon methyl group?

[@bonbonpony]

Zaitsev's rule. It can attack both, and it will in the beaker, but the major product will be the one in which the C=C is more substituted (more branched).

[@risingsun_gt]

@egyzzzz70 Trust me, you can never repeat this stuff enough. I've been reading/studying these reactions for weeks and still dont have a full grasp on them.

[@pchenester]

why is that H next to the carbon with the cholrine being knocked out?

[@bonbonpony]

Because it's more acidic due to its close vicinity to an electron-withdrawing atom (chlorine).

[@unit39fight]

can we say that Hydrogen replaced Sodium in NaOCH3 ????????????

[@robodoctor]

@aimezmoi13 ms paint

[@Soviless99]

reagents get so friggin confusing

[@prateeksharma291]

hey can you be a bit slow plzzz

[@prateeksharma291]

but nice

[@rdarmstrong100]

wtf