E2 E1 Sn2 Sn1 Reactions Example 3
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- Опубликовано: 17 сен 2010
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E2 E1 Sn2 Sn1 Reactions Example 3
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Here if u want to decide whether sn1 or e1 will dominate...... it depends on the temperature..... higher temperature favours e1(generally endothermic) while lower temperature favours sn1(generally endothermic)!
i mean exothermic** in the last
rearrangement can also occur right? as ders a carbocation intermediate
Khan, thank you. You save my life before every exam!! I'd donate to your academy, but I can barely feed myself hahahaha
you are so informative and I love your didactic tone. Keep em coming!
Finally, something that makes sense! It wasn't that clear to me what & why certain conditions favour Sn2, E2 or Sn1, E1 but this video made all the difference ^^
you are a great teacher. thanks for posting this video.
Thank you so much Sal!!!!
fantastic! really cleared things up!!!! love you sal! xo
thanks a lot! Very helpful
you are amazing!! thank you
Tnx for this useful video really appreciate it
In Sn1, wouldn't the Hydrogen in Methanol react with Iodide instead?
Naming, Orbital-system/atom model, electron configurations, electronegativity, valence-electrons, enthalpy, (strong/weak) base and acid, Sn1, Sn2, E1, E2 - reactions and some "rules" to orientate oneself... than you a base in (organic) chemistry. Maybe the key is the understanding of the orbital-system/atom model and the ability to juggling with electron configurations and electronegativity. - Thank you. Keep on going.
YAY chemistry! And yes, it is. Sometimes solvents are part of the reaction.
that is correct, b/c if it wasn't a planar species then the rxn mechanism can't take place due to not being able interact.
why rearrangement of C+ does not occurred (due to ring strain) it had chance of forming 6 member ring ?
FYI guys, at 9:20 (the 3rd step of SN1), Sal removes (deprotonates) the H+ from the newly formed molecule because all alkyloxonium ions are strong acids and will remove (dissociate) all the H+.
you always narrow it down between two (sn2/e2 or sn1/e1) how can you narrow it even more? bc that won't work on my exam… like I can't differentiate between strong nu/strong base and weak nu/weak base. based on that I'm supposed to be able to know what reaction occurs.
ily guise omg. I can't learn from an online etext and homework from WileyPlus, it's impossible. You guys are the best.
Slight correction? When the LG leaves, and the carbocation is formed, I believe it is a planar species, where as he has one Methyl group coming out.
I have a question because I haven been told that in sn1 ring expansion from pentane to cyclohexane will even been more stable than a tertiary carbocation?
yess you are right there will be a ring expanion from 5 carbon ring to 6 carbon ring because 5 member ring have more angle strain than 6 member ring and 6 membered ring is highly stable
Why ring doesn't expand ? There was a possibility of converting cyclo pentyl to cyclo hexyl, by methyl shift..?
for the e1 product is there any specifics on cis/trans?
I think that for E1 reaccion you will have two products; double bond with the C monosust and with the C none sust.
I was told that in an sn1, the leaving group, in this case iodide would attack the H and would form HI. How accurate is that?
It might in some reactions, depending on what electron removing groups you are working with, but it shouldn't (in the scope of this class) ever be the major product. Generally your professor wants _that_. Hydrogens, generally speaking, aren't good leaving groups. That's why we often have to introduce pyridine when we're doing regiospecific substitutions.
A good leaving group is a highly electronegative atom that, when it leaves, makes a relatively stable anion. The halogens tend to make good leaving groups.
ePsycho67 or if it had a charge
So what do you do with the Iodide ion? nothing? is it just another side product? because I can't help but think that the Iodide ion, with it's negative charge, can act as pretty good nucleophile to use in the Sn1 reaction to remove the extra H from the methanol that just bonded to the Carbocation. So instead of having another Methanol attacking a previously added methanol, isn't it better to use the Iodide ion as a nucleophile? Because I think Oxygen would prefer to have 2 bonds and 2 pairs of electrons and not 3 bonds and only 1 pair of election along with a positive charge--Oxygen is pretty electronegative, so I don't think having a positive charge on Oxygen is a good idea. Am I wrong?
It is stabilised by the protic solvent
Why it didn't go through ring expansion?
Im going to say it like mark hoppous: THANK YOU....THANK YOU.
for those who are watching this: Don't u think the the product get from E1 reaction should be a Zaitsev product instead of the Hoffmann product as it is given in the video? think carefully about the regiochemistry of the product for E1.
Also, in the Sn2 reaction, how do we take into account stereochemistry? Do we end up with a racemic mixture of R and S?
Yes, and also cis and trans
the carbocation intermediate would undergo a methyl shift, so the positive charge would be on the previously 4° carbon on the ring, in e1 the double bond would therefore be located in the ring, in sn1 the MeO would be attached to the previously 4°carbon making it 4° again
methyl/hydride shifts only occur when it produces a more stable carbonation. In this case, the carbonation is 3*, and the methyl shift would result in a 3* carbonation as well. So no methyl shift would occur because it would not produce a better stable carbonation, both are equal in stability.
@@robertlopez616wouldn't the stability of the entire molecule increase with the methyl shift? If you saw an electrostatic potential map of the molecule it would be quite different
@@max3eey Both are equally stable. But methyl and hydride shifts have a small activation energy requirement. So it depends on the thermodynamic and kinetic stability of each product, and is based on the experimental conditions. It "regular" cases, the thermodynamic product prevails, and in this case is produced as the majority product. Small amounts of kinetic are produced. If that makes sense. Its been a year so I don't remember what part in the video this comment refers to.
@luoshaoxiongmicky i think he did give the zaitsev
Why would the hydrogen from the positively charged methanol bond to another uncharged methanol when there is now a negatively charged iodide anion floating around? Does it have to do with the amount of methanol in comparison to iodide?
HI isnt a weak acid so it disassociates in solution
why not give away the H in the weak base in the sn1 reaction? why did you give away the atom from oxygen?
You explain everything really well but sometimes I wish that you wouldn't repeat yourself every 2 seconds and just move on to the next step. But great work though!
at 12:20, doesn't he mean the alpha carbon and not the primary carbon?
I don't know what a primary carbon is, but you're correct in saying he was referring to the alpha carbon.
Why doesnt the iodine bond with the methanol ion now?
+shavit34ss kinda late but anyways the methanol ion already has 3 bonds and a positive charge.
If anything one of the hydrogen's will break off the ion giving the charge back to the methonal and making it neutral.
Then the Hydrogen will bond with the iodide anion as the anion gives an electron to H making hydrogen iodide (HI)
i was told by my teacher that Sn1 is the only one that occures when you have the all the qualifications you mention plus polar prodic solvent ....she said that E1 would not occur in prodic solvent but u said it did so im kind of confused?
Ana Murillo sn reactions happen in polar APROTIC solvent.
@faceclutch why not 2-Cyclopentyl-3-Iodo-3-Methyl Pentane?????
#39
Haishhhh this organic chemistry ruind life
Naming compounds:
2-cyclopentyl-3-iodo-3-methylpentane (before the reaction)
4-cyclopentyl-3-methylpent-2-ene (product)
Anyone else agree?
Why didn't the H+ combine with I- ?
Because HI isnt a weak acid
The molecule looks like a stick figure
"..it doesn't have a neutron"