E1 Reaction Mechanism With Alcohol Dehydration & Ring Expansion Problems

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[@prettybird865]

You're actually the best dude. If I don't understand something about chem, you're the first place I look. Perfect voice for explaining things too!

[@tommyguir3820]

I watched 6 ads during the course of this video..... youtube is really pushing it

[@miriamnganga2561]

I know right 😪

[@ShreyanshGupta-cz6lr]

You can use revanced extended if you use android
For browser and adblock should easily work

[@SuloveKaushik]

I sometimes think what would happened of a person like me if a tutor like you was never there on TH-cam😢❤❤

[@sarahholland5980]

thank you for going through major/minor products! my professor rushes through it without explaining and I was so confused.

[@munch6814]

Thanks sooo much. It really helped. Please upload more like E1,E2,SN1,SN2, Stereoisomer etc.,plssss

[@randhirkotwal9738]

We are very thankful for this...

[@randhirkotwal9738]

Sir please upload more videos on organic chemistry

[@phil1pd]

A playlist on migratory rearrangements would be a helpful addition.

[@omagaomaga6313]

This dude is simply good at what he does.

[@mrmusaa2260]

I was looking for the major and minor products. Surprisingly, I found the hydride and methyl shifts!

[@tlili3990]

You're so much better than both of my professors

[@indianchika123]

i seriously have a crush on you

[@walterwhite4699]

awkward...

[@Gboy163]

Ayo 🤔

[@shaqaman1862]

Pause.

[@inexp1]

@@shaqaman1862 feel free to attempt this question

[@esliaddaccount1]

Thats like having a crush on ur professor

[@blinkybit]

11:00 you are never going to break that c-c sigma bond ... There'll be a 1,2 hydride shift to rearrange the carbo-cation into the ring and then the double bond will form. It will end up being an exocyclic alkene on the 2-carbon

[@geethacreations8915]

At 12:04 can the double bond be formed at methyl? Like outside the ring...

[@preetyluli7617]

Thanks so much🤗🤗
May God bless you😇😇

[@somgautam-yi4wk]

7:41 i would like to make a correction - although you got the correct majar product but since you made the cabocation you should had also done the hydride shift from left alpha H . by this you would get a more stable carbocation and 2 alkenes..HOPE YOU WILL DO THE CORRECTION

[@vihaanchandane8607]

u are wrong here, cuz the right one has +I effect of ethyl and a methyl group, whereas the left carbon has +I effect of two methyl groups, hence there will be no hydride shift as it decreases stability of carbocation

[@tangohdalij2590]

Very good and understandable video

[@user-bu6sp7dw8f]

PHARMACY STUDENT IS HERE
THX

[@theresa.gerges]

what is the difference between E1 and E2 mechanims? Like when do we use E1 and when do we use E2?

[@Future_iitian_]

​@xianhezhang3070nooo you are so wrong
E1 is used when good leaving group and weak base
E2 is used when bad leaving group and strong base attacking

[@sahitiputcha]

Your voice is awesome......:)

[@nostro1940]

1:00 is the subproduct a H30 (-) ? cause it loses a proton?

[@zaminahmed297]

You are great dude. I wish all chem professors were like you. A suggestion I would give though is could you make some of the videos a little shorter?

[@arditehrani1186]

So many thanks for your great simple educational videos!

[@GG-co2tb]

When do you use EtOH vs KOtBu/-OtBu?

[@a.i1137]

11:52 "basically" make the double bond..... I see what you did there

[@SophieSong-vf7bz]

Thank you for the clear explanations! For the last example, can EtOH attack the H on the left side and give the same product?

[@classicalcole7312]

Yes there is a line of symmetry down the center, so that should give you the same product.

[@shainablanco4051]

This is so Helpful hehehe😁

[@nostro1940]

10:17
1 carbon missing

[@chipp255]

9:00 where did that hydrogen come from?

[@bjswamy3193]

But isn't dehydrohalogenation E2 reaction? Why is it showing E1 mechanism in the video (for that ring expansion question)

[@msadjei9147]

I think it's because of the solvent, it's not strong enough for E2

[@camy563]

Thaaaank youuuu!!!!

[@talybodzhakow2641]

Thank you

[@randhirkotwal9738]

I know this week is busy for you in uploading videos..

[@nikodg194]

I bet that hso4- is too weak base to deprotonate anything

[@akshanshuanand9493]

How is H2O acting like a weak base here? Shouldn't it donate an electron pair to the carbocation and follow Sn1 instead of E1?

[@aylora7033]

are u saying because the strong acid reacted with the R-OH which is a weak base and donated a proton that H2O+ is now is a conjugate strong acid? H2O+ left as H2O and became a good stable leaving base. besides, Sn1 wouldnt happen because there is heat involved

[@Future_iitian_]

H20 is neutral so it's conjugate base is also weak

[@alexia7748]

Hello, why does the leaving group leave?

[@Moonlight-xl8wh]

as in e1 carbon becomes bulky and 3 degree carbocation is stable enough so halogen has to leave

[@Future_iitian_]

​@@Moonlight-xl8whkindly don't give false information
If u don't know just shut up and sit down

[@Future_iitian_]

Dear Alexia , leaving group leaves because the solvent used is polar protic
Polar solvents tends to break the bond which are ionic by dipole dipole interactions
If you have any other doubt you can ask me
I am from IIT (the institute which you get after giving the toughest exam of world)

[@TheHolyQuran2212]

​@@Future_iitian_what's protic I can understand polar but what's polar and non polar protic

[@Future_iitian_]

@@TheHolyQuran2212 the solvent which has a proton to share is called protic (H+)

[@shayneomale2776]

Really good, but cant say do this question but then introduce new stuff. Get's quite frustrating.

[@ngjackie7213]

Omg so many vids

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what is the plus icon?

[@RandomMadina]

that's your positive charge

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I love you

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flop, no me gusto

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Any Indian ?

[@Aman-b5k]

Yup, wassup though?