Why didn't you just rotate your methyls (the one on the plane, second on the dash) counter-clockwise. You'd end up with H on the dash, methyl on wedge and the other methyl on plane?
Can you please explain why in the first problem with E2 the double bond wasn't formed between the more substituted carbon? I wasn't told by my professor when you do the hoffman product
also at 26:00 why wouldn't you do E1 instead since the Beta carbon is more substituted. I was told that if you're picking between sn1 and E1, if a beta carbon is substituted you should side with E1.
SN1 vs E1 analogy can be subjective sometimes. Typically E1 is preferred at higher temperatures over SN1. But some books may have that E1 is always the minor product when dealing with alkyl halides regardless of temperature and would be major when dealing with alcohols dehydration at higher temperature. But at the end of day, with alkyl halides E1 and SN1 always tag alongs with one another and major deciding factor is usually temperature.
In problem 14:24, let's say that the CH3 was on a solid wedge, would Zaitsev's rule then apply and the double bond would attach to the CH3? (Since the hydrogen would now be behind and anti-periplanar to the Br)
Hello sir. Thank you for the perfect explanation for every reaction. At 12:40, after you placed both chlorine and hydrogen in "anti" positions on the plane, should we consider the factor of "carboncation rearrangement"
@@visuallearners1832 Got it! Thank you sir for your quick reply! I totally got it coz what confused me before is that I kept doing the same thing as we did over SN1 and E1 :(
@@max3eey there is nothing wrong with the process on paper. Mechanistically, you can use any CB or water to deprotonate at the last step. Practically in labs, these reactions involves a weak base work up as last step so remove the proton.
At 23:05, ring contraction should occur.
A major product is not a six membered ring but a five membered ring.
i thought so too
Ring contraction will result in a primary carbocation which is very unstable
@@g3itnalRing contraction will result in a tertiary carbocation by hydride shift, which is more stable and very stable cation.
At 25:00 I dont understand what you said about chiral centres. Can you please explain?
Why didn't you just rotate your methyls (the one on the plane, second on the dash) counter-clockwise. You'd end up with H on the dash, methyl on wedge and the other methyl on plane?
Can you please explain why in the first problem with E2 the double bond wasn't formed between the more substituted carbon? I wasn't told by my professor when you do the hoffman product
Bulky base prefer Hoffman elimination as major product.
Use this video for breakdown.
th-cam.com/video/8xYtRCnOr3w/w-d-xo.htmlsi=xttw_41bh15j4idu
tysm@@visuallearners1832
Thank you so much 🙏🏻
also at 26:00 why wouldn't you do E1 instead since the Beta carbon is more substituted. I was told that if you're picking between sn1 and E1, if a beta carbon is substituted you should side with E1.
SN1 vs E1 analogy can be subjective sometimes. Typically E1 is preferred at higher temperatures over SN1. But some books may have that E1 is always the minor product when dealing with alkyl halides regardless of temperature and would be major when dealing with alcohols dehydration at higher temperature. But at the end of day, with alkyl halides E1 and SN1 always tag alongs with one another and major deciding factor is usually temperature.
In problem 14:24, let's say that the CH3 was on a solid wedge, would Zaitsev's rule then apply and the double bond would attach to the CH3? (Since the hydrogen would now be behind and anti-periplanar to the Br)
OH- and OR- , acts as Nu for 1* and acts as base for 2* ?
Thanks it is so helpful❤❤
Hello sir. Thank you for the perfect explanation for every reaction. At 12:40, after you placed both chlorine and hydrogen in "anti" positions on the plane, should we consider the factor of "carboncation rearrangement"
No. E2 do not make carbocation.
@@visuallearners1832 Got it! Thank you sir for your quick reply! I totally got it coz what confused me before is that I kept doing the same thing as we did over SN1 and E1 :(
On the very first one, why would you not hydride shift from the carbon on the left since it would form a more stable product ?
Or not hydride shift but move the hydrogen from the left carbon
A bulky base was used so we use Hoffman's rule and hence the less stable alkene becomes the more stable alkene
Halide anions are unlikely to deprotonate due to their negative disassociation constants.
Halide ions (once formed) don’t even have proton left on them so not sure how would they deprotonate.
@@visuallearners1832 24:20 other than this mistake really helpful video
@@max3eey there is nothing wrong with the process on paper. Mechanistically, you can use any CB or water to deprotonate at the last step. Practically in labs, these reactions involves a weak base work up as last step so remove the proton.
why is NaSH strong nuc but KO isnt
what software do you use bro to make such videos?
Notability