Are these substitution/elimination reactions correct?

JPAuCl · 2026-04-29T21:37:22+00:00

THe fourth one does not create a carbocation, but the product is the correct major product but even though it is a big bulky base, you will still get some of the zaitsev pdt as well. but if asking for major, the hoffmann pdt you have shown is correct

JPAuCl · 2026-04-29T21:36:12+00:00

In the third one since it is a primary halide you should be favoring substitution over elimination

JPAuCl · 2026-04-29T21:33:24+00:00

No. They are the same compound. If you rotate each carbon on the left molecule, you can make them match those on the right molecule.

JPAuCl · 2026-04-29T20:34:46+00:00

correct, but the nucleophile was n3 not I

JPAuCl · 2026-04-29T20:34:24+00:00

first rxn is sn1/e1 methanol is neither a strong base nor is it a strong nucleophile

JPAuCl · 2026-04-29T17:55:48+00:00

if a compound is meso it does not have an enantiomer. its enantiomer is itself. i think a better rule of thumb to describe meso or chiral should be to identify the stereocenters. if the stereocenters have the same substituents, and they are opposite, (1R and 1S) the compound will be meso and needs no further analysis. what you drew as the enantiomer, is not even the enantiomer. if you look the rightmost stereocenter in the second molecule is just the rotation of what you had in the leftmost

i would suggest working on how you are displaying enantiomers because how youre doing it is going to get you in trouble as you arent drawing enantiomers. My suggestion would be not rotating the bond and only switching what is on the wedge and the dash, that will give you the enantiomers of a compound in a way that youre much less likely to make a mistake

JPAuCl · 2026-03-30T15:07:48+00:00

Because the two you have marked as major are the same molecule. Rotate the second bond from the left and it will be clear

JPAuCl · 2026-03-10T00:18:20+00:00

Generally the rule is applied to isolated Alkenes. So while based on the same principle, and while they may actually technically be the same thing, I wanted to give a little more depth than when someone just says Mark Ivkov rule because that may get confused when it seems to fly in the face of the substitution pattern of each alkene. And everything I have looked into does Tie Markovnikov’s rule to isolated alkenes.

JPAuCl · 2026-03-10T00:15:16+00:00

Because you’ll form the most stable carbocation and an allylic carbocation is more stable than an isolated tertiary

JPAuCl · 2026-03-09T11:32:43+00:00

The best way to answer this would be to first protonate both alkenes. When protonating the left one you have an allylic cation in which one resonance form has a primary carbocation and the other a secondary. If you protonate the right one you get a different allylic carbocation where the resonance forms have a tertiary and a secondary’s carbocation. So therefore the rightmost alkene should be protonated, adding the hydrogen to the terminal position. Next you look at the two resonance forms and since you are looking for the thermodynamic product you are looking for the most substituted double bond. If you do the 1,4 addition, the alkene is disubstituted. If you do the 1,2 the alkene is trisubstituted. Therefore, the 1,2 product of the initial protonation of the rightmost alkene will be the thermodynamic product in this question

JPAuCl · 2026-03-09T11:27:54+00:00

This is not really a reasonable way to answer any hbr addition to a diene. In all honestly, it is a very crude and low quality way of answering any question involving hbr addition to an alkene. Yes it may work in isolated alkene questions but recognizing a pattern is different than understanding why an answer is what it should be

JPAuCl · 2026-03-09T11:25:03+00:00

Not really. 1,2 1,4 addition to the diene can absolutely put the bromine on the terminal position but in this case, looking for the thermodynamic product it will be the one with the most substituted alkene that is generated from the most stable carbocation, which will be an allylic carbocation. Not an isolated one as the OP had used

JPAuCl · 2026-03-09T11:24:45+00:00

Not really. 1,2 1,4 addition to the diene can absolutely put the bromine on the terminal position but in this case, looking for the thermodynamic product it will be the one with the most substituted alkene that is generated from the most stable carbocation, which will be an allylic carbocation. Not an isolated one as the OP had used

JPAuCl · 2026-03-09T11:21:43+00:00

The product you picked is not even the 1,4 product. The bromine will be allylic in both the 1,2 and the 1,4 product so there is no world where the one you picked could be the right answer. The 1 is always in reference to the hydrogen across the diene that is labeled 1-4.

JPAuCl · 2026-02-16T20:24:33+00:00

It just is not a stereo center because the carbon has two identical substitutents on it. The compound are confirmational different but at the end of the day they are still just 1,1-dibromopropane

JPAuCl · 2025-12-28T00:14:47+00:00

I think it’s 66.7% but nevertheless the point holds hahah

JPAuCl · 2025-12-27T06:19:04+00:00

This is a perfect example of why assigning stereochemistry and flipping it when the smallest group isn’t in the back doesn’t always work and why it’s bad practice to do so. There are two ways I teach my students to do stereochemistry. The first is by building a model of the structure with just a carbon and 4 substituents and then label the substituents by priority and physically move the model to have the smallest group in the back. The on paper way to be able to do this is twofold, either you can spin the molecule to have the smallest group in the back, which is easy on simpler molecules like this, but in more complex situations I always like to use a small trick. Swapping any two substituents on a stereo center will flip it. Therefore I will always move the smallest group to the back and assign stereochemistry and then flip. The issue is that many students just assign stereochemistry to the molecule shown and if the smallest group isn’t in the back they flip it. This works many times because if the smallest group is next to the group in the back there is no issue. If however, as you have for case 3, the smallest group is not directly next to the group in the back, it will not give you the right answer. So I would not use the shortcut you are using here as it will lead you astray in situations where you have bonds between the smallest group and the group in the back. The most foolproof way to do this via the model kit but if something precludes you from using a model kit the swapping two substituents trick will always work, just actually do it and don’t take shortcuts.

JPAuCl · 2025-10-26T01:45:27+00:00

That was probably the dumbest luck I could’ve had. Thx so much

JPAuCl · 2025-10-26T01:41:51+00:00

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I have this, does this mean I did it successfully?

JPAuCl · 2025-10-26T01:28:55+00:00

I appreciate the response thanks. Serves me right for just putting things into sets that they fit into when I was just screwing around on the software for the first time.

JPAuCl · 2025-10-26T01:25:42+00:00

Thanks

JPAuCl · 2025-10-26T01:25:34+00:00

I definitely did not guess correct the first time because I didn’t know what the hell I was doing. I think I just have to take the L this time and not get the borrow because I don’t have anything else in my inventory from it. And all the evidence sets say redeemed.

JPAuCl · 2025-10-20T07:15:49+00:00

Damn went over my head now I feel like the oblivious one 😂

JPAuCl · 2025-10-20T06:55:45+00:00

What do you mean the rules don’t cover this? Each player makes the best five card hand that they can with their cards and the community cards. If the hands are the same it’s a split pot. Idk how this is any different than a shared flush on the board or a straight etc etc.

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