Substitution and Elimination Practice Questions | SN1 SN2 E1 E2

[@VictortheOrganicChemistryTutor]

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[@williammiketon5413]

please, I need more of these, this was soooooo helpful

[@VictortheOrganicChemistryTutor]

I’ve got a part 2 to this video 😉

[@jeffybrailin5377]

Finally completed this playlist 😊✌️

[@ramashakaroun2863]

Very good Job and great examples. Thank you :D

[@triple_gem_shining]

Feels good to get most of these correct after thinking deeply on it. I love these types of puzzles.

[@VictortheOrganicChemistryTutor]

This is a tricky topic. I’m glad you’re getting a lot of these questions correctly!

[@milliesmith8783]

Thank you! This was so helpful :)

[@VictortheOrganicChemistryTutor]

I’m glad you found it helpful 👍

[@rezalatifi1905]

Excellent job! Which software do you use for writing?

[@mwangalamusialelaamwalana3689]

Hello Victor! Hope you are well.. kindly help me with this question outline a synthetic scheme showing how you can efficiently synthesize busulfan from 1,4-dibromobutane. show all the steps clearly and assume you have all the needed reagents.

[@kavinduharshana740]

awesome explanations 😍

[@VictortheOrganicChemistryTutor]

Glad you liked it

[@himabindu5928]

Good video sir question no 2 what is the stereochemistry of the product

[@joshuamitchell5530]

Question 5, why wouldn’t you lose Br- for an SN1 mechanism? The cation is benzylic, and the resulting cation could also rearrange to the tertiary cation? I guess because you have a strong base E2 is just faster?

[@VictortheOrganicChemistryTutor]

For the carbocation formation, we need to have protic conditions which would help stabilize the leaving group upon dissociation. Here we have a strongly basic solution, so any protic properties that the ethanol solvent would exhibit are completely “negated” by the base. Thus, the formation of a carbocation is less likely.

[@joshuamitchell5530]

@@VictortheOrganicChemistryTutor makes sense. Would SN1 be more likely if we had a more Lewis acidic counter ion though?

[@VictortheOrganicChemistryTutor]

If you have a metal which strongly binds/coordinates to your anionic species, the anion is no longer going to be very basic or nucleophlic. Which then brings the question of why bother adding it at all 😉

[@joshuamitchell5530]

@@VictortheOrganicChemistryTutor That’s very true, thank you

[@SaifaSiddique]

Very welll explned

[@HexaneChemistry]

Very good

[@alexe6242]

For question 7 would SN1 proceed as the minor product of the reaction? If so, what would take the place of the leaving group?

[@VictortheOrganicChemistryTutor]

SN1 and E1 are always in the competition with each other, so yes, you can expect some amount of it depending on how exactly you do the reaction in the lab.

[@sportyyy5905]

Question 3: why doesn´t the OCH3 within the molecule react and form a 5 member ring?

[@VictortheOrganicChemistryTutor]

It's not a particularly good nucleophile, why would it? Besides, hypothetically, even if it did, what would your product be? How would it get rid of the + charge on oxygen?

[@carolinelecter9663]

Wonderful video, thank u, question 2 why there's no E1

[@VictortheOrganicChemistryTutor]

SN1 and E1 are always in a competition with each other. So, when you have one, you'll also have the other one. The question is which pathway is going to be major. And in this case, it's the SN1 reaction that will give you the major product, so we're not emphasizing the E1.

[@dwirevanityas6834]

why question 7, don't happened rearrangement carbon results like siklo?

[@TheThorRs]

Can someone provide more info as to why question 3 is E2? My thoughts is that ch3o- is a STRONG BASE and it's not bulky, and the substrate gives a secondary. so it can attack the molecule directly which makes me think SN2. MY thoughts as to why it's E2 is because maybe the polar protic solvent is occupying the base and so the base has no choice but to do E2 because of that. thoughts?

[@VictortheOrganicChemistryTutor]

Not as strong as you make it sound. Also, while secondary allyl halides can do both SN2 and E2 they tend to give an E2 major product when reacting with alkoxides regardless of the size.

[@devnarayanmaharshi773]

same doubt I have because HEAT is not given there, so that reaction must be SN2.

[@VictortheOrganicChemistryTutor]

@@devnarayanmaharshi773 See my comment above. You might also want to go back to the predictive model and my other videos on the topic. Heat has nothing to do with the outcome of bimolecular reactions.

[@michaelowogowog8518]

Why question 3 chose E2 Pathway instead of Sn2

[@VictortheOrganicChemistryTutor]

Based on the predictive model (I have a video on that too). 2° leaving group + base/nucleophile as a reagent = E2 major, SN2 minor.

[@madelineduckworth3413]

I'm going to fail my test:(

[@VictortheOrganicChemistryTutor]

I'm not accepting this attitude! You can do it! 💪