For question 17.30, how did you know which double bond to keep? What is the explanation behind it? I understand that there are two carbons that are not reduced. But which double bonds get to stay? Why not the carbon double bond with the methyl stays? and why can we not draw a parallel bond to that one? and the same goes for problem c, why was the carbon double attached to the ethyl kept and a parallel double bond drawn?
Hi Meg, very valid question! I think you are referring to the last problem (17.20 a and c). Let's start with 17.20a. The ring bears one substituent (an alkyl group), which is electron donating. The carbon atom next to the alkyl group will not be reduced. In a Birch reduction, only two positions are reduced, and they must be 1,4 to each other. This gives the product drawn (actually, you can draw two structures, but they are the same). For 17.20c, the ring bears 2 substituents which are both electron donating. The carbon atoms next to the alkyl group not be reduced, and they must be 1,4 to each other, which gives the product. Hope that helps but feel free to lmk if you have any follow up questions!:) have a great day
Yes, using H₃O⁺ instead of H₂O would also work for an SN1 reaction. Both will lead to the same product. H₂O and H₃O⁺ act as nucleophiles in SN1 reactions, with H₃O⁺ providing a stronger acidic environment, which can help with the departure of the leaving group. So, either method is correct.
Hello! I apologize for the questions, but I am confused because in 28:03, how is the reaction an SN2 when SN2 is only for primary and secondary carbon, but the alpha C is attached to 3 carbons.
Hello! No need to apologize; it’s a great question! Normally, SN2 reactions occur at primary or secondary carbons due to the lower steric hindrance, making a backside attack easier. However, the benzyl position (the carbon directly attached to the benzene ring) is an exception. Even though the alpha carbon in this case is attached to three groups, the benzyl group provides resonance stabilization to the transition state, making the SN2 mechanism possible here. This unique stabilization is why SN2 reactions can occur at benzyl positions, even when it seems sterically hindered. I hope this clarifies things!
Your videos have helped me out out so much!
I am so happy to hear that Coryna! Let me know if you need help with anything at all!
Your videos are so helpful, thank you!
Glad you like them!💖💖
GIRL you are amazing Mashallah . thank you ! ❤
I am so happy you are enjoying my videos!! LMK if you ever need anything
For question 17.30, how did you know which double bond to keep? What is the explanation behind it? I understand that there are two carbons that are not reduced. But which double bonds get to stay? Why not the carbon double bond with the methyl stays? and why can we not draw a parallel bond to that one? and the same goes for problem c, why was the carbon double attached to the ethyl kept and a parallel double bond drawn?
Hi Meg, very valid question! I think you are referring to the last problem (17.20 a and c). Let's start with 17.20a. The ring bears one substituent (an alkyl group), which is electron donating. The carbon atom next to the alkyl group will not be reduced. In a Birch reduction, only two positions are reduced, and they must be 1,4 to each other. This gives the product drawn (actually, you can draw two structures, but they are the same). For 17.20c, the ring bears 2 substituents which are both electron donating. The carbon atoms next to the alkyl group not be reduced, and they must be 1,4 to each other, which gives the product. Hope that helps but feel free to lmk if you have any follow up questions!:) have a great day
Thank you so much this video is amazing❤❤
You are so welcome!! I am so happy you liked it :)
Helpful this session
Glad to hear that!!!:)
C and D of synthesis were actually the two problems that I was most confused about
Hi! Do you need help with those? If so, just email me and I am happy to help you!!
Hey for minute 31 when you added NBS and then went to add H2O using an SN1 reaction, could you have just done H3O with a SN1 reaction instead?
Yes, using H₃O⁺ instead of H₂O would also work for an SN1 reaction. Both will lead to the same product. H₂O and H₃O⁺ act as nucleophiles in SN1 reactions, with H₃O⁺ providing a stronger acidic environment, which can help with the departure of the leaving group. So, either method is correct.
Hello! I apologize for the questions, but I am confused because in 28:03, how is the reaction an SN2 when SN2 is only for primary and secondary carbon, but the alpha C is attached to 3 carbons.
Hello! No need to apologize; it’s a great question! Normally, SN2 reactions occur at primary or secondary carbons due to the lower steric hindrance, making a backside attack easier. However, the benzyl position (the carbon directly attached to the benzene ring) is an exception. Even though the alpha carbon in this case is attached to three groups, the benzyl group provides resonance stabilization to the transition state, making the SN2 mechanism possible here. This unique stabilization is why SN2 reactions can occur at benzyl positions, even when it seems sterically hindered. I hope this clarifies things!
@@professoreman2289 thank you so much! I really appreciate your help
minute 28 did you mean SN1 for the H3O+ reagent? SN2 was written
Hi, YES! You are right! Great catch Abid!
@@professoreman2289 yeah all good lol. thanks for your videos
first
Winner!!