Sebgaudette Posted August 19, 2018 Share Posted August 19, 2018 Could anyone tell me why this reaction is sn1 and not sn2? ps: the OH shouldnt be on the phenyl. thanks Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted August 20, 2018 Share Posted August 20, 2018 What factor generally distinguishes SN1 from SN2? In other words, why do certain substrates go via SN2 and not SN1? Link to comment Share on other sites More sharing options...
studiot Posted August 20, 2018 Share Posted August 20, 2018 I don't know what text you are using and not all textbooks bring out the differences btween Sn1 and Sn2. This extract from Geissman is particularly clear. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted August 20, 2018 Share Posted August 20, 2018 It doesn’t give a particularly clear path to understanding the basis of the OP’s question, but it does hint at it if read correctly. As a further clue for the OP, it is important to consider the position of the carbon undergoing substitution relative to the aromatic ring, and how that ring might be able to participate. Link to comment Share on other sites More sharing options...
BabcockHall Posted August 21, 2018 Share Posted August 21, 2018 @OP, In general please show your work or give your thoughts in your first post. Substitutions reactions at a benzylic carbon might go one way or the other, depending on additional factors. The degree of substitution of the carbon and the hydroxy substituent on the aromatic ring (assuming it is present) are both factors here. Why don't you provide your thoughts now? Link to comment Share on other sites More sharing options...
BabcockHall Posted August 22, 2018 Share Posted August 22, 2018 @OP, What makes a substrate a good candidate for the SN1 process? for the SN2 process? Once you are sure of these answers, look at the substrate of this problem. Link to comment Share on other sites More sharing options...
USBSTAR Posted June 18, 2019 Share Posted June 18, 2019 (edited) So Br- is a good leaving group and the electrophilic carbon is secondary. Then??? Edited June 18, 2019 by USBSTAR Link to comment Share on other sites More sharing options...
BabcockHall Posted June 19, 2019 Share Posted June 19, 2019 Can the aromatic ring stabilize a putative carbocation (assuming it goes SN1)? A drawing may be helpful. Link to comment Share on other sites More sharing options...
cmatos Posted October 24, 2022 Share Posted October 24, 2022 This reaction occurs with a secondary compound. We can consider the leaving group, the Pka of the conjugated acid of Br is -9, making it a weak base. Nucleophiles for SN1 reactions are weak. The substitution will require two steps because a carbocation is formed. Link to comment Share on other sites More sharing options...
Jaymari Ayala Posted October 27, 2022 Share Posted October 27, 2022 This reaction is SN1 because is unimolecular. The rate- defining step involves only one molecule. Link to comment Share on other sites More sharing options...
jonathano23 Posted October 28, 2022 Share Posted October 28, 2022 The reaction that occurs is unimolecular, hence it is SN1 reaction. Also, this product can be possible in both reactions SN1 and SN2, but the SN1 reaction is more favorable. Link to comment Share on other sites More sharing options...
Rene Moreno Posted October 28, 2022 Share Posted October 28, 2022 For the reaction it will require two steps since it is a secondary compound. Link to comment Share on other sites More sharing options...
Gabriela Padilla Posted October 29, 2022 Share Posted October 29, 2022 It is SN1 because the shown reaction is known to be a unimolecular and it has a weak base due to Br-. Link to comment Share on other sites More sharing options...
Sebastian Figueroa Posted October 29, 2022 Share Posted October 29, 2022 The Br- made the base weak and the reaction is a unimolecular one, therefore the reaction shown is SN1. The product can be possible in both the SN1 and SN2. Link to comment Share on other sites More sharing options...
s.villanueva10 Posted October 29, 2022 Share Posted October 29, 2022 It is unimolecular and it has a weak base this is because of the Br- and meaning that the reaction is SN1. Link to comment Share on other sites More sharing options...
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