Could anyone tell me why this reaction is sn1 and not sn2?

 

ps: the OH shouldnt be on the phenyl.

 

thanks

What factor generally distinguishes SN1 from SN2? In other words, why do certain substrates go via SN2 and not SN1?

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.

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.

@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?

@OP, What makes a substrate a good candidate for the S_N1 process?  for the S_N2 process?  Once you are sure of these answers, look at the substrate of this problem.

So Br- is a good leaving group and the electrophilic carbon is secondary. Then???

Edited June 18, 20196 yr by USBSTAR

Can the aromatic ring stabilize a putative carbocation (assuming it goes S_N1)?  A drawing may be helpful.

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.

This reaction is SN1 because is unimolecular. The rate- defining step involves only one molecule. 

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. 

For the reaction it will require two steps since it is a secondary compound.

It is SN1 because the shown reaction is known to be a unimolecular and it has a weak base due to Br-.

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.

It is unimolecular and it has a weak base this is because of the Br- and meaning that the reaction is SN1.