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I wanted to see how to make the product on the right so my thought was to have an SN1 reaction to remove water and have the bromine ion attack making the product on the right hand side.

Is this theoretically possible?

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    $\begingroup$ yes, the benzyl cation is very stable which facilitates the process $\endgroup$
    – Waylander
    Sep 21 at 13:44
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    $\begingroup$ Yes, this can be done readily facilitated by the stable benzyl radical. This patent [ here](patents.google.com/patent/CN102329192A/en) shows an example. $\endgroup$
    – Waylander
    Sep 22 at 12:36
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The reaction would theoretically occur through an SN2 mechanism because the hydroxyl (OH) group is bound to a primary carbon, and SN1 will usually only occur under tertiary (steric) conditions with a weak nucleophile.

What you are trying to do is referred to as "activation of alcohols". The OH- is not a good leaving group because OH- is a stronger base compared to Br-. There are several ways to activate an alcohol group (such as the formation of tosylates, protonation with strong acids, and reacting with PBr3 etc.) Since your goal is to substitute the alcohol with a bromide ion, theoretically H-Br could be used, where alcohol will accept a hydrogen atom from H-Br and leave as H20 as you have noted. PBR3 would be a better choice compared to H-Br.

This is because you need to watch out for side products from any potential carbocation rearrangements. There is some steric bulk with the presence of the benzene ring which could arguably allow the mechanism to form a stable carbocation through SN1, in which case the carbocation could rearrange with the adjacent carbon in the benzene ring. Using PBr3 also avoids "harsh conditions".

Look up activation of alcohols with hydrogen halides and activation with PBr3 for more information.

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  • $\begingroup$ yes PBr3 and CBr4 would have been another option but I just wanted to see if the proposed reaction was also a theoretical possibility. Thanks $\endgroup$
    – bobsburger
    Sep 22 at 19:14
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