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Please help me understand synthesis sequence shown below:

Benzylbromide is used for protection, probably from action of $\ce{CuBr2}$, but it's reaction mechanism is unknown for me. The bromination of the methyl group is some kind of radical reaction? What would it do to the rest of the molecule weren't it for benzylbromide? Next, 18-crown-6 is clearly used as catalyst for halogen exchange — but how does it function? Does it enclose bromine making $\ce{Br-}$ a better leaving group or $\ce{F-}$ better nucleophile? Finally, how does $\ce{CF3COOH}$ manage to remove that protecting group? Maybe it protonate benzilic $\ce{O}$ leaving the stabilized carbocation behind? But this seems quite unlikely since this would probably attacked fluorine or carbonyl.

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    $\begingroup$ Please use the \ce{...} command for chemical formulae. $\ce{CuBr2}$, for example, automatically gives you $\ce{CuBr2}$. Crown ethers are generally used to solvate the cations not the anions; the reaction on the second arrow can be thought of as a simple $\mathrm{S_N2}$ reaction. Also, on the first arrow you have first and second steps, meaning benzyl bromide is applied first (likely with a base), that product is isolated and then subjected to $\ce{CuBr2}$. The latter must be the reason for bromination, but I can’t give you a mechanism. $\endgroup$
    – Jan
    Commented Oct 14, 2015 at 20:04
  • $\begingroup$ Thanks for suggestion Jan, it is indeed more practical. I see, so this crown ethers sequester from $\ce{F-}$ their $\ce{K+}$ countercations which make them better nucleophiles? $\endgroup$
    – wuschi
    Commented Oct 14, 2015 at 20:43
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    $\begingroup$ I would guess so, 18-crown-6 is known for its particularly high affinity for $\ce{K+}$ if I am not wrong. Each of the crown ethers has their own "favourite" cation, for 12-crown-4 it's $\ce{Li+}$, 15-crown-5 is $\ce{Na+}$, and so on. On top of that, if you are using an organic solvent, I think the addition of the crown ether is what makes KF soluble (an important practical consideration). $\endgroup$
    – orthocresol
    Commented Oct 14, 2015 at 20:52
  • $\begingroup$ Not sure about the particular case of KF. I know the solubility thing is important for some other salts. $\endgroup$
    – orthocresol
    Commented Oct 14, 2015 at 20:58
  • $\begingroup$ @orthocresol I think you’re right there. However, it is impossible to know whether the crown is needed without knowing the reaction solvent. $\endgroup$
    – Jan
    Commented Oct 15, 2015 at 9:14

1 Answer 1

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I have no idea what the addition of benzyl bromide accomplishes. There doesn't seem to be any need for a protecting group in any of the subsequent reactions. The rest of the reactions are relatively easy to explain.

This paper reports essentially 100% yield for bromination of 4-hydroxyacetophenone with $\ce{CuBr2}$, refluxed in chloroform-ethyl acetate for 15-20 mins. There is no mention of any need for a protecting group for the phenol. No mechanism is given but you are correct in your assumption that radicals will be involved.

The second reaction is a regular $S_N2$ reaction. The transition state of the reaction is highly stabilized by the adjacent carbonyl group and so the reactivity is dramatically increased compared to a regular primary halide. 18-crown-6 is used to solvate the $\ce{K+}$ cations, for which it has a very high affinity. This leaves the fluoride free to react, making it far more nucleophilic than it would otherwise be. The reaction will go to completion easily because bromide is a far more stable leaving group than fluoride.

The final reaction is a standard ether cleavage, catalysed by a strong acid. The oxygen is protonated by the acid and leaves to reform the phenol and the stabilized benzyl cation.

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