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Would his be a valid synthesis? I'm just unsure about the oxidative clevage step. I know that there is a benzylic hydrogen. So should oxidative clevage work here? What happens to the fragment that is lopped off? Is it also oxidized?

enter image description here

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    $\begingroup$ Your synthesis will work in theory, but it can be shortened. I think ron's method below is very nice, but it might be possible to do it all in one step with phosgene ($\ce{COCl2}$) in the presence of $\ce{AlCl3}$. $\endgroup$ – Greg E. Sep 29 '14 at 16:46
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    $\begingroup$ @GregE. That's a better approach then what I suggested, especially using something like triphosgene instead of phosgene. Why not post it as an answer? I'd vote for it. $\endgroup$ – ron Sep 29 '14 at 17:31
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    $\begingroup$ @ron, thank you. I'll expand it into a fuller answer a bit later when I have more time. Thanks also for the reminder about triphosgene; its existence had completely slipped my mind. $\endgroup$ – Greg E. Sep 29 '14 at 17:36
  • $\begingroup$ There is another nice synthesis, you can hydrolyze dicholodiphenylmethane. The latter can be synthesized from (a double Freidel-Crafts) benzene and carbon tetrachloride.. And carbon tet is a bit less toxic to work with than phosgene. $\endgroup$ – AlaskaRon Apr 5 '18 at 6:42
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Your basic strategy to involve a Friedel-Crafts acylation is correct, but your route is a bit circuitous. You could acylate benzene directly to produce benzaldehyde in one step using carbon monoxide and hydrochloric acid to generate formyl chloride in situ.

enter image description here

While it is "just another" Friedel-Crafts acylation, it has it's own name, the Gatterman-Koch reaction, after its discoverers. Once in hand, you could oxidize benzaldehyde to benzoic acid.

An alternate approach would be to 1) brominate benzene to form bromobenzene, 2) form the Grignard reagent with magnesium and 3) react it with carbon dioxide to produce benzoic acid.

In either case, with benzoic acid in hand, you can procede as you've indicated, on to benzophenone.

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I disagree with Ron on this one. First off, he got the Gatterman reaction wrong. The reaction he showed does not work, since the formyl chloride $\ce{(HC=OCl)}$ breaks down under the conditions required for Friedel–Crafts to occur, giving you $\ce{HCl}$ and $\ce{CO}$. The actual Gatterman reaction would work, since it uses a $\ce{C=N}$ instead of a $\ce{C=O}$ for the reaction. enter image description here

Simply oxidize this compound (benzaldehyde) with something like LiAlH4 (edit: this is a reducing agent, use KMnO4 in aqueous base or Chromic Acid) to get to benzoic acid, then continue with your thionyl chloride and then the second Friedel–Crafts reaction with another benzene.

Ron's second suggestion of using a Grignard to synthesize benzoic acid is also good, just note that you need to add an acid (even just water is a strong enough acid) to remove the Grignard reagent afterwards. Here's an image of the reaction he is referring to, except they used $\ce{HCl}$ to remove the Grignard.

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    $\begingroup$ The Gattermann-Koch reaction is not the same as the Gattermann reaction, they are two distinct reactions, see the link in my answer for clarification. The G-K reaction was specifically designed to generate formyl chloride in situ so that aromatic substrates could be formylated. The reason why the G-K approach is preferred over the Gatterman is that the latter route involves the use of $\ce{HCN}$. Also, I'm not sure what you mean by "oxidize this compound (Benzaldehyde) with something like $\ce{LiAlH4}$." $\ce{LiAlH4}$ is a reducing agent. $\endgroup$ – ron Apr 5 '18 at 17:33
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    $\begingroup$ Well done @ron. Two equiv. of phenyl Grignard added to 1 equiv. ethyl formate gives benzhydrol, readily converted to benzophenone by a myriad of oxidants. $\endgroup$ – user55119 Apr 5 '18 at 22:41
  • $\begingroup$ @user55119 Nice, only 3 steps and certainly more convenient to carry out. Take a look at Greg E.'s suggested synthesis up above - only 1 step! $\endgroup$ – ron Apr 5 '18 at 23:41
  • $\begingroup$ Oh yeah, thanks Ron, I messed that up with LiAlH4, it's quite obviously a reducing agent. KMnO4 in aqueous base or Chromic Acid would do the trick $\endgroup$ – Alex Apr 7 '18 at 0:12

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