What is the product of the reaction of Bromobenzene, 1-chloropropene with AlCl3?

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Explanation / Answer

The straight-forward synthesis of benzoic acid from benzene would utilise a Grignard reaction: benzene --- Br2 / FeBr3 ---> bromobenzene --- Mg in dry ether ---> phenyl magnesium bromide --- 1. CO2(s) 2. H+/H2O ---> benzoic acid C6H6 --- Br2 / FeBr3 ---> C6H5Br --- Mg in dry ether ---> C6H5MgBr --- 1. CO2(s) 2. H+/H2O ---> C6H5COOH Friedel-Craft reactions provide other means for introducing the additional carbon atom required and forming a carbon-carbon bond. Using an alkylation is awkward in that the required alkyl halides (such as CH3Cl) are gases at room temperature, so an acylation is the more natural approach: benzene --- HCOCl / AlCl3 ---> benzaldehyde --- H+ / MnO4^- ---> benzoic acid C6H6 --- HCOCl / AlCl3 ---> C6H5CHO --- H+ / MnO4^- ---> C6H5COOH There are numerous oxidising agents suitable for the benzaldehyde to benzoic acid conversion. Note that some are sufficiently strong to oxidise the side chain beyond the first carbon atom. So, if you were to use ethanoyl chloride instead of formyl chloride and thus formed acetophenone, it would still oxidise to benzoic acid. Sandmeyer reactions are really not at all suited to this synthesis. The prototypical reaction is used to convert aniline to an aryl halide via a diazonium salt intermediate, but the diazonium is actually much more useful for making phenols, nitriles and thioethers. Nevertheless, a couple of schemes can be imagined: benzene --- HNO3 / H2SO4 ---> nitrobenzene --- Sn / HCl ---> aniline --- NaNO2 / HCl / 0 °C ---> phenyl diazonium chloride --- CuCl / 60 °C ---> chlorobenzene --- Mg in dry ether ---> phenyl magnesium chloride --- 1. CO2(s) 2. H+/H2O ---> benzoic acid C6H6 --- HNO3 / H2SO4 ---> C6H5NO2 --- Sn / HCl ---> C6H5NH2 --- NaNO2 / HCl / 0 °C ---> [C6H5N2]Cl --- CuCl / 60 °C ---> C6H5Cl --- Mg in dry ether ---> C6H5MgCl --- 1. CO2(s) 2. H+/H2O ---> C6H5COOH Note that numerous reducing agents are capable of producing the nitrobenzene to aniline reduction. This synthetic pathway is, of course, an absurdly complex approach to making chlorobenzene, given it can be produced from benzene in a single step. A better approach if you use a modified Sandmeyer reaction would be to go to the nitrile rather than to an aryl halide: benzene --- HNO3 / H2SO4 ---> nitrobenzene --- Sn / HCl ---> aniline --- NaNO2 / HCl / 0 °C ---> phenyl diazonium chloride --- KCN ---> benzonitrile --- H+/H2O ---> benzoic acid C6H6 --- HNO3 / H2SO4 ---> C6H5NO2 --- Sn / HCl ---> C6H5NH2 --- NaNO2 / HCl / 0 °C ---> [C6H5N2]Cl --- KCN ---> C6H5CN --- H+/H2O ---> C6H5COOH However, even this is an inefficient synthesis, in that benzonitrile (cyanobenzene) can be formed by reaction of bromobenzene with copper(I) cyanide. In fact, the substitution to the nitrile and subsequent hydrolysis can be done in one-pot using a Rosenmund reaction: bromobenzene --- KCN in H2O / EtOH with CuCN (cat.) / 200 °C ---> benzoic acid C6H5Br --- KCN in H2O / EtOH with CuCN (cat.) / 200 °C ---> C6H5COOH In fact, having made bromobenzene, the major reason for prefering a one-pot Grignard synthesis over the Rosenmund synthesis is that the Grignard approach avoids the need to deal with highly toxic cyanide salts. Neither a Friedel-Craft approach nor (certainly) a Sandmeyer reaction is as simple as the benzene to bromobenzene to benzoic acid paths afforded by Grignard or Rosenmund methods.

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