I have this project that I've been struggling with. I'm already given the reactant and product, so basically what I have to do is to figure out the steps needed to get the product.


I figured that some of the steps would include the Friedel-Crafts acylation and nitration, but after that, I'm not too sure what other steps I'd have to take. Any help will be very much appreciated!

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    $\begingroup$ Welcome to ChemSE. With some real effort on your part, we will be pleased to help. Get back to us. $\endgroup$ – user55119 Apr 15 at 1:15
  • $\begingroup$ How many steps are you considering? $\endgroup$ – Waylander Apr 15 at 6:59
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    $\begingroup$ Hint #1: Identify suitable precursors to install any of the three substitutents on a benzene ring. E.g., $\ce{-NO2}$ is introduced by ... . #2: Not all remaining $\ce{C-H}$ e.g., in nitro benzene are equal. Some substitutents direct the next entrant substitutent to ortho / para position, or in meta position when running an electrophilic aromatic substitution. See how these patterns may align with the target of a triple substitution. #3: Some of the groups shown may the result of a substitution and subsequent group transformation (e.g., redox). #4: Don't limit to electrophilic reaction. $\endgroup$ – Buttonwood Apr 15 at 10:26

@Waylander has provided a concise route to the nitro compound in question. I am also in accord with his comments regarding the OP's effort and his assessment that benzene, as a starting material, is not a choice by a practicing chemist. Now that he has opened the door to a solution, I offer a more traditional approach.

Classical nitration of ketoacid 7 should be reasonably selective in that the introduction of the nitro group in 8 is both para and meta directed. Keto acid 7 arises from indene 6 via exhaustive oxidation. To access the indene, I would start with benzaldehyde (2), whose preparations from benzene (1) are numerous. The Perkin condensation 2 $\rightarrow$ 3 is known. Hydrogenation of 3 forms carboxylic acid 4, which can be cyclized to indanone 5 with polyphosphoric acid. Friedel-Crafts techniques may also be employed. Reduction of this ketone followed by dehydration leads to the critical indene 6.

  • $\begingroup$ The indene cleavage is a nice option. I would be tempted to carbonylate benzene to make benzaldehyde. $\endgroup$ – Waylander Apr 16 at 15:34
  • $\begingroup$ @Waylander: I agree with direct carbonylation except that the ones I found on Chem. Absts. require more sophisticated techniques. I opted for something the OP would understand although it is several steps. There is also Cl2CHOCH3/TiCl4 as a CO equivalent but, again, too sophisticated. The whole post is an academic exercise, not something you or I would have executed from benzene. $\endgroup$ – user55119 Apr 16 at 16:12

While I think the OP should make more of an effort at answering this I am going to suggest a route because I think this is an interesting challenge - if a little unreal. In the real world of preparative organic chemistry you would not be starting from benzene. There are many possible approaches to this, but here is mine

Step 1 - carbamoylation of benzene with dimethyl carbamoyl chloride/TMSOTf/Zn(OTf)2 as described here

Step 2 - directed lithiation of the dimethylbenzamide with sBuLi/TMEDA followed by quenching with the commercially available propylene oxide to install the C3 sidechain at the alcohol level.

Step 3 - oxidation with Dess-Martin periodinane to give the benzyl methyl ketone

Step 4 - nitration under standard HNO3/H2SO4 conditions as here should give the desired 1,2,5 substitution pattern as the major component.

Step 5 - amide hydrolysis under strong acid conditions with cHCl. I have chosen this over basic hydrolysis as think the formation of the anion of the starting benzyl methyl ketone activated by 2 EWGs is likely to be an unwanted complication.

  • $\begingroup$ The benzene rings in the carbamoylation paper appear to be activated. While toluene is the least activated, benzene is conspicuously absent as an example. $\endgroup$ – user55119 Apr 15 at 21:34
  • $\begingroup$ It is absent but think it will work with the addition of Zn(OTf)2 $\endgroup$ – Waylander Apr 15 at 21:36

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