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Problem Reaction

The following is a reaction from a school provided lab manual. However, I am unable to find anything similar to the cyclization reaction taking place. All of the reactions I have found online make use of either aldehydes or carboxylic acids for the ring closure. It is obvious from the product that the carbon connected to the 2 OMe substituents is the carbon that participates in the reaction. I just can't find anything that shows the possible mechanism that results in the product. Any help would be appreciated in how I should tackle this problem.

Edit: Mechanism according to Waylander?

I've taken Waylander's advice and come up with the above mechanism.

However, before I read his comment I also managed to come up with this mechanism:

Another possible mechanism?

Would either mechanism work? Or would one be more correct than the other?

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    $\begingroup$ Look up Bradsher reaction. For your particular reaction, doi.org/10.1055/s-1974-23372 and Wikipedia — Bradsher-Cyclisierung (in German) appear to be relevant. $\endgroup$
    – andselisk
    Commented Mar 21 at 7:57
  • $\begingroup$ Hello, thank you for the reply! Could I trouble you for further clarification? I took a look at the resources you linked and they use involve HBr. In this case would the H2SO4 act in the same way HBr does? The journal article you linked has compounds that seem to undergo the same reaction (2) --> (3), but there is no suggested mechanism so I am finding it hard to apply to my particular problem. $\endgroup$
    – Marethyu
    Commented Mar 21 at 8:26
  • $\begingroup$ The key step here is the protonation of one of the -OMe groups. This gives a cation which is which is stabilised by the other OMe group and captured by the aromatic ring. Further protonation/elimination of the remaining -OMe and the -OH gives aromatisation of system. It should be noted that there will most likely be a second isomer formed by cyclisation into the ortho position $\endgroup$
    – Waylander
    Commented Mar 21 at 11:11
  • $\begingroup$ I think your mechanism in pic 2 (apart from the unfortunate wandering alcohol group) is more correct. The site of first protonation is imho one of the acetal oxyens $\endgroup$
    – Waylander
    Commented Mar 21 at 13:51
  • $\begingroup$ Well, you do have there a protected aldehyde. BTW you'd do better not putting every proton exchange into separate step. $\endgroup$
    – Mithoron
    Commented Mar 21 at 14:24

1 Answer 1

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Terminal to the alkyl chain, you have the motif of an acetal. In presence of aqueous acid, it can pass to hemiacetal and free aldehyde.

enter image description here

From there, the aldehyde possibly is activated further by protonation. The arene (a pi-bond nucleophile) then reacts with the carbonyl-C (the electrophile) to establish the new ring.* The intermediate carbocation rearomatizes. In principle the subsequent dehydrations of the the alcohols pass equilibria, but the eventual gain of aromaticity (again) drives the reaction forward. (Maybe the sequence of which alcohol is dehydrated first, a tertiary vs. benzylic alcohol is a different one, than depicted; the result were the same.)

enter image description here

* At this stage, the pi-bond nucleophile can be recognized easier by moving the electron sextet, see the second structure. (We all know that neither one of the two mesomeric formulae represents the true state fully.)

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    $\begingroup$ I don't think you need to postulate cleavage of the dimethyl acetal to aldehyde. Simple protonation of the acetal is enough. $\endgroup$
    – Waylander
    Commented Mar 21 at 13:48
  • $\begingroup$ @Waylander Possibly influenced by ortho esters' hydrolysis in Claisen reactions, does the dehydration of a tertiary alcohol (as in the starting material) typically outrun the (re)generation of an aldehyde from an acetal? The example of a Bodroux Chichibabin reaction in Organic Synthesis couples this with steam distillation to harvest n-hexaldehyde and (hence) does not state the reaction time. $\endgroup$
    – Buttonwood
    Commented Mar 25 at 19:04

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