I take venlafaxine (Effexor) as medication, and having wrapped up 2 semesters of Organic Chemistry I thought it might be fun to design a synthesis for it. A picture of venlafaxine:


I came up with the following preparation (beginning with phenol) and was hoping someone could check it for me:

  1. React phenol with $\ce{NaOH}$ to form phenolate, followed by methyl iodide to form methoxybenzene. This forms a moderately activated ring with ortho/para direction.

  2. Friedel-Crafts alkylation: allyl bromide (3-bromo-1-propene) and $\ce{AlBr3}$ adds propene to the para position.

  3. Ozonolysis with DMS to give p-(acetaldehyde)methoxybenzene.

  4. Reductive amination with dimethylamine and triacetoxyborohydride.

  5. Radical bromination should be selective for the benzylic position.

  6. Finally, Grignard reaction with cyclohexanone.

  • 2
    $\begingroup$ Don't really need your first step as anisole is readily available $\endgroup$
    – Waylander
    Aug 3, 2019 at 8:39
  • $\begingroup$ I suspect the product from 5 is going to ring close to the aziridinium. $\endgroup$
    – Zhe
    Aug 3, 2019 at 12:22
  • $\begingroup$ @Zhe Thanks! I take it that means the nitrogen would attack and displace the bromide? I'll try and change the order to make amination follow the Grignard reaction if I can. Would the oxygen in the carbonyl compound do the same if I brominated the benzylic position after its formation in ozonolysis? $\endgroup$
    – David Reed
    Aug 3, 2019 at 18:29
  • $\begingroup$ Unlikely. Yours is the basic problem of trying to figure out functional groups (and protecting groups) that have orthogonal reactivity. $\endgroup$
    – Zhe
    Aug 3, 2019 at 18:56
  • $\begingroup$ @Zhe Thinking of that now though I would then have the problem of it interfering with the Grignard. What if I changed the order to 1,2,5,6,3,4? $\endgroup$
    – David Reed
    Aug 3, 2019 at 19:03

1 Answer 1


You may not necessarily need to start your procedure beginning with phenol. You may start it with readily available (for cheap price compared to the trouble you are going to have by handling methyl iodide, allyl bromide, and $\ce{AlBr3}$ or $\ce{AlCl3}$) 4-anisaldehyde (4-methoxybenzaldehyde). That is the exact product you are getting after your step 3. Although, steps 4-6 are some very thoughtful organic synthesis steps, I doubt it will give you the expected product. Therefore, I suggest following steps to follow:

  1. Buy 4-anisaldehyde (4-methoxybenzaldehyde) from Sigma-Aldrich (available in 98% purity.

  2. Prepare Grignard reagent using chlorocyclohexane/$\ce{Mg}$ in diethyl ether.

  3. React Grignard reagent (cyclohexyl magnesium chloride) with 4-anisaldehyde.

  4. Perform dehydration of resultant $2^\circ$-alcohol to give double bond.

  5. Perform epoxydation of double bond with 3-chloroperbenzoic acid.

  6. React epoxide with nitromethane.

  7. Reduce nitro group to $1^\circ$-amino group.

  8. Finally, convert $1^\circ$-amino group to dimethyl amino group by reductive amination with formaldehyde.

  • 3
    $\begingroup$ Cyanide may be a better nucleophile to use than nitromethane. Certainly the reduction will be easier, reductions of alkyl nitro are notoriously fickle $\endgroup$
    – Waylander
    Aug 4, 2019 at 6:17
  • $\begingroup$ I agreed. That is an one excellent suggestion. I suggested nitromethane because I have an experience with that during synthesis for inhibitors for DBM (dopamine-$\beta$-monooxygenese). $\endgroup$ Aug 4, 2019 at 6:18
  • 1
    $\begingroup$ @Waylander my thoughts exactly which is where the anisaldehyde suggestion originated $\endgroup$
    – Beerhunter
    Aug 4, 2019 at 10:04

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