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The production of bakelite using this method, but with an acid as the catalyst is simple enough; a protonation of the carbonyl oxygen, followed by an electrophilic substitution to the ring. The subsequent cross-linkage formation is the same for both cases, but I can't figure out how the initial reaction steps would work with a base as the catalyst. Google searches and the texts I have yield nothing.

How would the mechanism proceed with a base as the catalyst here?

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    $\begingroup$ Presuming that the reaction is base-catalyzed, "push" electrons toward the carbonyl by forming the anion of a phenol. In the acid-catalyzed reaction, the protonated formaldehyde is a "hot" electrophile that can react with a neutral phenol nucleophile. The base mechanism is similar to an aldol condensation. $\endgroup$
    – user55119
    Apr 29 at 14:35
  • $\begingroup$ @user55119: okay, so there's electromerism on the formaldehyde, yielding an electrophilic carbon, and deprotonation of the phenol yields a phenoxide ion that joins with the carbonyl carbon. But that doesn't result in anything like an elementary unit of Novolac. Where did I go wrong? $\endgroup$
    – harry
    Apr 29 at 14:46
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    $\begingroup$ Keep going once the addition occurs. Make the phenoxide again, beta-eliminate hydroxide. Now you have the equivalent of a conjugated unsaturated ketone. Do a Michael addition of another phenoxide. Now you have a methylene between two aromatic rings. Keep doing this until all ortho and para positions are substituted or you get bored. $\endgroup$
    – user55119
    Apr 29 at 14:52
  • $\begingroup$ @user55119: I got a deprotonated phenoxymethanol doing that, but that isn't a conjugated unsaturated ketone. Moreover, Novolac has hydroxyls ortho to the methylene bridges, which don't seem to come up here. $\endgroup$
    – harry
    Apr 30 at 9:38
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@Harry Holmes: Here is a simplified example of how the formation of Bakelite may occur under base-catalyzed conditions to incorporate formaldehyde as -CH2- at the ortho and para positions of phenol. There are a myriad of permutations as to the order of condensations. I have selected one that illustrates the Michael-like addition that seems to have eluded you.

Phenoxide 1 adds to formaldehyde at the para position in what amounts to the first step in an aldol addition. The addition is likely adding to the ortho position of phenoxide as well. Species 3 may condense with phenoxide at this point but I have chosen to incorporate two more equivalents of formaldehyde at the ortho positions of 3 to form 4. Now elimination of hydroxide from any of the three positions of 4 leads to a Michael acceptor, such as 5, which reacts with phenoxide or any substituted phenoxide having an open ortho or para position. Structure 6 simply illustrates the mechanism for incorporating formaldehyde. The o-methylenedienone 5 is the product of an aldol condensation wherein phenoxide is the enolate.

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