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The ATDBio book on nucleic acids describes tetrazole as an activator required to couple nucleotides.

The diisopropylamino group of the nucleoside phosphoramidite is protonated by the activator, and is thereby converted to a good leaving group. It is rapidly displaced by attack of the 5′-hydroxyl group of the support-bound nucleoside on its neighbouring phosphorus atom.

enter image description here

The first molecule in the scheme they provide is clearly already protonated; a second step in which the diisopropyl amino group is replaced by a tetrazole molecule is displayed.

What is the purpose of this step; why doesn't the support bound nucleotide attack the phosphorous, directly? Are the arguments purely kinetic (i.e. sterics and availability)?

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  • $\begingroup$ Er no, the first molecule in the scheme is not protonated - it is protonated by the tetrazole. The alcohol will not, on its own, displace the di-isopropylamino group from the phosphorus centre. $\endgroup$ – Waylander Nov 14 '18 at 18:07
  • $\begingroup$ In the image in the question, the leftmost amine is already drawn in its protonated form as $\ce{HNR3+}$ and is nucleophilically substituted for tetrazole, no? $\endgroup$ – Jacob Nov 14 '18 at 18:11
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    $\begingroup$ The language here is not quite clear. That is a substitution and there is a nucleophile involved, but the arrow-pushing shown for step 1 (or step 2, for that matter) are almost certainly not what mechanistically happens in this process. Phosphorus compound do not under $S_{\mathrm{N}}2$ reactions normally; they undergo addition-elimination-type substitutions. $\endgroup$ – Zhe Nov 14 '18 at 18:35
  • $\begingroup$ @jacob No, it is shown as protonated by the tetrazole $\endgroup$ – Waylander Nov 14 '18 at 18:45
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    $\begingroup$ Each step of catalysed reaction is much easier then uncatalysed, because activation energy is lower. Your tetrazole is both better nucleophile then alcohol and leaving group then amine. $\endgroup$ – Mithoron Nov 14 '18 at 20:21

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