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Can someone please elucidate how the reaction in the attached picture would proceed?[1] The products of the reaction are carboxylates. What exactly would cause the alkyne to cleave?

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In J. Org. Chem. 1979, 44 (15), 2726–2730, the authors Lee and Chang have investigated the permanganate oxidation of alkynes to α-diketones. In addition, they investigated the oxidative cleavage of several α-diketones to carboxylic acids. In the case of hexadecane-8,9-dione, permanganate oxidation under at least three sets of conditions provided n-octanoic acid 8 (major) and n-heptanoic acid 7 (minor). They suggested that the oxidation of the dione 1 leads only to octanoic acid 8 via intermediate 4 while oxidation of enol 2 affords 2-oxononanoic acid 6 and ultimately octanoic acid 8 via oxidative decarboxylation. In addition, oxidation of heptanal 5 affords heptanoic acid 7. The three solvent systems employed were acetone/aqueous HOAc, aqueous H2SO4 and water. Starting material was recovered in each experiment. Tetradecane-7,8-dione showed similar behavior.

For the mechanism of permanganate oxidation of alkynes to α-diketones, see the accompanying post by @Raul Luciano.
enter image description here

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  • $\begingroup$ Pinging with @ in posts does nothing. In comments it's quite limited. In chat is more useful, still sometimes "superping" from mod would be needed to actually ping someone. Just saying ;) $\endgroup$ – Mithoron Jul 20 '18 at 23:42
  • $\begingroup$ @Mithoron(ping?): I was trying to draw the attention of Raul Luciano. So how do I accomplish this goal ... in simple terms. $\endgroup$ – user55119 Jul 20 '18 at 23:47
  • $\begingroup$ You'd need to comment under his answer. BTW this time @.... wasn't necessary as there was no other users' comments under this post. Probably it wouldn't work otherwise as you didn't put space after user name and before "(" Here's related post chemistry.meta.stackexchange.com/questions/3889/… $\endgroup$ – Mithoron Jul 21 '18 at 0:10
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Alkynes react just as alkenes in the oxidative cleavage with $\ce{KMnO_4-}$ (the mechanism is the same). The difference is that since there is a triple bond, $\ce{MnO_4-}$ doesn't break the molecule apart like it does with alkenes (only if you use high temperature and high base/$\ce{MnO_4-}$ concentration), instead it forms a diketone. What causes the alkyne to cleave is the oxidant ion $\ce{MnO_4-}$ that is reduced to $\ce{MnO_2}$.

enter image description here

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  • $\begingroup$ This fails to answer the question asked. $\endgroup$ – A.K. Jul 18 '18 at 16:25

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