Reply 1
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It says that "the hydroxide acts as a nucleophile" in the question, so you know a curly arrow can be drawn starting from the lone pair on the hydroxide. To know for sure where it goes, you can see on the next box it has to go to the sulphur atom.
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Looking at the next box you can see that what was a neutral, double-bonded oxygen atom at the top of the molecule is now an anionic, single-bonded oxygen. This means that one of the bonds has broken and both electrons has gone to the oxygen. This also saves the sulphur from being bonded to more than 6 groups in the first step.
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Looking at the third box, the methoxide has split off, so the electrons in the S-O bond have gone to the oxygen, and the electrons on the anionic oxygen have come back down to reform the double bond.
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In the last step, the methoxide steals a hydrogen from the sulphonic acid, which can be shown as a curly arrow from the oxygen in the methoxide, and a curly arrow to show the breaking of the O-H bond.
Reply 2
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