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I recently made self-assembled monolayers (SAMs) of alkanethiols on a copper substrate. After scrubbing off the $\small{\ce{Cu2O}}$ top layer and rubbing the surface with ether and ethanol, I let my copper plates stand in a 5% (w/w) $\small{\ce{NaOH}}$ solution for 5 minutes, which I then throughly rinsed off with deionized water.

I'm wondering if I could skip this step. I'm not sure what it is needed for. My knowledge of chemistry tells me that $\small{\ce{Cu(0)}}$ should not react with sodium hydroxide. I'm also positive that any grease or other contaminants were washed off in the previous steps.

What do you think?

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Cu(0) will not be attacked by NaOH, and will not decompose in this aqueous environment at room temperature. So, this is not to create some reaction with the copper. Two uses I can think of:

  1. NaOH increases the solubility of Cu2O, so maybe this is a way of getting rid of the last remaining copper oxide.
  2. To bind the thiols with copper, you'd better be in alkaline conditions rather than acidic conditions, right? If so, then maybe this is the way of making sure you don't have any acidic groups or molecules remaining adsorbed on the copper surface, to improve bonding at the next step.
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Your first point might be the reason why... The protocol isn't clear on why we have to do this NaOH step. What I forgot to mention is that I thoroughly rinsed with distilled water after, so pH should be around 7. –  CHM Oct 29 '12 at 22:27
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