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I recently wrote a test in which I was asked:

What is this reaction $\ce{CH3CH=CH2 ->[CO + H2O][H+]CH3-CH(COOH)-CH3}$ known as?

  1. Wurtz Reaction
  2. Koch Reaction
  3. Clemmensen's Reduction
  4. Kolbe's Reaction

My thoughts on this:

  1. It's definitely not Wurtz Reaction, because that reaction eliminates terminal halogens and attaches together two alkyl groups. Also, that reaction needs sodium in dry ethereal medium.
  2. It's definitely not Clemmensen's reduction, because that needs zinc amalgam and acid. (Although acid is present, the reduction won't take place without the zinc, right?).
  3. Now, my options are Koch and Kolbe's reactions. As Koch's reaction (Gatterman-Koch Formylation) adds a carbonyl group, and not a carboxylic acid group, I decide to go with Kolbe's reaction. However, as per Wikipedia, this reaction requires carbon dioxide and sodium phenoxide. The carbon dioxide can be taken care of, by the water gas shift reaction that converts $\ce{CO + H2O->CO2 + H2}$. But now, there's no sodium phenoxide. What to do now?

Basically, my doubt is whether there's a mistake in my procedure of eliminating the options, or if there's something I'm missing out somewhere which may lead to make a different choice?

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    $\begingroup$ Obviously I agree with Klaus' reply below. I'm adding my own comment because it worries me that you were looking for the phenoxide when trying to 'fit' the reaction name to the scheme you had in your test. You should have recognised that the phenoxide, in the Kolbe reaction, is not a reagent but a reactant. What would your propene be doing there, if you were just reacting phenoxide and $\ce{CO2}$? You're clearly thorough in your searches, but you sometimes seem to take hasty shortcuts that lead you the wrong way. Which is dangerous, because details are everything in science. $\endgroup$ – user6376297 Dec 27 '17 at 16:52
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As Koch's reaction (Gatterman-Koch Formylation) adds a carbonyl group, and not a carboxylic acid group, I decide to go with Kolbe's reaction.

You started right, and then took the wrong turn ;-)

Under acidic conditions

  1. the proton adds to the $\ce{C=C}$ double bond and gives a secondary cation
  2. the cation reacts with carbon monoxide and yields an acylium cation
  3. the latter reacts with water to result in the formation of a carboxylic acid

The whole process is known as the Koch reaction (without the Gattermann).

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