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This is the answer to chapter 20, problem number 64 from the mcmurray organic chemistry textbook.
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This problem talks about the ritter reaction.
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So we have an alken, reacting with a nitrile, in the presence of strong aqueous sulfuric acid.
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And the end product is an amide.
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And so we are asked to draw a mechanism for this reaction.
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Okay.
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So we know that the first step is going to involve the acid, the strong aqueous sulfuric acid that we're told is in this reaction.
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And so that's going to be a protonation, as we might expect.
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And so that is going to look like this.
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So after the first step, we'll be at this intermediate.
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And so we've protonated our alken and as a result of that we have a carbocation right here at the carbon with the methyl substituent and this carbocation is then going to be attacked by the nitrile so we have our nitrile with a lone pair on the nitrogen and so that's going to be our nucleophile we'll attack the carbocation and so after the second step, we'll be here.
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Okay, so our nitrogen has four bonds now, so it's going to have a positive charge on it.
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And as a result of this, that's going to make this carbon susceptible to attack by a water molecule.
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And so this carbon will get attacked.
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One of these bonds of the carbon -nitrogen triple bond will break and the electrons will revert to the nitrogen.
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And so after that, you can kind of see our amide is starting to take shape.
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Okay.
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And so the next step that's going to happen, we're going to deprotonate this water that added here.
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So we have this...