This is a gold catalyzed alkyne hydration reaction: Phenylacetylene hydrated by

Question

This is a gold catalyzed alkyne hydration reaction:

Phenylacetylene hydrated by 2 mol% HAuCl4 + H2SO4 + MeOH/H2O yields acetophenone and/or phenylacetaldehyde

1. Gold is used in this experiment as a catalyst in 2 mol % or 0.02 equivalents relative to phenylacetylene. How many times does each gold catalyst have to perform the reaction on average to convert all of the phenylacetylene to the hydrated product?

2. How would you be able to identifiy which isomer, acetophenone or phenylacetaldehyde, was formed in your reactiton using IR spectroscopy? How would you know that all of your starting material, phenylacetylene, had been consumed?

Explanation / Answer

For the given gold catalyzed alkyne hydration reaction.

1. Each gold would have to run atleast 200 times fo complete hydrolysis of the starting materia.

2. The isomer acetophenone would show a sharp peak at around 1720 cm-1 for C=O stretch in the IR If formed. Similarly, phenylacetaldehyde would show along igth a peak at aound 1730 cm-1 for C=O stretch, two peaks at 2720 cm-1 and 2850 cm-1 corresonding to the aldehyde stretchings. Absence of this peak confirms only acetophenone formed in the reaction.

The starting material If present, phenylacetylene would show a peak at 3300 cm-1 for C-H stretch of alkyne and a peak at around 2200 cm-1 for CC stretch of alkyne. Absence of these peaks would confirm the consumption of starting material in the reaction.

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