One of my graduate students was measuring out samples of vanillin and acetamide and placed them...

One of my graduate students was measuring out samples of vanillin and acetamide and placed them on a watchglass for analysis. She did not label the samples and they both look very similar.How could she determine which sample is which? Briefly describe what you would do.

Expert Solution

Ans. The following tests can be conducted to distinguish the two compounds-

I. Solubility in water (at 25.0⁰C):

Solubility of vanillin = 0.01 g/ mL

Solubility of acetamide = 1.0 g/ mL

Procedure: Take 10 drops of water in two different Eppendorf tubes or clean glass tube. To one tube add 5 drops of sample from watch glass 1, to the other tube add 5 drops of sample from watch glass 2.

The sample that dissolves in water is acetamide.

The sample that does not dissolve in water is vanillin.

Note: any other similar ration of water to samples can be taken for assessment, given that the amount of organic sample does not exceed solubility of acetamide in water. Because, in that case, both samples become insoluble in water.

queen_honey_blossom answered 2 years ago

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