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Logic Puzzle #3: A Play on the Molecular Formulae

June 21, 2010

The goal of this puzzle is to determine how the molecular formulae of the intermediates may assist in reasoning out the final product.

In this puzzle, let’s consider the following one-pot synthetic reaction (solvents and additional reactants are not shown). The molecular formula (MF) and the RDBE information are also presented. The chemical reaction illustrates a C11 compound reacting to form a C9 compound and subsequently a C15 compound. The reaction continues and produces an unknown compound with a MF of C24 H23 N1 O2. Based on the given information, how can the MF of the unknown compound be explained?

LogicOnMF#3_1_Jun212010

The unknown comprises of 24 carbon atoms and so if you add the carbons from the 2nd intermediate with the carbons from the 3rd intermediate you arrive at the 24 carbons (9+15=24). This is also evident for the nitrogen atom count (0+1=1) and the RDBE count (5+9=14). This is not the case for the hydrogen (10+15=25) and oxygen (2+1=3) atoms. However, if the loss of a H2O molecule is considered, the unknown can be a combination of the 2nd and 3rd intermediate.

LogicOnMF#3_2_Jun212010

 

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