Logic Puzzle #3: A Play on the Molecular Formulae

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?

 The unknown comprises of 24 carbon atoms and so if you add the carbons from the 2^nd intermediate with the carbons from the 3^rd 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 2^nd and 3^rd intermediate.

Logic Puzzle #3: A Play on the Molecular Formulae

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?

 The unknown comprises of 24 carbon atoms and so if you add the carbons from the 2^nd intermediate with the carbons from the 3^rd 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 2^nd and 3^rd intermediate.

Blog on Hiatus from June 7 to 18, 2010

With the warm weather and some good football/soccer matches upon us, the blog P2C2E will be on hiatus for 2 weeks from June 7 to 18. Posts will resume the following week.

Stay tuned for more logic puzzles.

Blog on Hiatus from June 7 to 18, 2010

With the warm weather and some good football/soccer matches upon us, the blog P2C2E will be on hiatus for 2 weeks from June 7 to 18. Posts will resume the following week.

Stay tuned for more logic puzzles.

Logic Puzzle #2: How to link 3 Fragments

The goal of this puzzle is to logically combine a set of fragments using valence and NMR information.

In this puzzle, three fragments are correlated through 2-3J coupling responses (represented by a green arrow) that were extracted from a 1H-13C HMBC data (spectrum not shown). The carbon atoms with the 13C chemical shifts displayed in blue indicate the presence of an adjacent heteroatom. Based on these criteria, what 'complete' fragment(s) supports the data and is there anything missing?

In order to accommodate these restrictions, a logical fit is to consider a trivalent atom, e.g. nitrogen.

Logic Puzzle #2: How to link 3 Fragments

The goal of this puzzle is to logically combine a set of fragments using valence and NMR information.

In this puzzle, three fragments are correlated through 2-3J coupling responses (represented by a green arrow) that were extracted from a 1H-13C HMBC data (spectrum not shown). The carbon atoms with the 13C chemical shifts displayed in blue indicate the presence of an adjacent heteroatom. Based on these criteria, what 'complete' fragment(s) supports the data and is there anything missing?

In order to accommodate these restrictions, a logical fit is to consider a trivalent atom, e.g. nitrogen.

Logic Puzzle #1: The Missing Link

A great skill to master is the capability to conceptualize a fragment or structure directly off a spectrum without resorting to paper-and-pen work. This skill is learnt through lots of practice. Whenever partial information is available, an elucidator can conjure up a mental image of possibilities and should it be required instinctively hunt for any missing data.

In the following example, a set of fragments including 13C and 1H chemical shifts and long-range coupling information were extracted from an HMBC experiment (not shown). The green arrows represent the 2-3J coupling responses between the 3 equivalent methyl groups and the carbonyl’s quaternary carbon. Based on these restrictions, what fragment(s) support the data and is there anything missing?

To accommodate these restrictions, three potential fragments, assigned A, B and C, are shown below. Fragment A can be disregarded on the basis of the carbon valence. Fragment B is not a good candidate because the CH3 chemical shifts do not support the presence of an adjacent heteroatom. Fragment C seems to be the most logical choice. However, there is a missing quaternary carbon. The next step is to re-evaluate the NMR data in search of a weak 13C signal at ~40 ppm.