A chemist carried out a Fischer esterification using methanol that was isotopically labeled with 18O (indicated with an asterisk).
Which one of the esters below (A-D) was formed?
To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism.
In this mechanism, the nucleophile (methanol) becomes the -OCH3 group in the ester.
At least 80% of second semester organic chemistry is two mechnanisms: nucleophilic acyl addition and nucleophilic acyl substitution. It's worth your time to become familiar with these mechanisms. See problems 705 (basic conditions), 706 (acidic conditions), 707, and 708.
MendelSet practice problem # 725 submitted by Matt on July 24, 2011.
Use curved arrows to show the formation of the tetrahedral intermediate of a Fischer esterification reaction (shown below). There are three steps in total.
This is a good mechanism to know (nucleophilic acyl substitution). If you don't understand it, see problems 706 and 708.
MendelSet practice problem # 724 submitted by Matt on July 24, 2011.
The overall mechanism for Fischer esterification is shown below. This isn't a real mechanism, just an outline.
Methanol (the nucleophile) attacks the carbonyl carbon, forming a tetrahedral intermediate, which then loses a water to reform the carbonyl. This mechanism is called nucleophilic acyl substitution.
Use curved arrows to draw a full mechanism for this reaction. I've included structures for you to use as a guide.
This reaction takes place under acidic conditions, so the mechanism you draw will be similar to those in problem 706.
Acidic mechanisms only appear complicated because they contain several proton transfer steps.
Nucleophilic acyl substitution mechanisms have only three real steps- the "up, down, and kick."
First, the nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate (the "up").
Then the carbonyl reforms (the "down") and a leaving group leaves (the "kick").
MendelSet practice problem # 708 submitted by Matt on July 22, 2011.
Compound A (C5H12O) is oxidized using aqueous chromium (Jones reagent) to compound B (C5H10O2), which is then treated with methanol under acidic conditions to yield compound C (C6H12O2) and water.
The 1H NMR of compound C is shown below. Determine the structures of compounds A, B, and C.
Let's solve this NMR structure elucidation problem using steps similar to those used in problem 662.
1.Are there any hints?
Compound A has one oxygen and after treatment with aqueous chromium becomes compound B, which has two oxygens. This means A is probably an alcohol, B is probably a carboxylic acid.
Compound B is then treated with methanol under acidic conditions to form compound C. These are conditions for a Fischer esterification, so C is probably the methyl ester.
2.How many IHD are there?
Compound A: C5H12O = C5H12 should be C5H12 (CnH2n+2) so 0 IHD.
Compound B: C5H10O2 = C5H10 should be C5H12. Missing 2H, so 1 IHD.
Compound C: C6H12O2 = C6H12 should be C6H14. Missing 2H, so 1 IHD.
These IHD counts fit our assumptions from part 1).
3.Draw some structures and eliminate, learn, repeat.
Some clues from the NMR:
The isopropyl splitting pattern is present: d(6) (signal c at ~0.9 ppm) and multiplet(1) (signal b at ~2.4 ppm).
The s(3) at ~3.7 ppm is probably the methyl group from the methyl ester.
We know from before we have one IHD, and it's probably an ester.
So start drawing structures and eliminate those that don't fit the data!
MendelSet practice problem # 679 submitted by Matt on July 19, 2011.