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.
Draw out the mechanism for the addition of excess phenyl Grignard to the carbonyl compound below.
This carbonyl has two leaving groups attached to it- each of those oxygens can take part in a nucleophilic acyl substitution reaction and form a new carbonyl product.
First the Grignard attacks the oxidation state IV carbonyl carbon (4 oxygen bonds, so oxidation state 4). The carbonyl itself will act as a leaving group and form a tetrahedral intermediate. But tetrahedral intermediates don't last if there are any leaving groups attached to the carbon, so the -O will "come back down again", kick off an oxygen leaving group, and reform the carbonyl.
Then a second equivalent of Grignard will attack that carbonyl (an ester), and we will do another nucleophilic acyl substitution reaction to form yet another carbonyl.
Finally, the third carbonyl doesn't have any leaving groups built in (it's a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol.
Notice that as the reaction progresses the oxidation state of the carbonyl carbon (number of oxygen bonds attached to it) goes down form 4 to 3 to 2 and then to 1.
MendelSet practice problem # 670 submitted by Matt on July 18, 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.
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.
The ester below was dissolved in a solution of water, a small amount of which was isotopically labeled with O-18, denoted with an asterisk.
After a few hours, some isotopically labeled oxygen was found in the ester. Where was it found in the ester? Can you explain why?
To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism. (see problem 725)
Water reacts with an ester to form a carboxylic acid. This is what happens here.
If there is only trace amounts of water, it's possible that the water reacts with the ester to form a carboxylic acid, and then goes back to reform the ester. But, in this process, leaves an isotopically labeled oxygen in the carbonyl position.
MendelSet practice problem # 729 submitted by Matt on July 24, 2011.
N,N-dimethylformamide (DMF) is shown below. Based on its structure, you might expect to see only one -CH3 signal in the 1H NMR spectrum. But instead DMF shows two different -CH3 signals. Explain.
DMF appears to have two identical methyl groups. Since these six protons are all equivalent, its 1H NMR should only show one methyl signal (singlet, 6H).
So why is that the real life the 1H NMR of DMF shows two methyl signals? (two singlets with integration of 3H).
Because DMF is an amide.Recall that the "real" structure of molecule is the a mixture of its resonance forms. DMF doesn't look like either of the two resonance forms below. In real life, its somewhere in between.
For most carboxylic acid derivatives (such as esters), the resonance form is only a minor contributor and so the real "picture" looks very close to the carbonyl Lewis structure.
But for amides, its resonance form is fairly stable (it's common for nitrogen atoms to be positively charged), and so is a major resonance contributor.
In an amide, the bond between the carbonyl carbon and the nitrogen atom has a high degree of double bond character. (This also explains why it's harder to rotate the C-N "single bond" than you would expect from its Lewis structure- it's sort of like a double bond).
Because the C-N "single bond" is closer to a double bond, the two methyls are not equivalent. One methyl is cis, and the other is trans, and so they show two signals in the 1H NMR.
MendelSet practice problem # 730 submitted by Matt on July 24, 2011.
Show how each ketone below can be prepared from sodium cyanide and either ethylene or propene.
You may also use methyl Grignard and ethylene oxide.
Nitriles react with Grignard reagents to form ketones, so the carbonyl carbon on each product must have came from a nitrile.
You also have the restriction that the longest carbon chain building must be three carbons.
a) Working backwards, the left side must have come from a nitrile, so the left side came from a Grignard.
How can we make each starting from propene?
To make the Grignard, prepare propyl bromide from propene via anti-Markovnikov HBr addition (HBr/peroxides), and then add Mg0/ether.
To make the nitrile, add sodium cyanide (NaCN) to propyl bromide.
Then add the nitrile and Grignard together, which forms the imine intermediate, which will hydrolyze to the ketone after you add water (H3O+).
b) This problem is harder because you're limited to starting materials with three carbons or less, so we have to make three "cuts" instead of two, like in a).
Working backwards again, prepare propyl Grignard from propene as described in a), and then add ethylene oxide to form 1-pentanol. Form the 6-carbon nitrile by first converting the alcohol to a bromide using PBr3, and then adding NaCN. Finally, add methyl Grignard followed by H3O+ to form the desired ketone.
MendelSet practice problem # 735 submitted by Matt on July 25, 2011.