MS 919 - Nucleophilic Acyl Substitution and Addition
Submitted by
Matt on August 6, 2011.
Textbook and Chapters: Carey and Giuliano 8th Ed. (2010) , Chapters 17, 18, 19
Keywords: carbonyl hydrates, nucleophilic acyl addition, nucleophilic acyl substitution
Description: The most important mechanism in second semester organic chemistry. Goes over carbonyl addition/substitution mechanisms under both acidic and basic conditions.
Problem # 705
Carbonyls are in equilibrium with their hydrate forms. This equilibrium happens in both acid and base.
Let's go through this equilibrium under basic conditions . Draw a mechanism using curved arrows for each reaction below.
Remember that under basic conditions, most species are either neutral or negatively charged, and rarely positively charged. So your structures will contain either ROH or RO- , but not ROH2 + .
a) Carbonyl to Hydrate
Notice that no oxygen is ever positive during these basic mechanisms (always negative or neutral).
b) Hydrate to Carbonyl
Solution The mechanisms in this problem and problem 706 are the most common mechanisms you will draw during second semester organic chemistry, and so it's a good idea to draw them out a few times.
In a) , the nucleophile attacks the carbonyl carbon, and the double bond goes "up" to form a tetrahedral (sp 3 ) carbon.
In b) , one of the oxygen atoms acts as leaving group, and a lone pair on the other oxygen comes "down" to reform the double bond (and the sp 2 carbon).
You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!
a) Carbonyl to Hydrate (basic conditions) "UP"
b) Hydrate to Carbonyl (basic conditions) "DOWN"
Problem Details MendelSet practice problem #
705 submitted by
Matt on July 21, 2011.
Problem # 706
Carbonyls are in equilibrium with their hydrate forms. This equilibrium happens in both acid and base.
Let's go through this equilibrium under acidic conditions . Draw a mechanism using curved arrows for each reaction below.
Remember that under acidic conditions, most species are either neutral or positively charged, and rarely negatively charged. So your structures will contain either ROH or ROH2 + , but not RO- .
a) Carbonyl to Hydrate (acidic)
b) Hydrate to Carbonyl (acidic)
Solution The interconversion between a carbonyl (sp 2 carbon) and a tetrahedral intermediate (sp 3 carbon) is the most common mechanism you will encounter in second semester organic chemistry.
You should be familiar drawing it under both acidic (this problem) and basic (problem 705 ) conditions.
In a) , the carbonyl "goes up" to form a tetrahedral intermediate.
In b) , an oxygen "comes back down" to reform the carbonyl and kick off a leaving group.
You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!
a) Carbonyl to Hydrate (acidic) "UP"
Because this reaction takes place under acidic conditions, the carbonyl must protonate before the nucleophile attacks, to prevent oxygen from ever being negative (ROH instead of RO- ).
b) Hydrate to Carbonyl (acidic) "DOWN"
The leaving group must protonate before it leaves, so it doesn't leave as a negative molecule (H2 O instead of HO- ).
Problem Details MendelSet practice problem #
706 submitted by
Matt on July 22, 2011.
Problem # 707
The overall mechanism for imine formation is shown below. (This isn't a real mechanism, just an outline)
Use curved arrows to draw the full mechanism for imine formation under acidic conditions. (I've added outlines of the intermediate structures for you to use as a guide). This mechanism is similar to that in problem 706 (carbonyl hydrate equilibria).
Solution This reaction takes place under acidic conditions, so oxygen should never be negative (only neutral or positively charged).
Acidic mechanisms tend to be a bit of pain, because you have to include many proton transfer steps (protonation and deprotonation).
Problem Details MendelSet practice problem #
707 submitted by
Matt on July 22, 2011.
Problem # 724
Use curved arrows to show the formation of the tetrahedral intermediate of a Fischer esterification reaction (shown below). There are three steps in total.
Solution This is a good mechanism to know (nucleophilic acyl substitution). If you don't understand it, see problems 706 and 708 .
Problem Details MendelSet practice problem #
724 submitted by
Matt on July 24, 2011.
Problem # 708
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.
Solution 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" ).
Problem Details MendelSet practice problem #
708 submitted by
Matt on July 22, 2011.
Problem # 728
The acyl group is a protecting group for amines. Amines can be acylated using acetic anhydride, and deacylated with base.
Propose a mechanism for each reaction.
Solution These reactions are both nucleophilic acyl substitution under basic conditions. See problem 705 for the mechanism in detail.
Problem Details MendelSet practice problem #
728 submitted by
Matt on July 24, 2011.
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