MS 907 - Alkene Addition and Elimination Mechanisms
Submitted by
Matt on August 6, 2011.
Textbook and Chapter: Carey and Giuliano 8th Ed. (2010) , Chapter 6
Keywords: alkene addition, carbocation, dehydration, E1 mechanism
Description: This mendel set is a complement to Carbocation Drills . It is meant to prepare students on how to approach longer and more complicated mechanisms. Reaction mechanisms covered:
Acid-catalyzed elimination (dehydration, E1)
Acid-catalyzed addition to an alkene
Halohydrin formation (halonium ions, which result in anti-additions.)
Problem # 518
The alcohol below is protonated and contains an oxygen with a positive charge. Using curved arrows, show the two "legal moves" that result in a neutral oxygen.
Solution Protonation is a common first step in mechanisms. When you have a protonated alcohol, it's likely that either:
The alcohol will deprotonate to form a neutral alcohol (right), or
The alcohol will leave as a neutral molecule to form a carbocation.
Either one is possible. If you're not sure which one happens, remember that when writing out mechanisms, never go backwards - don't draw structures you have drawn before.
Problem Details MendelSet practice problem #
518 submitted by
Matt on June 30, 2011.
Problem # 519
Let's work through an elimination reaction. Draw the structures for each of the species in the three boxes below (protonated thiol, carbocation, and alkene). Also draw curved arrows to show electron movement.
Solution The first step in this reaciton, like many reactions, is an acid-base reaction- when you see an H+ (acid), the first step is usually something getting protonated. This starting material is a thiol (sulfur), but the same thing would happen with an alcohol or ether (oxygen).
When you have a protonated thiol (or alcohol, ether, etc.), you know the reaction isn't finished yet- products are rarely charged. There are two legal moves to get that positive charge off of the protonated thiol. Either the thiol deprotonates, or it takes off as a neutral leaving group. It can't deprontate because that would be going backwards, so your only "legal move" is for the HSCH2 CH3 to act as a leaving group (See problem 518 ).
But then you're left with a carbocation, which is definitely not the final product. How can we get rid of it? (See problem 335 for general ways carbocations react). If the HSCH2 CH3 attacked in an SN 1 type reaction that would be OK, but it would be going backwards! So the only way to get rid of the carbocation is to do a beta-elminiation (E1).
Problem Details MendelSet practice problem #
519 submitted by
Matt on June 30, 2011.
Problem # 520
Let's work through an alkene addition reaction. Draw the structures for each of the species in the three boxes below (3º carbocation, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Solution Note that this is the reverse of problem 519 . Instead of going from thiol (alcohol) to carbocation to alkene, we're going from alkene to carbocation to thiol. In each step ask yourself "what arrows can I draw?" and choose the step that doesn't go backwards.
The first step is the alkene picks up a proton to form the more stable carbocation. (See problem 334 if you don't understand why only the 3º carbocation is formed). To get rid of the carbocation, we can either do a beta-elimination (E1) to form an alkene, or an addition reaction (SN 1) to form a protonated thiol. Since forming the alkene would be going backwards, the only choice is addition.
Then you're left with a protonated thiol. How do we get rid of the positive charge on the sulfur? There are two legal moves- either the RSCH2 CH3 acts as a leaving group, or the sulfur deprotonates. Since the RSCH2 CH3 leaving would form a carbocation and take us backwards, the only option is for the sulfur to deprotonate.
Because the sulfur added to the more substituted carbon, this was a Markovnikov addition.
Problem Details MendelSet practice problem #
520 submitted by
Matt on June 30, 2011.
Problem # 521
Let's work through a halogenation reaction. Draw the structures for each of the species in the four boxes below (3º carbocation, halonium ion, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Solution Because the chlorine is more electronegative than iodine, the iodine will have a partial positive charge and will be attacked by the alkene. It forms the more stable carbocation as normal (as in problem 335 ). The nucleophile (HSCH2 CH3 ) will eventually attack the carbocation, but the iodine does something special first- it forms a cyclic intermediate.
The formation of the cyclic halonium ion is probably the most important concept you will learn in first semester organic chemistry. With iodine, it's called an iodonium ion. (with chlorine or bromine, it's called a chloronium or bromonium ion).
The HSCH2 CH3 will attack the carbon that was the carbocation, but it's forced to attack from the side opposite to that of the iodine. So the product will always have anti stereochemistry (you get the trans product).
Problem Details MendelSet practice problem #
521 submitted by
Matt on July 1, 2011.
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