Some clarifications--book vs. lecture
44 views
Skip to first unread message
Gregory Watkins
Jul 21, 2012, 1:25:22 AM7/21/12
to organic-chemist...@googlegroups.com
A few items:
1.) First, to be clear: Midterm II will NOT include the oxidation chemistry of alcohols with Cr reagents.
2.) The 6th edition of the book includes a number of examples of haloalkane chemistry that may confuse you, based on what I've said in lecture.
For instance: There is one example of ethoxide attacking a primary haloalkane in the presence of ethanol and heat. Your book calls this reaction Sn2, although you likely would label it as E2 based on what I said in class. My thoughts are as follows:
Let's assume for a second that your book is right. The logic is as follows: Although E2 is possible, primary haloalkanes usually react by Sn2. Ethoxide is a strong base and a good nucleophile, and can therefore do Sn2. Ethanol is a poor solvent for Sn2, so an impatient/sloppy chemist might be tempted to heat the reaction up in order to accelerate the forward reaction. Notice that your book does mention that some E2 products will also result under these conditions.
My response to your book: 1.) Yes, it is difficult to perform E2 chemistry on primary haloalkanes. However, because heat favors elimination, the reaction conditions will almost certainly give lots of E2 product *if you heat the reaction up to a high enough temperature.* 2.) This is a terrible example of haloalkane chemistry. If you really wanted to conduct an Sn2 reaction with ethoxide, you would use a polar aprotic solvent (e.g., DMSO) and avoid heat. If you really wanted to conduct an E2 reaction on a primary haloalkane, you would use a bulkier base in addition to a protic solvent and heat.
Another bad example from your book: The formation of ethanol from chloroethane and sodium hydroxide in methanol. Many of the above arguments hold here as well, but there is an additional problem: If sodium hydroxide is added to methanol, there will be an acid-base equilibrium between these reactants and methoxide + water. If methanol is the solvent, there will be thousands of methanol molecules for every hydroxide ion, and the equilibrium will be forced to the right. Hence, the dominant nucleophile will be methoxide (another small, strong base) and not hydroxide, and ethanol will not be formed as a major product. Couldn't follow all of that? Don't worry. We'll come back to this in the ether chapter.
Is all of this frustrating? Yes. Therefore, I will send out some additional haloalkane practice problems that are better than the ones in your book. Stay tuned.
Greg
1.) First, to be clear: Midterm II will NOT include the oxidation chemistry of alcohols with Cr reagents.
2.) The 6th edition of the book includes a number of examples of haloalkane chemistry that may confuse you, based on what I've said in lecture.
For instance: There is one example of ethoxide attacking a primary haloalkane in the presence of ethanol and heat. Your book calls this reaction Sn2, although you likely would label it as E2 based on what I said in class. My thoughts are as follows:
Let's assume for a second that your book is right. The logic is as follows: Although E2 is possible, primary haloalkanes usually react by Sn2. Ethoxide is a strong base and a good nucleophile, and can therefore do Sn2. Ethanol is a poor solvent for Sn2, so an impatient/sloppy chemist might be tempted to heat the reaction up in order to accelerate the forward reaction. Notice that your book does mention that some E2 products will also result under these conditions.
My response to your book: 1.) Yes, it is difficult to perform E2 chemistry on primary haloalkanes. However, because heat favors elimination, the reaction conditions will almost certainly give lots of E2 product *if you heat the reaction up to a high enough temperature.* 2.) This is a terrible example of haloalkane chemistry. If you really wanted to conduct an Sn2 reaction with ethoxide, you would use a polar aprotic solvent (e.g., DMSO) and avoid heat. If you really wanted to conduct an E2 reaction on a primary haloalkane, you would use a bulkier base in addition to a protic solvent and heat.
Another bad example from your book: The formation of ethanol from chloroethane and sodium hydroxide in methanol. Many of the above arguments hold here as well, but there is an additional problem: If sodium hydroxide is added to methanol, there will be an acid-base equilibrium between these reactants and methoxide + water. If methanol is the solvent, there will be thousands of methanol molecules for every hydroxide ion, and the equilibrium will be forced to the right. Hence, the dominant nucleophile will be methoxide (another small, strong base) and not hydroxide, and ethanol will not be formed as a major product. Couldn't follow all of that? Don't worry. We'll come back to this in the ether chapter.
Is all of this frustrating? Yes. Therefore, I will send out some additional haloalkane practice problems that are better than the ones in your book. Stay tuned.
Greg
Reply all
Reply to author
Forward
0 new messages