MS 4637 - Sacramento City College Chem 420 Unit 8 Alkenes (I)
Description: Please do the following Mendelset problems, and problems 1-38 in Ch. 11 of Second Language.
Problem # 335
Carbocations aren't very stable and so don't last very long after they are formed.
Use curved arrows to show:
a) how a carbocation reacts with a halide ions to form an alkyl halide.
b) how a carbocation reacts with water to form an alcohol.
c) how a carbocation reacts with a base to form an alkene.
Solution For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.
For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.
For c), there are two different types of beta hydrogens, and so two different alkenes are possible.
The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation : neighboring carbon atoms stabilize an alkene.
Problem Details MendelSet practice problem #
335 submitted by
Matt on June 7, 2011.
Problem # 348
For the reaction below, draw the structures of the carbocation intermediate and the final product.
Solution The delta (Δ) in the reaciton arrows means that heat is being added to this reaction, which tends to favor elimination over substitution. Also, the reaction is using a non-nucleophilic acid (H2 SO4 ), which tends to favor elimination reactions (H3 PO4 is another common reagent for E1 reactions, while HCl or HBr tend to go SN 1).
Because this reaction is taking place in acid, a carbocation is likely to form, so this is an E1 reaction. Since water is lost over the course of the reaction, this is a dehydration, which is a type of elimination reaction.
Problem Details MendelSet practice problem #
348 submitted by
Matt on June 7, 2011.
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 # 333
Let's go over how a carbocation can form from an alcohol.
Write in the curved arrows to show the formation of the protonated alcohol, and water acting as a leaving group to form a carbocation.
Solution
Problem Details MendelSet practice problem #
333 submitted by
Matt on June 7, 2011.
Problem # 319
For a molecule to undergo an E2 reaction, the leaving group and the beta-proton must be in an anti-coplanar conformation (one atom straight up, the other straight down). Based on this, which compound undergoes E2 reaction with KOt Bu faster? Why?
Solution
The methyl group is bulky and so is most stable in the equatorial position.
The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.
So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.
The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.
Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.
Problem Details MendelSet practice problem #
319 submitted by
Matt on June 7, 2011.
Problem # 341
Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.
Solution You know this is an elimination reaction because no nucleophile is present; H3 PO4 and H2 SO4 are non-nucleophilic acids. IF the reagent were HCl or HBr on the other hand, this would be a substitution reaction.
Since acid is present, a carbocation will probably form, so we know that this is an E1 mechanism.
The 2º carbocation that is initially formed will undergo a 1,2-hydride shift to become a 3° carbocation, but that doesn't affect the final product of the reaction; with either the 2° or 3° carbocation, the most stable alkene product is 2-methyl-2-butene.
Problem Details MendelSet practice problem #
341 submitted by
Matt on June 7, 2011.
Problem # 531
E2 elimination reactions require anti -coplanar geometry. (note: some textbooks call this anti -peri planar).
Let's work through an E2 reaction, and rotate the molecule eblow into an anti- coplanar geometry to predict the product of this E2 reaction.
Solution To predict the stereochemistry of the alkene product from an E2 reaction, we have to rotate the leaving group (Br) and the beta-proton into anti-coplanar geometry. It's easiest to put the H and Br in the plane of the page.
After we do this, we see that both the ethyl and phenyl groups are wedges, and the two methyls are dashes. So after the E2 reaction, the resulting alkene will have its two methyl groups cis to each other.
Problem Details MendelSet practice problem #
531 submitted by
Matt on July 2, 2011.
Follow MendelSet on , become a fan on
Facebook .
Printer-friendly version
