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.
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 HSCH2CH3 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 HSCH2CH3 attacked in an SN1 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).
MendelSet practice problem # 519 submitted by Matt on June 30, 2011.
When propyl bromine is treated with KF in benzene no reaction takes place. But when the crown ether 18-Crown-6 is added to the reaction mixture the desired propyl fluoride product is produced. Explain.
KF is insoluble in propyl bromide or benzene, so the two compounds never "touch" each other, and no reaction takes place. (F- and propyl bromide are in different phases and so don't come into contact).
But when the crown ether is added, it solvates the potassium ion, allowing the K+ and F- ions to dissolve in the solvent, so the fluoride ion and propyl bromide are able to "talk" to each other, and the rection can take place.
MendelSet practice problem # 698 submitted by Matt on July 21, 2011.
The acid-catalyzed condensation of alcohols to form ethers is reversable; ethers can be hydrolyzed back to alcohols. How can the direction of this equilibrium be controlled to preferentially form ethers?
To push an equilibrium to one side, add starting material and remove product. This is Le Chatelier's principle from general chemistry.
So to push this reaction to the right and form ether, add alcohol and remove ether and water as they form.
To push this reaction to the left and form alcohol, add water to ether and remove alcohol as it forms.
How do you "remove something as it forms?" Alcohols and ethers have (relatively) low boiling points, and can be removed by hooking up a vacuum line and condenser to your reaction. The ether (or alcohol) boils off under the reduced pressure, and then recondenses in a separate piece of glassware. (Sort of like in distillation.)
Water has a relatively high boiling point and so is difficult to remove under reduced pressure. To remove water, molecular sieves are used. They're like tiny sponges that only absorb water (and not other solvents), removing it from the reaction.
MendelSet practice problem # 701 submitted by Matt on July 21, 2011.
For the reaction below, draw the structures of the carbocation intermediate and the final product.
This is a substitution reaction. Because the reaction takes place in acid and the leaving group is on a 2º carbon, it will probably form a carbocation (SN1 mechanism).
-OH is a poor leaving group so the alcohol will protonate first, so it can leave as H2O, which is neutral.
MendelSet practice problem # 347 submitted by Matt on June 7, 2011.
Show how each compound can be prepared from an alkene containing 3 carbons (or less).
Each answer will involve the reaction of a Grignard with either a carbonyl or epoxide.
Note: epoxides are prepared from alkenes using a peroxy acid (epoxidation) such as mCPBA.
The trick to synthesis problems in second semester organic chemistry to recognize that alcohols ARE ketones ARE carboxylic acids. What do I mean? Alcohols, ketones/aldehydes, and carboxylic acids can all be easily converted using PCC or Jones Reagent (NaCr2O7/H2SO4).
For example, for a), the product is a ketone, but it may as well be an alcohol, because alcohols can be converted to ketones with PCC.
b) is similar, except the position of the alcohol (one away from the "bond cut", instead of directly connected to the cut as in a) ) indicates the starting material was an epoxide and not a carbonyl.
c) is just like b), except instead of PCC, use Jones Reagent to oxidize the alcohol all the way to a carboxylic acid.
MendelSet practice problem # 673 submitted by Matt on July 19, 2011.
Show how each compound can be prepared from the indicated starting material.
All carbon sources must contain three carbons or less.
a) When you see 2 carbons and 1 oxygen, that the tell-tale sign that you're adding ethylene oxide (the simplest epoxide).
But that would only leave you with an alcohol. How do you get to the ether? Using the Williamson ether synthesis.
b) As I mentioned in problem 673, when you see an alcohol you are also looking at a carbonyl, because you can interconvert the two (alcohol to aldehyde/ketone using PCC, aldehyde/ketone to alcohol using NaBH4).
To add the methyl group, convert the alcohol to a ketone (which is an electrophile), and then add methyl Grignard (a nucleophile). But once again, you are only left with an alcohol. How to convert it to an ether? The Williamson ether synthesis.
When you see an ether in a synthesis problem, remember the Williamson ether synthesis. It will come in handy.
MendelSet practice problem # 699 submitted by Matt on July 21, 2011.
Write out a mechanism for the reaction below using curved arrows. Be sure to include formal charges.
Ethers react with 2 equivalents of H-X to form water and two equivalents of alkyl halide.
In this case, the ether was cyclic, so the ring had to open up.
The reaction can go through either an SN1 or SN2 mechanism. Since this was a primary ether, it will go through an SN2 mechanism (the carbocation is too unstable for the reaction to go SN1).
MendelSet practice problem # 700 submitted by Matt on July 21, 2011.
Show two ways to prepare the ether below from a combination of an alcohol and an alkyl halide via the Williamson ether synthesis.
Is one way better than the other? Why?
The Williamson ether synthesis takes place in two steps. First an alcohol is deprotonated to form a strong nucleophile (RO-, this step isn't shown in the image below). Then the alkoxide (negative alcohol) attacks an alkyl halide in an SN2 reaction.
So this problem is really asking, which step of conditions is most favorable for an SN2 reaction?
Recall that SN2 reactions compete with E2 reactions. If the nucleophile is too basic, or if there is too much bulk, it will go E2 instead of SN2. (See problem 560 for a full explanation of these competition reactions)
Below, the top combination uses the less substituted (1º) alkyl halide, and so is the best for an SN2 reaction.
The bottom reaction uses a bulkier (2º) alkyl halide, and will probably give a higher percentage of E2 side reaction.
MendelSet practice problem # 703 submitted by Matt on July 21, 2011.
Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.
You know this is an elimination reaction because no nucleophile is present; H3PO4 and H2SO4 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.
MendelSet practice problem # 341 submitted by Matt on June 7, 2011.
Compound A (C5H12O) is oxidized using aqueous chromium (Jones reagent) to compound B (C5H10O2), which is then treated with methanol under acidic conditions to yield compound C (C6H12O2) and water.
The 1H NMR of compound C is shown below. Determine the structures of compounds A, B, and C.
Let's solve this NMR structure elucidation problem using steps similar to those used in problem 662.
1.Are there any hints?
Compound A has one oxygen and after treatment with aqueous chromium becomes compound B, which has two oxygens. This means A is probably an alcohol, B is probably a carboxylic acid.
Compound B is then treated with methanol under acidic conditions to form compound C. These are conditions for a Fischer esterification, so C is probably the methyl ester.
2.How many IHD are there?
Compound A: C5H12O = C5H12 should be C5H12 (CnH2n+2) so 0 IHD.
Compound B: C5H10O2 = C5H10 should be C5H12. Missing 2H, so 1 IHD.
Compound C: C6H12O2 = C6H12 should be C6H14. Missing 2H, so 1 IHD.
These IHD counts fit our assumptions from part 1).
3.Draw some structures and eliminate, learn, repeat.
Some clues from the NMR:
The isopropyl splitting pattern is present: d(6) (signal c at ~0.9 ppm) and multiplet(1) (signal b at ~2.4 ppm).
The s(3) at ~3.7 ppm is probably the methyl group from the methyl ester.
We know from before we have one IHD, and it's probably an ester.
So start drawing structures and eliminate those that don't fit the data!
MendelSet practice problem # 679 submitted by Matt on July 19, 2011.
Show how to prepare each compound starting from propylene oxide.
(Propylene oxide image below courtesy of Wikipedia.)
Epoxides can open up in two different ways.
To add a nucleophile to the less substituted side of an epoxide, use basic conditions. This is done in #2 below.
To add a nucleophile to the more substituted side of an epoxide, use acidic conditions. This is done is #1 below.
Why do the conditions matter? Epoxides have two electrophilic carbons. Normally nucleophiles will preferentially attack the less substituted carbon, as they do in SN2 reactions. Recall that SN2 reactions usually happen with strong nucleophiles- that is, negative charges (basic conditions).
When an epoxide reacts under acidic conditions, the transition state has carbocation character, and so it's sort of like an SN1 reaction. That is, instead of less substituted carbons being favored due to less steric bulk, more substituted carbons are favored do to a more stable carbocation. So acidic conditions cause an epoxide to open up on the more substituted side.
MendelSet practice problem # 702 submitted by Matt on July 21, 2011.