Show two ways of preparing the alkene below via the Wittig reaction starting from triphenyl phosphine (PPh3).
Is one route better than the other? Why?
Alkenes can be prepared from a combination of ylide and aldehyde or ketone. This is the Wittig reaction.
One carbon on the alkene comes from the carbon bonded to the PPh3 on the ylide, and the other carbon comes from the carbonyl.
There are usually two different ways to make an alkene via the Wittig reaction. So is one way better than the other?
Yes. Ylides come from the SN2 reaction of PPh3 with an alkyl halide. Because it's an SN2 reaction, you want to use the least substituted alkyl halide! (1º is better than 2º). So of the two reactions below that would result in the desired alkene, the top method is better because it involves the less bulky alkyl halide.
This principle also holds true for the Williamson ether synthesis (problem 703). Less substituted alkyl halides are better for SN2 reactions.
MendelSet practice problem # 711 submitted by Matt on July 22, 2011.
When a carbonyl is treated with semicarbazide under acidic conditions an "imine" is produced called a semicarbazone.
Which of the two products below is the correct structure for a semicarbazone? Explain.
What this problem is really asking is "Which nitrogen on semicarbazide is more nucleophilic?"
The nitrogens bonded to the carbonyl on semicarbazide all have resonance forms, so their lone pairs aren't as available for nucleophilic attack.
The other nitrogen (on the left) is the most nucleophilic because it isn't involved in any resonance forms, so its lone pair isn't shared with another atoms.
MendelSet practice problem # 713 submitted by Matt on July 22, 2011.
Complete each synthesis below. All carbon sources must come from alkenes.
Each synthesis will involve protecting groups.
The protecting group most commonly used for aldehydes and ketones (in undergraduate orgo) is ethylene glycol.
It is put on using HOCH2CH2OH/H+ and removed with H2O/H+.
For a), the reaction calls for the use of acetylide with an alkyl halide. But acetylides also react with carbonyls.
So before the alkyne is deprotonated using a strong base (such as NaNH2), the carbonyl must be protected with ethylene glycol.
For b), the reaction calls for the addition of two equivalents of phenyl Grignard to the ester. The problem is that esters aren't as reactive as ketones (or aldehydes), so the Grignard would react with the ketone before it ever touched the ester! To prevent this, the ketone must be protected before Grignard is added.
MendelSet practice problem # 714 submitted by Matt on July 22, 2011.
a) Is α-D-glucose an acetal, hemiacetal, ketal, or hemiketal?
b) Draw the carbonyl form of α-D-glucose.
a) You can spot the carbonyl carbon because it's the one with two oxygens bonded to it (oxidation state II).
The carbonyl carbon has an -OH (alcohol) bonded to it, and not an -OR (ether), so it's a hemiacetal or a kemiketal.
The carbonyl carbon also has a hydrogen bonded to it, so it must have come form an aldehyde (instead of ketone). So it's a hemiacetal. (remember: aldehyde -> acetal, ketone -> ketal).
b) The hemiacetal carbon's -OH (or -OR in the case of a ketal or acetal) is the oxygen that used to be the carbonyl C=O.
MendelSet practice problem # 715 submitted by Matt on July 23, 2011.