Show how each alcohol can be prepared from a combination of a carbonyl and a Grignard reagent.
The trick to these retrosynthesis problems is to determine where the connections or "cuts" were made.
The carbonyl carbon becomes an alcohol after a Grignard reaction, so that's where the "cut" must be.
Note that there can often be more than one correct answer to these types of problems.
For a), adding propyl Grignard to acetone or methyl Grignard to 2-pentanone will result in the product.
We can also use two equivalents of methyl Grignard with 4-carbon ester, such as ethyl butanoate. Esters contains a build in leaving group (-OR) and so react twice with Grignards.
For b), adding phenyl Grignard to cyclopentanone will do the job.
MendelSet practice problem # 668 submitted by Matt on July 18, 2011.
You may have noticed that the "solvent of choice" for many organometallic compounds such as Grignard reagents is ether (short for diethyl ether).
Why is it that for Grignard reactions this solvent is used over ethyl acetate, or protic solvents such as ethanol?
Grignards behave as though they are carbanions (negatively charged carbons), and so are very basic. There are so basic that they will deprotonate any O-H or N-H bond. So protic solvents such as water or ethanol aren't suitable for Grignard reactions; the Grignard reagent will react with the alcohol in an acid-base reaction.
In fact, water is used after a Grignard reaction to quench the Grignard reagent.
Grignards are also nucleophilic, and so react with carbonyls (which are electrophiles).
Ethyl acetate contins a carbonyl and would get attacked by a Grignard reagent, and so also isn't a suitable solvent for a Grignard reaction.
Diethyl ether doesn't have any acidic protons and isn't electrophilic and so won't react with a Grignard reagent, so it makes a good solvent.
MendelSet practice problem # 671 submitted by Matt on July 18, 2011.
Draw out the mechanism for the addition of excess phenyl Grignard to the carbonyl compound below.
This carbonyl has two leaving groups attached to it- each of those oxygens can take part in a nucleophilic acyl substitution reaction and form a new carbonyl product.
First the Grignard attacks the oxidation state IV carbonyl carbon (4 oxygen bonds, so oxidation state 4). The carbonyl itself will act as a leaving group and form a tetrahedral intermediate. But tetrahedral intermediates don't last if there are any leaving groups attached to the carbon, so the -O will "come back down again", kick off an oxygen leaving group, and reform the carbonyl.
Then a second equivalent of Grignard will attack that carbonyl (an ester), and we will do another nucleophilic acyl substitution reaction to form yet another carbonyl.
Finally, the third carbonyl doesn't have any leaving groups built in (it's a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol.
Notice that as the reaction progresses the oxidation state of the carbonyl carbon (number of oxygen bonds attached to it) goes down form 4 to 3 to 2 and then to 1.
MendelSet practice problem # 670 submitted by Matt on July 18, 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 a mechanism for the reduction of butyrolactone using LiAlH4.
Hydride reagents such as LiAlH4 and NaBH4 behave like hydride nucleophiles (H-), so that's what I used as shorthand. The real mechanism is very similar but involves aluminum coordinating to the oxygen.
Notice that the the ester will reform the carbonyl after the first hydride attacks. This is because esters have a built in leaving group, and so undergo nucleophilic acyl substitution reactions. The aldehyde that forms then undergoes a nucleophilic acyl addition reaction with the second equivalent of hydride.
Also note that you can't stop the reaction halfway at the aldehyde- LiAlH4 will take an ester all the way down to an alcohol.
MendelSet practice problem # 674 submitted by Matt on July 19, 2011.
Compound A has molecular formula C6H12O and shows a sharp peak at 1,710 cm-1 in its IR spectrum.
Treatment with 1 equivalent of phenyl Grignard yields compound B, which has formula C12H18O and whose IR shows a broad peak at 3,350 cm-1.
Compound B's 1H NMR spectrum is shown below. Determine the structures of compounds A and B.
Let's use steps similar those outlined in problem 662 to solve this NMR structure elucidation problem.
1.Are there any hints?
Yes, from the IR peaks. The starting material is a carbonyl (sharp IR peak at ~1,700 cm-1) and the product is an alcohol (broad IR peak ~3,300 cm-1).
2.How many IHD are in each compound? (also known as degrees of unsaturation or DBE).
C6H12O is the same as C6H12 (ignore oxygen) which should be C6H14 if fully saturated (CnH2n+2). It's missing 2 hydrogens so C6H12O has 1 IHD. This 1 IHD must be the carbonyl.
C12H18O is the smae as C12H18 which should be C12H26 if fully saturated. It's missing 8 hydrogens so it has 4 IHD. The benzene from the Grignard reagent must account for all 4 IHD.
3.Draw some C6H12O and C12H18O structures and elminate those that don't fit the data, then learn and repeat.
Things we know from the problem and NMR:
The starting material is a carbonyl and we're adding a Grignard, so we expect the product to be an alcohol. The NMR shows a peak that disappears with D2O addition, which also confirms that the product is an alcohol.
The product must have a benzene ring because the reagent was phenyl Grignard. That accounts for the aromatic signals (~7 ppm) and the 4 IHD in the product.
the NMR shows a doublet with an integration of 6 and a multiplet with an integration of 1. This is the splitting pattern on an isopropyl group.
The NMR also shows a quartet with an integration of 2 and a triplet with an integration of 3. This is the splitting pattern of an ethyl group.
Once you have a few clues from the NMR, start drawing structures! And then elminiate those that do not fit the data (too many signals, wrong multiplicity/integrations etc.).
MendelSet practice problem # 672 submitted by Matt on July 18, 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.