MS 917 - Alcohol and Carbonyl Redox
Submitted by
Matt on August 6, 2011.
Textbook and Chapter: Carey and Giuliano 8th Ed. (2010) , Chapter 15
Keywords: alcohols, hydride reduction, oxidation
Description: Goes over how alcohols can be oxidized to form aldehydes/ketones and carboxylic acids, which can be transformed further using Grignard reagents, hydride reagents such as NaBH4, and by performing Fischer esterification. Includes several synthesis problems involving carbonyls and epoxides. Also includes an NMR based problem.
Problem # 673
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 m CPBA.
Solution 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 (NaCr2 O7 /H2 SO4 ).
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.
Problem Details MendelSet practice problem #
673 submitted by
Matt on July 19, 2011.
Problem # 674
Show a mechanism for the reduction of butyrolactone using LiAlH4 .
Solution 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.
Problem Details MendelSet practice problem #
674 submitted by
Matt on July 19, 2011.
Problem # 677
Show a mechanism for the acid-catalyzed cyclization (condensation) of 1,4-butanediol.
Solution This is just an SN 2 reaction.
The first step is to turn one of the -OH groups into a better leaving group by having it pick up a proton. (I choose the hydroxyl on the right side).
Then the other -OH group can attack carbon 1, and the protonated hydroxyl group leaves as water.
Finally, the proton on the -OH group that acted as a nucleophile will "fall off" (deprotonate).
Problem Details MendelSet practice problem #
677 submitted by
Matt on July 19, 2011.
Problem # 678
Draw the structure of the major organic product from each reaction sequence.
Solution The purpose of this problem is to illustrate that there are often several routes to prepare a single molecule.
For the left route:
Br2 /FeBr3 adds a bromine to benzene via EAS.
Mg/ether converts bromobenzene to phenyl Grignard (a nucleophile).
Phenyl Grignard attacks the aldehyde.
A protic workup gets rid of all negative charges (and quenches the Grignard reagent).
For the right route:
Benzene undergoes Friedel-Crafts acylation (an EAS reaction) to form the ketone.
The ketone is then reduced to an alcohol using NaBH4 .
Note that both methods add a carbon chain to benzene, a step that is quite common in synthesis problems!
Problem Details MendelSet practice problem #
678 submitted by
Matt on July 19, 2011.
Problem # 679
Compound A (C5 H12 O) is oxidized using aqueous chromium (Jones reagent) to compound B (C5 H10 O2 ), which is then treated with methanol under acidic conditions to yield compound C (C6 H12 O2 ) and water.
The 1 H NMR of compound C is shown below. Determine the structures of compounds A, B, and C.
Solution 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: C5 H12 O = C5 H12 should be C5 H12 (Cn H2n+2 ) so 0 IHD .
Compound B: C5 H10 O2 = C5 H10 should be C5 H12 . Missing 2H, so 1 IHD.
Compound C: C6 H12 O2 = C6 H12 should be C6 H14 . 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!
Problem Details MendelSet practice problem #
679 submitted by
Matt on July 19, 2011.
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