Reduction of Fluorenone How do you calculate the oxidation state of a carbon? Ho
ID: 835963 • Letter: R
Question
Reduction of Fluorenone
How do you calculate the oxidation state of a carbon? How does it change during a reduction?
What are the advantages of using LiAlH4 over NaBH4? What are the advantages of using NaBH4 over LiAlH4?
What are the stereochemical consequences of reducing carbonyls? How many products (enantiomers, diastereomers, etc.) would you expect for a certain compound?
What is the stoichiometry of a borohydride or aluminum hydride reduction? How does that affect how much of the hydride source you use?
2. Cannizzaro Reaction
Why did the aldehyde used undergo the Cannizzaro reaction and not some other kind of reaction under basic conditions?
Which product comes from the organic layer and which comes from the aqueous layer?
Why is one product soluble in water but the other soluble in CH2Cl2?
What does acidification of the aqueous layer do? How does this help isolate the
product?
Explanation / Answer
How do you calculate the oxidation state of a carbon
First, identify the carbon of interest - assign numbers to its attachments (+1 for H, 0 for C, and -1 for halogens/N/O/S). The sum of these numbers + the oxidation number of the carbon = the charge on the carbon. (So if the numbers add up to 3 and the carbon is neutral, the oxidation number is -3)
How does it change during a reduction?
Oxidation = increased oxidation number/state (gain in electrons/hydrogens). Reduction = decreased oxidation number/state (loss of an oxygen/other electronegative atoms, or replacement with a less electronegative atom)
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What are the advantages of using LiAlH4 over NaBH4?
LiAlH4 will reduce all carbonyls as it is a strong nucleophile, while NaBH4 will only reduce aldehydes/ketones and NOT carboxylic acids or esters.
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What are the advantages of using NaBH4 over LiAlH4?
NaBH4 is safer than LiAlH4 (more stable reducing agent, doesn't react as violently with air/moisture and water). LiAlH4 reacts with protic solvents so it MUST be used in anhydrous ethereal solutions (ie. diethyl ether), whereas NaBH4 can be used in alcoholic and some aqueous solutions (will react with the C=O faster than the solvent)
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What are the stereochemical consequences of reducing carbonyls?
Due to carbonyls having a sp2 nature, attacking nucleophiles/reducing agents can attack from either the R or SI face
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How many products (enantiomers, diastereomers, etc.) would you expect for a certain compound?
Carbonyl reduction changes a carbon from sp2 to sp3
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What is the stoichiometry of a borohydride or aluminum hydride reduction?
4 9-fluorenone + 1 NaBH4 + MeOH solvent --> 4 [intermediate borate salt] --> 4 9-fluorenol
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How does that affect how much of the hydride source you use?
There is a hydride ion (H-) transfer from the BH4- to the carbonyl carbon; since BH4- has 4 different hydrogens, it can donate all of its hydrogens and thus we only need 1 molecule of BH4- for 4 molecules of 9-fluorenone. (Functions as having "4 times" as many moles.)
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Why did the aldehyde used undergo the Cannizzaro reaction and not some other kind of reaction under basic conditions
Cannizzaro reactions happen where there are no hydrogens on the alpha-carbon; aldol condensation reactions happen when there are hydrogens on the alpha-carbon (typically a base would pull off the alpha-hydrogen and form an enolate). In this case, the aldehyde does not have hydrogens on the alpha-carbon, so it undergoes Cannizzaro. Attack at the carbonyl carbon of the aldehyde is the most favorable reaction because this carbon is very polarized and electrophilic. OH is a nucleophile, thus it would like to attack this site. It would not react as a base because no proton in the molecule is particularly acidic. Finally, it is not a strong enough nucleophile to react at the carbon attached to Cl in a nucleophilic aromatic substitution reaction.
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Which product comes from the organic layer and which comes from the aqueous layer in the Cannizzaro reaction?
4-Chlorobenzoic acid comes from the aqueous layer. 4-Chlorobenzyl alcohol comes from the organic layer
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Why is one product soluble in water but the other soluble in CH2Cl2?
4-Chlorobenzoic acid (carboyxlic acid) exists in its carboxylate ion form, so the aqueous water can stabilize/protonate it into the carboyxlic acid form. Also, carboxylic acids are highly polar and H-bond with water, so it dissolves in water. 4-Chlorobenzyl alcohol is only slightly polar and organic, like dichloromethane (CH2Cl2). CH2Cl2 is a typical organic solvent, and "like dissolves like".
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What does acidification of the aqueous layer do? How does this help isolate the product?
Acidification of the aqueous layer protonates potassium 4-chlorobenzoate (the carboyxlic acid ion) to form the actual carboxylic acid. This helps isolate the product because it will make it precipitate out of the aqueous layer.
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