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LiAlH4
- Very powerful reducing agent. Aldehydes are reduced very easily.
- When a carboxylic acid is treated with this it is reduced to 1o Alcohol
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Acyl chlorides can be reduced to aldehydes by...
 Acyl chloride
treating them with LiAlH [OC(CH3)3]3, lithium tri-tert-butoxyaluminum hydride, at -78°C.
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Carboxylic acids can be converted to acyl chlorides by...
Using SOCl2
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Reducing agents
- Ni/H2
- Pd/H2
- NaBH4 Sodium borohydride
- Grignard Reagent Mg/ diethyl ether
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1o Alcohol ---------->Aldehyde
PCC (C5H5NH+CrO3Cl-)
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Aldehydes by Reduction of Acyl Chlorides, Esters, and Nitriles
When any aldehyde is treated with LiAlH4 It is reduced all the way to 1o alcohol
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Two derivatives of aluminum hydride that are less reactive than LAH (in part because they are much more sterically hindered and, therefore, have difficulty in transferring hydride ions) are:
Lithium tri-tert-butoxyaluminum hydride and diisobutylaluminum hydride (DIBAL-H)
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Acul chrolide -----> aldehyde
ester -----> Aldehyde
Nitrile -----> aldehyde
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What type of molecule is ethyl formate?
Ester
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Reduction of an Acyl Chloride to an Aldehyde
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Reduction of an Ester to an Aldehyde
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Reduction of a Nitrile to an Aldehyde
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synthesize propanal from each of the following:(a) 1-propanol (CH3CH2CH2OH) (b) propanoic acid (CH3CH2CO2H)
- (a) CH3CH2CH2OH + PCC-----> CH3CH2CHO
- (b) CH3CH2COHO --SOCl2--> CH3CClO [---(1)LiAlH(o-t-Bu)3 -78℃(2)H2O-->] CH3CH2CHO
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Treating a nitrile (R—CN) with either a Grignard reagent or an organolithium reagent followed by hydrolysis yields a...
Ketone
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KMnO4
KMnO4 is an oxidizing agent. It will oxidatively cleave the alkene to give an aldehyde, which further oxidizes to the carboxylic acid. If it were a ketone, however, it would not further oxidize.
Removes CH3 adds an aldehyde (adds double bond to O and an OH group)
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SOCl2
removes OH adds a Cl
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LiAlH(OCCH3)3
Removes halogen adds a H
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Difference between LiAlH4 and (CH3)2 Cu Li is...
- Li Al adds a H
- Cu Li adds a methyl group
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DIBAL-H
Reduces Esters into aldehydes
O-R into H
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(CH3)2 CuLi
Removes halogen adds methyl group becoming a ketone
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Hg2+/H2O
Removes triple bond and insert a ketone (Remember that number of carbons stays the same) Always count the carbons.
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Imine formation—reaction with a primary amine
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Oxime formation—reaction with hydroxylamine
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Wittig reaction
 Overall result is of this synthesis is:
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Horner–Wadsworth–Emmons modification
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