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.
Indicate the major organic product of the reaction below. Include stereochemistry.
This is a hydrogenation reaction. The product is an alkane. But this product has two stereocenters.
Normally, the product would be a racemic mixuture of enantiomers (equal amounts of each stereoisomer). But in this case, the starting material is optically active (chiral but not racemic), so the products won't be equal and opposite as usual.
Since there is a substitutent blocking the top face of the molecule (isopropyl is a wedge), the hydrogens will preferentially add to the opposite face of the molecule (dashes).
MendelSet practice problem # 529 submitted by Matt on July 2, 2011.
Let's work through anti and syn additions to alkenes.
Show the product for each reaction below, and indicate whether the product will be a racemic mixture of enantiomers, or a meso compound (which is achiral).
The reaciton of an alkene with Br2 is an anti-addition, and hydrogenation (H2 or D2) is a syn addition. From this we can figure our the relative stereochemistry of each product.
Each starting material is achiral, and therefore not optically active, so the products cannot be optically active.
There are three ways for products not to be optically active:
products can be achiral
products can be a racemic mixture
products can be meso
In both a) and b) each product has a sterocenter, so the products can't be achiral.
In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.
In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.
MendelSet practice problem # 530 submitted by Matt on July 2, 2011.