Question

1. Outline the syntheses indicated in Solved Problem 18-2, beginning with aldehydes and alkyl halides.
2. Both of these syntheses of 1-phenylbuta-1,3-diene form the central double bond. Show how you would synthesize this target molecule by forming the terminal double bond.

Short answer

The given alkene forms a four-membered transition state on reaction with triphenyl phosphine and butyl lithium which is used as base, and then on simultaneous bond-breaking and forming, we get the product in which negative charge is on carbon and positive charge on phosphorous, and in the next step, this negatively charged carbon or carbanion attacks on the carbonyl carbon of aldehyde, forming the alkene product with two double bonds not in conjugation.

The product formation can also occur starting from alkyl halide as the reactant, and using same reagents such as triphenyl phosphine and butyl lithium as base. Butyl lithium abstracts the proton from alkyl carbon forming witting ylide, and using different aldehyde, we can form the same product. Carbanion of wittig ylide attacks at carbonyl carbon of aldehyde forming the required product.

Formation of the alkene product

Step-by-step solution

Explanation of part (a):

The given alkene forms a four-membered transition state on reaction with triphenyl phosphine and butyl lithium which is used as base, and then on simultaneous bond-breaking and forming, we get the product in which negative charge is on carbon and positive charge on phosphorous, and in the next step, this negatively charged carbon or carbanion attacks on the carbonyl carbon of aldehyde, forming the alkene product with two double bonds not in conjugation.

The product formation can also occur starting from alkyl halide as the reactant, and using same reagents such as triphenyl phosphine and butyl lithium as base. Butyl lithium abstracts the proton from alkyl carbon forming witting ylide, and using different aldehyde, we can form the same product. Carbanion of wittig ylide attacks at carbonyl carbon of aldehyde forming the required product.

Formation of the alkene product

Explanation of part (b):

The given alkene forms a four-membered transition state on reaction with triphenyl phosphine and butyl lithium which is used as base, and then on simultaneous bond-breaking and forming, we get the product in which negative charge is on carbon and positive charge on phosphorous, and in the next step, this negatively charged carbon or carbanion attacks on the carbonyl carbon of aldehyde, forming the alkene product with two double bonds not in conjugation.

The product formation can also occur starting from alkyl halide as the reactant, and using same reagents such as triphenyl phosphine and butyl lithium as base. Butyl lithium abstracts the proton from alkyl carbon forming witting ylide, and using different aldehyde, we can form the same product. Carbanion of wittig ylide attacks at carbonyl carbon of aldehyde forming the required product, that is, 1-phenyl-buta-1,3-diene.

Formation of 1-phenyl-buta-1,3-diene