Overall: 2-iodobutane + (Ph)3P: w/DMSO solvent ------> C6H5Li w/THFsolvent ------> acetone w/ether solvent -----> in 2,3,-methyl-pent-2-ene (end product)
First Step:
2-Iodobutane undergoes SN2 with (Ph)3P: in aprotic DMSO solvent to give an ylide with the -P(Ph)3 on the 2-carbon (replacing the Iodine).
1)Why does the "bulkyness" of (Ph)3P: not prevent an SN2 reaction?
Second Step:
The ylide reacts with the strong base C6H5Li in aprotic THF solvent to give a carbanion on the 2-carbon attached to the -P(Ph)3 (triphenylphosphine).
2) How is the carbanion formed? I thought the ylide form creates a P+ and a C- already? Why is C6H5Li needed to form a carbanion in order to react with a ketone (later in the 3rd step)?
Third Step:
The ylide reacts with acetone in an ether solution to form 2,3-methyl-pent-2-ene.
I understand the mechanism regarding the zwitterionic betaine intermediate where the carbanion from the ylide attacks the partially positive carbonyl of acetone to form the intermediate, then the -O-P(Ph)3 bond is strong causing it to break away and create O=P(Ph)3 while the electrons between the 2-carbon and the P form the double bone between the 2-carbon and the 3-carbon but why is step 2 with C6H5Li necessary??
First Step:
2-Iodobutane undergoes SN2 with (Ph)3P: in aprotic DMSO solvent to give an ylide with the -P(Ph)3 on the 2-carbon (replacing the Iodine).
1)Why does the "bulkyness" of (Ph)3P: not prevent an SN2 reaction?
Second Step:
The ylide reacts with the strong base C6H5Li in aprotic THF solvent to give a carbanion on the 2-carbon attached to the -P(Ph)3 (triphenylphosphine).
2) How is the carbanion formed? I thought the ylide form creates a P+ and a C- already? Why is C6H5Li needed to form a carbanion in order to react with a ketone (later in the 3rd step)?
Third Step:
The ylide reacts with acetone in an ether solution to form 2,3-methyl-pent-2-ene.
I understand the mechanism regarding the zwitterionic betaine intermediate where the carbanion from the ylide attacks the partially positive carbonyl of acetone to form the intermediate, then the -O-P(Ph)3 bond is strong causing it to break away and create O=P(Ph)3 while the electrons between the 2-carbon and the P form the double bone between the 2-carbon and the 3-carbon but why is step 2 with C6H5Li necessary??
