Fill in the product for each reaction below. Indicate stereochemistry where appropriate.
NaNH2 is a very strong base (the pKa of ammonia, its conjugate acid, is about 35) and will easily depronate a terminal alkyne (pK ~ 25), producing a negatively charged alkyne carbon (an acetylide).
The acetylide is a strong nucleophile and will undergo an SN2 reaction with the 1ºalkyl halide. (acetylide is also a strong base, so with 2º or bulkier alkyl halides, it will go E2 instead).
Finally, Lindlar's catalyst will reduce the alkyne to a cis alkene.
MendelSet practice problem # 561 submitted by Matt on July 8, 2011.
Fill in the product for each reaction below. Indicate stereochemistry where appropriate.
-OtBu is a big, bulky base, and so will do an E2 reaction with the alkyl halide. The elimination product can be either a cis or trans alkene, but trans is usually the major product (cis alkenes have steric strain).
The alkene then reacts with Br2 to form 2,3-dibromobutane, which can undergo two consecutive E2 reactions with -OtBu to form an alkene and then the alkyne (2-butyne).
Finally, the alkyne reacts with the first equivalent of Br2 to form the trans 2,3-dibromo-2-butene, which reacts with a second equivalent of bromine to form the tetrabromo alkane.
MendelSet practice problem # 562 submitted by Matt on July 8, 2011.