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.
Rank the group of molecules below in in order of decreasing basicity. (1 = most basic)
Explain your reasoning.
The alkyne (triple bond) is the most stable and therefore the least basic. This is because it is sp hybridized.
An s-orbital is closer to the nucleus and more electronegative than a p-orbital, and an sp hybridized atom has 50% s-character.
Alkenes are sp2 hybridized and have 33% s-character, and alkanes are sp3 hybridized and have 25% s-character.
Because sp hybridized carbons have the most s-character, they are more electronegative and are better at stabilizing negative formal charges than sp2 or sp3 carbons are.
Therefore, the alkane (sp3) is the least stable/strongest base, while the alkyne (sp) is the most stable/weakest base.
MendelSet practice problem # 307 submitted by Matt on June 7, 2011.