Textbook: Brown 6th Ed. (2011)

Chapter 7: Alkynes

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Individual Problems

Individual Problems

Problem # 561

Fill in the product for each reaction below. Indicate stereochemistry where appropriate.

Solution

NaNH₂ is a very strong base (the pK_a 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 S_N2 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.

Problem Details

MendelSet practice problem # 561 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkynes and Alkyne Addition Reactions, Terminal Alkyne Acidity and Acetylide, Hydrogenation Reactions (Alkenes, Alkynes, Carbonyls)*

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): Lindlar catalyst, terminal alkyne

Problem # 562

Fill in the product for each reaction below. Indicate stereochemistry where appropriate.

Solution

^-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.

Problem Details

MendelSet practice problem # 562 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkynes and Alkyne Addition Reactions

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): alkyne formation

Problem # 307

Rank the group of molecules below in in order of decreasing basicity. (1 = most basic)

Explain your reasoning.

Solution

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.