All Practice Problems

Problem # 561

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

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 # 569

Draw all resonance forms for each species.

For the anion and cation species, used curved arrows. For the radical species, use hooks.

Solution

Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.

(Radical resonance uses hooks instead of arrows and so is a little different.)

Problem Details

MendelSet practice problem # 569 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

Question Type: Concept (explain why this is so..)

Keyword(s): resonance, allylic position

Problem # 570

Rank the following dienes in order of decreasing reactivity with a dienophile in a Diels-Alder reaction. (1 = most reactive).

Solution

To undergo a Diels-Alder reaction, a dienophile must be in s-cis conformation.

The bicylic compound below is locked into s-trans conformation; it can never rotate into s-cis conformation and so can't undergo a Diels-Alder reaction.

Of the other two compounds, the middle compound most easily rotates into s-cis conformation, and so will undergo a Diels-Alder reaction the fastest. The compound on the right has steric strain when in s-cis conformation, and so won't do Diels-Alder as easily.

Problem Details

MendelSet practice problem # 570 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Diels-Alder

Question Type: Concept (explain why this is so..)

Keyword(s): Diels-Alder

Problem # 571

Write the structure of the major organic product of each reaction.

Solution

a) is a free-radical bromination. The radical will be formed at the most stable position, which will be the allylic position due to resonance (see problem 569). There are two allylic positions in this molecule, so the more substituted one will be the more stable; the bromine will add to the 3º allylic carbon.

b) is an E2 reaction. There are two types of beta protons and so two possible alkene products. One is an isolated diene and the other is a conjugated diene. Conjugated dienes are more stable, and so that will be the major product.

Problem Details

MendelSet practice problem # 571 submitted by Matt on July 8, 2011.

Subject: Organic Chemistry

Subject Topic(s): Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)

Question Type: Concept (explain why this is so..)

Keyword(s): allylic position

Problem # 572

Let's work through a conjugate addition problem.

Write out the mechnanism for the formation of the 1,2 and 1,4 products of the reaction below.

Solution

This is an S_N1 reaction, but the carbocation is special because it's allylic.

If the Br^- nucleophile attacks the 3º resonance form, the 1,2 product is formed. (This is the product that would have formed if resonance was not involved.)

If the Br^- nucleophile attacks the 2º resonance form, the 1,4 product is formed.