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MS 1317 - Practice 1

MS 1317 - Practice 1

Submitted by rpongdee on October 17, 2011.

Description: Practice 1

Total Problems: 10

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Problem # 318

Two stereoisomers of trimethylcyclohexane are shown below (compounds A and B). Compare cyclohexane chair forms to determine which isomer has a lower heat of combustion. Explain your reasoning.
Solution
A has a chair form that avoids having any substituents in the axial position.

B's most stable chair form has one axial substituent.

So A is more stable than B. Therefore, B is higher eneregy and has a higher heat of combustion than A.

Problem Details
MendelSet practice problem # 318 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms
Question Type: Concept (explain why this is so..)
Keyword(s): cyclohexane chair forms, heats of combustion
Problem # 317

α-D-Glucose is shown below. Draw its two chair forms. Which conformation is more stable? Explain.

I recommend using the common convention wedge = "up" and dash = "down."
Solution
The chair form on the left is more stable because it has no axial substituents, which lead to 1,3-diaxial interactions and higher energies. Glucose is the most stable of the hexose carbohydrates because it is the only one that has al substituents at C2-C5 in the equatorial position.

It's probably no coincidence that the most abundant sugar in the world (glucose) is also the sugar that is most stable!

Problem Details
MendelSet practice problem # 317 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms, Carbohydrates
Question Type: Concept (explain why this is so..)
Keyword(s): cyclohexane chair forms
Problem # 334

Let's go over how a carbocation can form from an alkene.

Use curved arrows to show the two carbocations that can from from 1-methylcyclohexene.
Solution
3° carbocations are more stable than 2º carbocations, so only the 3º carbocation is formed.

Problem Details
MendelSet practice problem # 334 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Alkene Reactions and Markovnikov's Rule
Question Type: Mechanism (show a mechanism using curved arrows..)
Keyword(s): carbocation formation, Markovnikov's rule
Problem # 336

Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.
Solution
Due to Markovnikov's rule, only the more substituted alkyl halide is formed.

Problem Details
MendelSet practice problem # 336 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Alkene Reactions and Markovnikov's Rule
Question Type: Mechanism (show a mechanism using curved arrows..)
Keyword(s): Markovnikov's rule, alkene addition
Problem # 342

For the reaction below, draw the structures of the carbocation intermediate and the final product.
Solution
This is a acid-catalyzed hydration of an alkene, which is a Markovnikov addition. The product is an alcohol.

Problem Details
MendelSet practice problem # 342 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Alkene Reactions and Markovnikov's Rule
Question Type: Reactions (predict the product, show the starting material..)
Keyword(s): Markovnikov's rule, alkene addition
Problem # 343

For the reaction below, draw the structures of the chloronium ion intermediate and the final product.
Solution
Addition of a chlorine, bromine, or iodine to an alkene goes through a halonium ion intermediate, which results in an anti-addition (trans product).

Problem Details
MendelSet practice problem # 343 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Alkene Reactions and Markovnikov's Rule
Question Type: Reactions (predict the product, show the starting material..)
Keyword(s): alkene addition, anti stereochemistry, halonium ion
Problem # 344

For the reaction below, draw the structures of the borane intermediate and the final product.
Solution
This is a hydroboration reaction (also called hydroboration-oxidation in some textbooks). The product is an alcohol.

Boron bonds to the less substituted carbon, and is replaced with an oxygen after the borane intermediate is treated with peroxide, resulting in an anti-Markovnikov addition.

Note that one equivalent of BH₃ reacts with three equivalents of alkene; the borane intermediate is not BH₂(alkyl) but rather B(alkyl)₃.

Problem Details
MendelSet practice problem # 344 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Alkene Reactions and Markovnikov's Rule
Question Type: Reactions (predict the product, show the starting material..)
Keyword(s): alkene addition, hydroboration, anti-Markovnikov
Problem # 520

Let's work through an alkene addition reaction. Draw the structures for each of the species in the three boxes below (3º carbocation, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Solution
Note that this is the reverse of problem 519. Instead of going from thiol (alcohol) to carbocation to alkene, we're going from alkene to carbocation to thiol. In each step ask yourself "what arrows can I draw?" and choose the step that doesn't go backwards.

The first step is the alkene picks up a proton to form the more stable carbocation. (See problem 334 if you don't understand why only the 3º carbocation is formed). To get rid of the carbocation, we can either do a beta-elimination (E1) to form an alkene, or an addition reaction (S_N1) to form a protonated thiol. Since forming the alkene would be going backwards, the only choice is addition.

Then you're left with a protonated thiol. How do we get rid of the positive charge on the sulfur? There are two legal moves- either the RSCH₂CH₃ acts as a leaving group, or the sulfur deprotonates. Since the RSCH₂CH₃ leaving would form a carbocation and take us backwards, the only option is for the sulfur to deprotonate.

Because the sulfur added to the more substituted carbon, this was a Markovnikov addition.

Problem Details
MendelSet practice problem # 520 submitted by Matt on June 30, 2011.
Subject: Organic Chemistry
Subject Topic(s): Alkene Reactions and Markovnikov's Rule
Question Type: Mechanism (show a mechanism using curved arrows..)
Keyword(s): Markovnikov's rule, alkene addition
Problem # 529

Indicate the major organic product of the reaction below. Include stereochemistry.
Solution
This is a hydrogenation reaction. The product is an alkane. But this product has two stereocenters.

Normally, the product would be a racemic mixuture of enantiomers (equal amounts of each stereoisomer). But in this case, the starting material is optically active (chiral but not racemic), so the products won't be equal and opposite as usual.

Since there is a substitutent blocking the top face of the molecule (isopropyl is a wedge), the hydrogens will preferentially add to the opposite face of the molecule (dashes).

Problem Details
MendelSet practice problem # 529 submitted by Matt on July 2, 2011.
Subject: Organic Chemistry
Subject Topic(s): Alkene Reactions and Markovnikov's Rule, Stereochemistry, Chirality, and Optical Activity, Hydrogenation Reactions (Alkenes, Alkynes, Carbonyls)*
Question Type: Concept (explain why this is so..)
Keyword(s): alkene addition, stereoselectivity
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

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