Textbook: Smith 3rd Ed. (2010)

Chapter 12: Oxidation and Reduction

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Individual Problems

Individual Problems

Problem # 678

Draw the structure of the major organic product from each reaction sequence.

Solution

The purpose of this problem is to illustrate that there are often several routes to prepare a single molecule.

For the left route:

Br₂/FeBr₃ adds a bromine to benzene via EAS.
Mg/ether converts bromobenzene to phenyl Grignard (a nucleophile).
Phenyl Grignard attacks the aldehyde.
A protic workup gets rid of all negative charges (and quenches the Grignard reagent).

For the right route:

Benzene undergoes Friedel-Crafts acylation (an EAS reaction) to form the ketone.
The ketone is then reduced to an alcohol using NaBH₄.

Note that both methods add a carbon chain to benzene, a step that is quite common in synthesis problems!

Problem Details

MendelSet practice problem # 678 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution, Alcohol Oxidation Reactions (PCC, Jones Reagent)

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

Keyword(s): hydride reduction

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 # 673

Show how each compound can be prepared from an alkene containing 3 carbons (or less).

Each answer will involve the reaction of a Grignard with either a carbonyl or epoxide.

Note: epoxides are prepared from alkenes using a peroxy acid (epoxidation) such as mCPBA.

Solution

The trick to synthesis problems in second semester organic chemistry to recognize that alcohols ARE ketones ARE carboxylic acids. What do I mean? Alcohols, ketones/aldehydes, and carboxylic acids can all be easily converted using PCC or Jones Reagent (NaCr₂O₇/H₂SO₄).

For example, for a), the product is a ketone, but it may as well be an alcohol, because alcohols can be converted to ketones with PCC.

b) is similar, except the position of the alcohol (one away from the "bond cut", instead of directly connected to the cut as in a) ) indicates the starting material was an epoxide and not a carbonyl.

c) is just like b), except instead of PCC, use Jones Reagent to oxidize the alcohol all the way to a carboxylic acid.

Problem Details

MendelSet practice problem # 673 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alcohol Oxidation Reactions (PCC, Jones Reagent)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): epoxides, oxidation

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 # 530

Let's work through anti and syn additions to alkenes.

Show the product for each reaction below, and indicate whether the product will be a racemic mixture of enantiomers, or a meso compound (which is achiral).

Solution

The reaciton of an alkene with Br₂ is an anti-addition, and hydrogenation (H₂ or D₂) is a syn addition. From this we can figure our the relative stereochemistry of each product.

Each starting material is achiral, and therefore not optically active, so the products cannot be optically active.

There are three ways for products not to be optically active:

products can be achiral
products can be a racemic mixture
products can be meso

In both a) and b) each product has a sterocenter, so the products can't be achiral.

In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.

In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.

Problem Details

MendelSet practice problem # 530 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): stereoselectivity

Problem # 679

Compound A (C₅H₁₂O) is oxidized using aqueous chromium (Jones reagent) to compound B (C₅H₁₀O₂), which is then treated with methanol under acidic conditions to yield compound C (C₆H₁₂O₂) and water.

The ¹H NMR of compound C is shown below. Determine the structures of compounds A, B, and C.

Solution

Let's solve this NMR structure elucidation problem using steps similar to those used in problem 662.

1. Are there any hints?

Compound A has one oxygen and after treatment with aqueous chromium becomes compound B, which has two oxygens. This means A is probably an alcohol, B is probably a carboxylic acid.

Compound B is then treated with methanol under acidic conditions to form compound C. These are conditions for a Fischer esterification, so C is probably the methyl ester.

2. How many IHD are there?

Compound A: C₅H₁₂O = C₅H₁₂ should be C₅H₁₂ (C_nH_2n+2) so 0 IHD.

Compound B: C₅H₁₀O₂ = C₅H₁₀ should be C₅H₁₂. Missing 2H, so 1 IHD.

Compound C: C₆H₁₂O₂ = C₆H₁₂ should be C₆H₁₄. Missing 2H, so 1 IHD.

These IHD counts fit our assumptions from part 1).

3. Draw some structures and eliminate, learn, repeat.

Some clues from the NMR:

The isopropyl splitting pattern is present: d(6) (signal c at ~0.9 ppm) and multiplet(1) (signal b at ~2.4 ppm).
The s(3) at ~3.7 ppm is probably the methyl group from the methyl ester.
We know from before we have one IHD, and it's probably an ester.

So start drawing structures and eliminate those that don't fit the data!