Organic Chemistry

Problem # 743

Enolates are nucleophiles and react with a variety of electrophiles.

Carbonyls are electrophiles. But aldehydes/ketones and esters/acid chlorides often form different products.

Use curved arrows to draw a mechanism for each reaction below. How do the two products differ?

When enolates attack, the arrow is drawn from the double bond to the electrophile, and NOT from the negatively charged oxygen! The enolate's carbonyl reforms, and a new bond is created at the alpha position.

But Aldehydes and ketones react differently than ester and acid chlorides.

a) Aldehydes and ketones are oxidation state II carbonyls (the carbonyl has carbon-heteroatom bonds- the double bonded oxygen counts twice).

They do not have a built-in leaving group, and so undergo nucleophilic acyl addition reactions. The carbonyl on the electrophile becomes an alcohol.

So the product from this reaction is a beta hydroxyl carbonyl. If this β-hydroxyl carbonyl is heated up, the hydroxyl will eliminate to from an alpha, beta unsaturated carbonyl.

The mechanism for a) is very similiar to that of an aldol condensation reaction.

b) Esters (and acid chlorides) are oxidation state III carbonyls (the carbonyl carbon has three carbon-heteroatom bonds). They have a built-in leaving group (^-OR in the case of an ester, and Cl^- for an acid chloride), and so undergo nucleophilic acyl substitution reactions. ("up, down, kick"). The carbonyl on the electrophile is reformed.

So the product from this reaction is a beta keto carbonyl.

The mechanism for b) is very similiar to that of a Claisen condensation reaction.

Problem Details

MendelSet practice problem # 743 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork), Carbonyl Condensation Reactions (Aldol, Claisen, Dieckmann, Robinson)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition, enolates, carbonyl alpha substitution, aldol condensation, Claisen condensation

Problem # 742

Enamines are similar to enols and enolates in that they also undergo alpha substitution reactions.

The process of performing a carbonyl alpha substitution reaction via an enamine intermediate is called the Stork enamine synthesis. Let's work through this reaction.

Draw in the structures for the enamine and show it attacking the alkyl halide to form the "3º imine." Also draw the structure of the final and carbonyl product.

Solution

Enamines attack alkyl halides just as enolates do.

The trick to the Stock enamine synthesis is to remember how enamines (and imines) are related to carbonyls:

Enamines are formed from carbonyls and 2º amines.
Enamines (and imines) are converted back to carbonyls with H₃O⁺ (acidic hydrolysis).

Problem Details

MendelSet practice problem # 742 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): carbonyl alpha substitution, Stork enamine synthesis

Problem # 741

Like enolates (Q740), enols can also act as nucleophiles and make new bonds at the alpha position.

Let's go through the mechanism for how this happens, using alpha bromination as our example.

In the reaction below, the carbonyl form is in equilibrium with its enol form (Q738), which can then attack molecular bromine.

Complete the structure of the protonated carbonyl, and show the curved arrows that transform it into the final carbonyl product.

Solution

Enols are just like enolates in that they attack electrophiles and add to the alpha position.

Problem Details

MendelSet practice problem # 741 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enols, carbonyl alpha substitution, alpha halogenation

Problem # 740

Enolates are formed from carbonyls by adding a strong base, such as lithium diisopropyl amide (LDA), to deprotonate the alpha position. The enolate can then act as a nucleophile and attack an electrophile (such as an alkyl halide), to form a new bond at the alpha position. This is called a carbonyl alpha substitution reaction.

Let's go through the mechanism of how enolates are formed and how they react with electrophiles.

Draw in the curved arrows to show the formation of the enolate (middle compound), and draw the structure of the carbonyl product (right compound)

Solution

Unlike in problem 739 in which enolates were formed in two steps, enolate formation is usually drawn in one step, as shown below.

Enolates then can attack electrophiles (alkyl halides, carbonyls, etc.) to make new bonds at the alpha position.

Notice that after an alpha substitution reaction the carbonyl is reformed, and so can react again.

Problem Details

MendelSet practice problem # 740 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enolates, carbonyl alpha substitution

Problem # 739

Carbonyls are in equilibrium with their enol forms. An enolate is the deprotonated form of an enol.

Enolates are formed from carbonyls under basic conditions.

Let's go through this equilibrium under basic conditions. Draw a mechanism using curved arrows for each reaction below.

Remember that under basic conditions, most species are either neutral or negatively charged, and rarely positively charged. So your structures will contain either ROH or RO^-, but not ROH₂⁺.

a) Carbonyl to Enolate (basic)

b) Enolate to Carbonyl (basic)

Solution

These mechanisms are similar to those in problem 738, only under basic conditions.

a) Carbonyl to Enolate (basic)

In the mechanism below the carbonyl was deprotonated at the alpha position, and the enolate was then drawn as a resonance form. You can also go straight from carbonyl to enolate in one reaction arrow. (rip off the proton, form the new double bond, and send the carbonyl "up" to form a negative charged oxygen).

b) Enolate to Carbonyl (basic) "DOWN"

Like a), this reaction can be done in one reaction arrow (negative charge comes "down" and double bond attacks a proton)

It's important that you become familiar with the mechanisms for enol and enolate formation in both acid (Q738) and in base (this problem), as well as carbonyl hydrate formation in both acid (Q706) and base (Q705).

If you can draw these four mechanisms you will be able to figure most of the reactions in second semester organic chemistry!

Problem Details

MendelSet practice problem # 739 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enols, enolates, keto-enol tautomerization

Problem # 738

Carbonyls are in equilibrium with their enol forms. This process is called keto-enol tautomerization.

This equilibrium happens in both acid and base.

Let's go through this equilibrium under acidic conditions. Draw a mechanism using curved arrows for each reaction below.

Remember that under acidic conditions, most species are either neutral or positively charged, and rarely negatively charged. So your structures will contain either ROH or ROH₂⁺, but not RO^-.

a) Carbonyl to Enol (acidic)

b) Enol to Carbonyl (acidic)

Solution

Enol and enolate mechanisms always involve the alpha position of the carbonyl (one away from the carbonyl carbon). This the alpha hydrogen is acidic, and so can be deprotonated.

To form an enol (or enolate) from a carbonyl, the alpha position of the carbonyl must be deprotonated, and the double bond (pi electrons) travels "goes up" to the oxygen to become a lone pair.

To form a carbonyl from an enol (or enolate), the alpha position of the carbonyl must be protonated, and a lone pair on the oxygen "comes down" to reform the carbonyl.

This "up and down" mechanism is similar to the Nucleophilic Acyl Addition/Substitution mechanisms discussed in problems 705 and 706, with the difference being that the carbon carbonyl is not where bond formation takes place. In enol/enolate mechanisms, new bond formation occurs at the alpha position.

a) Carbonyl to Enol (acidic) "UP"

Because this mechanism takes place under acidic conditions, the carbonyl oxygen must be protonated before its double bond "goes up" to form a lone pair.

b) Enol to Carbonyl (acidic) "DOWN"

Most people remember that a lone pair from the enol oxygen "comes down." But many forget that you have to add something back to the alpha position! In this case, it's just a hydrogen- you reprotonate the alpha position to form the carbonyl.

Problem Details

MendelSet practice problem # 738 submitted by Matt on July 26, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): enols, enolates, keto-enol tautomerization

Problem # 735

Show how each ketone below can be prepared from sodium cyanide and either ethylene or propene.

You may also use methyl Grignard and ethylene oxide.

Solution

Nitriles react with Grignard reagents to form ketones, so the carbonyl carbon on each product must have came from a nitrile.

You also have the restriction that the longest carbon chain building must be three carbons.

a) Working backwards, the left side must have come from a nitrile, so the left side came from a Grignard.

How can we make each starting from propene?

To make the Grignard, prepare propyl bromide from propene via anti-Markovnikov HBr addition (HBr/peroxides), and then add Mg⁰/ether.

To make the nitrile, add sodium cyanide (NaCN) to propyl bromide.

Then add the nitrile and Grignard together, which forms the imine intermediate, which will hydrolyze to the ketone after you add water (H₃O⁺).

b) This problem is harder because you're limited to starting materials with three carbons or less, so we have to make three "cuts" instead of two, like in a).

Working backwards again, prepare propyl Grignard from propene as described in a), and then add ethylene oxide to form 1-pentanol. Form the 6-carbon nitrile by first converting the alcohol to a bromide using PBr₃, and then adding NaCN. Finally, add methyl Grignard followed by H₃O⁺ to form the desired ketone.

Problem Details

MendelSet practice problem # 735 submitted by Matt on July 25, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nitriles

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

Keyword(s): carboxylic acids, nitriles

Problem # 730

N,N-dimethylformamide (DMF) is shown below. Based on its structure, you might expect to see only one -CH₃ signal in the ¹H NMR spectrum. But instead DMF shows two different -CH₃ signals. Explain.

Solution

DMF appears to have two identical methyl groups. Since these six protons are all equivalent, its ¹H NMR should only show one methyl signal (singlet, 6H).

So why is that the real life the ¹H NMR of DMF shows two methyl signals? (two singlets with integration of 3H).

Because DMF is an amide.Recall that the "real" structure of molecule is the a mixture of its resonance forms. DMF doesn't look like either of the two resonance forms below. In real life, its somewhere in between.

For most carboxylic acid derivatives (such as esters), the resonance form is only a minor contributor and so the real "picture" looks very close to the carbonyl Lewis structure.

But for amides, its resonance form is fairly stable (it's common for nitrogen atoms to be positively charged), and so is a major resonance contributor.

In an amide, the bond between the carbonyl carbon and the nitrogen atom has a high degree of double bond character. (This also explains why it's harder to rotate the C-N "single bond" than you would expect from its Lewis structure- it's sort of like a double bond).

Because the C-N "single bond" is closer to a double bond, the two methyls are not equivalent. One methyl is cis, and the other is trans, and so they show two signals in the ¹H NMR.

Problem Details

MendelSet practice problem # 730 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives), Amino Acids, Peptides, and Proteins

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

Keyword(s): proton NMR, amides

Problem # 729

The ester below was dissolved in a solution of water, a small amount of which was isotopically labeled with O-18, denoted with an asterisk.

After a few hours, some isotopically labeled oxygen was found in the ester. Where was it found in the ester? Can you explain why?

Solution

To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism. (see problem 725)

Water reacts with an ester to form a carboxylic acid. This is what happens here.

If there is only trace amounts of water, it's possible that the water reacts with the ester to form a carboxylic acid, and then goes back to reform the ester. But, in this process, leaves an isotopically labeled oxygen in the carbonyl position.