Textbook: Wade 7th Ed. (2009)

Chapter 10: Structure and Synthesis of Alcohols

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Individual Problems

Individual Problems

Problem # 669

In your own words, what is the major difference in the addition of a Grignard reagent to an oxidation state III carbonyl (ester/acid chloride) versus an oxidation state II carbonyl? (aldehyde/ketone)

Solution

Oxidation state III carbonyls (esters, acid chlorides) contain a built-in leaving group (such as ^-OR or ^-Cl) and so undergo nucleophilic acyl substitution reactions. The product is another carbonyl.

Oxidation state II carbonyls (aldehydes and ketones) do not contain a built-in leaving group and so undergo nucleophilc acyl addition reactions (instead of substitution). The product is an alcohol.

Because Grignards react with all carbonyls- esters and aldehydes/ketones- esters and acid chlorides will react twice with Grignards: once in a Nuc Acyl Sub mechanism to form a ketone, which will then react with another equivalent of Grignard in a Nuc Acyl Add mechanism to form an alcohol.

Problem Details

MendelSet practice problem # 669 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition

Problem # 668

Show how each alcohol can be prepared from a combination of a carbonyl and a Grignard reagent.

Solution

The trick to these retrosynthesis problems is to determine where the connections or "cuts" were made.

The carbonyl carbon becomes an alcohol after a Grignard reaction, so that's where the "cut" must be.

Note that there can often be more than one correct answer to these types of problems.

For a), adding propyl Grignard to acetone or methyl Grignard to 2-pentanone will result in the product.

We can also use two equivalents of methyl Grignard with 4-carbon ester, such as ethyl butanoate. Esters contains a build in leaving group (^-OR) and so react twice with Grignards.

For b), adding phenyl Grignard to cyclopentanone will do the job.

Problem Details

MendelSet practice problem # 668 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

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

Keyword(s): Grignard, retrosynthesis

Problem # 671

You may have noticed that the "solvent of choice" for many organometallic compounds such as Grignard reagents is ether (short for diethyl ether).

Why is it that for Grignard reactions this solvent is used over ethyl acetate, or protic solvents such as ethanol?

Solution

Grignards behave as though they are carbanions (negatively charged carbons), and so are very basic. There are so basic that they will deprotonate any O-H or N-H bond. So protic solvents such as water or ethanol aren't suitable for Grignard reactions; the Grignard reagent will react with the alcohol in an acid-base reaction.

In fact, water is used after a Grignard reaction to quench the Grignard reagent.

Grignards are also nucleophilic, and so react with carbonyls (which are electrophiles).

Ethyl acetate contins a carbonyl and would get attacked by a Grignard reagent, and so also isn't a suitable solvent for a Grignard reaction.

Diethyl ether doesn't have any acidic protons and isn't electrophilic and so won't react with a Grignard reagent, so it makes a good solvent.

Problem Details

MendelSet practice problem # 671 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls)

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

Keyword(s): Grignard

Problem # 670

Draw out the mechanism for the addition of excess phenyl Grignard to the carbonyl compound below.

Solution

This carbonyl has two leaving groups attached to it- each of those oxygens can take part in a nucleophilic acyl substitution reaction and form a new carbonyl product.

First the Grignard attacks the oxidation state IV carbonyl carbon (4 oxygen bonds, so oxidation state 4). The carbonyl itself will act as a leaving group and form a tetrahedral intermediate. But tetrahedral intermediates don't last if there are any leaving groups attached to the carbon, so the ^-O will "come back down again", kick off an oxygen leaving group, and reform the carbonyl.

Then a second equivalent of Grignard will attack that carbonyl (an ester), and we will do another nucleophilic acyl substitution reaction to form yet another carbonyl.

Finally, the third carbonyl doesn't have any leaving groups built in (it's a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol.

Notice that as the reaction progresses the oxidation state of the carbonyl carbon (number of oxygen bonds attached to it) goes down form 4 to 3 to 2 and then to 1.

Problem Details

MendelSet practice problem # 670 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition

Problem # 674

Show a mechanism for the reduction of butyrolactone using LiAlH₄.

Solution

Hydride reagents such as LiAlH₄ and NaBH₄ behave like hydride nucleophiles (H^-), so that's what I used as shorthand. The real mechanism is very similar but involves aluminum coordinating to the oxygen.

Notice that the the ester will reform the carbonyl after the first hydride attacks. This is because esters have a built in leaving group, and so undergo nucleophilic acyl substitution reactions. The aldehyde that forms then undergoes a nucleophilic acyl addition reaction with the second equivalent of hydride.

Also note that you can't stop the reaction halfway at the aldehyde- LiAlH₄ will take an ester all the way down to an alcohol.

Problem Details

MendelSet practice problem # 674 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Hydride Reductions (NaBH4, LiAlH4)

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

Keyword(s): hydride reduction

Problem # 672

Compound A has molecular formula C₆H₁₂O and shows a sharp peak at 1,710 cm^-1 in its IR spectrum.

Treatment with 1 equivalent of phenyl Grignard yields compound B, which has formula C₁₂H₁₈O and whose IR shows a broad peak at 3,350 cm^-1.

Compound B's ^₁H NMR spectrum is shown below. Determine the structures of compounds A and B.

Solution

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

1. Are there any hints?

Yes, from the IR peaks. The starting material is a carbonyl (sharp IR peak at ~1,700 cm^-1) and the product is an alcohol (broad IR peak ~3,300 cm^-1).

2. How many IHD are in each compound? (also known as degrees of unsaturation or DBE).

C₆H₁₂O is the same as C₆H₁₂ (ignore oxygen) which should be C₆H₁₄ if fully saturated (C_nH_2n+2). It's missing 2 hydrogens so C₆H₁₂O has 1 IHD. This 1 IHD must be the carbonyl.

C12H18O is the smae as C12H18 which should be C12H26 if fully saturated. It's missing 8 hydrogens so it has 4 IHD. The benzene from the Grignard reagent must account for all 4 IHD.

3. Draw some C₆H₁₂O and C₁₂H₁₈O structures and elminate those that don't fit the data, then learn and repeat.

Things we know from the problem and NMR:

The starting material is a carbonyl and we're adding a Grignard, so we expect the product to be an alcohol. The NMR shows a peak that disappears with D₂O addition, which also confirms that the product is an alcohol.
The product must have a benzene ring because the reagent was phenyl Grignard. That accounts for the aromatic signals (~7 ppm) and the 4 IHD in the product.
the NMR shows a doublet with an integration of 6 and a multiplet with an integration of 1. This is the splitting pattern on an isopropyl group.
The NMR also shows a quartet with an integration of 2 and a triplet with an integration of 3. This is the splitting pattern of an ethyl group.

Once you have a few clues from the NMR, start drawing structures! And then elminiate those that do not fit the data (too many signals, wrong multiplicity/integrations etc.).