All Practice Problems

Problem # 670

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

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 # 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 # 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.).

Problem Details

MendelSet practice problem # 672 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Grignard and Masked Carbanions (alcohols from carbonyls)

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

Keyword(s): Grignard, proton NMR

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 # 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 # 677

Show a mechanism for the acid-catalyzed cyclization (condensation) of 1,4-butanediol.

Solution

This is just an S_N2 reaction.

The first step is to turn one of the -OH groups into a better leaving group by having it pick up a proton. (I choose the hydroxyl on the right side).

Then the other -OH group can attack carbon 1, and the protonated hydroxyl group leaves as water.

Finally, the proton on the -OH group that acted as a nucleophile will "fall off" (deprotonate).

Problem Details

MendelSet practice problem # 677 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Ethers and Epoxides, Alcohol Reactions (PBr3, SOCl2, HX, Sulfonate ester formation)

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

Keyword(s): SN2 mechanism, alcohol condensation

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 # 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!

Problem Details

MendelSet practice problem # 679 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Alcohol Oxidation Reactions (PCC, Jones Reagent)

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

Keyword(s): proton NMR, oxidation, Fischer esterification

Problem # 697

Rank the following compounds in order of decreasing boiling point.

Also, make a guess about their relative solubilities in water. Explain your reasoning.

Solution

Water is very polar, so the more polar a molecule is, the more soluble it will be in water.

Alcohols can hydrogen bond, and so will be the most polar, have the highest boiling point, and be the most soluble in water.

Aldehydes (carbonyls) can't hydrogen bond but the C=O double bond is very polar, so aldehydes are somewhat soluble in water.

The ether is the least polar of the three functional groups, and so will have the lowest boiling point, and be the least soluble in water.

So the overal trend for polarity, water solubility, and boiling point is:

alcohol > aldehyde > ether

Problem Details

MendelSet practice problem # 697 submitted by Matt on July 20, 2011.

Subject: Organic Chemistry

Subject Topic(s): Ethers and Epoxides, Intermolecular Forces and Functional Groups

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

Keyword(s): boiling point, polarity, solubility

Problem # 698

When propyl bromine is treated with KF in benzene no reaction takes place. But when the crown ether 18-Crown-6 is added to the reaction mixture the desired propyl fluoride product is produced. Explain.

Solution

KF is insoluble in propyl bromide or benzene, so the two compounds never "touch" each other, and no reaction takes place. (F^- and propyl bromide are in different phases and so don't come into contact).

But when the crown ether is added, it solvates the potassium ion, allowing the K⁺ and F^- ions to dissolve in the solvent, so the fluoride ion and propyl bromide are able to "talk" to each other, and the rection can take place.