Organic Chemistry

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.

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.).

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.

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.

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.

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.

This is similar to problem 665- a very difficult NMR problem because we're not given the unknown's molecular formula, only its molecular mass (122 from the molecular ion peak on the mass spec).

So the first thing we have to do is guess the molecular formula based on this mass. Here's my thought process:

• There are signals in the aromatic region on the ¹H NMR (~7 ppm). So we probably have at least 6 carbon's with at least 4 IHD's (a benzene ring has 4 IHD).
• There are two other signals on the ¹H NMR, so there are probably at least 2 more carbons, for 8 carbons in total.
• 8 carbons with 4 IHD would give a formula of C₈H₁₀ (C₈H₁₈ from C_nH_2n+2 minus 8 H's for the 4 IHDs). which only gives a mass of 106. But we need a mass of 122- we're 16 mass units short. That's exactly one oxygen!
• Based on this, C₈H₁₀O is a good guess for the formula.

Now we can follow the steps laid out in problem 662:

1. Are there any hints? & 2. How many IHD are there?

Mass spec NMR problems require us to have done these two steps by now.

3. Draw some C₈H₁₀O structures with 4 IHD and eliminate, learn, repeat.

Some things we know form the NMR spectra:

• Because there are only four protons in the aromatic region (total integration of 4), the benzene ring is disubstituted. The two peaks in the aromatic region (~7 ppm) are doublets with integrations of 2, so this ring is probably para substituted.
• The benzene ring "uses" 6 carbons which leaves 2 carbons for the other peaks. They're probably both methyls because they both have integrations of 3. They're also both singlets, so there's not adjacent to any other carbons that have a hydrogen. The peak that's further downfield (¹H ~3.8 ppm) is probably near the oxygen.

Draw a few structures based on these clues, and eventually you will come to the correct structure. 

This is a difficult problem because we are never given the compound's molecular formula, only its mass (the molecular ion is the mass of the compound).

So first, let's make a guess of this compound's molecular formula.

Do we have any clues about what atoms are present in this molecule? Yes. The 13C NMR peak at ~210 ppm indicates a carbonyl (specifically an aldehyde or ketone). So there must be at least one oxygen (and 1 IHD.)

So let's do some math:

1O is 16. Mass is 86. 86-16 - 70. That's the total mass of C's and H's we have to work with.

How many C's can we "fit" in 70? 6 C's would be 72, so let's try 5 C's.

5 C's is 60, so we need 10 H's to make 70.

So a formula of C₅H₁₀O has the correct mass. It also has the correct number of IHD (1). So let's go with it.

Now we follow the series of steps laid out in problem 662:

1. Are there any hints?

We sort of did this step already, but yes- this molecule contains an aldehyde or ketone. Since we don't see a singlet at ~10 ppm on the 1H NMR, it can't be an aldehyde. So it must be a ketone.

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

C₅H₁₀O is the same as C₅H₁₀ (ignore oxygen). C5 would be C₅H₁₂ if fully saturated (because of C_nH_2n+2), so this molecule is missing 2 hydrogens, which corresponds to 1 IHD.

3. Draw some C₅H₁₀O structures with 1 IHD and eliminate, learn, repeat.

Some things we know from the NMR spectra:

• There's a doublet with an integration of 6, and a multiplet with an integration of 1. This screams isopropyl group.
• That singlet with an integration of 3 is probably a methyl group.
• As mentioned above, since we don't see the aldehyde proton anywhere, the carbonyl must be a ketone.

Now draw a few candidate structures based on these clues and eliminate those structures that don't fit the data. If anything doesn't fit, you must elminate the entire structure.

These (no molecular formula, only MS data) are the hardest kinds of NMR problems you'll get in sophomore organic chemistry, so don't worry if you find it challenging- you're supposed to. 

For a detailed explanation of the general strategy for solving NMR structure elucidation problems, see problem 662.

Here are the steps I would take to solve this problem:

1. Are there any any hints?

Yes. The sharp IR peak at 1,700 cm^-1 tells you this molecule contains a carbonyl (C=O).

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

C₄H₆O₂ is the same as C₄H₆ (oxygens can be ignored) which if fully saturated would be C₄H₁₀ (from C_nH2_n+2).

So this compound is missing 4 H's (C₄H₁₀ - C₄H₆), which corresponds to 2 IHD.

3. Draw some C₄H₆O₂ structures with 2 IHD and eliminate, learn, repeat.

Some things we know so far:

• From the IR we know this molecule must have a carbonyl group. This "uses" 1C, 1O, and 1 IHD so we have 3C, 1O and 1 IHD remaining to work with).
• What is the the remaining 1 IHD? It's probably not another carbonyl since the ¹³C NMR shows only one carbonyl peak (~170 ppm). It's probably not an alkene since we don't see any vinyl hydrogen peaks on the ¹H NMR (~5-6 ppm). That leaves a ring.
• Another clue that this molecule contains a ring is that we only see CH₂'s in the ¹H NMR spectrum (all peaks have an integration of 2).
• The carbonyl peak on the ¹³C NMR spectrum is around ~170 ppm, which means it's an oxidation state III carbonyl (ester) rather than an oxidation state II carbonyl (aldehyde or ketone).

Draw a few structures based on these clues and you'll arrive at the correct answer.

Let's go through the steps you should take to solve any NMR structure elucidation problem.

1. Are there any hints?

Yes. A broad IR peak around 3,300 cm^-1 tells you this compound contains an alcohol.

2. How many IHD are there?

(indices of hydrogen deficiency are also called degrees of unsaturation or double bond equivalents, depending on the textbook.)

The formula is C₁₀H₁₄O. That's the same as C₁₀H₁₄ (oxygens don't change IHD count). A fully saturated compound has formula C_nH_2n+2, so a C₁₀ molecule should have 22 hydrogens (2 x 10 + 2). The difference between C₁₀H₂₂ and C₁₀H₁₄ is 8 hydrogens, which corresponds to 4 IHD.

3. Draw some C₁₀H₁₄O structures with 4 IHD and eliminate, learn, repeat.

This isn't shown on the image below, but if you don't have an idea of the structure of the molecule, just start drawing structures! Never look at a blank page- just start drawing structures with the correct number of IHD (in this case, 4 IHD).

The peaks around ~7 ppm on the ¹H NMR tell you its probably aromatic (also, benzene has 4 IHD) so that's a good place to start. Draw a few benzene candidate structures with formula C₁₀H₁₄O, but then eliminate structures that don't fit the data. How? I like to go through this check list:

1. Eliminate by number of signals: Do the candidate structures you drew give the proper number of ¹H and ¹³C NMR signals? If not, eliminate!
2. Eliminate by multiplicity: Do your structures have the correct ¹H NMR multiplicities? If not, eliminate!
3. Eliminate by integration: Do your structures have the correct ¹H NMR integrations?If not, eliminate!
4. Eliminate by chemical shifts: Would the structures you drew have chemical shifts that are about the same as the chemical shifts in the ¹H spectrum given in the problem?

That's the order in which I usually eliminate candidate structures- fastest method (determining the number of expected NMR signals) to slowest method (predicting chemical shifts).

• Some things we know so far about this molecule:
• Form the IR we know It contains an alcohol.
• The gigantic singlet with integration of 9 screams tert-butyl
• The total integration of the aromatic region (~7 ppm) is 4, which means this molecule is disubstituted benzene. They're both doublets with integration of 2 which points to a para-substitution pattern.

As you elminate incorrect structures you will learn what fits the data, and be able to draw better candidate structures. Then you can repeat this process. Once you have a structure that passes these four problems, you probably have the correct structure.

Notice how we didn't even really need the ¹³C NMR to answer this problem. The ¹H NMR is usually enough.

I can't stress how important it is to just draw something! Even if you have no idea what the answer might be, don't even leave a blank page for an NMR problem. Just starting drawing out structures with the proper formula and IHD count!