Questions from the 1998 Test 2.

• the third multiple choice, asking about reactivity with water.
  The most reactive of these compounds is choice b, which is the anhydride. It doesn't really matter WHAT they are reacting with. Anhydrides will always be more reactive than esters and amides. Acid chlorides are even more reactive, but there wasn't an acid chloride choice in this question.
• the fifth "provide the major product" question, using the LDA/EtBr(?)
  That question relates to chapter 22, which we haven't covered yet. (Big relief, I suspect.)
• the sixth provide the major product question, using Br2 and HOAc
  This question also relates to chapter 22, which we'll start on next week.
• Synthesis question b, I don't know what the starting materials are
  This synthesis involves yet another reaction from chapter 22, so you can ignore it for the time being.
• synthesis question c, I don't have a clue.
  Same answer here. This question is not from the chapters we have been studying.

I know in class we discussed a number of reactions that we didn't have to know the mechanism for. Still, I wrote them down anyway. Could you tell me which ones we are going to be responsible for on our exam? Thanks!
The reactions you're responsible for fall into two categories: aromatic substitution reactions and nucleophilic acyl substitution reactions. The aromatic substitution reactions include: electrophilic aromatic substitution and nucleophilic aromatic substitution (by two different mechanisms). The nucleophilic acyl substitution reactions include most of the reactions in chapter 21--both the forward and the reverse reactions where relevant. The list might seem overwhelming, but if you can focus on the general type of the reaction it should make the task much more doable for you.

In the lab portion, what are some of the other qual tests we are going to have to know besides the soot test and the copper wire test?
You should know the DNP test (for aldehydes and ketones) and the Jones test (for aldehydes, primary alcohols, and secondary alcohols).

I've been reviewing lab information for the test tomorrow, and I had a question about the organic qual. Is there somewhere where I should've read/studied about the best derivatives to make for each functional group? I know that you are supposed to choose the one with the greatest variation between the melting points, but other than that I'm at a loss! Is there something I could read/ look at this?

• If you have an alcohol, you'll be making an ester derivative. There's more than one kind of ester derivative. You only need to worry about what the functional group is. You make that ester by reacting your unknown alcohol with an acid chloride.
• If you have an amine (primary or secondary), you'll be making an amide derivative by reacting your unknown amine with an acid chloride. Again, there's more than one kind of amide derivative--you only need to know the functional group.
• If you have a carboxylic acid, you'll be making an amide derivative. You do this by first transforming your carboxylic acid into an acid chloride with thionyl chloride (SOCl₂). Then you turn the acid chloride into an amide by reacting it with an amine (or ammonia).
• If you have an aldehyde or ketone, you make an imine (general functional group name) by reacting the carbonyl group with an amine. The specific amine you typically use is 2,4-dinitrophenylhydrazine, commonly referred to as DNP.

Ok, so in response to the last info you sent me:
It says that the derivative for carboxylic acids is an amide. For my organic qual wet unknown (w-11) I think that I have an aromatic carboxylic acid. I tried to make an anilide because I thought that it was with this derivitive that had the greatest mp variation. But,with this test I was unable to obtain crystals. If I were to go back and make an amide I would run into the same problem as DeShawn and Cassie with the burning tar-like goo...what advice do you have for getting crystals?
An anilide is simply a specific type of amide. It's pretty much a crap shoot whether your stuff turns into tarry goo. It doesn't happen always. In the real world you'd have more time and would be required to do the reaction over and over until you got it to work. In the meantime you'd get more experience and would develop your skills in the art form of making crystals. It really is an art form, in case I didn't tell you that before.
The short answer to your question is that you don't need to do the reaction over again to get crystals.

I used the alternative method to make my ketone derivative which Dr. Workman named a semicarbazone. Is this an answer I could use on the test instead of DNP?
Certainly. When I answered that question, I was trying to be brief and give the gist of it. For each of the functional groups I described, there's more than one derivative that could be made. You would normally choose which derivative to make based on how far apart the mps of the derivatives are for the compounds you're considering as possibilities.

What does the Jones test tell you about your unknown. Negitive = ketone positive = ROH and aldehyde???
You have it almost right, assuming you are running the Jones test because you know your compound is neutral (aldehyde, ketone, or alcohol). Positive Jones means either aldehyde or a primary alcohol or a tertiary alcohol. We can't always tell very much from negative results. It's possible that you don't have a neutral compound at all but were just running all the tests. If it's definite that the compound is neutral, then a negative Jones test could also be a tertiary alcohol, which isn't readily oxidized (in addition to the ketone which you mentioned).

What does the DNP test tell you about your unknown?
Positive DNP test means either an aldehyde or ketone is present.

Could you please show me an example of a situation in which you would use sulfonation as a protecting group?
Sulfonation is used as a protecting group when you want the ortho isomer of something without any of the para isomer present. That's because it's easier to get the para isomer pure than the ortho isomer pure. Due to the regular shape of the para compounds, they crystallize more readily. Here's an example:

is there any way you could give me the reverse mechanism for sulfonation.
Here's the mechanism for the sulfonation reverse mechanism. I should point out that the SO₃ could also be a neutral leaving group if it exists on the ring as SO₃^- (without the H⁺).

can you give me a good example of a mechanism for nucleo. aromatic substitution
Question 16.58 is a good example of an explanation question involving nucleophilic aromatic substitution. So is Question 16.57. So is 16.54. They don’t all go by the same mechanism. In other words, one or more of them involves the benzyne mechanism and the other(s) involve the stepwise mechanism that is analogous to the electrophilic aromatic substitution.

Problem 16.33. Part b. I'm not sure what those reagents do.
16.33b The HNO₃/H₂SO₄ is a nitrating reagent. It will add a nitro group to the ring. That group can go ortho to one Br and meta to the other. Or it could be para to one Br and meta to the other. Which would you guess would be preferred? You have some similar experience with the nitration of veratrole that we did in lab recently. The second step is a reducing reagent, which interacts with the newly added nitro group.

Problem 16.33 Part d. Does the propyl group rearrange?
16.33d The propyl group will indeed rearrange. The major product will have isopropyl added to the ring. There may be some n-propyl added to the ring, but that unrearranged product would be minor.

Problem 16.34. How does ICl iodinize an aromatic ring?
16.34 ICl is polarized with a partial negative on the Cl and partial positive on the I. So, it reacts with the aromatic ring to put an I+ on the ring and leave a Cl- as a leaving group.

16.41 - i understand the problem but i really don't understand how you get to the resonance structure where there is a double bond off the ring and the C-N bond is only a double bond with a negative charge on the nitrogen.
Problem 16.41. To get the resonance structure you’re talking about, you basically break one of the pi bonds between C and N. You give both of the electrons to the N, giving the N a negative charge and the C a positive charge. Then do the steps I describe below if you don’t see why that positive charge on the C matters.
Draw resonance structures for the electrophile attacking the nitrile with the saturated alkyl chain. You’ll have nine if you draw them for each of the possible attacks (ortho, meta, and para). Do the same thing for the electrophile attacking the aromatic nitrile with the UNsaturated alkyl chain. You’ll again have nine. Then maybe you can get another one where the charges on the ring interact with the charges on the nitrile DUE to the conjugation. Thus, the nitrile has a much bigger effect if it’s on a conjugated chain, almost making it act like a regular nitrile DIRECTLY attached to the ring.

Problem 16.42. Why do you only get one product for the reaction of HBr to 1-phenylpropene?
16.42 You only get one product for this reaction because the preferred intermediate has the positive charge on the benzylic position (next to the ring). If the H went next to the ring and put the positive charge on the other position, you’d have a secondary cation, which is much less stable than a secondary cation which is also benzylic.

Problem 16.51. I have no clue how to make any of the three products for a, b, or c.
16.51 The key step for synthesizing each of these products is to functionalize the alkyl group after you add it to the ring. Two different possible ways to functionalize the alkyl group involve reacting an alkyl substituted aromatic ring with Br₂ and light OR NBS, peroxide and light. Another approach would be to acylate the ring and then reduce the carbonyl group to an alcohol. This second approach gives you a meta director on the ring, so you probably wouldn’t want to use for the group that gets added first. For part c, consider reactions you learned last semester to give anti-Markovnikov addition of water to an alkene. The alkene can’t be directly added to the ring, but you can make it with other groups that you got using the approaches in parts a or b.

Problem 16.53. How do you get hexachlorophene from 2,4,5-trichlorophenol and formaldehyde?
16.53 Draw out the structure of formaldehyde and then protonate it with your sulfuric acid. Now you have a good electrophile that can react with one of the aromatic rings. After another protonation step (and maybe the loss of water), you can have another aromatic ring reacting with your handy multi-purpose electrophile. The reaction that’s described here is very useful industrially in the formation of plastics.

Problem 20.21. I'm confused on how you know when things are more basic than others.
20.21 Basicity is inversely related to acidity. In other words, the conjugate base of a strong acid is a negligible base (e.g., HCl and Cl^-). The conjugate base of a weak acid is a weak base (e.g., acetic acid and acetate). The conjugate base of something that’s not particularly acidic is a strong base (e.g., ethanol and ethoxide). Adding electron withdrawing groups or electron donating groups can influence the acidity of an acid. It can also influence the basicity of the conjugate bases. Things that are electron withdrawing make the acid more acidic (and the base less basic). Things that are electron donating make the acid less acidic (and the base more basic).

Problem 20.29. Part d. Does potassium permanganate do anything to a carboxylic acid?
20.29d Nope. But it might react with the other part of the molecule.

Problem 20.34. Part c. Do any reagents reduce a carboxylic acid to an alkane?
20.34c Not directly. You’d have to reduce the carboxylic acid to an alcohol, dehydrate the alcohol, and then hydrogenate the alkene.

Are DIBAL and LiAl(OtBu)₃H interchangeable as reducing agents? In other words can you use DIBAL to reduce an acid chloride to only the aldehyde or is DIBAL not sterically hindered enough to prevent this from happening?
They aren't interchangeable, but sometimes one will work as a substitute for the other. You can't use DIBAL to reduce an acid chloride to an alcohol. I'm not sure what would form, as there's no entry in my table for that combination. But you could use LiAl(OtBu)₃H for reducing an amide to an aldehyde. (Esters, however, will turn to alcohols with that reagent.)

What is the reaction for soaponification? I have several reactions in my notebook, however, I forgot to mark which one it was.
The actual spelling is saponification. The reaction is the base catalyzed hydrolysis of an ester. It doesn't have to be a fat that you're hydrolyzing, but that's where the name came from.

in prep of carb acids, what is the difference again between grig. and nitrile reagents? I am still a little unclear.
Sometimes Grignard reagents work better. Sometimes nitriles are better. It depends where the halogen is to begin with. It also depends what other functional groups are in the ring. Alcohols interfere with Grignards, but they don’t interfere with nitriles. Aldehydes interfere with both Grignards and nitriles, so if you have an aldehyde, you would need to protect it before doing either one of those reactions.

I'm having a hard time figuring out 21.41 part b.
Think about changing your bromine to a carbon group by making a Grignard and then adding a carbon group to it (aldehyde, ketone, or carbon dioxide).

I have another question for you: 21.43 part g.
One or two of the carboxylic acid derivatives will allow you to add a SINGLE equivalent of an organometallic reagent and stop at the ketone rather than adding two equivalents of organometallic reagent and going all the way to the alcohol. Think about using one of those two acid derivatives. Either will work. One definitely requires fewer steps.