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"; Bonds Formed In An Sn1 Or Sn2 Reaction Making Ether As A Product Lab Report - Chem Homework Help
Bonds Formed in An Sn1 or Sn2 Reaction Making Ether as A Product Lab Report

Need help with my Chemistry question – I’m studying for my class. Instructions for the Assignment (can be found on Lab 7 pt.2 document at the very bottom of the last page; only answer questions from this sheet and use the other document as reference if needed):Using the same approach as above, design a Pathway A and Pathway B for your target molecule. In some cases, Pathway A and B are the same, in which case you should note that; in some cases, the ether may be part of a ring, in which case the starting material will be one molecule which has both alkyl halide and alcohol functional groups on it.In your notebook, analyze your Pathway A and Pathway B to determine whether an SN1 or SN2 approach will be better for each pathway. After finishing this analysis for both pathways, choose the overall best pathway for synthesizing your ether. Alternatively, if two or more pathways are both equally valid, explain that instead. The above example is a good template to follow for this analysis.————————————————————————————————————————————————————————Please read through the lab carefully and complete all the work with detailed responses on the document if possible if you can. If not, please answer the questions accurately numbered on a separate sheet of paper. You should answer the questions on the document Experiment 7 Part 2 completely.

Experiment 7 Pt. 1: Thinking Backwards, an Introduction to Retrosynthesis Read this document, follow along, and complete the exercise at the end to complete this week’s lab. This work need not be turned in, but should be done by the end of the week. Select an ether from the chart provided as part of this lab. On scratch paper, draw your selected ether. In this example, I’ll do the following one: Figure 1: A complicated ether. Identify which bonds could be formed in an SN1 or SN2 reaction that makes this ether as a product. Assume that you will use either an alcohol or its conjugate base as a nucleophile. Figure 2: The C-O bonds in an ether, colored for emphasis. One of these bonds must have been formed in the reaction which created this molecule. The C-O bonds are the only bonds in this molecule which could be formed in a reaction that we know how to do because we’ve seen it in class or lab before. Do a formal retrosynthesis step by showing one of these C-O bonds breaking in a heterolysis step. We will choose the red bond for an example. Figure 3: We can figure out what pieces a molecule is made from by drawing a mechanism for it breaking into pieces. Notice that this is not a reasonable reaction step – the RO- group is not a good leaving group, since it is a strong base. We are just doing it as a trick to help us figure out what molecules we could have made this ether from. Thinking backwards is hard, so we are doing it one step at a time. Now we have broken our molecule into parts. We just need to figure out what kind of reactants could give those parts. Let’s do it one piece at a time. The RO- ion can be made from a ROH (alcohol) by pulling off the H with a base. This is what we did in the synthesis of phenacetin. We use that fact in Figure 4. The arrow with two lines means “can be made from.” So, the thing on the left of the arrow “can be made from” the thing on the right. Figure 4: This is how you use the “can be made from” (retrosynthesis) arrow. This alkoxide can be made from an alcohol. The retrosynthesis arrow looks like a smiley face on its side. =) This alcohol will be one of our possible starting materials for the reaction. What about the carbocation? We can also write: Figure 5: We remember that heterolysis of an alkyl halide can give a carbocation, so we draw this. We say that the bromopropane is a “carbocation equivalent”, since we can’t exactly have a bottle of carbocations. In words, Figure 5 says “this carbocation can be made from 1-bromopropane.” (We could use another leaving group just as well, such as Cl) Remember, we may use an SN1 or an SN2 reaction to do this reaction, so we may not actually go through a carbocation intermediate. (Good thing, too, because that’s a lousy carbocation) This whole process is just an intellectual exercise, a trick, to help us figure out what chemicals we might react to make our target molecule. Look what we have now (Figure 6). Figure 6: Our final retrosynthesis of the ether, assuming that we started with the bond which is red in Figure 1. We would say that the ether on the left in Figure 6 can be made from the alcohol and alkyl halide on the right. (I colored the carbon groups on each side to make it more obvious where they came from.) We can write this as a normal-person reaction like this. (Figure 7) Figure 7: Our final line reaction showing the reactants from which our ether is made. This reaction looks more familiar. It has two reactants and a product. Of course, it’s not done yet – it doesn’t have any solvents, other reactants like base or acid, or temperatures written on it. We will do those in a later step. It does tell us what organic starting materials we need to do the reaction. The hard part is finished. Every ether has two C-O bonds. So, we could have started with the green bond in Figure 1 and done the same kind of analysis. That would have given us different starting materials. As a warmup exercise, perform this analysis with the other bond (green, Figure 1), and figure out what the other combination of alcohol and alkyl halide could be that would give us the same product. You don’t need to turn this work in, it’s only for practice. If you want to work ahead, try to perform this kind of retrosynthetic analysis on your ether chosen at the beginning of the lab. In the second part of this lab next week, you will do that analysis and choose which one is the better option as well as learn to write out a complete reaction with solvents and reagents.

Experiment 7 Pt. 2: Retrosynthesis of an Ether Last week, you selected an ether from the chart provided as part of this lab. On scratch paper, draw your selected ether. In the previous week’s assignment, we performed a retrosynthetic analysis on an ether: Figure 1: 3,5-dimethylcyclohexyl propyl ether (left) can theoretically be synthesized from two different sets of alcohol/alkyl halide combinations, Pathway A or Pathway B (right). Today, we will go through the process of choosing a final reaction by which we could actually synthesize our ether. For each pathway, we will decide whether it’s better to use unimolecular (SN1) conditions or bimolecular (SN2) conditions, and then compare the two to decide which is overall more practical. Let’s discuss pathway A first. In A, the electrophile is a primary (1°) alkyl halide. The electrophilic carbon (the one with the leaving group) has only one alkyl group on it, so it has low steric hindrance but would make a poor carbocation (and would shift if it did form one) This is a perfect candidate for an SN2 reaction, similar to our synthesis of phenacetin. Figure 2: Pathway A clearly favors a SN2 approach. In order to perform an SN2 reaction between this alkyl halide and alcohol, we need a strong nucleophile, and the alcohol itself ain’t it. To make the alcohol a nucleophile, we must react it with a base to remove the proton, making the strong alkoxide nucleophile. Figure 3: To perform a SN2 reaction, an alkoxide nucleophile (RO-) is prepared from an alcohol using a base. In Experiment 5, we used KOH as our base to prepare the nucleophile. Figure 4: Our previous synthesis of phenacetin used KOH as base and ethanol as solvent. We could get away with using KOH, the conjugate base of water (pKa 14), because the alcohol with resonance with the aromatic ring (called a phenol) had a much lower pKa than a normal alcohol (pKa 15). In this case, we have no resonance (no double bonds in the ring) so we must use a stronger base. A typical choice would be sodium hydride rather than hydroxide. Hydride is the conjugate base of H2 (hydrogen) and is an extremely powerful base due to the H atom’s tiny size/low polarizability (CPER). We choose one of the polar aprotic, that is, non-H-bonding, solvents as our solvent for the reaction, both to help the SN2 reaction (as we saw in class, polar aprotic solvents favor SN2) and because most of them are stable enough to survive this very powerful base. So, we have chosen to run Pathway A under bimolecular conditions (promoting SN2). To do this, we chose a base (NaH) and a solvent (DMSO) for Pathway A. We write: Figure 5: The final version of our Pathway A line reaction. We did it! While we never really know until we do the experiment, we would expect that this is a reasonable approach to synthesizing the molecule. Now let’s look at Pathway B. The halide is secondary, which isn’t really ideal for either an SN1 reaction or an SN2 reaction. If we follow the same SN2 approach as above, we get this: Figure 6: This reaction could work, but with such a bulky alkyl halide it will produce a lot of E2 product instead. Let’s look at an alternative. This is not as good as the SN2 version of Pathway A, because it is better for the electrophile (alkyl halide) to be less sterically hindered. With so much steric hindrance on the large, bulky alkyl halide, the strong R-O- base is likely to cause elimination instead, forming the E2 product. (Can you draw what the product would be?) Let’s consider an SN1 approach instead. To make an ether from an alkyl halide and an alcohol, we simply mix the two together and heat them to force heterolysis of the C-X bond. Figure 7: The SN1 version of Pathway B. This reaction is a possibility as well! As always, we have to watch out for shifts when we make a carbocation. However, the carbons next to the leaving group are also 2°, so there is no neighboring location with a more stable carbocation. Figure 8: The alkyl halide in Pathway B is secondary, and so are its neighbors. We conclude that both Pathway A, the SN2 version, or Pathway B, the SN1 version, are valid syntheses of our target ether. Either choice is fine. Instructions Using the same approach as above, design a Pathway A and Pathway B for your target molecule. In some cases, Pathway A and B are the same, in which case you should note that; in some cases, the ether may be part of a ring, in which case the starting material will be one molecule which has both alkyl halide and alcohol functional groups on it. In your notebook, analyze your Pathway A and Pathway B to determine whether an SN1 or SN2 approach will be better for each pathway. After finishing this analysis for both pathways, choose the overall best pathway for synthesizing your ether. Alternatively, if two or more pathways are both equally valid, explain that instead. The above example is a good template to follow for this analysis. Scan your notebook pages and turn your final analysis in on the Blackboard assignment provided.

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