Which reactions will produce the desired product in good yield? Y...

Question

Which reactions will produce the desired product in good yield? You may assume that aluminum chloride is added as a catalyst in each case. For the reactions that will not give a good yield of the desired product, predict the major products. (e) toluene + HNO3, H2SO4, heat (Desired Product: 2,4,6-trinitrotoluene (TNT))

Accepted Answer

Hey everyone consider the following reaction. Assumed that iron three chloride is added as a catalyst. Will this reaction produce the indicated product in good yield? If not predict which product will form instead. Now looking at this reaction, we understand that the starting material is nitrobenzene, we're using our alcohol chloride. It says that we are assuming that irons we chloride is added as a catalyst. So we may essentially add iron three chloride that were here. Which indicates that we're going to use it as our catalyst for this reaction. We call that whenever we have an L kill hey light in the form of our acts. Right, It's okay to indicate that as primary one. Even though here we have a tertiary one that's perfectly fine. We can see that we essentially have our our kill chain and whenever we are using our alcohol, Hey light and combination with a Lewis acid. Right. Iron three chloride were aluminum chloride. That's perfectly fine. As long as we have pre orbital's that can be used for our Lewis acid catalyst. That's fine. We are going to perform our freedom crafts alki elation. Okay, so basically we understand that we are performing the Friedel crafts calculation, let's recall that the starting material contains a nitro group and this nitrile group is highly deactivating because it's withdrawing the electron density from the ring. Recall that whenever we think about the residents, the oxygen atoms President or basically bonded to nitrogen can participate in the residence and electron density can be which one towards oxygen. And that's why nitro is deactivating the activating essentially implies that nitro group will be meta director. Right? So if we think about our positions that be or thought to Ortho positions to meta positions, we have two meta positions and electrolytic aromatic substitution for this reaction will take place at the meta position. Let's recall that Friedel Crafts reactions allow us to add our alcohol chain to our benzene ring. This is already a tertiary substrate. So no carbon Caroline rearrangement will take place when we form our electro file so we can definitely add our alcohol substitution to any of the two equivalent meta positions. They are equivalent because we have a line of symmetry. So at first glance, it looks like the reaction is fine. Right? Nitro group is meta directing. This is our meta position. We're adding our third beetle group just as indicated by the freedom craft calculation and everything seems fine. But yes, we can say that this is the product that we expect. So we can say that this is the expected product theoretically. However, let's recall that Friedel crafts calculation or basically FCS calculation. Right, let's use a shorthand notation fails general, there are two cases, but let's analyze the most important one for this problem. If there is a highly the activating substitution. Now it's really important to make sure that our ring is not deactivated whenever we have a deactivating group, it's essentially decreasing the rate of an electrolytic aromatic substitution significantly. Nitro is strongly deactivating. So this reaction would fail. Overall, the yield of this product will be extremely negligible and not significant. In other words, the reaction would fail. That's sad. Knowing this, we can say that no Friedel Crafts reactions don't work with ranks having highly deactivating substitutions, even though this is the theoretically expected product, because we have a substitution at the meta position. In practice, this reaction would fail because Friedel Crafts calculation doesn't work with highly deactivating substations. So this reaction is not efficient because the yield of the product would be extremely negligible and based on our answer. Because this reaction would fail due to the fact that nitro group is highly deactivating, we can conclude that the correct answer to the multiple choice question is the However, if you have any questions, don't hesitate to leave a comment down below. Thank you for watching.

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