(Solved): PROBLEM 4.2 Consider the following reaction, in which an aromatic ring undergoes chlorination, rath ...

PROBLEM 4.2 Consider the following reaction, in which an aromatic ring undergoes chlorination, rather than bromination: $\stackrel{{\mathrm{Cl}}_2}{{\mathrm{AlCl}}_3}$ The mechanism is very similar to the mechanism for bromination. First, ${\mathrm{Cl}}_2$ reacts with ${\mathrm{AlCl}}_3$ to generate a complex that can serve as a source of ${\mathrm{Cl}}^{+}$. Draw a mechanism for formation of this complex. PROBLEM 4.3 Draw a mechanism for the electrophilic aromatic substitution reaction that occurs when benzene is treated with the complex from problem 4.2. The mechanism is exactly the same as the mechanism for installing a $\mathrm{Br}$ on the ring. But PLEASE, do not look back at that mechanism to copy it. Try to do it without looking back. Then, when you are finished, compare your answer to the answer in the back of the book (and compare every arrow to make sure that all of your arrows were drawn correctly). PROBLEM 4.4 Aromatic rings will also undergo iodination when treated with a suitable source of ${\mathrm{I}}^{+}$, There are many ways to form ${\mathrm{I}}^{+}$; you should look in your textbook and in your lecture notes to see if you are responsible for knowing how to iodinate benzene. If so, be aware that the mechanism is exactly the same as what we have seen. The only difference will be in the mechanism of how ${\mathrm{I}}^{+}$is formed. Draw a mechanism for the reaction between benzene and ${\mathrm{I}}^{+}$to form iodobenzene. In the first step of your mechanism, simply draw ${\mathrm{I}}^{+}$as the electrophile (rather than a complex which delivers ${\mathrm{I}}^{+}$), and make sure to draw all resonance structures of the resulting sigma complex. Then, in the last step of your mechanism, use ${\mathrm{H}}_2\mathrm{O}$ as the base that removes the proton to restore aromaticity $({\mathrm{H}}_2\mathrm{O}$ will be present for many of the methods that are used to prepare a source of ${\mathrm{I}}^{+}$). In this case, we are using water to remove the proton (instead of ${\mathrm{AlBr}}_4{}^{-}$), which should make sense because we don't have any ${\mathrm{AlBr}}_4{}^{-}$in this reaction. There is plenty of water, because nitric and sulfuric acids are both aqueous solutions. Notice that the mechanism is very similar to what we have already seen in the previous reactions. So far, we have seen how to install a halogen $(\mathrm{Cl},\mathrm{Br}$, or $\mathrm{I})$ on an aromatic ring, and we have seen how to install a nitro group. Before we move on, let's just make sure that you are familiar with the reagents necessary to perform these reactions. In each of the following cases, identify the reagents that you would use in order to achieve the desired transformation. 4.5 $⟶$ 4.6 4.7 $⟶$ 4.8 Without looking back at the previous section, try to draw the mechanism for the nitration of benzene. You will need a separate piece of paper to record your answer. Make sure to start by drawing the mechanism for formation of ${\mathrm{NO}}_2^{+}$, and then show the reaction of benzene with ${\mathrm{NO}}_2{}^{+}$.