What is 'Z' in the given set of reactions?
C$_6$H$_5$OCH$_3$ $\xrightarrow{HI}$ X + Y
Y $\xrightarrow[\text{Anhy. AlCl}_3]{\text{C}_6\text{H}_6}$ Z
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Cleavage of mixed ethers (R-O-R') with HX follows an S$_N$2 mechanism. The halide ion (X$^-$) attacks the smaller, less hindered alkyl group. If one of the groups is tertiary, it follows an S$_N$1 mechanism. Phenyl-oxygen bonds are very strong and do not break under these conditions.
Step 1: Analyze the first reaction.
C$_6$H$_5$OCH$_3$ (Anisole) reacts with HI. This is the cleavage of an ether by a hydrohalic acid (Zeisel's method).
The reaction mechanism involves protonation of the ether oxygen, followed by an S$_N$2 attack by the iodide ion (I$^-$). The attack occurs on the less sterically hindered alkyl group.
The two groups attached to the oxygen are phenyl (C$_6$H$_5$) and methyl (CH$_3$). The methyl group is much less hindered than the phenyl group. Also, the C(sp$^2$)-O bond in the phenyl group is strong and resistant to cleavage.
Therefore, I$^-$ will attack the methyl group.
C$_6$H$_5$OCH$_3$ + HI $\rightarrow$ C$_6$H$_5$OH (Phenol) + CH$_3$I (Methyl iodide).
So, the products X and Y are Phenol and Methyl iodide. The question does not specify which is X and which is Y. We have to deduce from the next step.
Step 2: Analyze the second reaction.
The reaction is Y + C$_6$H$_6$ (Benzene) in the presence of Anhydrous AlCl$_3$. This is a Friedel-Crafts alkylation reaction.
In Friedel-Crafts alkylation, an alkyl halide reacts with an aromatic ring to attach the alkyl group.
If Y is Phenol (C$_6$H$_5$OH), it would not typically be used in a Friedel-Crafts reaction this way as the -OH group reacts with the AlCl$_3$ catalyst.
If Y is Methyl iodide (CH$_3$I), it will react with benzene to form toluene.
C$_6$H$_6$ + CH$_3$I $\xrightarrow{\text{Anhy. AlCl}_3}$ C$_6$H$_5$CH$_3$ + HI.
Therefore, it is logical to assume that Y is Methyl iodide (CH$_3$I).
The product Z is C$_6$H$_5$CH$_3$, which is Toluene.
