Tulobuterol contains a benzene ring in its structure. Describe and explain the shape of benzene. In your answer, include: the bond angle between carbon atoms; the hybridisation of the carbon atoms; how orbital overlap forms $\sigma$ and $\pi$ bonds between the carbon atoms.
Write an equation showing how $\text{Cl}_2$ reacts with $\text{AlCl}_3$ to produce an electrophile.
Complete the mechanism in Fig. 5.3 for the reaction between benzene and the electrophile formed in b(i). Include every relevant curly arrow and charge, and draw the intermediate structure.
Name the mechanism for the reaction in step 2.
Draw an isomer of $Q$ that is formed as an organic by-product in step 2.
The reactants used in step 2 include acyl chloride, alkyl chloride and aryl chloride functional groups. State and explain how readily acyl chlorides, alkyl chlorides and aryl chlorides undergo hydrolysis.
Suggest reagents and conditions for steps 3 and 4, and draw the structure of $Z$ in the box.
Define enantiomers.
Suggest one disadvantage of making two enantiomers in this synthesis.
Suggest a method of adapting the synthesis to give a single enantiomer.
Predict how many peaks would be seen in the carbon-13 NMR spectrum of tulobuterol.
The proton ($^1\text{H}$) NMR spectrum of tulobuterol dissolved in $\text{D}_2\text{O}$ shows peaks from four different proton environments. The peak for the $-\text{CH}_2-\text{N}-$ environment is a doublet in the chemical shift range $\delta = 2.0$-$3.0\,\text{ppm}$. Give details for each of the other three peaks in the proton NMR spectrum of tulobuterol, including the chemical shift, proton environment, splitting pattern, and number of $^1\text{H}$ atoms responsible.