Write an equation for each of the following. (If no redaction occurs, indicate "no reaction.") (a) potassium tert-butoxide + ethyl iodide (b) tert-butyl iodide + potassium ethoxide (c) ethyl alcohol \(+\mathrm{H}_{2} \mathrm{SO}_{4}\left(140^{\circ}\right)\) (d) n-butyl ether + boiling aqueous \(\mathrm{NaOH}\) (e) methyl ethyl ether + excess HI (hot) (f) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}+\) hot conc. \(\mathrm{HBr}\) (g) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}+\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}\) -

Nucleophilic substitution reactions are at the heart of organic chemistry. They occur when an electron-rich nucleophile targets an electron-deficient carbon atom that is typically bonded to a leaving group, such as a halogen. The nucleophile donates an electron pair to the carbon atom, forming a new chemical bond, while the leaving group departs with the electrons that were previously part of the bond.

For example, in the reaction between potassium tert-butoxide and ethyl iodide, the tert-butoxide ion serves as the nucleophile and ethyl iodide as the substrate with an iodide leaving group. This reaction proceeds through an SN2 mechanism, which is characterized by a single step where the nucleophile attacks the substrate and the leaving group exits simultaneously. Such reactions are common with primary alkyl halides and result in the inversion of stereochemistry at the carbon atom where substitution occurs.

In the context of the provided exercise, tert-butyl iodide doesn't react with potassium ethoxide because tertiary alkyl halides are not suitable substrates for SN2 due to steric hindrance and poor leaving group departure.

A dehydration reaction involves the removal of a water molecule from a compound, and it is a common type of elimination reaction in organic chemistry. Alcohols, such as ethyl alcohol, can undergo dehydration to form alkenes. This process often requires the presence of an acid, like sulfuric acid, and heat.

During dehydration, the acid acts as a catalyst, first protonating the alcohol to turn the -OH group into a better leaving group. Following this, a hydrogen atom is removed from an adjacent carbon, resulting in the formation of a double bond and the loss of a water molecule. The temperature helps in favoring the elimination path over substitution. The reaction described in step 3 of the exercise, where ethyl alcohol reacts with sulfuric acid at 140°C, is a classic example of an acid-catalyzed dehydration.

Ether cleavage is the breaking of the C-O-C bond in ethers, which can be achieved under acidic conditions. This type of reaction is important due to ethers' resistance to a variety of chemical reactions, making their cleavage noteworthy. A reagent commonly used for cleaving ethers is hydroiodic acid (HI).

In the presence of excess HI, especially when hot, ethers like methyl ethyl ether are cleaved into the corresponding alkyl iodides and water. This reaction proceeds as the acidic conditions protonate the ether oxygen, increasing the positive character of the carbon atoms and making them more susceptible to nucleophilic attack by iodide ions. The outcome is the cleavage of the ether into two separate molecules, each with an iodine atom attached, which is beautifully demonstrated in step 5 of the exercise solution with methyl ethyl ether and excess hot HI.