Please show full mechanism **Exercise 72: Reaction Products****Instructions:** Draw the products of the following reactions.---**a.**Starting Material: A beta-keto ester with the structure consisting of an ester and a ketone group, with two identical molecules participating.Reagents: 1. Ethoxide ion (\( \text{CH}_3\text{CH}_2\text{O}^- \))2. Protonation with hydronium ion (\( \text{H}_3\text{O}^+ \))**b.**Starting Material: A reactant with a benzene ring containing two ketone groups and a separate cyclic ketone compound.Reagent: Hydroxide ion (\( \text{HO}^- \))**c.**Starting Material: A cyclic compound with bromine and hydrogen atoms bonded to the same carbon, along with two ketone groups adjacent in the ring.Reagent: Hydroxide ion (\( \text{HO}^- \))**d.**Starting Material: A compound with a cyclic ester containing a bromine atom, along with a linear nitrile group.Reagent: Hydroxide ion (\( \text{HO}^- \))---**Note:** For each reaction, determine the expected product after considering potential transformations such as nucleophilic attack, elimination, or rearrangement. The presence of catalysts or specific reaction conditions will direct the conversion of reactants into products.

A reaction involves the transformation of a reactant molecule to another molecule. The reaction involves the use of reagent molecules and often contains more than one reactant molecule. A reagent is a molecule of inorganic or organic origin and it can add, remove, or substitute a functional group with another group. Note: Since you have posted a question with multiple sub-parts, we will provide the solution only to the first three sub-parts as per our Q&A guidelines. Please repost the remaining sub-parts separately. The given reaction a is a Claisen condensation. The mechanism of the reaction is described in the following steps. The first step is the formation of the enolate. The hydrogen atoms alpha to the ester group are acidic and can be abstracted by a base. This results in the formation of enolate. The next step is the nucleophilic attack of enolate. The enolate ion attacks the carbonyl group of another ester molecule and results in the formation of an intermediate ion. The next step is the removal of the leaving group. The ethoxide ion leaves and results in the formation of a C-O pi bond. The next step is the deprotonationThe deprotonation results in the formation of a carbanion. This carbanion is stabilized by the adjacent keto group and ester group. The final step is protonation. The protonation of carbanion results in the formation of the final product. The reaction b is a base-catalyzed aldol addition. It is described in the following steps. The first step is the formation of an enolate. The alpha H of ketones is acidic and can be abstracted by a strong base. The enolate can be formed by a compound containing at least one alpha H atom. Hence, this diketone can form enolate whereas the given dialdehyde cannot. The next step is the nucleophilic attack. The nucleophilic attack of enolate results in the formation of an intermediate ion. The next step is the protonation. The intermediate ion is protonated by a water molecule. Since one more aldehydic group is present, one more aldol addition takes place. So, one more time all these steps take place. The next step is the formation of enolate. The alpha H is abstracted by base which results in the formation of enolate. The next step is the nucleophilic attack of enolate. This is an intramolecular nucleophilic attack. This results in the formation of an oxide ion. The next step is protonation. The protonation results in the formation of diol. Since heat is not mentioned, the reaction stops here. If heat was present, two molecules of water were eliminated resulting in the formation of an alpha-beta unsaturated ketone. The given reaction c is a substitution reaction. The given halide is a secondary halide and a strong nucleophile hydroxide ion is present. The reaction proceeds via the SN2 mechanism. In an SN2 mechanism, the leaving group leaves and the nucleophile attacks at the same time. The nucleophile attacks from the backside to the leaving group, thus an inversion of configuration is observed. Thus, an alkyl bromide is converted to an alcohol. The products are shown below.a. b. c.