Deuterium (D) is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly (5.0 kJ/mol, or 1.2 kcal/mol) stronger than the C―H bond. Reaction rates tend to be slower if a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step. This effect on the rate is called a kinetic isotope effect. (Review PROBLEM 4-57)
a. Propose a mechanism to explain each product in the following reaction.

Verified step by step guidance

1. Step 1: Analyze the reaction conditions. The presence of KOH in alcohol suggests that the reaction can proceed via two possible mechanisms: elimination (E2) and substitution (SN2). The solvent (alcohol) and strong base (KOH) favor elimination, but substitution is also possible depending on the structure of the substrate.
2. Step 2: For the elimination product (CH2=CH-CH3), propose an E2 mechanism. In this mechanism, the strong base (OH⁻) abstracts a proton from the β-carbon (the carbon adjacent to the carbon bonded to the bromine atom). This leads to the formation of a double bond between the α-carbon (the carbon bonded to bromine) and the β-carbon, while bromine leaves as a bromide ion (Br⁻).
3. Step 3: For the substitution product (CH3-CH(OH)-CH3), propose an SN2 mechanism. In this mechanism, the hydroxide ion (OH⁻) directly attacks the α-carbon, displacing the bromine atom in a single concerted step. This results in the substitution of the bromine atom with a hydroxyl group (-OH).
4. Step 4: Consider the stereochemistry and regioselectivity of the reaction. The elimination product forms via the most stable alkene (Zaitsev's rule), favoring the more substituted double bond. The substitution product forms via direct nucleophilic attack, which does not involve rearrangement.
5. Step 5: Summarize the reaction outcomes. The reaction yields two products: the elimination product (CH2=CH-CH3) via the E2 mechanism and the substitution product (CH3-CH(OH)-CH3) via the SN2 mechanism. The relative proportions of these products depend on the reaction conditions, with elimination generally favored in the presence of a strong base and alcohol solvent.

Key Concepts

Kinetic Isotope Effect

The kinetic isotope effect refers to the change in reaction rate that occurs when one atom in a molecule is replaced by one of its isotopes. In this case, the C―D bond is stronger than the C―H bond, leading to slower reaction rates when breaking C―D bonds. This concept is crucial for understanding how isotopes can influence reaction mechanisms and product formation.

Elimination vs. Substitution Reactions

In organic chemistry, elimination reactions involve the removal of atoms or groups from a molecule, resulting in the formation of a double bond, while substitution reactions involve replacing one atom or group with another. The reaction shown produces both an elimination product (alkene) and a substitution product (alcohol), highlighting the competing pathways that can occur under the influence of a strong base like KOH.

Reaction Mechanisms

A reaction mechanism is a step-by-step description of how reactants transform into products, detailing the bond-breaking and bond-forming processes involved. Understanding the mechanism for the reaction of bromopropane with KOH is essential for predicting the formation of the elimination and substitution products, as it provides insight into the intermediates and transition states that occur during the reaction.