Question

Identify the set of reagents/reaction conditions ' \(\mathrm{X}\) ' and ' \(\mathrm{Y}\) ' in the following set of transformation \(\mathrm{CII}_{3} \mathrm{CII}_{2} \mathrm{CIIBr} \stackrel{\mathrm{X}}{\longrightarrow}\) product (1) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{HBr} /\) acetic acid, \(20^{\circ} \mathrm{C}\) (2) \(\mathrm{X}=\) conc alcoholic \(\mathrm{NaOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\) HBr/acetic acid, \(20^{\circ} \mathrm{C}\) (3) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHC} l_{3}, 0^{\circ} \mathrm{C}\) (4) \(\mathrm{X}=\) conc. alcoholic \(\mathrm{KOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHC} l_{3}, 0^{\circ} \mathrm{C}\)

Short answer

X: conc. alcoholic KOH, 80°C; Y: Br2/CHCl3, 0°C

Step-by-step solution

Understand the Reaction Conditions

Examine the given reaction conditions listed under X and Y. Determine which combination can result in the desired elimination and nucleophilic substitution.

Possible Elimination Products

Consider dilute aqueous NaOH (option 1 and 3) and concentrated alcoholic NaOH/KOH (option 2 and 4) as reagents. Dilute conditions tend to favor substitution, while concentrated and alcoholic conditions favor elimination of HBr.

Evaluate the Final Functionalization

Check the second reagent Y which could lead to halogenation or other functional group additions. HBr/acetic acid conditions usually help in substitution, whereas Br2 in CHCl3 ensures halogenation at the alpha position.

Choose Correct Reagents

Combining the understanding from steps 2 and 3, concentrated alcoholic KOH at 80°C can lead to the elimination of HBr forming a double bond, and subsequent treatment with Br2 in CHCl3 at 0°C will give the appropriate bromination.

Key concepts

Elimination Reactions

Elimination reactions are fundamental in organic chemistry. They involve the removal of a leaving group and a hydrogen atom from a molecule, resulting in the formation of a double bond. Two common types of elimination reactions are E1 and E2. In E2 eliminations, both the leaving group and the hydrogen are removed simultaneously, usually requiring a strong base like concentrated alcoholic KOH. In the exercise, concentrated alcoholic KOH at 80°C is used to eliminate HBr from the molecule, forming a double bond. When considering elimination reactions, focus on the strength of the base and the reaction conditions:

• Strong base and high temperature favor E2 eliminations.
• Weaker bases or lower temperatures may lead to E1 eliminations.

Remember, the conditions of the reaction play a critical role in determining whether elimination or substitution predominates.

Nucleophilic Substitution

Nucleophilic substitution reactions are types of reactions where a nucleophile replaces a leaving group on a carbon atom. The most common types are SN1 and SN2. In SN1 reactions, the leaving group departs before the nucleophile attacks, creating a carbocation intermediate. SN1 is typically seen with weaker nucleophiles in polar solvents. In SN2 reactions, the nucleophile attacks the carbon at the same time the leaving group departs. These are usually facilitated by strong nucleophiles. Consider the steps from the exercise: dilute aqueous NaOH (options 1 and 3) favors substitution over elimination because of lower base strength and temperature. Key points to remember for nucleophilic substitution:

• SN1 favors stable carbocations and polar solvents.
• SN2 requires strong nucleophiles and occurs in a single step.
• Temperature and base strength are decisive factors.

An understanding of these can help in predicting the outcome when different reagents and conditions are applied.

Halogenation

Halogenation is the addition of halogen atoms (X₂) to a molecule, commonly used to add functionalities or to induce transformations in organic molecules. In the exercise, we see halogenation with Br₂/CHCl₃ at 0°C. This ensures specific bromination, likely at an alpha position. Below are the crucial points to know about halogenation:

• Halogenation often results in the addition of halogen atoms that can act as leaving groups for future reactions.
• Specific reagents and conditions, like Br₂/CHCl₃, help in the desired regioselective addition.
• Different halogenation reagents lead to different products depending on their reactivity and the conditions imposed during the reaction.