Chemical Reactions

To solve the problem, we need to arrange the compounds 2-Bromo-2-methylbutane, 1-Bromopentane, and 2-Bromopentane in increasing order of their reactivity towards Sₓ2 displacement.

1. Understand Sₓ2 Reaction Mechanism:
Sₓ2 reactions involve a backside attack by the nucleophile, requiring access to the carbon attached to the leaving group (Br⁻). Reactivity decreases with increasing steric hindrance: primary (1°) > secondary (2°) > tertiary (3°).

2. Classify the Compounds:
- 2-Bromo-2-methylbutane: The carbon with Br is bonded to three other carbons (two methyl groups and one ethyl group), making it a tertiary (3°) halide.
- 1-Bromopentane: The carbon with Br is bonded to one other carbon, making it a primary (1°) halide.
- 2-Bromopentane: The carbon with Br is bonded to two other carbons, making it a secondary (2°) halide.

3. Assess Steric Hindrance:
- Tertiary (2-Bromo-2-methylbutane): High steric hindrance due to three alkyl groups, least reactive in Sₓ2.
- Secondary (2-Bromopentane): Moderate steric hindrance, more reactive than tertiary but less than primary.
- Primary (1-Bromopentane): Low steric hindrance, most reactive in Sₓ2.

4. Arrange in Increasing Order:
Reactivity increases as steric hindrance decreases: tertiary < secondary < primary.

Final Answer:
The compounds in increasing order of reactivity towards Sₓ2 displacement are: 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane.

To solve the problem, we need to determine which halogen compound undergoes an Sₓ1 reaction faster and explain why.

1. Identify the Compounds:
The first compound is a tertiary halide (3°), with Cl attached to a carbon bonded to three other carbons. The second compound is a secondary halide (2°), with Cl attached to a carbon bonded to two other carbons.

2. Understand Sₓ1 Reaction Mechanism:
Sₓ1 reactions proceed via a carbocation intermediate. The rate depends on the stability of the carbocation formed after the leaving group (Cl⁻) departs. The stability order is: tertiary carbocation > secondary carbocation > primary carbocation.

3. Compare Carbocation Stability:
For the first compound (tertiary halide), the carbocation formed is tertiary, which is highly stable due to hyperconjugation and inductive effects from three alkyl groups. For the second compound (secondary halide), the carbocation is secondary, which is less stable than a tertiary carbocation.

4. Determine Reaction Rate:
Since Sₓ1 reaction rate depends on carbocation stability, the tertiary halide will react faster because its carbocation intermediate is more stable.

Final Answer:
The first compound (tertiary halide) undergoes Sₓ1 reaction faster because it forms a more stable tertiary carbocation intermediate compared to the secondary carbocation formed by the second compound.

To solve the problem, we need to evaluate the Assertion (A) and Reason (R) regarding the preparation of aromatic primary amines using Gabriel Phthalimide synthesis and determine their validity and relationship.

1. Analyzing the Assertion (A):
Assertion (A) states that aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis. The Gabriel Phthalimide synthesis is a method to prepare primary aliphatic amines. It involves the reaction of potassium phthalimide with an alkyl halide to form N-alkyl phthalimide, which is then hydrolyzed to yield a primary aliphatic amine. For aromatic primary amines (e.g., aniline, C H NH ), the starting material would be an aryl halide (e.g., C H X). However, aryl halides typically do not undergo nucleophilic substitution reactions (like SN2) required for this synthesis under standard conditions, due to the stability of the aromatic ring and poor leaving group displacement. Thus, Gabriel Phthalimide synthesis is not suitable for preparing aromatic primary amines, making the assertion true.

2. Analyzing the Reason (R):
Reason (R) states that aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide. In Gabriel Phthalimide synthesis, the phthalimide anion (a nucleophile) reacts with an alkyl halide via an SN2 mechanism to displace the halide. Aryl halides, however, have a carbon-halogen bond strengthened by resonance with the aromatic ring, making them resistant to nucleophilic attack in SN2 reactions. Additionally, SN1 mechanisms are unfavorable for aryl halides due to the instability of aryl carbocations. Only under harsh conditions (e.g., high temperature or specific catalysts) or via alternative mechanisms (e.g., benzyne formation with strong bases) do aryl halides react, but these are not applicable in standard Gabriel synthesis conditions. Thus, the reason is true, as aryl halides do not readily undergo the required nucleophilic substitution with the phthalimide anion.

3. Evaluating the Relationship:
The reason explains why the assertion is true. The inability of aromatic primary amines to be prepared via Gabriel Phthalimide synthesis (A) is directly due to the fact that aryl halides do not undergo nucleophilic substitution with the phthalimide anion (R). The stability of the aryl halide bond prevents the formation of the N-aryl phthalimide intermediate, a necessary step in the synthesis. Therefore, R is a correct explanation for A.

4. Conclusion:
- Assertion (A) is true because aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis.
- Reason (R) is true because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
- The reason correctly explains the assertion, as the failure of aryl halides to react is the cause of the limitation stated in the assertion.

Final Answer:
Assertion (A) is true, Reason (R) is true, and R is the correct explanation for A.

Step 1: Identify longest chain:
6 carbon atoms → hexane
Step 2: Locate substituent:
Methyl group at C-3
Step 3: Apply IUPAC rules: Final Answer: