Haloalkanes And Haloarenes

Hence, to obtain propanenitrile, should be chloroethane.

To solve this question, we need to identify the reagents (X, Y, Z) used in the transformation of into different compounds. Let's analyze each part of the reaction scheme:

1. X: Transformation to
   • The conversion of to can be carried out using a fluoride source. The most suitable reagent from the options provided is , which facilitates the halogen exchange reaction.
2. Y: Transformation to
   • This step involves dehydrohalogenation of to form ethylene . Alcoholic KOH is commonly used for such elimination reactions.
3. Z: Transformation to
   • The conversion of to butane can occur via the Wurtz reaction, involving the coupling of alkyl halides. This reaction uses sodium in dry ether as the reagent.

Based on the above analysis, the correct choice of reagents for X, Y, and Z are , alcoholic KOH, and Na in dry ether, respectively.

Conclusion: The correct answer is , alcoholic KOH, Na in dry ether.

The question asks which factors determine if a haloalkane undergoes an or reaction. Here's a step-by-step explanation of the relevant concepts:

1. Understanding and Reactions:
   • (Substitution Nucleophilic Bimolecular) reaction is a single-step process where the nucleophile attacks the substrate and the leaving group departs simultaneously. This type of reaction involves a transition state where the carbon atom is simultaneously bonded to the nucleophile and the leaving group.
   • (Substitution Nucleophilic Unimolecular) reaction, on the other hand, is a two-step process. First, the leaving group departs to form a carbocation intermediate, followed by the attack of the nucleophile.
2. Influence of Solvent on Reaction Type:
   • reactions are favored in polar aprotic solvents because these solvents do not form strong interactions with the nucleophile, thus making it more nucleophilic.
   • reactions are favored in polar protic solvents which stabilize the carbocation intermediate and the leaving group through hydrogen bonding.
3. Analyzing Other Options:
   • Low temperature: While temperature can influence reaction rates, it does not directly determine whether an or mechanism is preferred.
   • The type of halogen atom: The leaving ability of a halogen atom can affect the rate but does not decide the type of nucleophilic substitution (though better leaving groups favor both types).
   • Stability of the haloalkane: While more stable carbocations (tertiary) favor , stability alone does not determine the mechanism without considering solvent effects.
4. Conclusion: The most influential factor determining whether a haloalkane undergoes an or reaction is the solvent used in the reaction, as it directly influences the mechanisms described above.

The question requires us to determine which pair of haloalkanes, when used in the Wurtz reaction with metallic sodium and dry ether, produces 2-methylpropane. The Wurtz reaction involves coupling two alkyl halides in the presence of sodium metal to form a higher alkane.

2-Methylpropane has the molecular formula of . To achieve this via the Wurtz reaction, the two haloalkanes chosen must form the correct structure upon coupling. Let's analyze the options:

1. Chloromethane and 2-chloropropane:
   - Chloromethane and 2-chloropropane couple to form 2-methylpropane .
2. Chloroethane and chloromethane:
   - Chloroethane and chloromethane will not correctly form 2-methylpropane.
3. Chloroethane and 1-chloropropane:
   - This option will yield alkanes different from 2-methylpropane.
4. Chloromethane and 1-chloropropane:
   - Combination of these would form isomers but not specifically 2-methylpropane.

The correct pair is therefore Chloromethane and 2-chloropropane. When these are reacted with sodium in dry ether, they successfully couple to form 2-methylpropane due to the favorable position of the functional groups facilitating this particular isomer formation.

Thus, the correct answer is:

• Chloromethane and 2-chloropropane

Given the sequence of reactions, we need to identify the reagents 'P', 'Q', and 'S'. The reaction sequence is:

1. The first step involves converting an alkene to a dibromo compound. This reaction is typically performed using bromine in the presence of an inert solvent. Hence, P is .
2. The second step involves dehydrohalogenation to form a terminal alkyne. This requires a strong base. Thus, Q involves , which is used for elimination to form an alkyne from a dibromo compound.
3. The alkyne reacts further to form the aromatic compound benzene. This step (S) generally requires a strong base followed by a red-hot iron tube for the conversion of alkynes to benzene; hence, S is and a red-hot iron tube.

Therefore, the correct sequence of reagents is: Br₂, Alc, KOH, NaNH₂, Red hot iron tube.

This matches with the correct answer choice provided: Br₂, Alc, KOH, NaNH₂, Red hot iron tube.