Haloalkanes And Haloarenes

Step 1: Understanding the reaction.
Alkyl halides can react with to form ethers. The reaction typically occurs when the alkyl halides react with moist under heat, facilitating the formation of the ether. Step 2: Analysis of options.
(A) Dry : Dry silver oxide reacts with alkyl halides, but it does not typically lead to ether formation.
(B) Moist : Moist silver oxide can react with alkyl halides to form ethers in some cases, but moist is more commonly used.
(C) Dry : Dry zinc oxide does not typically participate in ether formation.
(D) Moist : Correct — Moist reacts with alkyl halides to form ethers.
Step 3: Conclusion.
The correct answer is (D) because moist is used to form ethers from alkyl halides.

Step 1: Understand the reaction.
To prepare ethane (C H ) from methyl iodide (CH I), we need to use a nucleophilic substitution reaction. This can be done by using a strong nucleophile, such as a cyanide ion (CN ) or an alkoxide ion (RO ), in a reaction called the Wurtz reaction.
Step 2: Wurtz Reaction.
The Wurtz reaction involves two molecules of methyl iodide reacting in the presence of sodium metal (Na). Sodium provides electrons, which initiate the formation of a new carbon-carbon bond, resulting in the formation of ethane. The reaction proceeds as follows:
Step 3: Mechanism.
- Sodium (Na) donates electrons to methyl iodide (CH I), resulting in the formation of two methyl radicals (CH •). - These methyl radicals then combine to form ethane (C H ), while iodine (I ) is displaced.
Final Answer: Ethane can be prepared from methyl iodide through the Wurtz reaction by using sodium metal.
Final Answer:

Step 1: Understand the reaction.
To prepare iodoform (CHI ) from acetylene (C H ), we use the reaction between acetylene and iodine in the presence of a base. The process involves the halogenation of the acetylene molecule.
Step 2: Reaction mechanism.
The reaction is a halogenation reaction, where acetylene reacts with iodine (I ) in the presence of a base, such as sodium hydroxide (NaOH), leading to the formation of iodoform. The steps are: 1. Acetylene ( ) reacts with iodine to form a diiodo compound. 2. Further halogenation leads to the formation of triiodomethane (iodoform, CHI ). The reaction is as follows:
Step 3: Explanation.
In this reaction, the iodine atoms replace the hydrogen atoms on the acetylene molecule, leading to the formation of iodoform.
Final Answer: Iodoform can be prepared from acetylene by halogenating acetylene with iodine in the presence of sodium hydroxide.
Final Answer:

Step 1: Understand the reaction.
When n-butyl chloride (C H Cl) reacts with alcoholic potassium hydroxide (KOH), a dehydrohalogenation reaction occurs. In this reaction, the hydroxyl group from KOH removes a hydrogen atom from the carbon adjacent to the carbon bonded to the chlorine atom, resulting in the elimination of hydrogen chloride (HCl).
Step 2: Mechanism of the reaction.
Alcoholic KOH induces an elimination (E2) reaction, where the chlorine atom is replaced by a double bond, forming butene. The reaction can proceed as follows: In this case, n-butyl chloride undergoes elimination to form butene, which is an alkene.
Step 3: Conclusion.
Thus, when n-butyl chloride reacts with alcoholic KOH, it undergoes an elimination reaction to form butene and hydrogen chloride.
Final Answer: When n-butyl chloride reacts with alcoholic KOH, butene is formed through an elimination reaction, releasing HCl.
Final Answer:

Step 1: Define chloroform.
Chloroform is a chemical compound with the formula . It is a type of halogenated hydrocarbon where three chlorine atoms are bonded to a single carbon atom, along with one hydrogen atom.
Step 2: Understand the classification of chloroform.
Since chloroform contains three chlorine atoms attached to a carbon, it is classified as a trihalogen derivative. A trihalogen derivative contains three halogen atoms attached to a carbon atom.
Step 3: Eliminate the incorrect options.
- Option (A) Primary halide: Incorrect, as chloroform is not a primary halide (it has three halogen atoms). - Option (B) Tertiary halide: Incorrect, a tertiary halide has three alkyl groups attached to the carbon atom, which is not the case with chloroform. - Option (C) Trihalogen derivative: Correct, chloroform is a trihalogen derivative because it contains three chlorine atoms attached to one carbon. - Option (D) Tetrahalogen derivative: Incorrect, chloroform has three halogen atoms, not four.
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