Haloalkanes And Haloarenes

The Wurtz-Fittig reaction is a coupling reaction that involves the condensation of different types of halides. Let's go through the options and determine which one correctly describes this reaction:

1. Two molecules of aryl halides: The Wurtz reaction typically involves coupling two alkyl halides, not aryl halides. Therefore, this option is incorrect for the Wurtz-Fittig reaction, which specifically includes both an aryl and an alkyl component.
2. One molecule of each of aryl-halide and alkyl-halide: This option describes the Wurtz-Fittig reaction correctly. In this reaction, an aryl halide and an alkyl halide are condensed in the presence of sodium to form an aryl-alkyl compound.
3. One molecule of each of aryl-halide and phenol: Phenols are not involved in the Wurtz-Fittig reaction, as it only involves halides. Hence, this option is incorrect.
4. Two molecules of aralkyl-halides: This is not the mechanism of the Wurtz-Fittig reaction. The Wurtz reaction, when involving aralkyl halides, differs and does not specifically refer to the Wurtz-Fittig reaction.

In conclusion, the correct answer is one molecule of each of aryl-halide and alkyl-halide. The Wurtz-Fittig reaction takes advantage of the reactivity of both types of halides to form a new carbon-carbon bond in the resulting compound. This reaction is particularly useful in organic synthesis where mixed alkyl-aryl products are needed.

The correct answer is 1

CH₃CH₂-Br + NaI (in acetone) → CH₃-CH₂-I + NaBr

Correct Statement:

NaI is soluble in acetone, but NaBr precipitates in acetone.

Explanation:

• The reaction occurs via the Finkelstein reaction, which is used to prepare alkyl iodides from alkyl bromides or chlorides.
• Acetone is a polar aprotic solvent that helps dissolve NaI and facilitates the reaction.
• When NaI reacts with CH₃CH₂-Br:
  • NaI is soluble in acetone.
  • NaBr, the byproduct, is not soluble in acetone and precipitates out.
• The precipitation of NaBr drives the equilibrium of the reaction toward the formation of the product, CH₃-CH₂-I.

Conclusion:

The correct statement is: NaI is soluble in acetone, but NaBr precipitates in acetone.

Correct answer is (c)
Three compounds meet the criteria for both tests, indicating the presence of specific functional groups reactive to these tests. These compounds contain functional groups that react with ceric ammonium nitrate and the iodoform test:

Ceric ammonium nitrate is used to detect alcohols and phenols, where the reaction forms a color change.

The iodoform test is specific for methyl ketones, where the formation of a yellow precipitate (iodoform) indicates the presence of the functional group.

Thus, the three compounds must contain alcohols, phenols, or methyl ketones to show a positive result for both tests.
 

The correct answer is (C) :

The reaction with is used to test for halide ions. A precipitate of (pale yellow) will form if a bromide ion is available to react with .

Analysis of the Options:

Option 1: Cyclopropyl bromide

• The bond in cyclopropyl bromide is not easily broken.
• The carbocation formed upon the removal of bromide is stabilized by aromatic nature.
• Hence, it does not readily form a precipitate with .

Option 2: Bromobenzene

• The bond in aryl halides is strong due to resonance and the hybridization of carbon.
• This bond does not readily break to release bromide ions.
• The carbocation is not formed as it is highly unstable.
• Thus, bromobenzene does not react with to form a precipitate.

Option 3: Benzylic bromide

• In benzylic bromide, the benzylic carbocation formed after the removal of bromide ion is stabilized by resonance with the benzene ring.
• This makes the bromide ion readily available to react with , forming a precipitate.

Option 4: Allylic bromide

• In allylic bromide, the allylic carbocation formed is stabilized by resonance.
• This allows the bromide ion to react with , forming a precipitate.

Conclusion:

Final Answer: Bromobenzene will not produce a precipitate with .

Step 1: Reaction with and

The alcohol reacts with and to form the corresponding iodoalkane via a nucleophilic substitution reaction. In this process:

• acts as an acid catalyst.
• from acts as the nucleophile, replacing the hydroxyl group with an iodine atom.

The reaction is as follows:

Step 2: Formation of the Grignard Reagent

The iodoalkane reacts with magnesium in dry ether to form a Grignard reagent:

Here:

• Dry ether is used as a solvent because it stabilizes the Grignard reagent and prevents it from reacting with water.

Step 3: Reaction with

The Grignard reagent reacts with deuterated water ( ) to form the deuterated alkane. The bond is replaced by a bond:

Final Answer:

The product formed is .

Freons are chlorofluorocarbons (CFCs) or hydrochlorofluorocarbons (HCFCs), which are hydrocarbons where some or all of the hydrogen atoms have been replaced by chlorine and fluorine atoms.

Key Characteristics of Freons:

• They contain carbon (C), chlorine (Cl), and fluorine (F).
• Freons are widely used as refrigerants and propellants due to their stability and low toxicity.

Analysis of the Given Options:

Among the given options, only (dichlorodifluoroethylene) contains both chlorine and fluorine along with carbon, making it a Freon.

Final Answer:

is an example of a Freon.

1. Lassaigne’s Test: This test detects nitrogen, sulfur, and halogens in organic compounds.
- The compound is fused with sodium, converting nitrogen into NaCN and halogen into NaX ( ).
2. Compound Analysis:
: Contains nitrogen but no halogen.
: Contains both nitrogen (from the group) and halogen ( from ).
: Contains nitrogen and halogen but is not an organic compound.
: Contains nitrogen and halogen but does not form the required products in Lassaigne’s test.
3. Conclusion: Only gives a positive Lassaigne’s test for both nitrogen and halogen.
Final Answer: (2) .

Step 1: Understanding nucleophilicity.
Nucleophilicity refers to the ability of a nucleophile to donate electrons to form a new bond with an electrophile (in this case, CH –Br). The stronger the nucleophile, the faster the nucleophilic substitution reaction will occur.

Step 2: Comparing the nucleophilicity of the given nucleophiles.
- I is the most nucleophilic due to its larger size and lower electronegativity, which makes it more willing to donate electrons.
- C H O (ethoxide) is a good nucleophile, but not as strong as I , because oxygen is more electronegative, making it less willing to donate electrons.
- PhO (phenoxide) is also a strong nucleophile, but the resonance stabilization of the phenoxide ion reduces its nucleophilicity compared to C H O and I .
- F is the least nucleophilic because fluorine is highly electronegative, making it reluctant to donate electrons.

Step 3: Conclusion.
The correct order of nucleophilicity is I >C H O >PhO >F , which corresponds to option (3).