Structural Representations Of Organic Compounds

Answer (c)

Analyze the application of alkyl iodides in various reactions.

The Finkelstein reaction typically produces alkyl iodides. We need to consider where these products are utilized:

• The Wurtz reaction and Wurtz-Fittig reaction both use alkyl halides, including iodides, to form new carbon-carbon bonds.
• The Williamson reaction uses alkyl halides to synthesize ethers and can utilize alkyl iodides.
• The Fittig reaction, however, primarily involves aryl halides, and typically alkyl iodides are not used here.

Step 2: Determine the reaction where alkyl iodides are not typically employed.

While alkyl iodides are versatile in many reactions, they are not a common choice for the Fittig reaction, which focuses on coupling aryl halides, not alkyl halides. Thus, the Fittig reaction does not commonly utilize the products of the Finkelstein reaction.

Step 1: Examine the reaction conditions and requirements.

The conversion of a nitrile to a methyl group involves significant reduction, typically requiring strong reducing agents and conditions that facilitate both reduction and hydrolysis.

Step 2: Analyze the selected option.

SnCl in the presence of HCl (reagent X) is used initially to reduce the nitrile group partially, likely to an imine.

H O (part of reagent Y) then hydrolyzes the imine to an amine.

The final step involves the use of Zn-Hg amalgam in concentrated HCl (additional part of reagent Y), which is a classic Clemmensen reduction, reducing the amine further to a methyl group. This sequence of reactions is particularly effective for fully reducing nitrile to a primary alkyl group without over-reduction or unwanted side reactions.

The given reaction shows the formation of an aromatic compound. - The reaction involves the conversion of n-propyl bromide to an alkene via the reaction with sodium. - The subsequent reaction with chromium oxide ( ) and high temperature leads to a Friedel-Crafts type reaction, converting the product to an aromatic compound. Therefore, the compound Y is aromatic in nature.

1. Reaction of C H with Hg /H O:

C H is propyne (CH C-CH ).
Hydration of alkynes in the presence of Hg /H O leads to the formation of a ketone.
X is propanone (CH COCH ).

2. Reaction of propanone with MeMgBr followed by H O:

Propanone reacts with methyl magnesium bromide (MeMgBr) to form a tertiary alcohol.
Y is 2-methylpropan-2-ol ((CH ) COH).

3. Reaction of C H with (BH ) followed by H O /OH :

C H is propene (CH CH=CH ).
Hydroboration-oxidation of alkenes leads to the formation of an alcohol.
B is propan-1-ol (CH CH CH OH).

4. Dehydration of alcohols:

Tertiary alcohols (Y) are easily dehydrated with mild conditions like 20% H PO /358 K.
Primary alcohols (B) require stronger conditions like Conc. H SO /443 K for dehydration.

Therefore, B is dehydrated with Conc. H SO /443 K and Y with 20% H PO /358 K.

Final Answer:
B is dehydrated with Conc. H SO /443 K and Y with 20% H PO /358 K.

1. Reaction of acetophenone with Br /OH followed by H :

This is a haloform reaction.


A is benzoic acid ( ).

2. Reaction of benzoic acid with NH and heat:

B is benzamide ( ).

3. Reaction of benzamide with Br /NaOH:

This is Hoffmann bromamide degradation.

C is aniline ( ).

4. Reaction of aniline with CHCl /alc. KOH:

This is the carbylamine reaction.

D is benzene isocyanide ( ).

Therefore, D is benzene isocyanide.

Final Answer:
Benzene isocyanide.

1. Reaction of propan-1-ol with X:

Propan-1-ol (CH CH CH OH) is a primary alcohol.
Primary alcohols can be oxidized to aldehydes or carboxylic acids.
CrO /H SO (Jones reagent) is a strong oxidizing agent that will oxidize primary alcohols to carboxylic acids.
PCC (pyridinium chlorochromate) is a mild oxidizing agent that will oxidize primary alcohols to aldehydes.
Since the product Y reacts with benzene in the presence of anhydrous AlCl , it must be an acid chloride.
Therefore, X must be CrO /H SO , which will oxidize propan-1-ol to propanoic acid (CH CH COOH).

2. Reaction of propanoic acid with SOCl :

Propanoic acid reacts with thionyl chloride (SOCl ) to form propanoyl chloride (CH CH COCl).
Y is propanoyl chloride.

3. Reaction of propanoyl chloride with benzene in presence of anhydrous AlCl :

This is a Friedel-Crafts acylation reaction.
Propanoyl chloride reacts with benzene to form propiophenone (C H COCH CH ).
Z is propiophenone.

Therefore, X is CrO /H SO and Z is propiophenone.

Final Answer:
CrO /H SO ; Propiophenone.

Step 1: Analyze the solubility of each compound based on their chemical structure.

is a sodium salt of a sulfonated fatty alcohol, which is typically water-soluble due to the ionic nature of the sulfate group and the sodium ion.

represents a quaternary ammonium salt, which is generally soluble in water due to the ionic nature and the hydrophilic quaternary nitrogen.

is the sodium salt of a fatty acid, and like most soap molecules, it is soluble in water due to the carboxylate group's ionic nature.

, being a calcium salt of a fatty acid, tends to be less soluble in water compared to sodium salts due to the calcium ion's ability to form more stable, insoluble compounds with fatty acids.

Step 2: Conclude which compound is least likely to be soluble in water.

The calcium salt of a fatty acid, , is typically insoluble in water due to its divalent calcium ion, which enhances the lattice energy of the salt, making it less soluble compared to its monovalent counterparts.

Step 1: Determine the reactants and conditions for each transformation. X is initially chlorobenzene.
Y involves the conversion of chlorobenzene using strong base and heat, likely aiming for a substitution reaction that introduces a hydroxyl group, leading to phenol.
Z would then be the final product, phenol, under the given reaction conditions.
Step 2: Confirm the specific conditions and product. The conditions described (NaOH at 623 K and 300 atm) are indicative of a high-pressure, high-temperature environment suitable for converting chlorobenzene to phenol, a process that involves breaking the aromatic ring's stability to introduce the hydroxyl group.

Explanation: - The basic structural unit of silicates is actually the tetrahedron, not . So statement (i) is incorrect.
- Silicones are biocompatible and widely used in medical implants and prosthetics, making statement (ii) correct.
- Producer gas is a mixture of carbon monoxide (CO) and nitrogen (N_2), which is a result of the partial combustion of carbon-based fuels in a limited oxygen supply. Thus, statement (iii) is also correct.
- Since statements (ii) and (iii) are correct, the correct answer is option (B).

Statement I: Benzanamine (aniline) cannot be prepared directly from phthalimide. Phthalimide is typically used in the Gabriel synthesis to prepare primary aliphatic amines, not aromatic amines like benzanamine. Therefore, Statement I is not correct.

Statement II: Benzanamine (aniline) is less basic than phenyl methanamine (benzylamine). This is because the lone pair of electrons on the nitrogen in aniline is delocalized into the benzene ring, reducing its availability for protonation, whereas in benzylamine, the nitrogen's lone pair is more available for protonation. Thus, Statement II is correct.

Final Answer:
Statement I is not correct, but statement II is correct.

1. Identify the reactant:

The reactant is a tertiary alkyl halide, 2-bromo-2-methylpropane.

2. Identify the reaction type:

The reaction involves an alkyl halide reacting with a strong nucleophile (OH ).
Since the reactant is a tertiary alkyl halide, the reaction will proceed via an S 1 mechanism.

3. S 1 mechanism:

In an S 1 reaction, the leaving group (Br ) departs first, forming a carbocation intermediate.
The nucleophile (OH ) then attacks the carbocation.
The product formed will have the OH group attached to the carbon that was bonded to the bromine.

4. Determine the product:

The carbocation formed is a tertiary carbocation.
The OH attacks the carbocation, resulting in the formation of 2-methylpropan-2-ol.

Therefore, X is 2-methylpropan-2-ol.

Final Answer:
2-methylpropan-2-ol.