Organic Reactions

Step 1: Gabriel Phthalimide Synthesis Overview.
Gabriel phthalimide synthesis is a method used to prepare primary amines by reacting phthalimide with an alkyl halide in the presence of a strong base, followed by hydrolysis. This synthesis is typically effective for the preparation of primary amines, but there are limitations.

Step 2: Analysis of Options.
- (A) n-Butylamine: n-Butylamine can be easily synthesized using Gabriel phthalimide synthesis as the alkylation step proceeds with primary alkyl halides.
- (B) Alanine: Alanine contains both an amino group and a carboxyl group, which complicates its synthesis via Gabriel phthalimide, as it involves a secondary amino group. Thus, alanine cannot be synthesized via this method.
- (C) Aniline: Aniline is an aromatic amine, and Gabriel phthalimide is not effective for preparing aromatic amines like aniline. Hence, aniline cannot be prepared by this method.
- (D) t-Butylamine: t-Butylamine involves a bulky tertiary-butyl group, which also prevents the alkylation step from proceeding effectively in the Gabriel synthesis. Thus, t-butylamine cannot be synthesized by this method.

Step 3: Conclusion.
Therefore, the correct answer is (4) (B), (C) and (D) as these compounds cannot be prepared by Gabriel phthalimide synthesis.

Final Answer:

Step 1: Borane Reduction.
The reaction starts with diborane ( ) and hydrogen peroxide ( ) in the presence of NaOH, which reduces the ethoxy group to a hydroxyl group. This results in the formation of a diol at positions 1 and 6 of the hexane chain.

Step 2: Hydrolysis of the Organoborane.
The organoborane intermediate is hydrolyzed to form the desired diol product.

Step 3: Conclusion.
The product formed is Hexane-1,6-diol, making option (1) the correct answer.

Final Answer:

Step 1: Kolbe Electrolysis.
In Kolbe's electrolysis of an ester like ethyl propionate, two molecules of the ester undergo decarboxylation at the anode, leading to the formation of alkanes. The reaction does not form ethylene as ethylene is an unsaturated compound, and Kolbe's electrolysis typically forms saturated alkanes.

Step 2: Formation of Alkanes.
During the electrolysis, products such as n-butane, n-propane, and ethane can form due to the decarboxylation of the ester, but ethylene does not appear.

Step 3: Conclusion.
Hence, ethylene is not formed during Kolbe's electrolysis of ethyl propionate. The correct answer is (3) Ethylene.

Final Answer:

Step 1: Oxidation with V O .
Vanadium pentoxide (V O ) and air oxidize the starting compound, initiating the process by forming a benzyl alcohol derivative.

Step 2: Bromination.
Bromination with Br and KOH proceeds to form a brominated intermediate that undergoes dehydrobromination.

Step 3: Final Step.
The final step involves heating with sodium hydroxide (NaOH) to induce a coupling reaction, leading to the formation of naphthalene.

Step 4: Conclusion.
Thus, the final product (D) is naphthalene, which matches option (1).

Final Answer:

Step 1: Friedlander's Synthesis.
Friedlander's synthesis involves the reaction of o-aminobenzaldehyde with a suitable electrophilic reagent. Hence, it matches with (I) o-Aminobenzaldehyde.

Step 2: Doebner-Miller's Synthesis.
Doebner-Miller's synthesis involves the reaction of -phenylethylamide with a suitable electrophile, so it matches with (II) -Phenylethylamide.

Step 3: Hantzsch's Synthesis.
Hantzsch's synthesis involves the reaction of aniline with a carbonyl compound, so it matches with (III) Aniline.

Step 4: Bischler-Napieralski's Synthesis.
Bischler-Napieralski's synthesis involves the reaction of a -ketoneester with an amine, so it matches with (IV) -Ketoneester.

Step 5: Conclusion.
Thus, the correct matching is (A) - (I), (B) - (II), (C) - (III), (D) - (IV), which corresponds to option (1).

Final Answer:

Step 1: Cyclic Polymerization.
The starting material undergoes a cyclic polymerization reaction. This type of reaction typically forms a cyclic structure that is a precursor for further transformations.

Step 2: Gatterman-Koch's Reaction.
This reaction involves the formylation of an aromatic compound. In this case, it adds a formyl group to the aromatic compound, which is a critical intermediate step in the transformation.

Step 3: Pomeranz-Fritsch's Reaction.
The Pomeranz-Fritsch's reaction leads to the formation of isquinoline from the reaction intermediate formed in step 2.

Step 4: Conclusion.
Thus, the correct sequence involves cyclic polymerization, Gatterman-Koch's reaction, and the Pomeranz-Fritsch's reaction. Hence, option (4) is correct.

Final Answer:

Step 1: Base Catalysis.
Both the Benzil-Benzylic acid rearrangement and Cannizzaro's reaction are base-catalyzed reactions. In the Benzil-Benzylic acid rearrangement, the base is required to deprotonate the intermediate. Similarly, in Cannizzaro's reaction, hydroxide ions facilitate the transfer of hydride.

Step 2: Anion Shifting.
Both reactions involve the shifting of an anion in their mechanisms. In the Benzil-Benzylic acid rearrangement, an anion is formed during the formation of the intermediate, and in the Cannizzaro reaction, an anion of the aldehyde is involved in the hydride transfer.

Step 3: Inter and Intra-molecular Reactions.
Both reactions can occur inter-molecularly and intra-molecularly. However, this is not specifically true for both reactions and doesn't apply to all cases. Thus, (C) is not necessarily true for both reactions.

Step 4: Redox Reactions.
Both reactions involve redox changes. In the Cannizzaro reaction, one molecule undergoes reduction while another undergoes oxidation. The Benzil-Benzylic acid rearrangement also involves a redox process.

Step 5: Conclusion.
Thus, the correct answer is (2) (A), (B) and (D) only.

Final Answer:

Step 1: Hydrolysis Rates.
The rate of hydrolysis generally increases as the electrophilicity of the carbonyl carbon increases. Acetyl chloride (B) is the most reactive due to the presence of a highly electrophilic chlorine atom. Next in reactivity is acetic anhydride (D), followed by ethyl acetate (C) and acetamide (A). Acetamide is the least reactive as it involves a more stable amide group.

Step 2: Conclusion.
Thus, the correct order is (A), (C), (D), (B). Hence, the correct answer is (1).

Final Answer:

Step 1: Acid Chloride Formation.
The first step of the reaction is the conversion of 2-pentanone into an acyl chloride using HCl and EISH. This forms a reactive acyl chloride intermediate.

Step 2: Hydrogenation.
Next, the acyl chloride undergoes reduction in the presence of nickel (Ni) and ethanol (EtOH), which leads to the formation of a thiol group (–SH) in the final product.

Step 3: Conclusion.
The final major product is propane-2-thiol, which corresponds to option (4).

Final Answer:

Step 1: Rosenmund's Reduction.
The reduction of benzoyl chloride using hydrogen ( ) over palladium on barium sulfate ( ) is called Rosenmund's reduction. This reaction selectively reduces the acyl group to an aldehyde, yielding benzaldehyde.

Step 2: Claisen-Schmidt's Reaction.
The reaction of benzaldehyde with a base (like dilute NaOH) leads to the Claisen-Schmidt condensation, where it reacts with acetaldehyde to form cinnamaldehyde.

Step 3: Conclusion.
Thus, the correct sequence involves Rosenmund's reduction followed by the Claisen-Schmidt reaction.

Final Answer:

Step 1: KOH (alc.).
Potassium hydroxide in alcohol (KOH (alc.)) induces elimination of the bromine atom, producing an alkene.

Step 2: Oxymurcuration-demercuration.
The alkene then undergoes oxymurcuration-demercuration to form an alcohol at the position where the bromine was originally attached. Hence, option (1) is correct.

Final Answer: