Organic Chemistry

Grignard Reaction with Nitriles: Ketone Formation

This transformation involves the nucleophilic addition of a Grignard reagent to a nitrile group, followed by acidic hydrolysis to yield a ketone.

1. Nucleophilic Addition of Grignard Reagent

R-C≡N + CH₃MgBr → R-C(-N=MgBr)(CH₃)

• The nucleophilic methyl group from CH₃MgBr attacks the electrophilic carbon of the nitrile triple bond
• The magnesium bromide coordinates with the nitrogen lone pair
• Forms an imine anion intermediate (ketimine salt)

2. Acidic Hydrolysis

R-C(-N=MgBr)(CH₃) + H₃O⁺ → R-CO-CH₃ + NH₄⁺ + MgBrOH

• Protonation of the nitrogen initiates hydrolysis
• Water adds to the imine carbon, forming an unstable amino alcohol
• Elimination of ammonia yields the final ketone product

Important Characteristics:

• Nitrile Reactivity: The C≡N triple bond is highly polarized, making the carbon electrophilic
• Stoichiometry: 1:1 ratio of Grignard reagent to nitrile is required
• Product Structure: The new ketone has one additional carbon (from the Grignard) compared to the starting nitrile
• Limitation: Cannot be used to prepare ketones with tertiary alkyl groups adjacent to carbonyl

Practical Considerations:

1. Reaction must be performed under anhydrous conditions initially
2. Hydrolysis is typically done with dilute acid at 0°C to room temperature
3. Common solvents: dry ether or THF

Final Product: The reaction converts the nitrile (R-C≡N) to a methyl ketone (R-CO-CH₃)

Therefore, the correct answer is D (a ketone with structure R-CO-CH₃).

Gabriel's synthesis is a chemical reaction used to synthesize primary amines ( amines). This process typically involves the following steps:

1. Phthalimide is treated with a base to form the phthalimide anion.

2. The phthalimide anion then undergoes nucleophilic substitution with an alkyl halide. This step introduces the alkyl group to the nitrogen atom.

3. Hydrolysis (with acid or base) of the alkylated phthalimide leads to the cleavage of the phthalimide group, resulting in the formation of a primary amine.

Gabriel's synthesis is specifically designed to prepare primary amines, avoiding the formation of secondary and tertiary amines, which can occur in other synthetic methods. Therefore, among the given options, Gabriel's synthesis is used for the preparation of:

• Primary amines ( amines)

Acetophenone Aldol Condensation Mechanism

1. Enolate Formation (Initiation)

When acetophenone (C₆H₅COCH₃) reacts with sodium ethoxide (C₂H₅ONa):

C₆H₅COCH₃ + C₂H₅O^- → C₆H₅COCH₂^- + C₂H₅OH

The base abstracts an acidic α-hydrogen (pK_a ≈ 20), forming a resonance-stabilized enolate anion.

2. Nucleophilic Addition (Aldol Formation)

The enolate attacks the carbonyl carbon of another acetophenone molecule:

C₆H₅COCH₂^- + C₆H₅COCH₃ → C₆H₅COCH₂CH(OH)C₆H₅

This forms a β-hydroxy ketone (aldol) intermediate.

3. Dehydration (Elimination)

Under basic conditions, the aldol undergoes elimination:

C₆H₅COCH₂CH(OH)C₆H₅ → C₆H₅CH=CHCOC₆H₅ + H₂O

Water is eliminated to form the α,β-unsaturated ketone (chalcone derivative).

Key Observations:

• The reaction follows standard aldol condensation mechanism
• Product stabilization occurs through conjugation (C=C-C=O)
• Sodium ethoxide serves both as base and nucleophile source
• The final product is more stable than the aldol intermediate

Final Product: The stable α,β-unsaturated ketone (1,3-diphenylprop-2-en-1-one)

Therefore, the correct answer is A.

This reaction describes a two-step transformation where an alkene is converted to a cyclic compound through hydroxyl addition and subsequent ring closure.

Reaction Mechanism:

1. Hydroxyl Addition:
The hydroxyl group ( ) adds to the most substituted carbon of the alkene double bond, following Markovnikov's rule. This regioselectivity occurs because:
- The more substituted carbon stabilizes the partial positive charge in the transition state
- The resulting carbocation intermediate is more stable at the more substituted position

2. Ring Formation:
The added hydroxyl group then participates in an intramolecular reaction:
- The oxygen's lone pair attacks an electrophilic carbon center in the same molecule
- This nucleophilic substitution results in water elimination ( )
- Simultaneous formation of a new carbon-oxygen bond creates the cyclic structure

Key Features:
- Regioselectivity: OH always adds to the more substituted carbon
- Stereochemistry: The reaction typically proceeds with anti-addition
- Driving Force: Ring formation provides thermodynamic stability

Final Product:
The transformation yields a cyclic ether (if forming C-O bond) or alcohol (if forming C-C bond), with the ring size determined by the original molecule's structure.

• D-glucose: At C-1 and C-2, D-glucose has the configuration HO-CH₂-CHOH-CHOH-.
• D-fructose: As a ketose, it shares the C-2 configuration with D-glucose after rearranging. It has HO-CH₂-CO-CHOH-.
• D-mannose: It has a C-2 epimer of D-glucose with the same configuration: HO-CH₂-CHOH-CHOH-.

To analyze the decarboxylation behavior of the given compounds, we'll examine both gem-dicarboxylic acids and β-keto acids.

1. Gem-Dicarboxylic Acids:
These are dicarboxylic acids where both carboxyl (-COOH) groups are attached to the same carbon atom (geminal position). The general decarboxylation reaction is:
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Product: Acetone (a ketone) and carbon dioxide gas.

4. Conclusion:
Both types of compounds undergo facile decarboxylation when heated:
- Gem-dicarboxylic acids lose one CO₂ to form monocarboxylic acids
- β-Keto acids lose CO₂ to form ketones

Final Answer:
The β-keto acid shown will decarboxylate upon heating to yield a ketone (acetone) and carbon dioxide gas.

The given compound is , which represents an amine group. The reaction described involves the formation of a compound with a foul smell upon treatment with chloroform ( ) in the presence of alcoholic KOH, known as the Hofmann degradation reaction.
This reaction typically involves the removal of one carbon atom from the amine chain, leading to the formation of an amine with one less carbon atom.
The Hofmann reaction usually occurs with primary amines, where a foul-smelling product such as an amine with fewer carbon atoms (like a lower alkyl amine) is formed after the removal of one carbon from the parent chain.
Now, let’s evaluate the given options:
- (A) 2-Methylpropan-1-amine: This compound has a branched structure and would likely produce a different set of products, thus it's not the correct answer.
- (B) Butan-1-amine: This is a primary amine, but it will undergo the Hofmann reaction to produce ethylamine. However, it is not optically active, so it cannot be the correct answer.
- (C) Butan-2-amine: This compound is a secondary amine and will show the reaction with chloroform and alcoholic KOH, resulting in a foul-smelling product. This compound is also optically active due to the presence of a chiral center at carbon 2.
- (D) N-Methylpropan-1-amine: This is a substituted amine and will not undergo the Hofmann degradation reaction.
Thus, the correct answer is (C) Butan-2-amine, as it is the optically active isomer and undergoes the described reaction.

When compounds are reacted with chloroethane and anhydrous AlCl₃, they undergo Friedel-Crafts alkylation or acylation reactions. The reactivity in these reactions is primarily influenced by the electronic nature of the substituents on the benzene ring. The reactivity of the compounds follows this pattern:
- Electron-donating groups (such as CH₃) increase the electron density on the benzene ring, making it more reactive in electrophilic aromatic substitution reactions (such as Friedel-Crafts reactions). Hence, Benzene with CH₃ (compound [B]) will be highly reactive.
- Electron-withdrawing groups (such as NO₂) decrease the electron density on the benzene ring, making it less reactive. Hence, Benzene with NO₂ (compound [D]) will be less reactive. - Br is a halogen, and it has a slightly electron-withdrawing effect through induction but also has a lone pair that can participate in resonance donation, making it slightly less deactivating than NO₂.

Thus, Benzene with Br (compound [A]) will have medium reactivity. - NO₂ is a strongly electron-withdrawing group, significantly reducing the electron density on the benzene ring, making the compound C (Benzene with NO₂) much less reactive compared to other compounds. Thus, the correct order of increasing reactivity is: C

The given Assertion and Reason can be analyzed as follows:

1. Assertion: - This statement is correct. In the Williamson ether synthesis, an alkoxide ion (such as ) reacts with an alkyl halide (such as ) to form an ether, in this case, n-propyl tert-butyl ether.

2. Reason: The reaction occurs by attack of Primary alkoxide on Tert-alkyl halide to give a good yield of the product, - This statement is incorrect. The reaction mechanism for the Williamson synthesis generally follows an mechanism, not . In an mechanism, the nucleophile (alkoxide) directly displaces the leaving group (the halide) in a single step. The mechanism, on the other hand, involves the formation of a carbocation intermediate, which is not favored in this case because of the primary nature of the alkyl halide.

Thus, the Assertion is correct, but the Reason is incorrect. Therefore, the correct option is (C).

The given reactions are:
1. Reaction with / Sunlight: Toluene ( ) undergoes a free radical substitution reaction with chlorine in the presence of sunlight. This results in the formation of chlorotoluene (where a hydrogen atom is replaced by chlorine in the methyl group, forming a benzyl chloride). The methyl group ( ) gets chlorinated to form a benzyl chloride intermediate ( ).
2. Reaction with : In this step, the benzyl chloride undergoes a hydrolysis reaction, resulting in the formation of the corresponding alcohol. The benzyl chloride is converted to a benzyl alcohol.
3. Reaction with Acetophenone / at 293 K: In the final step, the product reacts with acetophenone in the presence of a base ( ) at 293 K. This leads to the formation of a conjugated enone structure, where the product is 1, 3-Diphenylprop-2-en-1-one.
Thus, the final product is 1, 3-Diphenylprop-2-en-1-one. Hence, the correct answer is .

Let's analyze the statements:
1. (A) Phenol has lower value than -cresol: This statement is true because -cresol is a weaker acid compared to phenol.
The methyl group in -cresol is an electron-donating group, which increases the electron density on the aromatic ring, making phenol less acidic than -cresol.
2. (B) 2-Chlorophenol is more acidic than phenol: This statement is true. The chlorine atom in 2-chlorophenol is an electron-withdrawing group that stabilizes the negative charge on the oxygen atom after deprotonation, thus making 2-chlorophenol more acidic than phenol.
3. (C) Ortho and para nitrophenols can be separated by steam distillation since -Nitrophenol is more steam volatile than -Nitrophenol: This statement is incorrect. -Nitrophenol has lower steam volatility compared to -Nitrophenol due to the stronger intermolecular hydrogen bonding in -nitrophenol, which makes it less volatile than -nitrophenol. Therefore, they cannot be separated by steam distillation based on their volatility.
4. (D) Phenol on reaction with yields a conjugated diketone: This statement is true. The reaction of phenol with results in oxidation of phenol, and the product formed is a conjugated diketone, which is a product of the oxidation of phenol.
Thus, the incorrect statement is (C), and the correct answer is .

The molecular formula of the hydrocarbon is , which indicates that it is propene. Let's analyze the reactions step by step:
- Step 1: (C₃H₆) reacts with , a halogenation reaction that leads to the formation of 1,2-dibromopropane. Therefore, compound is 1,2-dibromopropane.
- Step 2: When (1,2-dibromopropane) is heated with alcoholic KOH, a dehydrohalogenation occurs, which leads to the formation of propene. Hence, compound is propene.
- Step 3: When 3 moles of (propene) are passed through red-hot iron tubes, a benzene ring is formed through dehydrogenation. In this process, benzene is formed by the elimination of hydrogen atoms from propene.

Now, the reaction proceeds to form a specific substituted benzene. The correct substitution pattern on the benzene ring is 1, 3, 5-trimethylbenzene, where the methyl groups are located at the 1st, 3rd, and 5th positions of the benzene ring.

Thus, the correct compound is 1, 3, 5-Trimethylbenzene.