Organic Chemistry

To determine the acidity order, we need to consider the effects of the hydroxyl groups on the acidity of the benzoic acid.

Step 1: Draw the Structures

Let's draw the structures of the three compounds:

• Salicylic acid (2-hydroxybenzoic acid): The hydroxyl group is in the ortho position.
• 4-hydroxybenzoic acid: The hydroxyl group is in the para position.
• 2,6-dihydroxybenzoic acid: There are two hydroxyl groups in the ortho positions.

Step 2: Factors Affecting Acidity

Several factors influence the acidity of organic acids:

• Inductive effect: Electron-withdrawing groups increase acidity, while electron-donating groups decrease acidity.
• Resonance effect: Groups that stabilize the conjugate base increase acidity.
• Ortho effect: Substituents in the ortho position can have a significant effect on acidity due to steric and electronic interactions. This is complex and not predictable just by looking at inductive and resonance effect.
• Intramolecular hydrogen bonding: Can stabilize the conjugate base (increase acidity) or the acid itself (decrease acidity).

Step 3: Acidity Analysis of the Given Compounds

1. 4-hydroxybenzoic acid:

The hydroxyl group in the para position has a weaker effect on acidity because it is farther from the carboxyl group. The -OH group exhibits both +R (resonance) and -I (inductive) effects. Usually, the +R effects counteract each other, and the -I effect slightly decreases the acidity of the compound relative to benzoic acid.

1. Salicylic acid (2-hydroxybenzoic acid):

The hydroxyl group in the ortho position has a stronger effect than in the para position. The ortho effect and intramolecular hydrogen bonding influence acidity. Salicylic acid benefits from intramolecular hydrogen bonding which stabilises the conjugate base, increasing its acidity.

1. 2,6-dihydroxybenzoic acid:

The presence of two hydroxyl groups in the ortho positions further increases the acidity. The effect of one hydroxyl group in ortho is stabilising the conjugate base through hydrogen bonding. With two OH groups, more hydrogen bonding is possible, stabilising the anion more than in salicylic acid.

Step 4: Putting It All Together

Based on the analysis, we can conclude the following acidity order:

2, 6-dihydroxybenzoic acid > salicylic acid > 4- hydroxybenzoic acid

Conclusion

The correct order of acidity is: 2, 6-dihydroxybenzoic acid > salicylic acid > 4- hydroxybenzoic acid.

To determine the correct enol form of ethyl-3-oxobutanoate, we analyze the compound's tendency to undergo keto-enol tautomerism.

Step-by-step Analysis:

1. Structure of Ethyl-3-oxobutanoate: This is a β-keto ester with a ketone at the 3-position and an ester at the 1-position. Such compounds readily exhibit keto-enol tautomerism.

2. Enol Formation: The hydrogen at the alpha carbon (between the ester and ketone groups) is acidic and can be removed to form a double bond with the carbon adjacent to the ketone. The carbonyl oxygen of the ketone becomes an OH group.

3. Stability of Enol: The most stable enol is typically the one where the OH group is hydrogen bonded to the ester oxygen, forming a six-membered ring. Additionally, conjugation between the C=C bond and the ester group enhances stability.

4. Evaluating the Options: Among the given images, the correct enol form is the one where:
- A C=C bond is formed between the alpha and carbonyl carbon
- The ketone oxygen is converted into an OH group
- Intramolecular hydrogen bonding and conjugation are possible

Conclusion: The correct enol form is shown in the third image.

To determine the relative stability of the given conformers of n-butane, we analyze the Newman projections about the central C–C bond.

Conformer II: This is the anti-conformation where the two methyl groups (CH₃) are 180° apart. It is the most stable due to minimal steric hindrance.

Conformer I and III: Both are gauche conformations, where the methyl groups are 60° apart. These are less stable than anti due to gauche interactions (steric repulsion between bulky groups), but they are equally stable with respect to each other.

Hence, the correct order of relative stability is:

II > I = III

The reaction is the addition of HBr to an alkene:



This follows the electrophilic addition mechanism. Since there is no peroxide present, the addition follows Markovnikov’s rule.

According to Markovnikov’s rule, the hydrogen (H) adds to the carbon of the double bond that already has more hydrogens, and the bromine (Br) adds to the other carbon.

In the given compound:

- The terminal carbon ( ) gets the H
- The middle carbon (adjacent to ) gets the Br

However, the group is strongly electron-withdrawing, making the formation of a carbocation at the adjacent carbon (near CF₃) highly unstable. Therefore, instead of a Markovnikov product, the reaction favors anti-Markovnikov addition due to the destabilization of the carbocation by the electron-withdrawing group.

Thus, the H⁺ adds to the middle carbon and Br⁻ adds to the terminal one, forming:


Answer: F₃C–CH₂–CH₂Br

The correct order of stability is II > I > III.

Here's why:

• Free Radical Stability: Free radical stability is primarily determined by the degree of substitution. More substituted free radicals are more stable. This is due to hyperconjugation (overlap of the singly occupied p-orbital with sigma bonds on adjacent carbon atoms) and inductive effects (electron donation from alkyl groups).
• Radical I: Radical I is a secondary allylic free radical. The allylic stabilization from the adjacent double bond makes it more stable than simple secondary radicals.
• Radical II: Radical II is a tertiary free radical. It is more stable than secondary or primary radicals due to hyperconjugation. Since it's also benzylic, with the unpaired electron delocalized across the benzene ring, it has greater stability. Benzyl radicals are more stable than allyl radicals. Hence this is most stable.
• Radical III: Radical III is a primary free radical. The primary radicals are the least stable because there are fewer alkyl groups that can stabilize the carbon atom with the unpaired electron through hyperconjugation or inductive effects.

Because radical II benefits from tertiary and benzylic stabilization, is most stable; radical I is only stabilized through allylic and secondary carbons, and radical III is only stabilized through a primary carbon, it follows that the order of stability is II > I > III.

Correct Answer: II > I > III

The correct product is Option 3.

Here's why:

• Hydrogenation: Hydrogenation is the addition of hydrogen (H₂) to a molecule, typically in the presence of a metal catalyst (like Pt, Pd, or Ni). It's commonly used to reduce multiple bonds (alkenes and alkynes) to single bonds (alkanes).
• Catalytic Hydrogenation and Stereochemistry: In catalytic hydrogenation with a metal catalyst, hydrogen adds to the *same face* of the double bond (syn addition). This is because both hydrogen atoms are adsorbed onto the surface of the metal catalyst, and the alkene approaches the catalyst surface from one side.
• Analyzing the Reactant and Products: The reactant contains a double bond in a cyclic system. Option 3 shows the addition of two hydrogen atoms on the *same side* of the ring (syn addition). Therefore, only **Option 3** corresponds to this stereochemical outcome. Other products show anti addition that doesn't match the process.

Therefore, the product of the hydrogenation reaction is the isomer in which both hydrogen atoms have been added to the *same face* of the original double bond, corresponding to the structure in Option 3.

Correct Answer: Option 3

• The Reaction: The reaction is an intramolecular aldol condensation. The starting material is a dialdehyde. Under basic conditions, one aldehyde can act as a nucleophile and attack the other aldehyde in the same molecule, forming a cyclic aldol product. Heat then causes dehydration, leading to the formation of a cyclic , -unsaturated aldehyde.
• Mechanism: The base removes a proton from an -carbon of one of the aldehyde groups, creating an enolate. This enolate then attacks the other aldehyde group intramolecularly. After protonation, this is followed by dehydration (loss of water) to form a stable , -unsaturated carbonyl compound.
• Product Formation: The specific structure of the starting material dictates the ring size of the cyclic product. In this case, a six-membered ring is formed, with the double bond in conjugation with the carbonyl group.

Therefore, the aldol condensation reaction leads to the formation of the cyclic enone product shown in Option 1.

Correct Answer: Option 1

• Grignard Reaction: Methyl magnesium iodide (CH₃MgI) is a Grignard reagent, which is a strong nucleophile. It reacts with carbonyl compounds (aldehydes and ketones) to form alcohols.
• Chirality: A molecule is chiral if it is non-superimposable on its mirror image (i.e., it has a stereocenter with four different substituents).
• Benzaldehyde (C₆H₅CHO): Reaction with CH₃MgI followed by protonation gives 1-phenylethanol (C₆H₅CH(OH)CH₃). The carbon atom bonded to the OH group is attached to a phenyl group, a methyl group, a hydrogen atom, and an OH group. These are four different groups, so it is a chiral center, creating a pair of enantiomers.
• Propiophenone (C₆H₅C(=O)CH₂CH₃): Reaction with CH₃MgI followed by protonation gives 2-phenylbutan-2-ol (C₆H₅C(OH)(CH₃)CH₂CH₃). The carbon atom bonded to the OH group is attached to a phenyl group, a methyl group, an ethyl group, and an OH group. These are four different groups, so it is a chiral center, creating a pair of enantiomers.
• Acetone (CH₃C(=O)CH₃): Reaction with CH₃MgI followed by protonation gives 2-methylpropan-2-ol ((CH₃)₃COH). The carbon bonded to the OH group is attached to three methyl groups and an OH group. It has two or more identical groups attached to it, so it is *not* chiral.
• Acetaldehyde (CH₃CHO): Reaction with CH₃MgI followed by protonation gives propan-2-ol (CH₃CH(OH)CH₃). The carbon bonded to the OH group is attached to two methyl groups, a hydrogen, and an OH group. It has two or more identical groups attached to it, so it is *not* chiral.

Therefore, only benzaldehyde and propiophenone produce chiral alcohols (and thus enantiomers) upon reaction with methyl magnesium iodide.

Correct Answer: Benzaldehyde, Propiophenone

Stability Order of the Given Carbocations:

The stability of a carbocation depends on:

• Resonance effects: Delocalization of the positive charge increases stability.
• Inductive effects: Electron-donating groups stabilize the carbocation, while electron-withdrawing groups destabilize it.

Given species:

1. H₂C⁺–CH=CH–CH₃: Allylic carbocation, resonance stabilized.
2. ⁺CH₂–CH=CH–BMe₂: Allylic, but boron is an electron-withdrawing group → destabilizes.
3. H₂C⁺–CH=CH–NMe₂: Allylic, NMe₂ is an electron-donating group via resonance → highly stabilized.
4. H₂C⁺–CH=CH–OMe: Allylic, OMe is also electron-donating → stabilizes, but not as strongly as NMe₂.

Stability order: III > IV > I > II

Correct option: (C)