Organic Chemistry Problems - Detailed Solutions

Problem 1: Novocaine (Procaine) - IUPAC Nomenclature

Given Structure:

The compound contains: benzene ring with NH₂ group (para position), ester linkage (-COO-), ethylene bridge (-CH₂CH₂-), and diethylamino group (-N(C₂H₅)₂)

Step-by-Step IUPAC Naming:

Step 1: Identify the Principal Functional Group

The ester group (-COO-) has the highest priority. The parent structure will be named as an ester.

Ester nomenclature: [Alkyl group from alcohol] [alkanoate from acid]

Step 2: Identify the Acid Component

The acid part is benzoic acid with an amino group at position 4 (para):

Parent: benzoic acid (benzene ring + carboxylic acid)
Substituent: amino group (-NH₂) at carbon 4
Name: 4-aminobenzoic acid (or p-aminobenzoic acid)

Step 3: Identify the Alcohol Component

The alcohol part after removing the ester proton:

Chain: -CH₂CH₂-N(C₂H₅)₂
Two-carbon chain: ethyl
At position 2: diethylamino group -N(CH₂CH₃)₂
Name: 2-(diethylamino)ethyl

Step 4: Combine the Names

Following ester nomenclature rules: [Alcohol part] [Acid part with -oate ending]

Alcohol: 2-(diethylamino)ethyl
Acid (as ester): 4-aminobenzoate

IUPAC NAME: 2-(diethylamino)ethyl 4-aminobenzoate

Molecular Structure:

2D Structure

3D Structure

Problem 2: Ketoprofen - IUPAC Nomenclature

Given Structure:

The compound contains: two benzene rings, ketone group, carboxylic acid group, and methyl substituent

Step-by-Step IUPAC Naming:

Step 1: Identify the Principal Functional Group

The carboxylic acid (-COOH) has the highest priority. This will be the parent functional group.

The compound will be named as a substituted propanoic acid.

Step 2: Identify the Parent Chain

The carbon chain containing the carboxylic acid:

Three-carbon chain with COOH: propanoic acid
Numbering starts from the carboxylic acid carbon (C1)

Step 3: Identify Substituents on the Parent Chain

At carbon 2 (alpha carbon):

A phenyl group substituted with a benzoyl group at position 3
This is: 3-benzoylphenyl group

Step 4: Number and Name Substituents

Breaking down the 3-benzoylphenyl group:

Phenyl ring attached to C2 of propanoic acid
This phenyl ring has a benzoyl group (C₆H₅CO-) at position 3
Benzoyl = benzene ring + carbonyl (C₆H₅CO-)

Step 5: Assemble the Complete Name

Position of substituent: 2-(3-benzoylphenyl)

Parent chain: propanoic acid

IUPAC NAME: 2-(3-benzoylphenyl)propanoic acid

Molecular Structure:

2D Structure

3D Structure

Problem 3: Acetylcholine - IUPAC Nomenclature

Given Structure:

CH₃COO-CH₂-CH₂-N⁺(CH₃)₃ Cl⁻ (quaternary ammonium salt)

Step-by-Step IUPAC Naming:

Step 1: Identify the Principal Functional Group

The compound contains:

Ester group (-COO-)
Quaternary ammonium group (-N⁺(CH₃)₃)

Quaternary ammonium salts have higher priority than esters.

The compound will be named as an ammonium salt.

Step 2: Name the Cation (Ammonium Part)

The nitrogen-containing chain:

Two-carbon chain: ethane
Amino group becomes: ethanaminium (for the cation)
At position 2: acetyloxy group (CH₃COO-)
On nitrogen: three methyl groups (trimethyl)
Position of N: N,N,N-trimethyl

Step 3: Number the Chain

Numbering from the nitrogen end (gives lowest numbers to substituents):

N⁺ is at position 1
Acetyloxy group (-OCOCH₃) is at position 2

Name: 2-(acetyloxy)ethanaminium

Step 4: Add N-Substituents

Three methyl groups on nitrogen: N,N,N-trimethyl

Complete cation name: 2-(acetyloxy)-N,N,N-trimethylethanaminium

Step 5: Name the Anion

Counterion: chloride (Cl⁻)

Step 6: Combine Cation and Anion

Cation + Anion (as separate words)

IUPAC NAME: 2-(acetyloxy)-N,N,N-trimethylethanaminium chloride

Molecular Structure:

2D Structure

3D Structure

Problem 4: Adiphenine - IUPAC Nomenclature

Given Structure:

Complex molecule with: two benzene rings, ester group, diethylamino group

Step-by-Step IUPAC Naming:

Step 1: Identify the Principal Functional Group

The ester group (-COO-) has the highest priority.

The compound will be named as an ester.

Step 2: Identify the Acid Component

The acid part contains:

Acetic acid (ethanoic acid) as the base
Two phenyl groups attached to carbon 2
Name: 2,2-diphenylacetic acid
As ester: 2,2-diphenylacetate

Step 3: Identify the Alcohol Component

The alcohol part:

Two-carbon chain: ethyl
At position 2: diethylamino group
Name: 2-(diethylamino)ethyl

Step 4: Combine the Names

Ester nomenclature: [Alcohol part] [Acid part]

Alcohol: 2-(diethylamino)ethyl
Acid: 2,2-diphenylacetate

IUPAC NAME: 2-(diethylamino)ethyl 2,2-diphenylacetate

Molecular Structure:

2D Structure

3D Structure

Problem 5: Formalin Preparation - Formaldehyde Hydration

Background:

Formalin is a 37% aqueous solution of formaldehyde used as a disinfectant and preservative.

Reaction:

HCHO + H₂O ⇌ CH₂(OH)₂
Formaldehyde + Water ⇌ Methanediol (Methylene glycol)

Mechanism - Nucleophilic Addition:

Step 1: Nucleophilic Attack

The oxygen atom of water (nucleophile) attacks the electrophilic carbonyl carbon of formaldehyde.

The carbonyl carbon is partially positive (δ+) due to the electronegative oxygen
Water's oxygen has lone pairs and acts as a nucleophile
A tetrahedral intermediate forms

Step 2: Proton Transfer

Proton transfer occurs from the positively charged oxygen to another water molecule or to the carbonyl oxygen.

This stabilizes the intermediate
Forms the geminal diol (methanediol)

Step 3: Equilibrium

The reaction is reversible and exists in equilibrium:

In aqueous solution, formaldehyde exists primarily as methanediol
The equilibrium strongly favors the hydrated form (>99.9%)
This is why formalin is stable

Why This Reaction Occurs:

Electrophilicity: Formaldehyde has a highly electrophilic carbonyl carbon
No Steric Hindrance: Formaldehyde is the smallest aldehyde (no alkyl groups)
Equilibrium: The hydrate is more stable in water due to hydrogen bonding

Molecular Structures:

Formaldehyde (HCHO)

Methanediol (CH₂(OH)₂)

Problem 6: Glutaraldehyde Cyclic Hemiacetal Formation

Background:

Glutaraldehyde cyclic hemiacetal is used for sterilization of medical equipment that cannot be autoclaved.

Reaction:

OHC-CH₂-CH₂-CH₂-CHO (4-hydroxybutanal form)
↓ Intramolecular cyclization
Cyclic hemiacetal (6-membered ring)

Mechanism - Intramolecular Hemiacetal Formation:

Step 1: Molecular Structure Analysis

Glutaraldehyde (pentanedial) exists in equilibrium with its enol form, which can tautomerize to 4-hydroxybutanal.

5-carbon chain with aldehyde groups at both ends
Can form a 6-membered cyclic hemiacetal

Step 2: Nucleophilic Attack (Intramolecular)

The hydroxyl group at C4 attacks the carbonyl carbon at C1:

Oxygen of -OH acts as nucleophile
Attacks the electrophilic carbonyl carbon
This is an INTRAmolecular reaction (within the same molecule)
Forms a 6-membered ring (thermodynamically favorable)

Step 3: Tetrahedral Intermediate Formation

A tetrahedral intermediate forms with:

Oxygen from the original hydroxyl now bonded to carbonyl carbon
Positive charge on the oxygen

Step 4: Proton Transfer

Proton transfer stabilizes the hemiacetal:

Proton moves to carbonyl oxygen
Forms the stable cyclic hemiacetal
Contains both -OH and -OR groups on the same carbon

Why 6-Membered Ring Forms:

Thermodynamic Stability: 6-membered rings have minimal ring strain
Kinetic Favorability: The distance between reacting groups is optimal
Entropy: Intramolecular reactions are entropically favored over intermolecular

Molecular Structures:

4-Hydroxybutanal

Cyclic Hemiacetal

Problem 7: Acetylcholine Synthesis - Esterification

Reaction:

CH₃COOH + HO-CH₂-CH₂-N⁺(CH₃)₃ → CH₃COO-CH₂-CH₂-N⁺(CH₃)₃ + H₂O
Acetic acid + Choline → Acetylcholine + Water

Mechanism - Fischer Esterification:

Step 1: Protonation of Carboxylic Acid

Acid catalyst (H⁺) protonates the carbonyl oxygen of acetic acid:

Makes the carbonyl carbon more electrophilic
Increases reactivity toward nucleophilic attack

Step 2: Nucleophilic Attack by Choline

The hydroxyl group of choline attacks the activated carbonyl carbon:

Oxygen of -OH acts as nucleophile
Forms a tetrahedral intermediate
This is the rate-determining step

Step 3: Proton Transfer

Proton transfer within the intermediate:

Proton moves from the choline oxygen to one of the hydroxyl groups
Prepares the leaving group (water)

Step 4: Elimination of Water

Water molecule is eliminated:

The protonated hydroxyl leaves as H₂O
Reforms the carbonyl double bond
Forms the ester linkage

Step 5: Deprotonation

Loss of proton regenerates the acid catalyst:

Forms the final acetylcholine product
H⁺ catalyst is regenerated (catalytic cycle)

Key Features:

Acid-catalyzed: Requires acid catalyst (H₂SO₄ or HCl)
Reversible: Equilibrium reaction (Le Chatelier's principle applies)
Water removal: Removing water drives equilibrium to the right

Molecular Structures:

Acetic Acid

Choline

Acetylcholine

Problem 8: Novocaine (Procaine) Synthesis - Esterification

Reaction:

NH₂-C₆H₄-COOH + HO-CH₂-CH₂-N(C₂H₅)₂ → NH₂-C₆H₄-COO-CH₂-CH₂-N(C₂H₅)₂ + H₂O
p-Aminobenzoic acid + Diethylaminoethanol → Procaine + Water

Mechanism - Acid-Catalyzed Esterification:

Step 1: Activation of Carboxylic Acid

Protonation of the carbonyl oxygen of p-aminobenzoic acid:

Acid catalyst (H⁺ from H₂SO₄) protonates C=O
Creates a more electrophilic carbonyl carbon
Resonance stabilization of the intermediate

Step 2: Nucleophilic Attack

The alcohol (diethylaminoethanol) attacks the activated carbonyl:

Hydroxyl oxygen donates electron pair to carbonyl carbon
Forms tetrahedral intermediate
Slow, rate-determining step

Step 3: Proton Transfer

Intramolecular proton transfer:

Proton moves from the alcohol oxygen to the carboxylic OH
Converts -OH into a better leaving group (-OH₂⁺)

Step 4: Water Elimination

Loss of water molecule:

Water leaves as the leaving group
Carbonyl π bond reforms
Forms protonated ester

Step 5: Deprotonation

Final deprotonation:

Base (water or alcohol) removes proton
Forms neutral procaine molecule
Regenerates acid catalyst

Reaction Conditions:

Catalyst: Concentrated H₂SO₄ or dry HCl gas
Temperature: Reflux conditions (heating)
Water removal: Dean-Stark apparatus or molecular sieves
Protection: The amino group may need protection to prevent side reactions

Molecular Structures:

p-Aminobenzoic Acid

Diethylaminoethanol

Procaine (Novocaine)

Problem 9: Novocaine Hydrolysis in the Body

Reaction - Acid Hydrolysis:

NH₂-C₆H₄-COO-CH₂-CH₂-N(C₂H₅)₂ + H₂O → NH₂-C₆H₄-COOH + HO-CH₂-CH₂-N(C₂H₅)₂
Procaine + Water → p-Aminobenzoic acid + Diethylaminoethanol

Mechanism - Acid-Catalyzed Ester Hydrolysis:

Step 1: Protonation of Ester Carbonyl

In the acidic environment of the body (or stomach):

H⁺ protonates the carbonyl oxygen of procaine
Makes carbonyl carbon more electrophilic
Activates the ester toward nucleophilic attack

Step 2: Nucleophilic Attack by Water

Water molecule attacks the activated carbonyl carbon:

Water acts as nucleophile (oxygen lone pair)
Forms tetrahedral intermediate
This is the rate-determining step

Step 3: Proton Transfer

Proton rearrangement in the intermediate:

Proton transfers from water oxygen to ester oxygen
Prepares the alcohol as leaving group

Step 4: Elimination of Alcohol

Cleavage of the ester bond:

Diethylaminoethanol leaves as the alcohol
Carbonyl double bond reforms
Forms protonated carboxylic acid

Step 5: Deprotonation

Final step to form products:

Loss of proton gives p-aminobenzoic acid
Diethylaminoethanol is released
Acid catalyst is regenerated

Biological Significance:

Metabolism: This hydrolysis occurs in the body via esterases (enzymes)
Duration: Hydrolysis limits the duration of anesthetic action
Detoxification: Breaks down the drug into less active metabolites
p-Aminobenzoic acid: Can cause allergic reactions in some patients

Molecular Structures:

Procaine

p-Aminobenzoic Acid

Diethylaminoethanol

Problem 10: Crotonic Acid (2-Butenoic Acid) Preparation

Background:

Crotonic acid is involved in protein and fat metabolism. It exists as cis and trans isomers (trans is more stable).

Method 1: Aldol Condensation followed by Dehydration

Step 1: 2 CH₃CHO → CH₃CH(OH)CH₂CHO (Aldol addition)
Step 2: CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O (Dehydration)
Step 3: CH₃CH=CHCHO + [O] → CH₃CH=CHCOOH (Oxidation)
Overall: Acetaldehyde → Crotonaldehyde → Crotonic acid

Detailed Mechanism:

Step 1: Enolate Formation

Base-catalyzed enolate formation from acetaldehyde:

Base (OH⁻) removes α-hydrogen from acetaldehyde
Forms resonance-stabilized enolate ion
Enolate is nucleophilic at the α-carbon

Step 2: Aldol Addition

Nucleophilic attack of enolate on another acetaldehyde:

Enolate attacks carbonyl carbon of second acetaldehyde
Forms alkoxide intermediate
Protonation gives 3-hydroxybutanal (aldol)

Step 3: Dehydration (E1cB Mechanism)

Base-catalyzed elimination of water:

Base removes another α-hydrogen
Forms carbanion/enolate
Elimination of OH⁻ gives α,β-unsaturated aldehyde
Product: crotonaldehyde (but-2-enal)

Step 4: Oxidation

Oxidation of aldehyde to carboxylic acid:

Oxidizing agent: KMnO₄, K₂Cr₂O₇, or Ag₂O
Aldehyde group converts to carboxylic acid
Product: crotonic acid (but-2-enoic acid)

Method 2: Perkin Reaction

CH₃CHO + (CH₃CO)₂O → CH₃CH=CHCOOH
Acetaldehyde + Acetic anhydride → Crotonic acid
(with sodium acetate catalyst)

Molecular Structures:

Acetaldehyde

Crotonaldehyde

Crotonic Acid

Problem 11: Magnesium 4-Aminobutanoate (GABA) Preparation

Background:

4-Aminobutanoic acid (GABA - gamma-aminobutyric acid) is an important neurotransmitter. Its magnesium salt has enhanced bioavailability.

Method: Acid-Base Neutralization

2 H₂N-CH₂-CH₂-CH₂-COOH + Mg(OH)₂ → (H₂N-CH₂-CH₂-CH₂-COO)₂Mg + 2 H₂O
2 × 4-Aminobutanoic acid + Magnesium hydroxide → Magnesium 4-aminobutanoate + Water

Detailed Mechanism:

Step 1: Dissociation of Base

Magnesium hydroxide dissociates in water:

Mg(OH)₂ → Mg²⁺ + 2OH⁻
Hydroxide ions are strong bases

Step 2: Deprotonation of Carboxylic Acid

Hydroxide removes proton from carboxylic acid:

OH⁻ attacks the acidic proton of -COOH
Forms carboxylate anion: -COO⁻
Water is produced: OH⁻ + H⁺ → H₂O
This is fast and exothermic

Step 3: Salt Formation

Magnesium ion coordinates with carboxylate anions:

Mg²⁺ has +2 charge
Two carboxylate anions (-COO⁻) balance the charge
Ionic bond formation: Mg²⁺ + 2 R-COO⁻ → (R-COO)₂Mg
Forms ionic salt (magnesium 4-aminobutanoate)

Step 4: Crystallization

Product isolation:

Evaporation of water
Crystallization of the magnesium salt
Purification by recrystallization if needed

Alternative Preparation Methods:

From succinic acid: Via Hofmann rearrangement of succinimide
From glutamic acid: Decarboxylation using glutamate decarboxylase enzyme
From γ-butyrolactone: Ring opening with ammonia, then neutralization

Biological Importance:

Neurotransmitter: GABA is the main inhibitory neurotransmitter in CNS
Magnesium salt: Better absorption and bioavailability
Medical uses: Anxiety, stress, sleep disorders

Molecular Structures:

4-Aminobutanoic Acid (GABA)

Magnesium 4-Aminobutanoate

Problem 12: Acetone Preparation by Decarboxylation

Background:

Acetone is a ketone body formed during starvation and diabetes. It can be prepared by decarboxylation of acetoacetic acid.

Reaction:

CH₃COCH₂COOH → CH₃COCH₃ + CO₂
Acetoacetic acid → Acetone + Carbon dioxide
(Heat or enzymatic decarboxylation)

Mechanism - Thermal Decarboxylation:

Step 1: Cyclic Transition State Formation

Six-membered cyclic transition state:

Carbonyl oxygen of ketone interacts with carboxylic acid proton
Forms a cyclic, hydrogen-bonded transition state
This is a concerted mechanism
Electrons rearrange in a cyclic fashion

Step 2: C-C Bond Cleavage

Simultaneous bond breaking and forming:

C-C bond between α-carbon and carboxyl carbon breaks
C=O π bond of carboxylic acid becomes C=O of CO₂
O-H bond breaks, proton transfers to α-carbon
All happens in one step (concerted)

Step 3: Product Formation

Formation of final products:

CO₂ is released as gas (drives reaction forward)
Enol form of acetone initially forms
Rapid tautomerization to keto form
Stable acetone molecule

Step 4: Tautomerization (if needed)

Enol to keto conversion:

Enol: CH₃C(OH)=CH₂
Keto: CH₃COCH₃ (more stable)
Keto form is thermodynamically favored (>99%)

Why Decarboxylation Occurs Easily:

β-Keto acid: The ketone at β-position stabilizes the transition state
Entropy: Gas evolution (CO₂) increases entropy, drives reaction
Resonance: Enol intermediate is stabilized by resonance
Low activation energy: Cyclic transition state lowers energy barrier

Biological Context:

Ketogenesis: In liver during fasting/diabetes
Acetoacetate decarboxylase: Enzyme that catalyzes this reaction
Ketone bodies: Acetone, acetoacetate, β-hydroxybutyrate
Clinical significance: Elevated in diabetic ketoacidosis

Molecular Structures:

Acetoacetic Acid

Acetone

Carbon Dioxide

🧪 Organic Chemistry Solutions

End of Solutions