2nd PUC Chemistry Named Reactions You MUST Know | Midterm Exam

2nd PUC Chemistry Named Reactions: Organic chemistry in the 2nd PUC syllabus requires a strong grasp of named reactions, reagents, and mechanisms. From nucleophilic substitutions to electrophilic aromatic reactions, standard transformations appear repeatedly in theory exams. This guide breaks down high-yield organic reactions from Haloalkanes, Haloarenes, Alcohols, Phenols, and Ethers to help you prepare effectively for your midterm exams.

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Haloalkane and Haloarene Coupling Reactions

Coupling reactions combine organic halides using metallic sodium in dry ether to form new carbon-carbon bonds.

Fittig Reaction

• Reactants: Two molecules of haloarene (aryl halide).

• Reagent: Metallic Sodium (Na) in dry ether.

• Transformation: Cleavage of C–X bonds to couple two phenyl rings.

• Product: Biphenyl (Diphenyl).

• Equation: 2 Ar–X + 2 Na (in dry ether) -> Ar–Ar + 2 NaX

Wurtz-Fittig Reaction

• Reactants: One haloarene (aryl halide) and one haloalkane (alkyl halide).

• Reagent: Metallic Sodium (Na) in dry ether.

• Transformation: Coupling between an aryl group and an alkyl group.

• Product: Alkylarene (such as Toluene / Methylbenzene).

• Equation: Ar–X + 2 Na + R–X (in dry ether) -> Ar–R + 2 NaX

• Example: C6H5Cl + 2 Na + CH3Cl (in dry ether) -> C6H5–CH3 + 2 NaCl

Wurtz Reaction

• Reactants: Two molecules of haloalkane (alkyl halide).

• Reagent: Metallic Sodium (Na) in dry ether.

• Transformation: Symmetrical coupling of alkyl groups.

• Product: Higher Alkane.

• Equation: 2 R–X + 2 Na (in dry ether) -> R–R + 2 NaX

• Example: 2 CH3Cl + 2 Na (in dry ether) -> CH3–CH3 + 2 NaCl

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Comparison of Sodium Coupling Reactions

(Memory Tip: Think of Fittig as Aryl + Aryl coupling to make Biphenyl, Wurtz-Fittig as Aryl + Alkyl coupling to make Alkylbenzene, and Wurtz as Alkyl + Alkyl coupling to make an Alkane).

Halogen Exchange Reactions

Halogen exchange methods offer a reliable route to prepare specific haloalkanes that cannot be synthesized easily by direct halogenation.

Finkelstein Reaction

• Purpose: Preparation of alkyl iodides.

• Reaction: Alkyl chlorides or bromides react with sodium iodide (NaI) in dry acetone.

• Equation: R–X + NaI (in dry acetone) -> R–I + NaX (where X = Cl, Br)

• Role of Dry Acetone: NaI dissolves in acetone, but the by-products NaCl and NaBr do not dissolve and precipitate out. Following Le Chatelier's principle, this precipitation shifts equilibrium to the forward direction, driving the reaction forward.

Swarts Reaction

• Purpose: Preparation of alkyl fluorides.

• Reaction: Alkyl chlorides or bromides are heated with heavy metal fluorides such as AgF, Hg2F2, CoF2, or SbF3.

• Equation: R–X + AgF -> R–F + AgX (where X = Cl, Br)

(Memory Tip: Remember F–I for Finkelstein producing Alkyl Iodide, and S–F for Swarts producing Alkyl Fluoride).

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Nucleophilic Substitution: SN1 vs. SN2 Mechanisms

Nucleophilic substitution in haloalkanes follows two distinct pathways depending on substrate structure and reaction conditions.

       Nucleophilic Substitution Pathways
                /          \
                /            \
          SN1 Pathway     SN2 Pathway
          (Two Steps)     (Single Step)
              |               |
      Carbocation Inter.  Transition State
              |               |
          3° > 2° > 1°    1° > 2° > 3°

Unimolecular Nucleophilic Substitution (SN1)

• Kinetics: First-order kinetics (Rate = k[Substrate]).

• Steps: Two-step mechanism.

• Substrate Reactivity: Tertiary (3°) alkyl halides react fastest due to the high stability of tertiary carbocations stabilized by hyperconjugation and +I effects.

• Solvent: Polar protic solvents (like H2O) stabilize the carbocation and leaving group.

Two-Step Mechanism for tert-Butyl Bromide:

1. Step 1 (Slow / Rate-Determining Step): Ionization forms a planar carbocation intermediate and a bromide ion:
   (CH3)3C–Br (slow) -> (CH3)3C(+) + Br(-)

2. Step 2 (Fast Step): Nucleophile (OH-) attacks the planar carbocation from either side to give tert-butyl alcohol:
   (CH3)3C(+) + OH(-) (fast) -> (CH3)3C–OH

Biomolecular Nucleophilic Substitution (SN2)

• Kinetics: Second-order kinetics (Rate = k[Substrate][Nucleophile]).

• Steps: Single-step (concerted) mechanism.

• Substrate Reactivity: Primary (1°) alkyl halides and methyl halides react fastest due to low steric hindrance.

• Stereochemistry: Occurs with complete inversion of configuration (Walden Inversion).

Mechanism for Methyl Chloride:

The nucleophile (OH-) attacks the central carbon from the rear side (backside attack) to avoid repulsion from the leaving group (Cl-). A simultaneous bond-forming and bond-breaking step creates a pentacoordinated transition state:

HO(-) + CH3Cl -> [HO…CH3…Cl](transition state) -> CH3OH + Cl(-)

Comparison of SN1 and SN2 Mechanisms

Sandmeyer Reaction

The Sandmeyer reaction converts primary aromatic amines into aryl halides.

• Step 1 (Diazotization): Aniline (Ar–NH2) reacts with nitrous acid (NaNO2 + HCl) at 0–5 °C to yield benzenediazonium chloride (Ar–N2(+)Cl(-)).

• Step 2 (Sandmeyer Substitution): Treating the diazonium salt with cuprous halides (Cu2Cl2 or Cu2Br2) gives chloroarene or bromoarene:
  Ar–N2(+)X(-) + Cu2X2 -> Ar–X + N2 (where X = Cl, Br)

• Aryl Iodide Synthesis: Cuprous halide is not needed. Warming the diazonium salt directly with aqueous potassium iodide (KI) yields iodobenzene:
  Ar–N2(+)Cl(-) + KI -> Ar–I + KCl + N2

Important Alkyl Halide Reactions and Rules

Saytzeff's (Zaitsev's) Rule

In dehydrohalogenation (beta-elimination) using alcoholic KOH, base removes a beta-hydrogen. The preferred product is the more highly substituted alkene having the greater number of alkyl groups attached to the doubly bonded carbon atoms.

• Example: CH3–CH2–CH(Br)–CH3 + Alc. KOH -> CH3–CH=CH–CH3 (Major, 2-Butene) + CH3–CH2–CH=CH2 (Minor, 1-Butene)

Friedel-Crafts Reactions of Haloarenes

Halogens on benzene rings are ortho/para-directing due to resonance electron donation (+M effect). The para-isomer is the major product due to less steric hindrance.

• Alkylation: Chlorobenzene + CH3Cl (Anhydrous AlCl3) -> 1-Chloro-4-methylbenzene (para, Major) + 1-Chloro-2-methylbenzene (ortho).

• Acylation: Chlorobenzene + CH3COCl (Anhydrous AlCl3) -> 4-Chloroacetophenone (para, Major) + 2-Chloroacetophenone (ortho).

Darzens Reaction

Alcohols react with thionyl chloride (SOCl2) to form alkyl chlorides:

R–OH + SOCl2 -> R–Cl + SO2(g) + HCl(g)

• Synthetic Advantage: The by-products (SO2 and HCl) are gaseous and escape easily, leaving behind pure alkyl chloride without complicated purification.

Fundamental Reaction Mechanisms in Alcohols and Ethers

1. Intermolecular Dehydration of Ethanol to Diethyl Ether (413 K)

• Overall Equation: 2 C2H5OH (H2SO4 at 413 K) -> C2H5–O–C2H5 + H2O

1. Protonation: CH3CH2OH + H(+) <=> CH3CH2–OH2(+) (Ethyl oxonium ion)

2. Nucleophilic Attack: CH3CH2–OH + CH3CH2–OH2(+) (slow) -> (CH3CH2)2OH(+) + H2O

3. Deprotonation: (CH3CH2)2OH(+) -> CH3CH2–O–CH2CH3 + H(+)

2. Acid-Catalyzed Hydration of Ethene to Ethanol

• Overall Equation: CH2=CH2 + H2O (H+) -> CH3–CH2OH

1. Protonation of Alkene: Electrophilic attack by H3O(+) creates a carbocation intermediate:
   CH2=CH2 + H3O(+) <=> CH3–CH2(+) + H2O

2. Nucleophilic Attack of Water: CH3–CH2(+) + H2O <=> CH3–CH2–OH2(+)

3. Deprotonation: CH3–CH2–OH2(+) + H2O <=> CH3–CH2OH + H3O(+)

3. Intramolecular Dehydration of Ethanol to Ethene (443 K)

• Overall Equation: CH3CH2OH (Conc. H2SO4 at 443 K) -> CH2=CH2 + H2O

1. Protonation: CH3CH2OH + H(+) <=> CH3CH2–OH2(+)

2. Carbocation Formation (Slow Step): CH3CH2–OH2(+) (slow) -> CH3CH2(+) + H2O

3. Deprotonation: CH3CH2(+) -> CH2=CH2 + H(+)

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Reactions of Phenol and Ether Synthesis

Reimer-Tiemann Reaction

Phenol reacts with chloroform (CHCl3) and aqueous NaOH to form Salicylaldehyde (2-Hydroxybenzaldehyde) via a dichlorocarbene (:CCl2) intermediate.

• Equation: C6H5OH + CHCl3 + 3 NaOH -> C6H4(OH)(CHO) + 3 NaCl + 2 H2O

Kolbe's Reaction

Phenol reacts with NaOH to form sodium phenoxide, which reacts with carbon dioxide (CO2) at 400 K and 4–7 atm, followed by acidification to produce Salicylic acid (2-Hydroxybenzoic acid).

• Equation: C6H5ONa + CO2 (400 K, 4-7 atm) -> then H(+) -> C6H4(OH)(COOH)

(Memory Tip: Reimer-Tiemann yields Salicylaldehyde containing -CHO, while Kolbe's reaction yields Salicylic Acid containing -COOH).

Industrial Preparation of Phenol from Cumene

1. Auto-oxidation: Cumene (isopropylbenzene) oxidizes in air to form Cumene Hydroperoxide.

2. Acid Cleavage: Dilute acid decomposes the peroxide into Phenol and Acetone (a valuable co-product).

• Equation: C6H5–CH(CH3)2 + O2 -> C6H5–C(CH3)2–O–O–H (H+/H2O) -> C6H5OH + CH3COCH3

Williamson Ether Synthesis

Reaction between a primary (1°) alkyl halide and a sodium alkoxide (R'–O(-)Na(+)) yields an ether through an SN2 pathway.

• Equation: R'–O(-)Na(+) + R–X -> R'–O–R + NaX

• Key Requirement: The alkyl halide must be primary to avoid elimination. Tertiary alkoxides can react with primary alkyl halides to yield unsymmetrical branched ethers.

Synthesis of Aspirin

Salicylic acid reacts with acetic anhydride in acid to acetylate the phenolic -OH group, producing acetylsalicylic acid (Aspirin), an important analgesic and antipyretic drug.

• Equation: C6H4(OH)(COOH) + (CH3CO)2O (H+) -> C6H4(OCOCH3)(COOH) + CH3COOH

Additional Key Phenol Transformations

• Reduction: Phenol heated with Zn dust gives Benzene.

• Bromination: Phenol with aqueous Br2 forms a white precipitate of 2,4,6-Tribromophenol.

• Nitration: Dilute HNO3 yields ortho- and para-nitrophenol; concentrated HNO3 with H2SO4 yields 2,4,6-Trinitrophenol (Picric Acid).

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