Unit 1 ✨️✨️ 1)What are acids and bases. 2)Define Tortomerism with example. 3)Define metaberism with exapmle. 4)What is lewis concept of acid and base. 5)Define intermolecular forces. 6)What do you mean by ion paring. 7)Give the examples of protic and aprotic solvents. 8)Define isomerism with example. (Unit 2)✨️✨️ 1)Define alcohols and acids. 2)Define cyloalkanes. 3)What are phenols. 4)Define relative reactivity. 5)What are free radicals. 6)Define organic chemistry. 7)Define ketones and Diane's. 8)Draw the functional group for esters and amide. Unit 3 ✨️✨️✨️ 1)Write down the formula to calculate angel staine. 2)What is bare staine theory. 3)Define nucleofiles. 4)What is leaving group. 5)Define SN2 reaction. 6)What is steric hinderance. 7)What do you mean by solvolysis. 8)What is stability order of carbo cations. Unit 4✨️✨️✨️ 1)What do u mean by halogenation. 2)Define E1 and E2. 3)Define Eelimination reaction. 4)Define substitution reaction. 5)Define catalysis. 6)What is acid catalysis. 7)Define kinetics. 8)What do you mean by dehydrogenation. Unit 5 ✨️✨️✨️ 1)Define electrofile. 2)Define free radical. 3)What is markonioff rule. 4)Define rearrangment. 5)What is alohydration. 6) what is peroxide effect. 7 Define hydrogen exchange. 8)Define free radical addition. Unit 6 ✨️✨️✨️ 1)Define nucleophilic substitution. 2)Define resonance. 3)What is hyper conjunction. 4)What is substrate. 5)What is alylic rearrangment. 6)Define free radical substitution. 7)Define equilibrium. 8)Define dienes. Unit 7 ✨️✨️✨️ 1)Define sulfonation. 2)Define electrophilic aromatic substitution. 3)Define halogenation. 4)Define activating group. 5)Define nitration. 6)Define fridal craft acylation. 7)Define deactivating group. 8)What are ortho pera meta directing group. Unit 8 ✨️✨️✨️ 1)Define nucleofiles. 2)Write down nucleophilic addition reaction. 3)Write down aldol condensation reaction. 4)Write down witting reaction. 5)Write down benzoic condensation. 6) Write down canizzaro reaction. 7)Write down perkine reaction. 8)Write down reform matsq reaction. Unit 9 ✨️✨️✨️ 1)Define diazotyration. 2)Write down fries rearrangement. 3)Write down kolbe reaction. 4)What id rimer timen reaction. 5)Define bimolecular displacement. 6)Write down sandmyers reaction. 7)Define electron migration. 8)Define nucleophilic aromatic substitution. Unit 10 ✨️✨️✨️ 1)What is oxidation. 2)What is reduction. 3)Define assay. 4)Draw the structure of sodium (SLS). 5)Draw the structure of SL acid. 6)What is use of lactic acid. 7)Write a purity test for dimercaprol. 8)Draw the structure if urea. Answers these as per 2 marks answer for pharm d 1st year student

Here are all 80 answers across 10 units, written as concise 2-mark answers for Pharm D 1st year:

UNIT 1

1. What are acids and bases? An acid is a substance that donates a proton (H⁺) or accepts an electron pair. A base is a substance that accepts a proton or donates an electron pair. Example: HCl (acid), NaOH (base). (Arrhenius, Bronsted-Lowry definitions)

2. Define Tautomerism with example. Tautomerism is a type of isomerism where two isomers are interconvertible by the migration of a proton (H⁺) and a shift of a double bond. These forms exist in dynamic equilibrium. Example: Keto-enol tautomerism of acetoacetic ester - the keto form (CH₃-CO-CH₂-COOC₂H₅) and enol form (CH₃-C(OH)=CH-COOC₂H₅) exist in equilibrium.

3. Define Metamerism with example. Metamerism is a type of isomerism where compounds have the same molecular formula but different alkyl groups on either side of the same functional group (like -O-, -S-, -NH-). Example: Diethyl ether (C₂H₅-O-C₂H₅) and methyl propyl ether (CH₃-O-C₃H₇) are metamers - both have molecular formula C₄H₁₀O.

4. What is Lewis concept of acid and base? According to G.N. Lewis:

A Lewis acid is a species that can accept an electron pair. Example: BF₃, AlCl₃
A Lewis base is a species that can donate an electron pair. Example: NH₃, H₂O This concept is broader than Bronsted-Lowry as it includes reactions without proton transfer.

5. Define intermolecular forces. Intermolecular forces are attractive or repulsive forces between neighbouring molecules. They include:

Van der Waals forces (London dispersion forces)
Dipole-dipole interactions
Hydrogen bonding These forces determine physical properties like boiling point, melting point, and solubility.

6. What do you mean by ion pairing? Ion pairing is the association of oppositely charged ions (cation + anion) in solution to form a neutral or lower-charged unit called an "ion pair." It occurs in solvents of low dielectric constant and affects conductivity and reactivity of ionic compounds. Example: Na⁺Cl⁻ ion pair in a non-polar solvent.

7. Give examples of protic and aprotic solvents.

Protic solvents: Have an O-H or N-H bond and can donate protons. Examples: Water (H₂O), ethanol (C₂H₅OH), methanol (CH₃OH), acetic acid.
Aprotic solvents: Do not have O-H or N-H bonds; cannot donate protons. Examples: Acetone, DMSO (dimethyl sulfoxide), DMF (dimethylformamide), acetonitrile.

8. Define isomerism with example. Isomerism is the phenomenon where two or more compounds have the same molecular formula but different structural arrangements, resulting in different physical or chemical properties. Such compounds are called isomers. Example: n-butane and isobutane both have formula C₄H₁₀ but different structures.

UNIT 2

1. Define alcohols and acids.

Alcohols: Organic compounds containing a hydroxyl group (-OH) attached to a saturated carbon atom. Example: Ethanol (C₂H₅OH).
Organic Acids (Carboxylic acids): Compounds containing the carboxyl group (-COOH). They are weak acids. Example: Acetic acid (CH₃COOH).

2. Define cycloalkanes. Cycloalkanes are saturated cyclic hydrocarbons with the general formula CₙH₂ₙ. The carbon atoms are joined in a ring structure with only single bonds. They are less reactive than alkenes. Examples: Cyclopropane (C₃H₆), Cyclohexane (C₆H₁₂), Cyclopentane (C₅H₁₀).

3. What are phenols? Phenols are organic compounds in which one or more hydroxyl groups (-OH) are directly attached to an aromatic (benzene) ring. General formula: C₆H₅OH (phenol). They are more acidic than alcohols due to resonance stabilization of the phenoxide ion. Example: Phenol (carbolic acid), cresol, catechol.

4. Define relative reactivity. Relative reactivity refers to the comparative ability of different compounds or functional groups to undergo a chemical reaction under similar conditions. It is used to compare how readily different substrates react with a reagent. Example: Tertiary alcohols > secondary > primary in SN1 reactions.

5. What are free radicals? Free radicals are highly reactive chemical species with one or more unpaired electrons. They are electrically neutral. They are formed by homolytic cleavage of a covalent bond. Example: Chlorine radical (Cl•), methyl radical (CH₃•). They play a key role in halogenation and polymerization reactions.

6. Define organic chemistry. Organic chemistry is the branch of chemistry that deals with the study of carbon-containing compounds, their structure, properties, composition, reactions, and synthesis. It also includes compounds of hydrogen, oxygen, nitrogen, sulfur, and halogens bonded to carbon. It forms the basis of biochemistry, pharmacology, and drug synthesis.

7. Define ketones and dienes.

Ketones: Organic compounds containing a carbonyl group (C=O) bonded to two carbon atoms (R-CO-R'). Example: Acetone (CH₃-CO-CH₃).
Dienes: Hydrocarbons containing two carbon-carbon double bonds (C=C). General formula CₙH₂ₙ₋₂. Example: 1,3-Butadiene (CH₂=CH-CH=CH₂) - a conjugated diene.

8. Draw the functional group for esters and amides.

Ester functional group:

    O
    ‖
 R—C—O—R'

(-COO- or -RCOOR')

Amide functional group:

    O
    ‖
 R—C—NH₂

(-CONH₂ or -RCONHR')

Esters are formed from acids + alcohols; amides are formed from acids + amines.

UNIT 3

1. Write down the formula to calculate angle strain. Angle strain (also called Baeyer strain) arises when the bond angles in a cyclic compound deviate from the normal tetrahedral angle (109.5°). Formula:

Angle strain = ½ × (109.5° - actual bond angle of the ring carbon)

For cyclopropane: ½ × (109.5° - 60°) = ½ × 49.5° = 24.75° per carbon.

2. What is Baeyer Strain Theory? Baeyer strain theory (1885) states that cyclic compounds are strained because their carbon atoms are forced to adopt bond angles different from the normal tetrahedral angle of 109.5°. Smaller rings (cyclopropane, cyclobutane) have greater angle strain and are more reactive. Cyclohexane is most stable as it can adopt a strain-free chair conformation.

3. Define nucleophiles. Nucleophiles are electron-rich species that have a lone pair of electrons or a negative charge and attack electron-deficient (positive) centers in a molecule. The word means "nucleus-loving." Examples: OH⁻, CN⁻, NH₃, H₂O, Cl⁻, RO⁻. They initiate nucleophilic substitution (SN1, SN2) reactions.

4. What is a leaving group? A leaving group is an atom or group of atoms that departs from the substrate along with a pair of electrons (as an anion or neutral species) during a substitution or elimination reaction. A good leaving group is a weak base (stable after leaving). Examples: Cl⁻, Br⁻, I⁻, OTs⁻ (tosylate), H₂O. F⁻ is a poor leaving group.

5. Define SN2 reaction. SN2 (Substitution Nucleophilic Bimolecular) is a one-step mechanism where the nucleophile attacks the carbon bearing the leaving group from the back side simultaneously as the leaving group departs. It involves a transition state and results in inversion of configuration (Walden inversion).

Rate = k[substrate][nucleophile]
Favoured by primary substrates and polar aprotic solvents.

6. What is steric hindrance? Steric hindrance is the slowing or blocking of a chemical reaction due to the bulky size of groups surrounding the reactive site in a molecule. Large substituents physically obstruct the approach of a reagent. Example: Tertiary alkyl halides undergo SN1 (not SN2) because three alkyl groups block the back-side attack of the nucleophile.

7. What do you mean by solvolysis? Solvolysis is a type of substitution reaction in which the solvent itself acts as the nucleophile. The solvent (water, alcohol, acetic acid) attacks the substrate and displaces the leaving group.

Hydrolysis (water as nucleophile)
Alcoholysis (alcohol as nucleophile)
Acetolysis (acetic acid as nucleophile) It follows SN1 mechanism.

8. What is the stability order of carbocations? Carbocations are positively charged carbon intermediates. Their stability increases with the number of alkyl groups attached (hyperconjugation and inductive effect):

Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (CH₃⁺)

Allylic and benzylic carbocations are especially stable due to resonance delocalization.

UNIT 4

1. What do you mean by halogenation? Halogenation is the introduction of a halogen atom (F, Cl, Br, or I) into an organic molecule. It can occur by:

Free radical halogenation (alkanes, UV light)
Electrophilic addition (alkenes)
Electrophilic aromatic substitution (benzene ring) Example: CH₄ + Cl₂ → CH₃Cl + HCl (chlorination of methane)

2. Define E1 and E2.

E1 (Elimination Unimolecular): Two-step reaction. The leaving group departs first to form a carbocation intermediate, then a proton is removed. Rate = k[substrate]. Favoured by tertiary substrates.
E2 (Elimination Bimolecular): One-step concerted reaction. A base removes a proton while the leaving group departs simultaneously. Rate = k[substrate][base]. Requires anti-periplanar geometry.

3. Define elimination reaction. An elimination reaction is a type of organic reaction in which atoms or groups are removed from adjacent carbons of a molecule, resulting in the formation of a double bond (alkene) or triple bond (alkyne). It is the reverse of addition. Example: CH₃CH₂Br + KOH (alc.) → CH₂=CH₂ + KBr + H₂O

4. Define substitution reaction. A substitution reaction is one in which an atom or group of atoms in a molecule is replaced by another atom or group. The overall molecular formula changes, but the carbon skeleton is retained. Types: Nucleophilic substitution (SN1, SN2), Electrophilic substitution (aromatic), Free radical substitution. Example: CH₃Br + OH⁻ → CH₃OH + Br⁻

5. Define catalysis. Catalysis is the process by which the rate of a chemical reaction is increased by a substance called a catalyst, which is not consumed in the reaction. A catalyst lowers the activation energy of the reaction. Types: Homogeneous (same phase), Heterogeneous (different phase), Enzymatic (biological catalyst).

6. What is acid catalysis? Acid catalysis is a type of catalysis in which an acid (proton donor or Lewis acid) speeds up a chemical reaction by donating a proton to the substrate, making it more reactive (more electrophilic).

Specific acid catalysis: H₃O⁺ acts as catalyst (e.g., hydrolysis of esters in H₂SO₄).
General acid catalysis: Any proton donor acts as catalyst. Example: Acid-catalyzed hydration of alkenes.

7. Define kinetics. Chemical kinetics is the branch of chemistry that studies the rate of chemical reactions and the factors that affect it (concentration, temperature, pressure, catalyst). It helps determine the mechanism of a reaction. The rate law: Rate = k[A]^m [B]^n, where k is the rate constant and m, n are orders.

8. What do you mean by dehydrogenation? Dehydrogenation is a chemical reaction in which hydrogen atoms are removed from an organic compound, resulting in the formation of a double or triple bond. It is the reverse of hydrogenation. Example: C₂H₆ → C₂H₄ + H₂ (ethane → ethylene) It is used industrially in the production of styrene, butadiene, etc.

UNIT 5

1. Define electrophile. An electrophile is an electron-deficient species that seeks electrons and attacks electron-rich centers (nucleophiles) in a molecule. The word means "electron-loving." They have a positive charge or partial positive charge (δ+). Examples: H⁺, Br⁺, NO₂⁺ (nitronium ion), carbocations (R⁺), BF₃, AlCl₃.

2. Define free radical. A free radical is a chemical species with one or more unpaired electrons. It is highly reactive and electrically neutral. Formed by homolytic bond cleavage. Examples: Methyl radical (•CH₃), Chlorine radical (Cl•), Hydroxyl radical (•OH). Free radicals participate in chain reactions (initiation, propagation, termination).

3. What is Markovnikov's rule? Markovnikov's rule states that in the addition of an unsymmetrical reagent (like HX) to an unsymmetrical alkene, the hydrogen adds to the carbon bearing the greater number of hydrogen atoms (more substituted carbon gets the negative part X⁻). Example: CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (not CH₃-CH₂-CH₂Br)

4. Define rearrangement. A rearrangement reaction is a chemical reaction in which the carbon skeleton or substituent of a molecule is reorganized to form a structural isomer, usually via migration of an atom, group, or bond. Example: 1,2-hydride shift or 1,2-methyl shift in carbocation intermediates. Example reaction: Wagner-Meerwein rearrangement.

5. What is halohydration? Halohydration is the addition of a halogen (X₂) and water (OH) across a double bond of an alkene in an aqueous solution of halogen. It produces a halohydrin (a compound with both -OH and -X on adjacent carbons). The reaction follows Markovnikov's rule (OH adds to more substituted carbon). Example: CH₂=CH₂ + Br₂/H₂O → BrCH₂-CH₂OH (bromohydrin)

6. What is the peroxide effect? The peroxide effect (also called Kharasch effect or anti-Markovnikov addition) is the reversal of Markovnikov's rule when HBr is added to an alkene in the presence of peroxides (ROOR). The reaction proceeds by a free radical mechanism. Example: CH₃-CH=CH₂ + HBr (peroxide) → CH₃-CH₂-CH₂Br (anti-Markovnikov product)

7. Define hydrogen exchange. Hydrogen exchange (also called hydrogen-deuterium exchange) is a reaction in which a hydrogen atom in an organic molecule is replaced by another hydrogen isotope (deuterium, D) or vice versa under acidic, basic, or enzymatic conditions. It is used to study reaction mechanisms and metabolic pathways. Example: C₆H₆ + D₂SO₄ → C₆D₆ (deuterobenzene) via electrophilic aromatic substitution.

8. Define free radical addition. Free radical addition is an addition reaction that proceeds through a free radical mechanism, initiated by light (hv) or peroxides. The reaction involves three steps: initiation, propagation, and termination. Example: Addition of HBr to alkene in the presence of peroxides - follows anti-Markovnikov rule. Cl• + CH₂=CH₂ → ClCH₂-CH₂• (propagation step)

UNIT 6

1. Define nucleophilic substitution. Nucleophilic substitution is a reaction in which an electron-rich nucleophile replaces a leaving group attached to a carbon atom. The nucleophile donates its electron pair to the electrophilic carbon. Two types: SN1 (unimolecular, stepwise) and SN2 (bimolecular, concerted). Example: CH₃Br + OH⁻ → CH₃OH + Br⁻

2. Define resonance. Resonance is a concept used to describe molecules whose electronic structure cannot be accurately represented by a single Lewis structure. The actual structure (resonance hybrid) is an average/blend of two or more contributing structures (resonance structures) connected by double-headed arrows (↔). Example: Benzene has two resonance structures (Kekulé structures).

3. What is hyperconjugation? Hyperconjugation is the delocalization of electrons from a C-H sigma bond (of an alkyl group) into an adjacent empty p-orbital or antibonding π* orbital. It stabilizes carbocations, free radicals, and alkenes. Example: Stability of tert-butyl carbocation - 9 C-H bonds contribute to hyperconjugation, giving maximum stability. More C-H bonds adjacent = greater hyperconjugation = greater stability.

4. What is a substrate? In organic chemistry, a substrate is the reactant molecule that undergoes a chemical reaction at the site of attack by a reagent (nucleophile, electrophile, or radical). It is the organic compound being transformed. Example: In CH₃Br + OH⁻ → CH₃OH + Br⁻, methyl bromide (CH₃Br) is the substrate.

5. What is allylic rearrangement? Allylic rearrangement is a type of rearrangement reaction occurring in allylic systems (CH₂=CH-CH₂-) where the double bond migrates and the substituent shifts from one end of the allylic system to the other via a resonance-stabilized allylic carbocation or radical. Example: 1-chlorobut-2-ene ⇌ 3-chlorobut-1-ene (via allylic cation intermediate).

6. Define free radical substitution. Free radical substitution is a reaction in which a hydrogen atom in an organic molecule is replaced by a halogen (or other group) via a free radical mechanism. It involves three stages: initiation (radical formation), propagation (chain reaction), and termination. Example: CH₄ + Cl₂ → CH₃Cl + HCl (UV light)

7. Define equilibrium. Chemical equilibrium is the state in a reversible reaction where the rate of forward reaction equals the rate of reverse reaction, so the concentrations of reactants and products remain constant. It is a dynamic equilibrium. Expressed by the equilibrium constant K = [products]/[reactants]. Example: N₂ + 3H₂ ⇌ 2NH₃

8. Define dienes. Dienes are hydrocarbons containing two carbon-carbon double bonds (C=C). General formula: CₙH₂ₙ₋₂. Types:

Isolated dienes: Double bonds separated by 2+ single bonds (CH₂=CH-CH₂-CH=CH₂)
Conjugated dienes: Alternating single and double bonds (CH₂=CH-CH=CH₂) - most stable
Cumulated dienes (allenes): Two double bonds on same carbon (CH₂=C=CH₂)

UNIT 7

1. Define sulfonation. Sulfonation is an electrophilic aromatic substitution reaction in which a sulfonyl group (-SO₃H) is introduced into an aromatic ring by reacting with fuming sulfuric acid (oleum, H₂SO₄·SO₃). The electrophile is SO₃. Example: C₆H₆ + H₂SO₄ (fuming) → C₆H₅SO₃H + H₂O (benzenesulfonic acid) The reaction is reversible.

2. Define electrophilic aromatic substitution. Electrophilic Aromatic Substitution (EAS) is a reaction in which an electrophile replaces a hydrogen atom on an aromatic ring, maintaining the aromaticity of the ring. It occurs in two steps: formation of arenium ion (sigma complex/Wheland intermediate), then loss of H⁺. Examples: Nitration, sulfonation, halogenation, Friedel-Crafts reactions.

3. Define halogenation (of aromatics). Halogenation of aromatic compounds is the introduction of a halogen atom (Cl or Br) into the aromatic ring via electrophilic aromatic substitution. It requires a Lewis acid catalyst (FeBr₃ or AlCl₃) to activate the halogen. Example: C₆H₆ + Br₂ → C₆H₅Br + HBr (in presence of FeBr₃)

4. Define activating group. An activating group (activating substituent) is a group already present on the benzene ring that increases the reactivity of the ring toward electrophilic aromatic substitution by donating electrons into the ring (by resonance or induction). They are ortho/para directors. Examples: -OH, -NH₂, -OCH₃, -CH₃, -NHCOCH₃.

5. Define nitration. Nitration is an electrophilic aromatic substitution reaction in which a nitro group (-NO₂) is introduced into the aromatic ring using a nitrating mixture of concentrated HNO₃ and H₂SO₄. The electrophile is the nitronium ion (NO₂⁺). Example: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (nitrobenzene)

6. Define Friedel-Crafts acylation. Friedel-Crafts acylation is an electrophilic aromatic substitution in which an acyl group (-COR) is introduced into the benzene ring using an acyl halide (RCOCl) and a Lewis acid catalyst (AlCl₃). It produces aryl ketones. Example: C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl (acetophenone) (No rearrangement, unlike Friedel-Crafts alkylation)

7. Define deactivating group. A deactivating group is a substituent on the benzene ring that decreases the electron density of the ring, making it less reactive toward electrophilic aromatic substitution. They withdraw electrons by induction or resonance and are meta directors. Examples: -NO₂, -CN, -COOH, -SO₃H, -CHO, -COR, halogens (weakly deactivating but ortho/para directing).

8. What are ortho, para, meta directing groups? These groups determine the position where the next substituent enters the benzene ring during EAS:

Ortho/Para directors: Activate the ring; direct incoming electrophile to ortho (1,2) and para (1,4) positions. Examples: -OH, -NH₂, -OR, -CH₃, -X (halogens).
Meta directors: Deactivate the ring; direct incoming electrophile to meta (1,3) position. Examples: -NO₂, -CN, -COOH, -CHO, -SO₃H.

UNIT 8

1. Define nucleophiles. Nucleophiles are electron-rich species with lone pairs or negative charges that attack electron-deficient (electrophilic) centers in a molecule. They donate electrons to form new covalent bonds. Examples: OH⁻, CN⁻, NH₃, H₂O, RO⁻, Br⁻. They are central to nucleophilic addition and substitution reactions.

2. Write down nucleophilic addition reaction. Nucleophilic addition is the addition of a nucleophile to the electrophilic carbonyl carbon (C=O) of aldehydes or ketones. General reaction:

      O                OH
      ‖                |
 R—C—H  + Nu⁻  →   R—C—H
                       |
                       Nu

Example: CH₃CHO + HCN → CH₃CH(OH)CN (cyanohydrin formation)

3. Write down aldol condensation reaction. Aldol condensation occurs between two carbonyl compounds (same or different) in the presence of a dilute base (NaOH) or acid. The product is a β-hydroxy aldehyde (aldol), which on heating dehydrates to an α,β-unsaturated carbonyl compound.

2 CH₃CHO → (NaOH) → CH₃CH(OH)CH₂CHO (aldol)
             → (heat) → CH₃CH=CHCHO + H₂O (crotonaldehyde)

4. Write down Wittig reaction. The Wittig reaction converts a carbonyl compound (aldehyde/ketone) into an alkene using a phosphorus ylide (Wittig reagent, R₂C=PPh₃).

R₁R₂C=O  +  R₃R₄C=PPh₃  →  R₁R₂C=CR₃R₄  +  O=PPh₃

Example: Benzaldehyde + Ph₃P=CH₂ → Styrene (Ph-CH=CH₂) + Ph₃P=O

5. Write down Benzoin condensation. Benzoin condensation is the coupling of two benzaldehyde molecules in the presence of a cyanide ion (CN⁻) catalyst (or thiamine) to form benzoin (a hydroxy ketone).

2 C₆H₅CHO → (CN⁻/ethanol) → C₆H₅-CH(OH)-CO-C₆H₅
                                    (Benzoin)

This is a nucleophilic addition reaction where CN⁻ acts as both nucleophile and leaving group (Umpolung).

6. Write down Cannizzaro reaction. The Cannizzaro reaction is a disproportionation reaction of aldehydes without α-hydrogen, in the presence of concentrated NaOH. One molecule is oxidized to a carboxylate salt and another is reduced to an alcohol.

2 HCHO → (conc. NaOH) → CH₃OH + HCOONa
(Formaldehyde)            (Methanol) (Sodium formate)
2 C₆H₅CHO → (NaOH) → C₆H₅CH₂OH + C₆H₅COONa

7. Write down Perkin reaction. The Perkin reaction is the condensation of an aromatic aldehyde with an acid anhydride in the presence of the sodium salt of the corresponding acid (base catalyst) to form an α,β-unsaturated aromatic acid (cinnamic acid type).

C₆H₅CHO + (CH₃CO)₂O → (CH₃COONa, heat) → C₆H₅CH=CHCOOH + CH₃COOH
(Benzaldehyde + Acetic anhydride → Cinnamic acid)

8. Write down Reformatsky reaction. The Reformatsky reaction is the reaction of an aldehyde or ketone with an α-halo ester in the presence of zinc metal to give a β-hydroxy ester after hydrolysis.

R-CHO + BrCH₂COOC₂H₅ → (Zn/ether) → R-CH(OH)-CH₂COOC₂H₅
(Aldehyde + Ethyl bromoacetate → β-hydroxy ester)

UNIT 9

1. Define diazotization. Diazotization is the reaction of a primary aromatic amine (ArNH₂) with nitrous acid (HNO₂ = NaNO₂ + HCl) at low temperature (0-5°C) to form a diazonium salt (ArN₂⁺Cl⁻).

C₆H₅NH₂ + NaNO₂ + HCl → (0-5°C) → C₆H₅N₂⁺Cl⁻ + NaCl + H₂O
(Aniline → Benzenediazonium chloride)

2. Write down Fries rearrangement. Fries rearrangement is the conversion of a phenol ester into a hydroxy aryl ketone (ortho or para) upon heating with a Lewis acid catalyst (AlCl₃).

C₆H₅-O-CO-CH₃ → (AlCl₃, heat) → o-OH-C₆H₄-CO-CH₃ + p-OH-C₆H₄-CO-CH₃
(Phenyl acetate → o- and p-hydroxyacetophenone)

Low temperature favors para product; high temperature favors ortho product.

3. Write down Kolbe reaction (Kolbe-Schmitt reaction). The Kolbe-Schmitt reaction is the reaction of sodium phenoxide (C₆H₅ONa) with CO₂ under high pressure and temperature (125°C) to form sodium salicylate, which on acidification gives salicylic acid.

C₆H₅ONa + CO₂ → (125°C, pressure) → C₆H₄(OH)COONa → (H⁺) → C₆H₄(OH)COOH
(Sodium phenoxide → Sodium salicylate → Salicylic acid)

4. What is Reimer-Tiemann reaction? The Reimer-Tiemann reaction is the formylation of phenol to introduce a -CHO group at the ortho position. Phenol reacts with chloroform (CHCl₃) and NaOH to form o-hydroxybenzaldehyde (salicylaldehyde) as the major product.

C₆H₅OH + CHCl₃ + NaOH → o-HO-C₆H₄-CHO (salicylaldehyde) + para product

5. Define bimolecular displacement. Bimolecular displacement (SN2) is a one-step nucleophilic substitution reaction in which both the substrate and nucleophile are involved in the transition state. The rate depends on both concentrations. Rate = k[substrate][nucleophile] The nucleophile attacks from the back and the leaving group departs simultaneously, causing inversion of configuration.

6. Write down Sandmeyer reaction. The Sandmeyer reaction is the replacement of the diazonium group (-N₂⁺) in an aryl diazonium salt with -Cl, -Br, -CN, or -OH using cuprous salts (CuCl, CuBr, CuCN) as catalysts.

C₆H₅N₂⁺Cl⁻ + CuCl → C₆H₅Cl + N₂ (chlorobenzene)
C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN + N₂ (benzonitrile)

7. Define electron migration. Electron migration is the movement of electrons (either a pair or single electron) from one part of a molecule to another during a chemical reaction. It includes:

1,2-hydride shift (H:⁻ migration in carbocation rearrangements)
1,2-methyl shift (CH₃:⁻ migration)
Electron delocalization through conjugated systems (resonance) It is a key concept in rearrangement reactions and reaction mechanisms.

8. Define nucleophilic aromatic substitution. Nucleophilic aromatic substitution (SNAr) is a reaction where a nucleophile replaces a leaving group on an aromatic ring. Unlike aliphatic systems, it requires strong electron-withdrawing groups (e.g., -NO₂) on the ring ortho/para to the leaving group to activate the ring. Mechanism: Addition-elimination (Meisenheimer complex intermediate). Example: 2,4-dinitrochlorobenzene + NaOH → 2,4-dinitrophenol

UNIT 10

1. What is oxidation? In organic chemistry, oxidation is a reaction involving the loss of hydrogen, gain of oxygen, or an increase in oxidation state of a carbon atom. Oxidizing agents include KMnO₄, K₂Cr₂O₇, H₂O₂, OsO₄. Example: Ethanol → Acetic acid (oxidation); Primary alcohol → Aldehyde → Carboxylic acid.

2. What is reduction? Reduction in organic chemistry is the gain of hydrogen, loss of oxygen, or decrease in oxidation state of a carbon atom. Reducing agents include LiAlH₄, NaBH₄, H₂/Pd, Zn/HCl. Example: Aldehydes → Primary alcohols; Ketones → Secondary alcohols; Nitrobenzene → Aniline.

3. Define assay. An assay is an analytical procedure used to determine the purity, potency, or concentration of a substance (drug, chemical, or biological agent). In pharmacy, it is performed to confirm that a drug meets prescribed quality standards. Types: Physical assay, Chemical assay (titrimetry, gravimetry), Biological assay (bioassay).

4. Draw the structure of sodium lauryl sulfate (SLS). SLS (Sodium Lauryl Sulfate / Sodium Dodecyl Sulfate) is an anionic surfactant.

CH₃-(CH₂)₁₁-O-SO₃⁻ Na⁺
or
C₁₂H₂₅-OSO₃Na

It consists of a 12-carbon alkyl chain (lauryl) attached to a sulfate group with a sodium counterion. Used as a detergent and emulsifying agent.

5. Draw the structure of salicylic (SA) acid. Salicylic acid is 2-hydroxybenzoic acid:

        COOH
        |
    [benzene ring]
        |
        OH   (ortho position)

Molecular formula: C₇H₆O₃ It has a benzene ring with -COOH at position 1 and -OH at position 2 (ortho). Used as a keratolytic and precursor to aspirin.

6. What are the uses of lactic acid? Lactic acid (2-hydroxypropanoic acid, CH₃-CH(OH)-COOH) is used in:

Pharmaceutical: As a pH adjuster in injections and topical preparations; treatment of skin conditions (keratolytic).
Food industry: Preservative and acidulant in fermented foods.
Cosmetics: Moisturizer, exfoliant (alpha-hydroxy acid).
Intravenous fluids: Component of Ringer's lactate solution.

7. Write a purity test for dimercaprol. Dimercaprol (BAL - British Anti-Lewisite; 2,3-dimercaptopropanol) purity tests include:

Specific gravity: 1.24 g/mL (measured by pycnometer).
Refractive index: 1.571-1.578.
Boiling range: 70-80°C at 1 mmHg.
Limit test for sulfur compounds: Purity confirmed by absence of excessive discoloration.
Assay: By iodimetric titration - dimercaprol is oxidized by iodine; not less than 98% purity required.

8. Draw the structure of urea.

        O
        ‖
  H₂N—C—NH₂

Molecular formula: CH₄N₂O (or CO(NH₂)₂) Urea is a diamide of carbonic acid. It has a central carbonyl group (C=O) flanked by two amino groups (-NH₂). Used as a diuretic (in pharmacology), skin moisturizer, and as a laboratory reagent for protein denaturation.

Quick Reference Summary for all 10 units:

Unit	Topics Covered
1	Acids/bases, Tautomerism, Metamerism, Lewis concept, Intermolecular forces, Ion pairing, Protic/Aprotic solvents, Isomerism
2	Alcohols/acids, Cycloalkanes, Phenols, Relative reactivity, Free radicals, Organic chemistry, Ketones/Dienes, Functional groups
3	Angle strain formula, Baeyer strain theory, Nucleophiles, Leaving groups, SN2, Steric hindrance, Solvolysis, Carbocation stability
4	Halogenation, E1/E2, Elimination, Substitution, Catalysis, Acid catalysis, Kinetics, Dehydrogenation
5	Electrophile, Free radical, Markovnikov's rule, Rearrangement, Halohydration, Peroxide effect, Hydrogen exchange, Free radical addition
6	Nucleophilic substitution, Resonance, Hyperconjugation, Substrate, Allylic rearrangement, Free radical substitution, Equilibrium, Dienes
7	Sulfonation, EAS, Halogenation, Activating groups, Nitration, Friedel-Crafts acylation, Deactivating groups, Directing groups
8	Nucleophiles, Nucleophilic addition, Aldol condensation, Wittig, Benzoin condensation, Cannizzaro, Perkin, Reformatsky
9	Diazotization, Fries rearrangement, Kolbe reaction, Reimer-Tiemann, Bimolecular displacement, Sandmeyer, Electron migration, SNAr
10	Oxidation, Reduction, Assay, SLS structure, Salicylic acid structure, Lactic acid uses, Dimercaprol purity, Urea structure

All answers are written for 2-mark Pharm D 1st year exam format - concise, accurate, with at least one example each. Good luck with your exams!

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Paper: OM Sterling Global University - Pharmaceutical Organic Chemistry (DPH104) - Pharm D Year 1

📋 SECTION - I (Attempt ALL) [10×2 = 20 Marks]

Q1. Define Intermolecular Forces

Definition: Intermolecular forces are the attractive or repulsive forces that act between neighbouring molecules. They are weaker than covalent bonds but play a major role in determining physical properties like boiling point, melting point, viscosity, and solubility.

Types:

Type	Example
Van der Waals / London Dispersion	Ar, CH₄
Dipole-Dipole	HCl, acetone
Hydrogen Bonding (strongest)	H₂O, NH₃, HF
Ion-Dipole	NaCl in water

Key Point: Hydrogen bonding is the strongest intermolecular force and is responsible for the unusually high boiling point of water (100°C).

Q2. Define Organic Chemistry

Definition: Organic chemistry is the branch of chemistry that deals with the study of carbon-containing compounds - their structure, properties, composition, reactions, and synthesis.

Scope includes:

Compounds of C, H, O, N, S, halogens
Natural products, drugs, polymers, dyes

Importance in Pharmacy:

All drugs are organic molecules
Drug synthesis, metabolism, and stability depend on organic chemistry principles

"Organic chemistry is the chemistry of life and the foundation of pharmaceutical science."

Q3. What is Stearic Hindrance?

Definition: Steric hindrance is the slowing or blocking of a chemical reaction due to the large size of substituent groups surrounding the reactive site (the carbon being attacked). Bulky groups physically block the approach of an incoming reagent.

Example:

Primary alkyl halide (CH₃CH₂Br): Less hindered - SN2 reaction occurs easily
Tertiary alkyl halide (CH₃)₃CBr: Highly hindered - three bulky methyl groups block attack, so SN1 occurs instead

Rule: More bulky groups around the reactive carbon = more steric hindrance = slower reaction

Q4. Define Kinetics

Definition: Chemical kinetics is the branch of chemistry that studies the rate (speed) of chemical reactions and the factors that influence it.

Rate Law: $$\text{Rate} = k[A]^m [B]^n$$

Where:

k = rate constant
m, n = order of reaction
[A], [B] = concentrations

Factors affecting rate:

Concentration of reactants
Temperature (higher T = faster rate)
Catalyst (lowers activation energy)
Pressure (for gases)
Surface area

Importance: Kinetics helps determine the mechanism of a reaction (SN1 vs SN2, E1 vs E2).

Q5. What is Peroxide Effect?

Definition: The peroxide effect (also called the Kharasch effect or anti-Markovnikov addition) is the reversal of Markovnikov's rule during the addition of HBr to an alkene when peroxides (ROOR) are present. The reaction proceeds by a free radical mechanism instead of ionic mechanism.

Mechanism (steps):

Step 1 - INITIATION:
ROOR → 2RO•  (peroxide breaks by heat/light)
RO• + HBr → ROH + Br•  (bromine radical formed)

Step 2 - PROPAGATION:
Br• + CH₂=CH₂ → BrCH₂-CH₂•  (Br adds to less substituted C)
BrCH₂-CH₂• + HBr → BrCH₂-CH₃ + Br•

Step 3 - TERMINATION:
Br• + Br• → Br₂

Result:

CH₃-CH=CH₂ + HBr  --[Peroxide]-->  CH₃-CH₂-CH₂Br  (anti-Markovnikov)
                                     (Br goes to terminal C)

WITHOUT peroxide:
CH₃-CH=CH₂ + HBr  -----------------> CH₃-CHBr-CH₃  (Markovnikov)

Note: Peroxide effect is observed ONLY with HBr, NOT with HCl or HI.

Q6. Define Free Radical Substitution

Definition: Free radical substitution is a reaction in which a hydrogen atom of an organic molecule is replaced by a halogen (or other radical) through a free radical chain mechanism. It is initiated by UV light or heat.

Mechanism - Three Stages:

STAGE 1: INITIATION
Cl₂  --hν-->  2Cl•  (homolytic cleavage)

STAGE 2: PROPAGATION (chain reaction)
Cl• + CH₄  →  •CH₃ + HCl
•CH₃ + Cl₂ →  CH₃Cl + Cl•  (cycle repeats)

STAGE 3: TERMINATION
Cl• + Cl• → Cl₂
•CH₃ + Cl• → CH₃Cl
•CH₃ + •CH₃ → C₂H₆

Overall reaction:

CH₄ + Cl₂  --hν-->  CH₃Cl + HCl
(Methane)  (Chloromethane)

Q7. Define Free Radical Addition

Definition: Free radical addition is the addition of a molecule (like HBr) across a double bond via a free radical mechanism, initiated by peroxides or UV light. It gives the anti-Markovnikov product.

Example:

CH₂=CH₂ + Br•  →  BrCH₂-CH₂•
BrCH₂-CH₂• + HBr → BrCH₂-CH₃ + Br•

Difference from Ionic Addition:

Feature	Ionic Addition	Free Radical Addition
Initiator	H⁺ (acid)	Peroxide / UV light
Rule followed	Markovnikov	Anti-Markovnikov
Intermediate	Carbocation	Free radical

Q8. Define Friedel-Craft Acylation

Definition: Friedel-Crafts acylation is an electrophilic aromatic substitution reaction in which an acyl group (-CO-R) is introduced into a benzene ring using an acyl halide (RCOCl) and a Lewis acid catalyst (AlCl₃).

Reaction:

        + CH₃COCl  --AlCl₃-->  [benzene-CO-CH₃] + HCl
Benzene  Acetyl chloride         Acetophenone

Mechanism:

Step 1: RCOCl + AlCl₃ → RCO⁺ (acylium ion, electrophile) + AlCl₄⁻
Step 2: RCO⁺ attacks benzene ring → sigma complex (arenium ion)
Step 3: Loss of H⁺ → aryl ketone product + aromaticity restored

Advantages over Friedel-Crafts Alkylation:

NO carbocation rearrangement
Gives single product
Product deactivates ring (no polyacylation)

Q9. Write Down Sandmeyer's Reaction

Definition: Sandmeyer's reaction is the replacement of the diazonium group (-N₂⁺) in an aryl diazonium salt with -Cl, -Br, -CN, -OH using cuprous salts (Cu⁺ salts) as catalysts.

Reactions:

C₆H₅-N₂⁺Cl⁻ + CuCl  → C₆H₅Cl + N₂↑  (Chlorobenzene)
C₆H₅-N₂⁺Cl⁻ + CuBr  → C₆H₅Br + N₂↑  (Bromobenzene)
C₆H₅-N₂⁺Cl⁻ + CuCN  → C₆H₅CN + N₂↑  (Benzonitrile)
C₆H₅-N₂⁺Cl⁻ + CuI   → C₆H₅I  + N₂↑  (Iodobenzene)

Starting material:

C₆H₅-NH₂ + NaNO₂ + HCl → C₆H₅-N₂⁺Cl⁻
(Aniline)  [0-5°C]        (Diazonium salt)

Importance: Sandmeyer's reaction is used to introduce halogens and cyano groups into benzene ring, which are otherwise difficult to introduce directly.

Q10. Draw the Structure of Salicylic Acid

Name: Salicylic acid = 2-Hydroxybenzoic acid Molecular Formula: C₇H₆O₃ Molecular Weight: 138.12 g/mol

Structure:

         COOH   ← carboxyl group at position 1
          |
    [  benzene  ]
          |
         OH    ← hydroxyl group at position 2 (ortho)

Full drawn structure:

        COOH
       /
  [1]
 /    \
[6]   [2]
|      |
[5]   [3]
 \    /
  [4]

Position 1: -COOH
Position 2: -OH (ortho to COOH)

Properties:

White crystalline solid
Melting point: 159°C
Slightly soluble in water

Pharmaceutical Uses:

Keratolytic agent (removes hard skin)
Precursor to Aspirin (acetylsalicylic acid)
Anti-fungal in topical preparations
Treatment of acne, psoriasis, warts

📋 SECTION - II (Attempt ANY TWO) [2×10 = 20 Marks]

Q (Marked *): Explain in detail about SN₂ Reaction. What are the factors affecting it?

SN2 Reaction - Full Detail

Full Name: Substitution Nucleophilic Bimolecular

Definition: SN2 is a one-step (concerted) nucleophilic substitution reaction in which the nucleophile attacks the carbon from the back side (180°) while the leaving group departs simultaneously. Both the substrate and nucleophile are involved in the rate-determining step.

Rate Law: $$\text{Rate} = k[\text{Substrate}][\text{Nucleophile}]$$ (2nd order kinetics - depends on BOTH concentrations)

Mechanism:

        Nu:⁻  +  C—LG   →   [Nu---C---LG]‡   →   Nu—C  +  LG⁻
                              Transition State
                              (pentacoordinate)

Key Feature - BACK SIDE ATTACK:

                        LG
                        |
  Nu:⁻ ----→ ----→ ----C----   Nu attacks exactly opposite to LG
                        |
                      (R groups)

Result - WALDEN INVERSION (Inversion of Configuration):

  Like an umbrella turning inside-out in wind!

  Before: R-configuration
  After:  S-configuration  (complete inversion)

Energy Profile Diagram:

Energy
  |        ‡ (Transition State - highest point)
  |       /\
  |      /  \
  |     /    \
  |----/      \----
  |  Reactants  Products
  |_______________________
            Reaction Progress

Only ONE energy peak = single transition state
NO intermediate formed

Factors Affecting SN2 Reaction:

1. Structure of Substrate (MOST IMPORTANT)

Reactivity order in SN2:
CH₃X > Primary (1°) > Secondary (2°) >> Tertiary (3°)
(Methyl = fastest)              (Tertiary = too hindered, does NOT do SN2)

Reason: More alkyl groups = more steric hindrance = back-side attack blocked

2. Nature of Nucleophile

Strong Nucleophile	Weak Nucleophile
OH⁻, CN⁻, I⁻, RS⁻	H₂O, ROH
Favors SN2	Does NOT favor SN2

Strong nucleophiles are needed. Large/bulky nucleophiles (e.g., tert-butoxide) slow SN2.

3. Nature of Leaving Group

Good leaving group = weak base = leaves easily

Best leaving groups: I⁻ > Br⁻ > Cl⁻ > F⁻
                     (I⁻ is BEST for SN2)

4. Solvent

Polar Aprotic Solvent	Polar Protic Solvent
Acetone, DMSO, DMF	Water, Ethanol
BEST for SN2	Slows SN2 (solvates nucleophile)

Polar aprotic solvents leave the nucleophile "naked" and more reactive.

5. Concentration of Nucleophile

High nucleophile concentration → faster SN2
Because rate = k[substrate][nucleophile]

Example:

CH₃CH₂Br + OH⁻  --DMSO-->  CH₃CH₂OH + Br⁻
(Ethyl bromide)  (SN2)  (Ethanol)

Back-side attack → Product has inverted configuration

Summary Table: SN1 vs SN2

Feature	SN1	SN2
Steps	2-step	1-step
Rate	k[substrate]	k[substrate][Nu]
Substrate	3° > 2°	CH₃ > 1° > 2°
Intermediate	Carbocation	None (transition state)
Stereochemistry	Racemization	Inversion
Solvent	Polar protic	Polar aprotic

Q2 (Marked 2): Explain the Kinetics and Mechanism of E1 in Detail

E1 Elimination - Full Detail

Full Name: Elimination Unimolecular (E1)

Definition: E1 is a two-step elimination reaction in which the leaving group departs first to form a carbocation intermediate, and then a base removes a proton (β-hydrogen) from the adjacent carbon to give an alkene.

Rate Law: $$\text{Rate} = k[\text{Substrate}]$$ (1st order - depends ONLY on substrate concentration, NOT on base)

Mechanism - Step by Step:

Step 1: Ionization (Slow - Rate Determining Step)

         CH₃                    CH₃
          |                      |⊕
  CH₃ -- C -- Br  →  CH₃ -- C     +  Br⁻
          |                      |
          H                      H
   (tert-butyl bromide)    (carbocation intermediate)
         SLOW STEP

Step 2: Proton Removal (Fast)

              CH₃                     CH₃
               |⊕                      |
  Base: +  CH₃-C   →  CH₃ -- C=CH₂  +  Base-H⁺
               |                   
               CH₂-H             
           (Base removes β-H)    (ALKENE formed)
               FAST STEP

Energy Profile Diagram for E1:

Energy
  |          ‡₁                ‡₂
  |         /\               /\
  |        /  \             /  \
  |       /    \           /    \
  |------/      ----------/      \------
  |  Reactants  Carbocation     Products
  |             Intermediate
  |_________________________________________
                Reaction Progress

TWO energy peaks = TWO transition states
ONE intermediate = Carbocation

Kinetics of E1:

Rate = k [R-X]

If [substrate] doubles → Rate doubles
If [base] doubles → Rate UNCHANGED (base not in rate equation)

This confirms the 2-step mechanism where step 1 (ionization) is rate-determining.

Factors Favoring E1:

1. Substrate Structure

Tertiary (3°) >> Secondary (2°) >> Primary (1°)
(3° carbocation is most stable → most favored)

2. Solvent

Polar protic solvents (water, ethanol) strongly favor E1
They stabilize the carbocation intermediate by solvation

3. Temperature

High temperature favors elimination (E1 and E2) over substitution

4. Leaving Group

Good leaving groups (I⁻, Br⁻, OTs⁻) favor E1

5. Base Concentration

E1 does NOT require a strong base
Weak bases (H₂O, ROH) are sufficient

Markovnikov's Rule for E1 Products (Zaitsev's Rule):

When multiple alkenes can form, the more substituted (more stable) alkene is the major product.

        CH₃                          
         |                           
  CH₃ - C - CH₂CH₃  --E1-->  CH₃-C=CHCH₃  (major - more substituted)
         |                  +  CH₂=C-CH₂CH₃ (minor)
         Br

Stereochemistry of E1:

E1 gives a mixture of E and Z (cis/trans) isomers
No strict stereospecificity (unlike E2 which needs anti-periplanar geometry)
The more stable E (trans) isomer is usually the major product

Comparison: E1 vs E2

Feature	E1	E2
Steps	2 (stepwise)	1 (concerted)
Rate law	k[substrate]	k[substrate][base]
Intermediate	Carbocation	None
Base needed	Weak base	Strong base
Substrate	3° > 2°	3° > 2° > 1°
Solvent	Polar protic	Polar aprotic
Stereochemistry	Mixture (E+Z)	Anti-periplanar (E)
Rearrangement	YES possible	NO

Q3: Explain Peroxide Effect and Markovnikov's Rule (Detail)

(Already covered in Section I Q5 - expanded version below)

Markovnikov's Rule:

Statement: When an unsymmetrical reagent (HX) adds to an unsymmetrical alkene, the hydrogen atom adds to the carbon with MORE hydrogen atoms (more hydrogen-rich carbon), and the X group adds to the carbon with FEWER hydrogen atoms.

Simple language: "The rich gets richer" - the carbon already having more H atoms gets one more H.

CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃  (Markovnikov product, MAJOR)
                   → CH₃-CH₂-CH₂Br (Anti-Markovnikov, MINOR)

Reason: The more substituted carbocation (2° or 3°) is more stable, so H⁺ adds to give the more stable intermediate.

Peroxide Effect reverses this rule (as explained in Section I Q5 above) via free radical mechanism.

📋 SECTION - III (Attempt ANY SIX) [6×5 = 30 Marks]

Q1 (Marked *): Explain Chain Isomerism and Functional Isomerism with Example

A) Chain Isomerism (Skeletal Isomerism)

Definition: Chain isomerism occurs when two compounds have the same molecular formula but different arrangements of the carbon skeleton (chain). The main carbon chain differs - one may be straight (normal) while another is branched.

Example: C₅H₁₂ (Pentane)

1. n-Pentane (straight chain):
   CH₃-CH₂-CH₂-CH₂-CH₃

2. Isopentane (one branch):
   CH₃-CH-CH₂-CH₃
       |
       CH₃

3. Neopentane (highly branched):
       CH₃
       |
   CH₃-C-CH₃
       |
       CH₃

All three have formula C₅H₁₂ but different carbon skeletons.

Properties differ: n-pentane (bp 36°C) vs neopentane (bp 9.5°C)

B) Functional Isomerism

Definition: Functional isomers have the same molecular formula but different functional groups, so they belong to different classes of organic compounds and have very different chemical properties.

Examples:

C₂H₆O:

1. Ethanol (Alcohol):     CH₃-CH₂-OH  (functional group: -OH)
2. Dimethyl ether:        CH₃-O-CH₃   (functional group: -O-)

C₃H₆O:

1. Propanal (Aldehyde):   CH₃-CH₂-CHO   (functional group: -CHO)
2. Acetone (Ketone):      CH₃-CO-CH₃    (functional group: -CO-)

C₂H₄O₂:

1. Acetic acid:           CH₃-COOH      (functional group: -COOH)
2. Methyl formate:        HCOO-CH₃      (functional group: -COO-)
3. Glycolaldehyde:        HOCH₂-CHO     (two functional groups)

Key Difference:

Feature	Chain Isomers	Functional Isomers
Same functional group?	YES	NO
Same carbon chain?	NO	Can be same or different
Chemical behavior	Similar	Completely different

Q2: Define Free Radicals, Explain its Mechanism with Example

Free Radicals - Complete Note

Definition: A free radical is a highly reactive chemical species containing one or more unpaired electrons. It is electrically neutral and is formed by homolytic cleavage of a covalent bond (each atom gets one electron).

Homolytic cleavage:    A : B  →  A•  +  B•
(Each atom gets 1 electron - shown as dot •)

Formation of Free Radicals:

1. By UV light (photolysis):

Cl : Cl  --hν-->  Cl•  +  Cl•
(Each Cl gets one electron)

2. By heat:

(CH₃)₃C-OO-C(CH₃)₃  --heat-->  2 (CH₃)₃C-O•
(Peroxide breaks to give oxy radicals)

Properties of Free Radicals:

Property	Description
Electrons	One unpaired electron
Charge	Neutral (no charge)
Stability	Very short-lived, highly reactive
Geometry	Planar (sp² hybridized)
Stability order	3° > 2° > 1° > methyl

Chain Mechanism of Free Radical Halogenation:

Example: Chlorination of Methane

Overall reaction:

CH₄ + Cl₂  --hν-->  CH₃Cl + HCl

Stage 1: INITIATION (starting the chain)

Cl₂  --hν-->  2 Cl•
(UV light breaks Cl-Cl bond homolytically)

Stage 2: PROPAGATION (chain carries forward)

Step 1: Cl• + CH₄ → •CH₃ + HCl
Step 2: •CH₃ + Cl₂ → CH₃Cl + Cl•
(Cl• formed in Step 2 goes back and repeats Step 1)
(Chain reaction - thousands of cycles!)

Stage 3: TERMINATION (chain stops)

Cl•  + Cl•  → Cl₂
•CH₃ + Cl•  → CH₃Cl
•CH₃ + •CH₃ → C₂H₆
(Two radicals combine - chain stops)

Stability Order of Free Radicals:

       CH₃             CH₃             CH₃
        |               |               |
   CH₃-C•   >   CH₃-C•-H   >   CH₃-C•-H₂   >   •CH₃
        |               |
       CH₃             H

   Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl
   (MOST stable)                              (LEAST stable)

Reason: More alkyl groups donate electrons by hyperconjugation, stabilizing the radical.

Importance in Pharmacy:

Free radicals cause oxidative damage to drugs and body cells
Antioxidants (Vitamin C, E) work by neutralizing free radicals
Used in polymerization of pharmaceutical polymers
Free radical reactions are used in drug synthesis

Q3: Discuss the Kinetics of 1st Order and 2nd Order Reactions

Chemical Kinetics - Reaction Orders

Kinetics Definition: The study of the rate of chemical reactions and factors affecting it.

First Order Reaction:

Definition: A reaction in which the rate depends on the concentration of ONE reactant raised to the first power.

Rate Law: $$\text{Rate} = k[A]^1 = k[A]$$

Integrated Rate Law: $$\ln[A] = \ln[A]_0 - kt$$ $$\text{or } [A] = [A]_0 \cdot e^{-kt}$$

Half-life (t₁/₂): $$t_{1/2} = \frac{0.693}{k}$$ (Half-life is CONSTANT and independent of concentration)

Graph: ln[A] vs time = straight line

ln[A]
  |\.
  | \.
  |  \.
  |   \.  slope = -k
  |    \.
  |_____________
       Time

Examples:

Radioactive decay
Drug decomposition in pharmacy (most drugs follow 1st order)
N₂O₅ → 2NO₂ + ½O₂

Second Order Reaction:

Definition: A reaction in which the rate depends on the concentration of TWO reactants (each raised to first power) or one reactant raised to the second power.

Rate Law: $$\text{Rate} = k[A]^2$$ $$\text{or Rate} = k[A][B]$$

Integrated Rate Law: $$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$$

Half-life: $$t_{1/2} = \frac{1}{k[A]_0}$$ (Half-life DEPENDS on initial concentration - doubles when concentration halves)

Graph: 1/[A] vs time = straight line

1/[A]
  |          /
  |         /
  |        /  slope = +k
  |       /
  |      /
  |_____________
       Time

Examples:

SN2 reactions: Rate = k[RX][Nu]
H₂ + I₂ → 2HI
Saponification of esters

Comparison Table:

Feature	1st Order	2nd Order
Rate law	k[A]	k[A]² or k[A][B]
Units of k	s⁻¹	L mol⁻¹ s⁻¹
Half-life	0.693/k (constant)	1/k[A]₀ (varies)
Linear plot	ln[A] vs t	1/[A] vs t
Examples	Drug degradation	SN2 reactions

Q4: Mechanism of Halogenation (Free Radical)

(Complete mechanism as written in Section I Q6 and Section III Q2 above - refer to those)

Additional Points for full marks:

Selectivity in Halogenation:

When propane reacts with Cl₂ or Br₂, different products form:

CH₃-CH₂-CH₃ + Cl₂ → CH₃-CHCl-CH₃ (2-chloropropane, major)
                    + ClCH₂-CH₂-CH₃ (1-chloropropane, minor)

Selectivity order:

Bromine (Br₂) is more selective than Chlorine (Cl₂)
Chlorine is more reactive but less selective

Reactivity of H atoms:

Tertiary H (3°) > Secondary H (2°) > Primary H (1°)
(Easiest to abstract)                (Hardest to abstract)

Q5: Write a Note on 1,4-Addition vs 1,2-Addition

1,2-Addition vs 1,4-Addition in Conjugated Dienes

Conjugated diene example: 1,3-Butadiene

CH₂=CH-CH=CH₂
 1    2   3   4

When HBr is added to 1,3-butadiene, TWO products can form:

1,2-Addition (Direct Addition):

HBr adds to carbons 1 and 2 (adjacent carbons of one double bond)
Product: 3-Bromobut-1-ene

CH₂=CH-CH=CH₂ + HBr → CH₃-CHBr-CH=CH₂
  1   2  3   4              (3-Bromobut-1-ene)
                       [double bond remains at C3-C4]

Favored at low temperature (-80°C)
Product is formed FASTER (kinetic control)
Kinetic product

1,4-Addition (Conjugate Addition):

HBr adds to carbons 1 and 4 (ends of the conjugated system)
Product: 1-Bromobut-2-ene

CH₂=CH-CH=CH₂ + HBr → CH₃-CH=CH-CH₂Br
  1   2  3   4              (1-Bromobut-2-ene)
                       [double bond shifts to C2-C3]

Favored at high temperature (40°C)
Product is more STABLE (thermodynamic control)
Thermodynamic product

Mechanism (via allylic carbocation):

Step 1: H⁺ adds to C1:
CH₂=CH-CH=CH₂ + H⁺ → CH₃-⁺CH-CH=CH₂
                       ↕ resonance
                       CH₃-CH=CH-⁺CH₂
                      (allylic carbocation)

Step 2a: Br⁻ attacks C2 → 1,2-addition product
Step 2b: Br⁻ attacks C4 → 1,4-addition product

Summary:

Feature	1,2-Addition	1,4-Addition
Position of addition	C1 and C2	C1 and C4
Temperature	Low (-80°C)	High (40°C)
Product type	Kinetic product	Thermodynamic product
Stability	Less stable	More stable
Double bond position	C3-C4	C2-C3 (internal)

Q6: Explain the Mechanism of Nitration

Nitration of Benzene - Complete Mechanism

Definition: Nitration is an electrophilic aromatic substitution (EAS) reaction in which a nitro group (-NO₂) replaces a hydrogen atom on the benzene ring using a mixture of concentrated HNO₃ and H₂SO₄ (nitrating mixture).

Overall Reaction:

C₆H₆ + HNO₃  --conc.H₂SO₄-->  C₆H₅NO₂ + H₂O
(Benzene)                       (Nitrobenzene)

Step 1: Generation of Electrophile (Nitronium ion, NO₂⁺)

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
              (Nitronium ion = electrophile)

H₂SO₄ protonates HNO₃, which then loses water to form the very reactive NO₂⁺.

Step 2: Electrophilic Attack on Benzene Ring

        NO₂⁺
         |
  [benzene] + NO₂⁺  →  [sigma complex / arenium ion]
                         (Wheland Intermediate)
                         positive charge delocalized on ring
                         AROMATICITY TEMPORARILY LOST

Structural representation:

      NO₂             NO₂             NO₂
       |               |               |
  ⊕[benzene] ↔  [benzene]⊕ ↔  [benzene with ⊕]
  (ortho +)     (para +)       (meta +)
  (3 resonance structures of sigma complex)

Step 3: Loss of Proton (Restoration of Aromaticity)

Sigma complex  →  C₆H₅NO₂ + H⁺
                  (Nitrobenzene)    (aromaticity restored)
H⁺ + HSO₄⁻ → H₂SO₄ (catalyst regenerated)

Energy Profile:

Energy
  |         ‡ (sigma complex formation - SLOW step)
  |        /\
  |       /  \
  |      /    \.........
  |-----/              \---
  |  Benzene+NO₂⁺     Nitrobenzene
  |_________________________________
         Reaction Progress

Effect of Substituents on Nitration:

Substituent Present	Effect	Second -NO₂ goes to
-OH, -NH₂, -CH₃ (activating)	Faster than benzene	Ortho + Para positions
-NO₂, -COOH, -CN (deactivating)	Slower than benzene	Meta position

Example:

Toluene (has -CH₃ group):
C₆H₅CH₃ + HNO₃/H₂SO₄ → o-Nitrotoluene + p-Nitrotoluene (major)
                                           (m-Nitrotoluene = minor)

Q7: Write a Note on Ionization of Carboxylic Acids

Ionization of Carboxylic Acids

Ionization: When a carboxylic acid dissolves in water, it partially dissociates (ionizes) to release a proton (H⁺) and form a carboxylate anion (RCOO⁻).

General equation:

RCOOH  + H₂O  ⇌  RCOO⁻  +  H₃O⁺
(Weak acid)      (Carboxylate ion)  (Hydronium ion)

Acid Dissociation Constant (Ka):

$$K_a = \frac{[\text{RCOO}^-][\text{H}_3\text{O}^+]}{[\text{RCOOH}]}$$

$$pK_a = -\log K_a$$

Lower pKa = Stronger acid

Factors Affecting Ionization (Acidity) of Carboxylic Acids:

1. Inductive Effect:

Electron-withdrawing groups (-Cl, -F, -NO₂) increase acidity (lower pKa)
They stabilize the carboxylate anion by pulling electron density away

Cl-CH₂-COOH > HCOOH > CH₃COOH > (CH₃)₃C-COOH
(Chloroacetic)  (Formic)  (Acetic)   (Pivalic)
  pKa 2.86       3.75      4.76       5.05
  (MOST acidic)              (LEAST acidic)

2. Number of electron-withdrawing groups:

CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH
pKa: 0.65    1.48         2.86        4.76
(More Cl = more acidic)

3. Resonance Stabilization of Carboxylate Ion: The carboxylate anion (RCOO⁻) is stabilized by resonance - the negative charge is spread over both oxygen atoms equally:

       O                    O⁻
       ‖                    |
  R—C—O⁻   ↔   R—C=O
  
Actual structure: Both C-O bonds are equal (1.5 bond order)
Negative charge equally distributed on both oxygens

This resonance stabilization makes RCOOH MORE acidic than alcohols (ROH).

4. Aromatic vs Aliphatic acids:

Benzoic acid (C₆H₅COOH) pKa = 4.2
Acetic acid  (CH₃COOH)  pKa = 4.76
(Benzoic acid is slightly stronger due to phenyl ring)

Comparison: Acidity of different compounds

Strongest acid                                    Weakest acid
     ↓                                                 ↓
  HCl > RCOOH > H₂CO₃ > ArOH > ROH > R-C≡C-H > RH
  (Mineral)  (Carboxylic) (Carbonic) (Phenol) (Alcohol)     (Alkane)

Q8: Write a Note on Basicity of Amines

Basicity of Amines

Amines as Bases: Amines are organic bases because the nitrogen atom has a lone pair of electrons that can accept a proton (H⁺) from acids.

General reaction:

R-NH₂  +  H₂O  ⇌  R-NH₃⁺  +  OH⁻
(Amine)              (Ammonium ion)

Basicity constant (Kb): $$K_b = \frac{[RNH_3^+][OH^-]}{[RNH_2]}$$ $$pK_b = -\log K_b$$

Higher Kb (lower pKb) = Stronger base

Factors Affecting Basicity of Amines:

1. Inductive Effect of Alkyl Groups: Alkyl groups push electrons toward nitrogen (electron-donating), making the lone pair MORE available for proton acceptance.

In gas phase: (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
              (Trimethyl)  (Dimethyl)  (Methyl) (Ammonia)

However, in aqueous solution, solvation complicates the order:

In water: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
          (Dimethylamine is MOST basic in water)

Reason: Trimethylamine has 3 methyl groups blocking solvation of N-H⁺, reducing stability of conjugate acid.

2. Aromatic Amines vs Aliphatic Amines:

Aliphatic amine (CH₃NH₂, pKb = 3.36) >> Aromatic amine (C₆H₅NH₂, pKb = 9.4)
(MUCH more basic)                         (MUCH less basic)

Reason: In aniline (C₆H₅NH₂), the lone pair on N is delocalized into the benzene ring by resonance. This makes the lone pair less available to accept H⁺, reducing basicity.

C₆H₅-NH₂ resonance structures:
   NH₂              ⊕NH₂            ⊕NH₂
    |           /      |        \      |
[benzene] ↔  [ring]  ↔  [ring]  ...
(Lone pair donated to ring - less available for protonation)

3. Effect of Substituents on Aromatic Amines:

Substituent on Ring	Effect on Basicity
Electron-donating (-OH, -OCH₃, -CH₃)	Increases basicity
Electron-withdrawing (-NO₂, -Cl, -CN)	Decreases basicity
Position effect: para > ortho > meta

Example:

p-Aminophenol > Aniline > p-Nitroaniline
(More basic)             (Least basic)

4. Hybridization of Nitrogen:

N hybridization	Example	Basicity
sp³	Alkylamines (R-NH₂)	Most basic
sp²	Aniline (Ar-NH₂)	Less basic
sp	R-C≡N (Nitriles)	Least basic

More s-character = lone pair held more tightly = weaker base

Summary: Basicity Order of Amines

Aliphatic amines > NH₃ > Aromatic amines > Amides
(RNH₂)         (Ammonia)  (ArNH₂)       (RCONH₂)
Most basic                              Least basic (N lone pair in resonance with C=O)

Exam Tip: For 5-mark questions, always include: (1) Definition, (2) Equation/Reaction, (3) Mechanism or explanation, (4) Examples with structures, (5) A comparison table. This structure guarantees maximum marks.

All answers are written in:

Simple, examiner-friendly language
Proper definitions with examples
Step-by-step mechanisms
Comparison tables for clarity
Logical flow that earns extra marks

Good luck with your exam! 🎯

(Unit 1) ✨️10 marks ✨️✨️ 1)Explain all the theorys of acid and bases in details. 2)Give a brief summary of intermolecular forces. 3)Define boiling point,melting point and solubility.give a detail note on dipolar moment. 4)Define the term isomerism. Give a detail note on structural isomerism in organic compound. (Unit 2) ✨️✨️✨️ 1) What is organic chemistry. Give a detail note on classification of organic compound. 2)Give a detail note on IUPAC nomenclature of aldehydes and amines with example. 3) Write a note on free radical chain. Reaction of alkene with mechanism. 4) Write a note on relative reactivity and stability of free radicals. (Unit3)✨️✨️✨️ 1)Explain in detail about SN2 reaction. What are the factor affecting the reaction. 2)Discuss the mechanism of SN1 reaction.what are the factor affecting the reaction. 3)Give a detail note on nucleophilis and leaving group. What is the role of steric hinderance. 4) Give a brief note on carbo cations their stability and rearrangment. (Unit 4)✨️✨️✨️ 1) Explain the kinetics and mechanism of E1 in detail. 2)Explain the kinetics and mechanism of E2 in detail. 3)Give a detail note on element effects orrientation and reactivity in E1 and E2. 4) Discuss a note on elemination v/s substitutio. Dehydration of alcohol and assay of dehydration. (Unit 5)✨️✨️✨️ 1)Write a note on mechanism of free radical addition. 2)Explain in detail about proxide effects and markonioff rule. 3)Discuss about mechanism of peroxide initiation addition of hydrogen bromide. 4)What is the mechanism of halogenation. Explain orientation about free radical additions. (Unit 6)✨️✨️✨️ 1)Explain yhe mechanism of free radical, halogenation of alkenes. 2)Discuss about nucleophilics substitution in alkalic substrate. 3)Explain the orientation and reactivity of free radical addition of conjucated dienes. 4)Compare free radical substitution with free radical additions. (Unit 7)✨️✨️✨️ 1)Explain the effect of halogen on electrophilics aromatics substituent in alkyl banzene. 2)Discuss about resonance stabilization of benzyl radical. 3)Write about freedial craft alkylation with mechanism. 4)Discuss about determination, orientation and relative reactivity of ESR (Unit 8 )✨️✨️✨️ 1)Write down cross aldol condensation reaction with mechanism. 2)Explain the mechanism of perkine condensation. 3)Explain the mechanism of ì canizzaro reaction. (Unit 9)✨️✨️✨️ 1)Write about villiam son synthesis. 2)Explain fries rearrangement. 3)Discuss about kolbe reaction. 4)Write a note on remer triman reaction. (Unit 10)✨️✨️✨️ 1)Write down the preparation purity test and medicinal use of urra. 2)Write down the preparation purity test and medicinal use of benzyl. 3)Write down the preparation purity test and medicinal of benzyl. 4)Write down the preparation purity test and assay of chlorobutol. Answer these with high yield marks and for university level checker give extra marks impressive format in easy language try to have same basic concepts and simple words through which examiner can easily understand what's written as per 10 marks answer

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