Samacheer Kalvi Class 11 Chemistry Solutions Chapter 12 Basic Concepts of Organic Reactions

Textbook Evaluation:

I. Choose the best answer:

Question 1. For the following reactions
(A) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{KOH} \rightarrow \text{CH}_3 - \text{CH} = \text{CH}_2 + \text{KBr} + \text{H}_2\text{O} \)
(B) \( (\text{CH}_3)_3\text{CBr} + \text{KOH} \rightarrow (\text{CH}_3)_3\text{COH} + \text{KBr} \)
(C)
Which of the following statement is correct?
(a) (A) is elimination, (B) and (C) are substitution
(b) (A) is substitution, (B) and (C) are elimination
(c) (A) and (B) are elimination and (C) is addition reaction
(d) (A) is elimination, (B) is substitution and (C) is addition reaction
Answer: (d) (A) is elimination, (B) is substitution and (C) is addition reaction
In simple words: Reaction (A) removes parts of the molecule to form a double bond. Reaction (B) replaces one atom with another. Reaction (C) adds atoms across a double bond.

🎯 Exam Tip: To identify reaction types, look for changes in bonding: elimination forms double/triple bonds, substitution exchanges groups, and addition breaks double/triple bonds to form single bonds.

Question 2. What is the hybridisation state of benzyl carbonium ion?
(a) sp²
(b) spd²
(c) sp³
(d) sp²d
Answer: (a) sp²
In simple words: The central carbon atom in a benzyl carbonium ion forms three bonds and has no lone pairs, making its hybridization sp². This causes it to have a flat shape.

🎯 Exam Tip: Remember that a carbocation carbon typically has sp² hybridization because it has three bonds and an empty p-orbital, resulting in a planar geometry.

Question 3. Decreasing order of nucleophilicity is
(a) \( \text{OH}^- > \text{NH}_2^- > -\text{OCH}_3 > \text{RNH}_2 \)
(b) \( \text{NH}_2^- > \text{OH}^- > -\text{OCH}_3 > \text{RNH}_2 \)
(c) \( \text{NH}_2^- > \text{CH}_3\text{O}^- > \text{OH}^- > \text{RNH}_2 \)
(d) \( \text{CH}_3\text{O}^- > \text{NH}_2^- > \text{OH}^- > \text{RNH}_2 \)
Answer: (b) \( \text{NH}_2^- > \text{OH}^- > -\text{OCH}_3 > \text{RNH}_2 \)
In simple words: Nucleophilicity is how strongly a molecule attacks a positive center. Ammonia (NH2-) is the strongest attacker here, followed by hydroxide (OH-), then methoxide (-OCH3), and finally a primary amine (RNH2). This is because NH2- is less stable and more reactive than OH-.

🎯 Exam Tip: Nucleophilicity generally decreases as electronegativity increases across a period and increases down a group in the periodic table, and negative charges usually make a species more nucleophilic.

Question 4. Which of the following species is not electrophilic in nature?
(a) \( \text{Cl}^+ \)
(b) \( \text{BH}_3 \)
(c) \( \text{H}_3\text{O}^+ \)
(d) \( \text{NO}_2^+ \)
Answer: (c) \( \text{H}_3\text{O}^+ \)
In simple words: Electrophiles are molecules that like electrons. \( \text{H}_3\text{O}^+ \) (hydronium ion) is a source of protons, but it's not looking for electrons to complete its octet, unlike the others which are electron-deficient or have empty orbitals. \( \text{H}_3\text{O}^+ \) mainly acts as an acid.

🎯 Exam Tip: Electrophiles are typically electron-deficient species (like positive ions or molecules with incomplete octets) that can accept an electron pair. Hydronium, while positively charged, can donate a proton but is not typically considered an electrophile in organic reactions as it doesn't accept electrons.

Question 5. Homolytic fission of covalent bond leads to the formation of
(a) electrophile
(b) nucleophile
(c) carbocation
(d) free radical
Answer: (d) free radical
In simple words: When a covalent bond breaks evenly, each atom gets one electron, forming a free radical. Free radicals are atoms or molecules with an unpaired electron.

🎯 Exam Tip: Homolytic fission is characteristic of reactions that produce free radicals, often initiated by heat or light, where bonds break symmetrically.

Question 6. Hyper conjugation is also known as
(a) no bond resonance
(b) Baker - nathan effect
(c) both (a) and (b)
(d) none of these
Answer: (c) both (a) and (b)
In simple words: Hyperconjugation is a special way electrons can spread out in a molecule, making it more stable. It's sometimes called "no bond resonance" because electrons move without forming a full bond, and also the "Baker-Nathan effect" after the scientists who studied it.

🎯 Exam Tip: Remember these alternative names for hyperconjugation, as they are often used interchangeably in exams and indicate the delocalization of sigma electrons into an adjacent pi or empty p-orbital.

Question 7. Which of the group has highest + I effect?
(a) \( \text{CH}_3- \)
(b) \( \text{CH}_3 - \text{CH}_2- \)
(c) \( (\text{CH}_3)_2 - \text{CH}- \)
(d) \( (\text{CH}_3)_3 - \text{C}- \)
Answer: (d) \( (\text{CH}_3)_3 - \text{C}- \)
In simple words: The \( +\text{I} \) effect (inductive effect) means a group pushes electrons away. A group with more alkyl branches (like three methyl groups around a carbon) pushes electrons the most. This makes the carbon it is attached to more electron-rich.

🎯 Exam Tip: The \( +\text{I} \) effect increases with the number of alkyl groups attached to the central carbon, as each alkyl group contributes to electron donation.

Question 8. Which of the following species does not exert a resonance effect?
(a) \( \text{C}_6\text{H}_5\text{OH} \)
(b) \( \text{C}_6\text{H}_5\text{Cl} \)
(c) \( \text{C}_6\text{H}_5\text{NH}_2 \)
(d) \( \text{C}_6\text{H}_5\text{NH}_3^+ \)
Answer: (d) \( \text{C}_6\text{H}_5\text{NH}_3^+ \)
In simple words: Resonance effect happens when electrons can move around a molecule through connected double bonds or lone pairs. In \( \text{C}_6\text{H}_5\text{NH}_3^+ \), the nitrogen atom has no available lone pair to share with the benzene ring, so it cannot participate in resonance. The positive charge on nitrogen makes it electron-deficient.

🎯 Exam Tip: For a group to show a resonance effect with a benzene ring, it must either have a lone pair of electrons (for \( +\text{R} \) effect) or be connected to a multiple bond (for \( -\text{R} \) effect).

Question 9. – I effect is shown by
(a) \( -\text{Cl} \)
(b) \( -\text{Br} \)
(c) both (a) and (b)
(d) \( -\text{CH}_3 \)
Answer: (c) both (a) and (b)
In simple words: The \( -\text{I} \) effect (inductive effect) means a group pulls electrons towards itself. Both chlorine (Cl) and bromine (Br) are electronegative atoms, so they pull electrons and show a \( -\text{I} \) effect.

🎯 Exam Tip: Remember that halogens (F, Cl, Br, I) are electron-withdrawing groups and exert a negative inductive effect ( \( -\text{I} \) effect).

Question 10. Which of the following carbocation will be most stable?
(a) \( \text{Ph}_3\text{C}^+ - \)
(b) \( \text{CH}_3 - ^+\text{CH}_2- \)
(c) \( (\text{CH}_3)_2 - ^+\text{CH} \)
(d) \( \text{CH}_2 = \text{CH} - ^+\text{CH}_2 \)
Answer: (d) \( \text{CH}_2 = \text{CH} - ^+\text{CH}_2 \)
In simple words: The stability of carbocations depends on how well the positive charge can be spread out. The allyl carbocation (CH2=CH-CH2+) is stabilized by resonance, where the positive charge can move between the two end carbons, making it very stable. \( \text{Ph}_3\text{C}^+ \) (triphenylmethyl carbocation) is also highly stabilized by resonance with three phenyl groups, often considered one of the most stable. In this list, the allyl carbocation is given as the answer choice, indicating strong resonance stabilization.

🎯 Exam Tip: Carbocation stability increases with more alkyl groups (due to inductive effect and hyperconjugation) and especially with resonance stabilization (e.g., allyl or benzyl carbocations).

Question 11. Assertion: Tertiary Carbocations are generally formed more easily than primary Carbocations ions. Reason: Hyper conjucation as well as inductive effect due to additional alkyl group stabilize tertiary carbonium ions.
(a) both assertion and reason are true and reason is the correct explanation of assertion.
(b) both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false.
(d) both assertion and reason are false.
Answer: (a) both assertion and reason are true and reason is the correct explanation of assertion.
In simple words: Tertiary carbocations form more easily because they are more stable. This extra stability comes from hyperconjugation and the electron-donating inductive effect of the alkyl groups attached to the positively charged carbon. These effects help spread out the positive charge.

🎯 Exam Tip: Always remember that the stability order of carbocations is tertiary > secondary > primary > methyl, which is directly explained by the combined effects of hyperconjugation and the inductive effect of alkyl groups.

Question 12. Heterolytic fission of C – Br bond results in the formation of
(a) free radical
(b) Carbanion
(c) Carbocation
(d) Carbanion and Carbocation
Answer: (d) Carbanion and Carbocation
In simple words: When a C-Br bond breaks unevenly, the more electronegative bromine atom takes both electrons, forming a bromide ion. This leaves the carbon with a positive charge (carbocation) and the bromide with a negative charge (anion). The question is asking what *forms*, and an anion (bromide) is a type of carbanion, and the carbon becomes a carbocation. So, both are produced.

🎯 Exam Tip: Heterolytic fission is an unsymmetrical bond breaking process where one atom retains both bonding electrons, leading to the formation of a cation and an anion.

Question 13. Which of the following represent a set of nucleophiles?
(a) \( \text{BF}_3, \text{H}_2\text{O}, \text{NH}_2^- \)
(b) \( \text{AlCl}_3, \text{BF}_3, \text{NH}_3 \)
(c) \( \text{CN}^-, \text{RCH}_2^-, \text{ROH} \)
(d) \( \text{H}^+, \text{RNH}_3^+, :\text{CCl}_2 \)
Answer: (c) \( \text{CN}^-, \text{RCH}_2^-, \text{ROH} \)
In simple words: Nucleophiles are electron-rich species that donate electrons. Cyanide ion \( (\text{CN}^-) \), carbanions \( (\text{RCH}_2^-) \), and alcohols \( (\text{ROH}) \) all have lone pairs or negative charges that they can donate. Water (H2O) and ammonia (NH3) are also nucleophiles, but options (a) and (b) include electrophiles.

🎯 Exam Tip: Nucleophiles are Lewis bases (electron-pair donors), and they can be negatively charged ions or neutral molecules with at least one lone pair of electrons.

Question 14. Which of the following species does not acts as a nucleophile?
(a) \( \text{ROH} \)
(b) \( \text{ROR} \)
(c) \( \text{PCl}_3 \)
(d) \( \text{BF}_3 \)
Answer: (d) \( \text{BF}_3 \)
In simple words: A nucleophile gives away electrons. Boron trifluoride \( (\text{BF}_3) \) is an electron-deficient molecule, meaning it wants to *accept* electrons, not give them away. It acts as a Lewis acid (an electrophile). Alcohols, ethers, and \( \text{PCl}_3 \) all have lone pairs of electrons to donate.

🎯 Exam Tip: Species with incomplete octets, like \( \text{BF}_3 \) or \( \text{AlCl}_3 \), are strong Lewis acids and therefore electrophiles, not nucleophiles.

Question 15. The geometrical shape of carbocation is
(a) Linear
(b) tetrahedral
(c) Planar
(d) Pyramidal
Answer: (c) Planar
In simple words: A carbocation has a carbon atom with a positive charge. This carbon forms three bonds and has an empty p-orbital, which means it arranges itself in a flat (planar) shape. This geometry is sp² hybridized.

🎯 Exam Tip: Remember that sp² hybridization always corresponds to a trigonal planar geometry, which is why carbocations are planar.

II. Write brief answer to the following questions:

Question 16. Write short notes on
(a) Resonance
(b) Hyper Conjugation
Answer:
(a) Resonance (or) Mesomeric effect:
Resonance is a chemical effect seen in some organic compounds that have double bonds in a suitable place. It means these compounds can be shown with more than one structure. These structures only differ in where the bonding and lone pairs of electrons are placed. These different forms are called resonance structures (or canonical structures). The actual molecule is a mix of these structures, called a resonance hybrid. This spreading out of electrons makes the molecule more stable. For example, the 1,3-butadiene molecule has delocalized pi electrons across its four carbon atoms, making its C-C bond lengths equal, which a single structure cannot explain.
Question 1. In organic reactions the reactant is called as _____
(a) intermediate
(b) product
(c) substrate
(d) by product
Answer: (c) substrate
In simple words: The main chemical that takes part in an organic reaction is called the substrate. It is the molecule that undergoes change.

🎯 Exam Tip: Identify the role of each component (reactant, product, intermediate, byproduct) in a reaction.

Question 2. Heterolysis of C - Cl bond produces
(a) Two free radicals
(b) Two carbonium ions
(c) Two carbanions
(d) One cation and one anion
Answer: (d) One cation and one anion
In simple words: When a C-Cl bond breaks heterolytically, it forms one positively charged ion and one negatively charged ion.

🎯 Exam Tip: Understand heterolysis as an unsymmetrical bond breaking process that forms ions, not neutral free radicals.

Question 3. Heterolysis of propane gives
(a) Methyl and ethyl free radicals
(b) Methyl carbocation and ethyl free radicals
(c) Methyl anion and ethyl carbocation
(d) Methyl free radical and ethyl carbocation
Answer: (c) Methyl anion and ethyl carbocation
In simple words: Heterolysis of propane can result in a methyl anion and an ethyl carbocation.

🎯 Exam Tip: Remember that heterolysis produces charged species (ions), not free radicals, and the specific ions formed depend on the bond that breaks.

Question 4. Removal of hydride ion from a methane molecule will give a
(a) Methyl radical
(b) Carbonium ion
(c) Carbanion
(d) Methyl group
Answer: (b) Carbonium ion
In simple words: Removing a hydride ion from methane creates a carbonium ion, which is a positively charged methyl group.

🎯 Exam Tip: A hydride ion (\( \text{H}^- \)) removal means the hydrogen leaves with its two electrons, leaving a positive charge on the carbon.

Question 5. The shape of carbonium ion is
(a) Planar
(b) Linear
(c) Pyramidal
(d) Tetrahedral
Answer: (a) Planar
In simple words: Carbonium ions have a flat, planar shape because the central carbon atom is sp² hybridized.

🎯 Exam Tip: Connect the hybridization state (sp²) directly to the molecular geometry (trigonal planar) for carbocations.

Question 6. The shape of methyl free radicals is
(a) Planar
(b) Pyramidal
(c) Tetrahedral
(d) Linear
Answer: (a) Planar
In simple words: Methyl free radicals are planar in shape because their central carbon atom is sp² hybridized.

🎯 Exam Tip: Understand that like carbocations, simple free radicals such as methyl radicals are often planar due to sp² hybridization.

Question 7. The reaction, \( \text{(CH}_3\text{)}_3\text{C} - \text{Br} \rightarrow \text{(CH}_3\text{)}_3\text{C}^+ + \text{Br}^- \) is an example of
(a) Homolytic fission
(b) Heterolytic fission
(c) Cracking
(d) All of the options
Answer: (b) Heterolytic fission
In simple words: The reaction is an example of heterolytic fission because the bond breaks unevenly, creating ions.

🎯 Exam Tip: Distinguish between homolytic (equal sharing, free radicals) and heterolytic (unequal sharing, ions) bond fission based on the products formed.

Question 8. Which of the following contains only three pairs of electrons
(a) Carbocation
(b) Carbanion
(c) Free radical
(d) All of these
Answer: (a) Carbocation
In simple words: A carbocation has a carbon atom that forms three bonds, meaning it has only three pairs of electrons around it.

🎯 Exam Tip: Remember that carbocations are electron deficient, having only six valence electrons, which makes them highly reactive.

Question 9. Which of the following species is paramagnetic
(a) A carbaonium ion
(b) A free radical
(c) A carbanion
(d) All of these
Answer: (b) A free radical
In simple words: A free radical is paramagnetic because it has at least one electron that is not paired with another.

🎯 Exam Tip: Recall that paramagnetism arises from unpaired electrons; carbocations and carbanions have all electrons paired (though incomplete octet for carbocations).

Question 10. In carbonium ion the carbon bearing the positive charge is
(a) sp hybridized
(b) sp² hybridized
(c) sp³ hybridized
(d) un hybridized
Answer: (b) sp² hybridized
In simple words: The positive carbon in a carbonium ion is sp² hybridized.

🎯 Exam Tip: Hybridization state dictates geometry and reactivity; sp² hybridization is key for carbocations.

Question 11. Among the following the true property about \( \text{H}_3\text{C}-\text{C}^+ \text{H}_2 \) is
(a) non - planar
(b) C+ is sp² - hybridized
(c) Electrophile can attack on C+
(d) Does not undergo hydrolysis
Answer: (b) C+ is sp² - hybridized
In simple words: The positive carbon atom in \( \text{H}_3\text{C}-\text{C}^+ \text{H}_2 \) is sp² hybridized.

🎯 Exam Tip: Always relate the charge on carbon to its hybridization and electron deficiency.

Question 12. The geometry of a methyl carbanion is likely to be
(a) Pyramidal
(b) Tetrahedral
(c) Planar
(d) Linear
Answer: (a) Pyramidal
In simple words: Methyl carbanions usually have a pyramidal shape because their central carbon is sp³ hybridized and has a lone pair.

🎯 Exam Tip: Remember that the lone pair in carbanions influences their geometry, often leading to a pyramidal shape due to sp³ hybridization.

Question 13. Which of the following statements are correct for nucleophile?
(a) All negatively charged species are nucleophiles
(b) Nucleophiles are Lewis bases
(c) Alkenes, alkynes, benzene and pyrrole are nucleophiles
(d) All are correct
Answer: (d) All are correct
In simple words: All the options are true: negatively charged things are nucleophiles, nucleophiles are Lewis bases, and molecules with double/triple bonds or lone pairs are also nucleophiles.

🎯 Exam Tip: A nucleophile is an electron-rich species that donates an electron pair to form a new bond, acting as a Lewis base.

Question 14. Which one of the following statement is not correct for electrophile?
(a) Electron deficient species are electrophiles
(b) Electrophiles are Lewis acids
(c) All positively charged species are electrophiles
(d) \( \text{AlCl}_3 \) and \( \text{SO}_3 \) are electrophiles
Answer: (c) All positively charged species are electrophiles
In simple words: Not all positively charged species are electrophiles; for example, some metal ions might not behave like them in organic chemistry.

🎯 Exam Tip: While most positively charged species are electrophiles, remember the core definition: an electrophile is an electron-deficient species that accepts an electron pair.

Question 15. Electrophiles are
(a) Lewis bases
(b) Lewis acids
(c) Amphoteric
(d) All of these
Answer: (b) Lewis acids
In simple words: Electrophiles are Lewis acids because they accept electrons from other molecules.

🎯 Exam Tip: Clearly differentiate between Lewis acids (electron acceptors/electrophiles) and Lewis bases (electron donors/nucleophiles).

Question 16. Electrophiles are
(a) Electron loving species
(b) Electron hating species
(c) Nucleus loving reagents
(d) Nucleus hating reagents
Answer: (a) Electron loving species
In simple words: Electrophiles love electrons, so they are called electron loving species.

🎯 Exam Tip: Understand the etymology of "electrophile" and "nucleophile" to remember their key characteristic: electron-loving vs. nucleus-loving.

Question 17. Nucleophiles are
(a) Electron loving
(b) Electron hating
(c) Nucleus loving
(d) Nucleus hating
Answer: (c) Nucleus loving
In simple words: Nucleophiles are nucleus loving because they are attracted to the positive parts of atoms.

🎯 Exam Tip: A nucleus is positively charged, so "nucleus loving" means being attracted to positive charges, implying the species itself is electron-rich.

Question 18. Which of the following statement is false about an electrophile?
(a) Electron - deficient species
(b) An acidic reagent
(c) A reagent which attacks an electron - deficient site in a molecule
(d) A species which seeks a pair of electrons
Answer: (c) A reagent which attacks an electron - deficient site in a molecule
In simple words: The false statement is that an electrophile attacks an electron-deficient site. Electrophiles actually attack places that have lots of electrons.

🎯 Exam Tip: Electrophiles are electron-deficient and are attracted to electron-rich areas, whereas nucleophiles are electron-rich and attracted to electron-deficient areas.

Question 19. Methyl carbanium ion is
(a) Electrophile
(b) Lewis acid
(c) Nucleophile
(d) Both (a) and (b)
Answer: (c) Nucleophile
In simple words: A methyl carbanion is a nucleophile because it has a negative charge and extra electrons it can share.

🎯 Exam Tip: Carbanions are electron-rich due to their negative charge and lone pair, making them powerful nucleophiles and Lewis bases.

Question 20. Ammonia molecule is
(a) A nucleophile
(b) An electron deficient
(c) A electrophile
(d) An acid
Answer: (a) A nucleophile
In simple words: Ammonia is a nucleophile because its nitrogen atom has an extra pair of electrons to donate.

🎯 Exam Tip: Any molecule with an accessible lone pair of electrons (like water, alcohols, amines) can act as a nucleophile.

Question 21. Methyl carbanion is
(a) Electrophile
(b) Lewis acid
(c) Both (a) and (b)
(d) Nucelophile
Answer: (d) Nucelophile
In simple words: Methyl carbanion is a nucleophile as it has a negative charge and can donate electrons.

🎯 Exam Tip: Reiterate the definition: nucleophiles are electron-rich species that donate electron pairs; carbanions fit this perfectly.

Question 22. Which of the following statements is correct about inductive effect?
(a) Implies the transfer of lone pair of electrons from more electronegative atom to lesser electronegative atom in a molecule.
(b) Implies the transfer to lone pair of electrons from lesser electronegative atom to the more electronegative atom in a molecule
(c) Increases with increase in distance
(d) Implies the atoms ability to cause bond polarization
Answer: (d) Implies the atoms ability to cause bond polarization
In simple words: The inductive effect is about how atoms can make bonds slightly uneven (polarized) by pulling or pushing electrons through them.

🎯 Exam Tip: Key features of inductive effect: permanent, operates through sigma bonds, causes polarization, decreases with distance.

Question 23. The displacement of electrons in a multiple bond in the presence of attacking reagent is called
(a) inductive effect
(b) electromeric effect
(c) resonance
(d) Hyper-conjugation
Answer: (b) electromeric effect
In simple words: When an attacking chemical is near, electrons in a double or triple bond can shift temporarily; this is called the electromeric effect.

🎯 Exam Tip: Remember that the electromeric effect is temporary and requires an attacking reagent, unlike the permanent inductive or resonance effects.

Question 24. The electromeric effect in organic compounds is a
(a) Temporary effect
(b) Permanent effect
(c) Temporary or permanent effect
(d) All of the options
Answer: (a) Temporary effect
In simple words: The electromeric effect is a temporary change that only happens when another chemical is attacking.

🎯 Exam Tip: Distinguish electromeric effect (temporary, reagent-dependent) from inductive and resonance effects (permanent).

Question 25. Which of the following is not the correct condition for resonance?
(a) The positions of all the atomic nuclei in the resonating structures may be differ
(b) The resonating structures must have the same number of unpaired or paired electrons
(c) The molecules exhibiting resonance must be planar in nature
(d) The resonating structures must have nearly the same energies
Answer: (a) The positions of all the atomic nuclei in the resonating structures may be differ
In simple words: The condition that positions of atomic nuclei can be different in resonating structures is not correct. Nuclei must stay in place; only electrons move.

🎯 Exam Tip: Remember that resonance involves electron delocalization, not atom movement. Nuclei positions are fixed in all resonance contributors.

Question 26. Resonance is due to _____
(a) delocalization of sigma electrons
(b) migration of H atoms
(c) migration of proton
(d) delocalization of pi electrons
Answer: (d) delocalization of pi electrons
In simple words: Resonance happens because pi electrons are not stuck in one place; they can spread out over several atoms.

🎯 Exam Tip: Pi electron delocalization is the hallmark of resonance, contributing to molecular stability.

Question 27. Resonance in benzene is accompanied by delocalization of \( \pi \) - electrons. Each \( \pi \) electron is attached with
(a) 4 carbons
(b) 2 carbons
(c) 3 carbons
(d) 6 carbons
Answer: (d) 6 carbons
In simple words: In benzene, the pi electrons are spread out and connected to all six carbon atoms in the ring.

🎯 Exam Tip: The continuous delocalization of six pi electrons over six carbon atoms in a planar ring is a key characteristic of benzene's aromaticity.

Question 28. Reaction mechanism describes
(a) Sequential account of each step describing the details of electron movement
(b) energy changes during bond breaking and bond formation
(c) kinetics of the reaction
(d) all of these
Answer: (d) all of these
In simple words: A reaction mechanism tells us everything about how a reaction occurs: step by step, how electrons move, energy changes, and how fast it happens.

🎯 Exam Tip: A reaction mechanism is a detailed, step-by-step pathway of a chemical reaction, including intermediates and transition states.

Question 29. Substitution reaction may be
(a) Free radical substitution
(b) Nucleophilic substitution
(c) Electrophilic substitution
(d) All are correct
Answer: (d) All are correct
In simple words: Substitution reactions can happen in three main ways: by free radicals, by nucleophiles, or by electrophiles. So, all the options are correct.

🎯 Exam Tip: Be familiar with the three main types of substitution reactions and their typical substrates.

Question 30. \( \text{CH}_3 - \text{CH}_2 - \text{X} + \text{KOH(alc)} \rightarrow \text{CH}_2 = \text{CH}_2 + \text{KX} + \text{H}_2\text{O} \) is
(a) addition reaction
(b) substitution reaction
(c) elimination reaction
(d) molecular rearrangement
Answer: (c) elimination reaction
In simple words: This reaction is an elimination reaction because atoms are removed from the starting molecule to create a double bond.

🎯 Exam Tip: Recognize elimination reactions by the formation of a multiple bond (double or triple) and the loss of smaller molecules, often favored by strong bases and heat.

Question 31. \( \text{CH}_4 + \text{Cl}_2 \xrightarrow{hv} \text{CH}_3\text{Cl} + \text{HCl} \) is an example for
(a) free radical substitution reaction
(b) electrophilic substitution reaction
(c) nucleophile substitution reaction
(d) nueclophilic addition reaction
Answer: (a) free radical substitution reaction
In simple words: This reaction where methane and chlorine react with light is a free radical substitution because hydrogen is replaced by chlorine through a free radical process.

🎯 Exam Tip: Sunlight (hv) or high temperatures are common indicators for free radical reactions, especially halogenation of alkanes.

Question 32. \( \text{C}_6\text{H}_6 + 3 \text{Cl}_2 \xrightarrow{hv} \text{C}_6\text{H}_6\text{Cl}_6 \) is
(a) free radical substitution reaction
(b) electrophilic addition reaction
(c) free radical addition reaction
(d) nueclophilic addition reaction
Answer: (c) free radical addition reaction
In simple words: This is a free radical addition reaction because chlorine atoms are added to the benzene ring, breaking its double bonds, with the help of light.

🎯 Exam Tip: Benzene usually undergoes substitution, but under UV light with halogens, it can undergo free radical addition across its double bonds.

Question 33. Which one of the following orders is correct regarding the - I effect of the substituents?
(a) \( \text{– NR}_2 < \text{– OR} > \text{– F} \)
(b) \( \text{– NR}_2 > \text{– OR} > \text{– F} \)
(c) \( \text{– NR}_2 < \text{– OR} < \text{– F} \)
(d) \( \text{– NR}_2 > \text{– OR} < \text{– F} \)
Answer: (c) \( \text{– NR}_2 < \text{– OR} < \text{– F} \)
In simple words: The correct order for how strongly these groups pull electrons through a single bond (negative inductive effect) is that Fluorine pulls the strongest, then -OR, and then -NR2 pulls the least among these.

🎯 Exam Tip: The -I effect generally increases with increasing electronegativity of the atom directly attached to the carbon chain.

Question 34. Decreasing - I power of given groups is
(1) \( \text{-CN} \)
(2) \( \text{-NO}_2 \)
(3) \( \text{– NH}_3^+ \)
(4) \( \text{-F} \)
(a) 2 > 1 > 4 > 3
(b) 2 > 3 > 4 > 1
(c) 1 > 4 > 3 > 2
(d) 3 > 2 > 1 > 4
Answer: (a) 2 > 1 > 4 > 3
In simple words: The groups pull electrons in this decreasing order: \( \text{– NO}_2 \) is the strongest puller, then \( \text{– CN} \), then \( \text{– F} \), and \( \text{– NH}_3^+ \) pulls the least among these.

🎯 Exam Tip: For -I effect, positively charged groups are generally very strong electron-withdrawing groups, often stronger than highly electronegative atoms like fluorine.

Question 35. Which of the following belongs to - I group?
(a) \( \text{– C}_6\text{H}_5 \)
(b) \( \text{– CH}_3 \)
(c) \( \text{– CH}_2\text{CH}_3 \)
(d) \( \text{– C (CH}_3\text{)}_3 \)
Answer: (a) \( \text{– C}_6\text{H}_5 \)
In simple words: The \( \text{– C}_6\text{H}_5 \) group pulls electrons through a single bond, making it a -I group, while the others donate electrons.

🎯 Exam Tip: Alkyl groups typically show a +I effect (electron-donating), while groups like phenyl ( \( \text{C}_6\text{H}_5 \) ) can show a weak -I effect, depending on the context.

Question 36. Which of the following functional group shows + R effect?
(a) \( \text{– CHO} \)
(b) \( \text{– NO}_2 \)
(c) \( \text{– CN} \)
(d) \( \text{– NR}_2 \)
Answer: (d) \( \text{– NR}_2 \)
In simple words: The \( \text{– NR}_2 \) group shows a positive resonance effect because it has lone pairs of electrons it can share into a connected system.

🎯 Exam Tip: Groups with lone pairs on the atom directly attached to the conjugated system (e.g., \( \text{-OH, -OR, -NH}_2 \)) usually show a +R effect, while groups with multiple bonds to an electronegative atom (e.g., \( \text{-CHO, -NO}_2 \)) usually show a -R effect.

Question 37. Hyper conjugation is
(a) \( \sigma – \pi \) conjugation
(b) Due to delocalization of \( \sigma \) and \( \pi \) bonds
(c) No bond resonance
(d) All of these
Answer: (d) All of these
In simple words: Hyperconjugation involves sigma electrons mixing with pi systems, it's a type of "no bond resonance," and it's also called sigma-pi conjugation. All these descriptions are correct.

🎯 Exam Tip: Hyperconjugation is a stabilizing interaction often seen in carbocations, free radicals, and alkenes, involving sigma-electron delocalization.

Question 38. + R power of the given groups (1) \( \text{-O}^- \) (2) \( \text{– NH}_2 \) (3) \( \text{– OH} \) (4) \( \text{– NHCOCH}_3 \) in decreasing order is
(a) 1 > 2 > 3 > 4
(b) 4 > 3 > 2 > 1
(c) 1 > 3 > 2 > 4
(d) 1 > 4 > 3 > 2
Answer: (a) 1 > 2 > 3 > 4
In simple words: The order of how well these groups can donate electrons through resonance, from strongest to weakest, is: a negatively charged oxygen, then an amino group, then a hydroxyl group, and lastly an acetamido group.

🎯 Exam Tip: A negative charge on the atom directly attached to the conjugated system greatly enhances its +R effect. Any electron-withdrawing group on the donating atom will reduce its +R effect.

Question 39. In pyridine: (diagram of pyridine) Number of conjugated electrons are:
(a) 6
(b) 8
(c) Zero
(d) 5
Answer: (a) 6
In simple words: Pyridine has three double bonds in its ring, and each double bond has two conjugated electrons, making a total of six conjugated electrons. The nitrogen's lone pair is not part of this.

🎯 Exam Tip: For aromatic compounds, count the pi electrons in the conjugated ring system, following Hückel's rule (4n+2). Nitrogen's lone pair only counts if it's necessary for aromaticity and in a p-orbital.

Question 40. In hyper conjugation, the atom involved is
(a) \( \beta – \text{H atom} \)
(b) \( \alpha – \text{H atom} \)
(c) \( \gamma – \text{H atom} \)
(d) All
Answer: (b) \( \alpha – \text{H atom} \)
In simple words: Hyperconjugation involves hydrogen atoms that are directly next to the carbon with a special electron system (alpha-H atoms).

🎯 Exam Tip: The number of alpha-hydrogens is directly proportional to the stability gained through hyperconjugation.

Question 41. Hyper conjugation involves overlap of the following orbtials
(a) \( \sigma- \sigma \)
(b) \( \sigma - \text{p} \)
(c) \( \text{p-p} \)
(d) \( \pi-\pi \)
Answer: (b) \( \sigma - \text{p} \)
In simple words: Hyperconjugation happens when a sigma bond's electrons share space with an empty p-orbital.

🎯 Exam Tip: The overlap of \( \sigma \) orbitals with empty or partially filled p or \( \pi \) orbitals is the fundamental basis of hyperconjugation.

Question 42. + I effect is shown by
(a) \( \text{– NO}_2 \)
(b) \( \text{– Cl} \)
(c) \( \text{– Br} \)
(d) \( \text{– CH}_3 \)
Answer: (d) \( \text{– CH}_3 \)
In simple words: The \( \text{– CH}_3 \) group pushes electrons away, which is called a positive inductive effect. Other options pull electrons.

🎯 Exam Tip: Remember that alkyl groups are generally electron-donating (+I), while electronegative atoms and groups with multiple bonds to electronegative atoms are electron-withdrawing (-I).

Question 43. Zero inductive effect is exerted by
(a) \( \text{C}_6\text{H}_5- \)
(b) \( \text{– H} \)
(c) \( \text{CH}_3- \)
(d) \( \text{– Cl} \)
Answer: (b) \( \text{– H} \)
In simple words: The hydrogen atom is considered to have no inductive effect, making it the neutral reference point.

🎯 Exam Tip: Hydrogen is the standard against which inductive effects are measured; its own effect is defined as zero.

Question 44. Hyper conjugation is most useful for stabilizing which of the following carbocations?
(a) neo - Pentyl
(b) tert - Butyl
(c) iso - Propyl
(d) Ethyl
Answer: (b) tert - Butyl
In simple words: Hyperconjugation helps to make carbocations more stable by sharing electrons from nearby bonds. Tertiary butyl carbocations have more such bonds, making them very stable.

🎯 Exam Tip: Remember that the more alkyl groups (like methyl, ethyl) are attached to the positively charged carbon, the more opportunities there are for hyperconjugation, leading to greater stability.

Question 45. Hyper conjugation phenomenon is possible in
(a) \( \text{H}_2\text{C} = \text{CH}_2 \)
(b) \( \text{CH}_3\text{CH}_2-\text{CH} = \text{CH}_2 \)
(c) \( \text{C}_6\text{H}_5\text{CH} = \text{CH}_2 \)
(d) \( \text{(CH}_3\text{)}_3\text{C} - \text{CH} = \text{CH}_2 \)
Answer: (b) \( \text{CH}_3\text{CH}_2-\text{CH} = \text{CH}_2 \)
In simple words: Hyperconjugation needs an alpha-hydrogen atom next to a double bond or a charged carbon. In this option, the \( \text{CH}_3\text{CH}_2 \) group has such hydrogens available to participate in hyperconjugation.

🎯 Exam Tip: To identify if hyperconjugation is possible, always look for C-H bonds directly attached to the carbon that is next to an unsaturated system (like a double bond) or a carbon with a positive charge.

Question 46. Among the following carbocations which is more stable
(a) \( \text{(C}_6\text{H}_5\text{)}_3\text{C}^+ \)
(b)
Answer: (a) \( \text{(C}_6\text{H}_5\text{)}_3\text{C}^+ \)
In simple words: The more benzene rings attached to the carbocation, the more places the positive charge can spread out, making it much more stable. Triphenylmethyl carbocation has three such rings.

🎯 Exam Tip: Benzene rings stabilize carbocations very effectively through resonance. The more aryl (benzene) groups directly attached to the carbocation, the greater the resonance stabilization, and thus the higher the stability.

Question 47. The compound which gives the most stable carbonium ion on dehydration
(a) \( \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2\text{OH} \)
(b) \( \text{H}_3\text{C}-\text{C}(\text{CH}_3)(\text{OH})-\text{CH}_3 \)
(c) \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH} \)
(d) \( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_3 \)
Answer: (b) \( \text{H}_3\text{C}-\text{C}(\text{CH}_3)(\text{OH})-\text{CH}_3 \)
In simple words: Dehydration removes water from an alcohol, usually forming a carbocation first. Tertiary alcohols like the one in option (b) form the most stable carbocations because the positive charge is surrounded by three alkyl groups, which help share the electron density.

🎯 Exam Tip: Remember the order of carbocation stability: Tertiary > Secondary > Primary > Methyl. Alcohols that can form tertiary carbocations upon dehydration will be the most reactive in this process.

Question 48. Which carbocation is more stable?
(a)
(b)
(c)
(d)
Answer: (a)
In simple words: Electron-donating groups like \( \text{CH}_3 \) (methyl) help stabilize a positive charge. When it's in the para position, it can donate electrons more effectively through resonance and induction, spreading out the positive charge and making the carbocation most stable.

🎯 Exam Tip: Electron-donating groups (like alkyls) stabilize carbocations, especially when they can interact through resonance (ortho/para positions). Electron-withdrawing groups (like \( \text{NO}_2 \)) destabilize them.

Question 49. Arrange the following groups in order of decreasing – R (Or – M) power (1) \( \text{NO}_2 \) (2) \( \text{SO}_3\text{H} \) (3) \( \text{CF}_3 \) (4) \( \text{CHO} \)
(a) 1 > 3 > 2 > 4
(b) 1 > 2 > 3 > 4
(c) 1 > 4 > 3 > 2
(d) 4 > 3 > 2 > 1
Answer: (a) 1 > 3 > 2 > 4
In simple words: The -R or -M effect shows how strongly a group pulls electrons through resonance. Some groups are better at this than others, making the electron-pulling strength decrease in the order shown.

🎯 Exam Tip: Memorize the general order of electron-withdrawing groups for resonance effect. Groups with positively charged atoms or multiple bonds where the electronegative atom is external usually have strong -R effects.

Question 50. The reaction intermediate produced by homolytic cleavage of bond is called
(a) carbocations
(b) carbanions
(c) free radicals
(d) carbenes
Answer: (c) free radicals
In simple words: When a bond breaks evenly, each atom gets one electron, forming a free radical. This is called homolytic cleavage.

🎯 Exam Tip: Homolytic cleavage always produces species with unpaired electrons (free radicals), while heterolytic cleavage produces ions (carbocations or carbanions).

Question 51. Most stable carbocation is
(a) \( \text{CH}_3 - \text{CH}_2^+ \)
(b) \( \text{CH}_2^+\text{CHCl}_2 \)
(c) \( \text{CH}_2^+\text{CH}_2\text{Cl} \)
(d) \( \text{CH}_2^+ - \text{CH}_2\text{NO}_2 \)
Answer: (a) \( \text{CH}_3 - \text{CH}_2^+ \)
In simple words: Alkyl groups next to a carbocation help stabilize it by donating electrons through hyperconjugation and induction. The more alkyl groups or electron-donating groups nearby, the more stable the carbocation. Here, \( \text{CH}_3 - \text{CH}_2^+ \) is the most stable among the given options because the other options have electron-withdrawing groups like chlorine or nitro, which destabilize the positive charge.

🎯 Exam Tip: Electron-donating groups stabilize carbocations, while electron-withdrawing groups destabilize them. Always look for groups that can best delocalize the positive charge.

Question 52. Carbocation is a reaction intermediate in which of the following reactions
(a) \( \text{E}_1 \) reactions
(b) Electrophilic addition reactions of alkenes and alkynes
(c) \( \text{S}_{\text{N}}1 \) reactions
(d) All of the options
Answer: (d) All of the options
In simple words: Carbocations are common temporary species that form during many organic reactions. They play a key role in elimination (E1), addition to double or triple bonds, and certain substitution (SN1) reactions.

🎯 Exam Tip: Understanding the role of carbocations as intermediates helps predict reaction pathways and major products in organic chemistry. They are typically formed in reactions where a good leaving group departs, or where a pi bond attacks an electrophile.

Question 53. Halogenation of an alkane takes place through the intermediate
(a) Carbocation
(b) carbanion
(c) carbon free radical
(d) carbene
Answer: (c) carbon free radical
In simple words: When alkanes react with halogens, the process happens in steps. It starts with the formation of free radicals, which are atoms or groups with unpaired electrons.

🎯 Exam Tip: Remember that free radical halogenation of alkanes proceeds through a chain mechanism involving initiation, propagation, and termination steps, with free radicals being key intermediates.

Question 54. Which free radical is the most stable?
(a) \( \text{C}_6\text{H}_5-\dot{\text{C}}\text{H}_2 \)
(b) \( \text{CH}_2=\text{CH}-\dot{\text{C}}\text{H}_2 \)
(c) \( \text{CH}_3-\dot{\text{C}}\text{H}-\text{CH}_3 \)
(d) \( \text{(CH}_3\text{)}_3\dot{\text{C}} \)
Answer: (a) \( \text{C}_6\text{H}_5-\dot{\text{C}}\text{H}_2 \)
In simple words: Benzyl radicals, like the one in option (a), are very stable because the unpaired electron can spread out over the benzene ring through resonance. This spreading of charge makes the radical less reactive and more stable.

🎯 Exam Tip: The stability of free radicals increases with resonance (e.g., benzyl, allyl) and also with increasing alkyl substitution (tertiary > secondary > primary), which provides hyperconjugation and inductive effects.

Question 55. The most stable carbonium ion is
(a) Methyl carbonium ion
(b) Primary carbonium ion
(c) Secondary carbonium ion
(d) Tertiary carbonium ion
Answer: (d) Tertiary carbonium ion
In simple words: Carbonium ions (carbocations) are more stable when they have more carbon atoms directly attached to the positive carbon. Tertiary carbocations have three such attachments, making them the most stable due to electron donation and hyperconjugation.

🎯 Exam Tip: The general stability order for carbocations is tertiary > secondary > primary > methyl, primarily due to hyperconjugation and the inductive effect of alkyl groups.

Question 56. Stability of which intermediate is not govern by hyperconjugation?
(a) Carbon cation
(b) carbon anion
(c) carbon free radical
(d) carboniumion
Answer: (b) carbon anion
In simple words: Hyperconjugation helps stabilize carbocations (carbon cations) and carbon free radicals by spreading out electron density or an unpaired electron. However, carbon anions have a negative charge, and hyperconjugation doesn't help stabilize them; instead, electron-withdrawing effects would.

🎯 Exam Tip: Hyperconjugation involves the interaction of \( \sigma \)-electrons with an adjacent empty p-orbital (in carbocations) or a singly occupied p-orbital (in free radicals). Carbanions have a filled orbital, so this type of stabilization does not apply.

Question 57. Which of the following is an electrophilic reagent?
(a) \( \text{H}_2\text{O} \)
(b) \( \text{OH}^- \)
(c) \( \text{NO}_2^+ \)
(d) All of the options
Answer: (c) \( \text{NO}_2^+ \)
In simple words: An electrophile is an electron-loving species, meaning it seeks electrons. \( \text{NO}_2^+ \) has a positive charge, making it electron-deficient and eager to accept electrons, so it acts as an electrophile.

🎯 Exam Tip: Electrophiles are typically positively charged ions or electron-deficient neutral molecules, while nucleophiles are negatively charged ions or electron-rich neutral molecules.

Question 58. \( \text{CH}_3 - \text{CH}_2 - \text{X} + \text{KOH(alc)} \rightarrow \text{CH}_2 = \text{CH}_2 + \text{KX} + \text{H}_2\text{O} \) is an example for
(a) addition reaction
(b) substitution reaction
(c) elimination reaction
(d) molecular rearrangement
Answer: (c) elimination reaction
In simple words: In this reaction, a smaller molecule (\( \text{HX} \) and \( \text{H}_2\text{O} \)) is removed from the starting compound, leading to the formation of a double bond. This type of reaction is called an elimination reaction.

🎯 Exam Tip: Elimination reactions often occur when an alkyl halide is heated with a strong base like alcoholic KOH, resulting in the removal of hydrogen and a halogen (HX) to form an alkene.

Question 59. \( \text{R-X} + \text{OH}^-_{\text{(aq)}} \rightarrow \text{R - OH} + \text{X}^+_{\text{(aq)}} \) is an example for
(a) addition reaction
(b) substitution reaction
(c) elimination reaction
(d) molecular rearrangement
Answer: (b) substitution reaction
In simple words: In this reaction, one group (X) in the starting molecule is replaced by another group (OH). This kind of exchange is known as a substitution reaction.

🎯 Exam Tip: Nucleophilic substitution reactions, like the hydrolysis of alkyl halides, involve a nucleophile replacing a leaving group. Pay attention to the reagent (aqueous vs. alcoholic base) as it often dictates substitution versus elimination.

Question 60. \( \text{C}_6\text{H}_6 + \text{Cl}_2 \overset{\text{Fe}}{\longrightarrow} \text{C}_6\text{H}_5\text{Cl} + \text{HCl} \) is an example for
(a) Free radical substitution reaction
(b) Electrophilic substitution reaction
(c) Nucleophilic addition reaction
(d) Nucelophilic substitution reaction
Answer: (b) Electrophilic substitution reaction
In simple words: Benzene is an electron-rich ring, and it prefers to react with electron-loving species (electrophiles). In this reaction, a hydrogen on the benzene ring is replaced by a chlorine atom, guided by an electrophile formed by \( \text{Cl}_2 \) and \( \text{Fe} \).

🎯 Exam Tip: Aromatic compounds like benzene typically undergo electrophilic substitution reactions because their delocalized pi-electron system acts as a good nucleophile, attracting electrophiles.

Question 61. Alkenes readily undergo
(a) Substitution reactions
(b) Addition reactions
(c) Elimination reactions
(d) Rearrangement reactions
Answer: (b) Addition reactions
In simple words: Alkenes have a carbon-carbon double bond, which is rich in electrons. This double bond can easily break to add new atoms or groups across it, leading to addition reactions.

🎯 Exam Tip: The presence of a \( \pi \)-bond in alkenes makes them highly reactive towards addition reactions, where the \( \pi \)-bond is broken and two new \( \sigma \)-bonds are formed.

Question 62. Nitration of benzene is
(a) nucleophilic substitution
(b) nucleophilic addition
(c) electrophilic substitution
(d) free radical substitution
Answer: (c) electrophilic substitution
In simple words: Nitration is a reaction where a nitro group (\( \text{NO}_2 \)) replaces a hydrogen atom on the benzene ring. This process involves an electrophile (an electron-seeking species) attacking the electron-rich benzene, hence it's an electrophilic substitution.

🎯 Exam Tip: Nitration of benzene is a classic example of Electrophilic Aromatic Substitution (EAS), where the electrophile is the nitronium ion (\( \text{NO}_2^+ \)), generated from nitric and sulfuric acids.

Question 63. The reaction \( \text{(CH}_3\text{)}_3\text{C} - \text{Br} \overset{\text{H}_2\text{O}}{\longrightarrow} \text{(CH}_3\text{)}_3\text{C} - \text{OH} \) is
(a) elimination reaction
(b) substitution reaction
(c) free radical reaction
(d) displacement reaction
Answer: (b) substitution reaction
In simple words: In this reaction, the bromine atom attached to the carbon is replaced by an hydroxyl group (\( \text{OH} \)) from water. This exchange of one atom or group for another is the definition of a substitution reaction.

🎯 Exam Tip: This reaction is specifically a nucleophilic substitution (likely \( \text{S}_{\text{N}}1 \) due to the tertiary carbocation intermediate) where water acts as a nucleophile to replace the bromide leaving group.

Question 64. Inductive effect involves
(a) displacement of \( \sigma \) - electrons
(b) delocalization of \( \pi \) - electrons
(c) delocalization of \( \sigma \) - electrons
(d) displacement of \( \pi \) - electrons
Answer: (b) delocalization of \( \pi \) - electrons
In simple words: This refers to how electrons in pi-bonds can spread out or move within a molecule. This spreading of electron density influences the overall charge distribution and reactivity of the molecule.

🎯 Exam Tip: Electron delocalization, whether of \( \pi \) or \( \sigma \) electrons, is a crucial concept for understanding the stability and reactivity of organic molecules and reaction intermediates.

Question 65. Compound which shows positive mesomeric effect
(a) \( \text{H}_2\text{C} = \text{CH} - \text{Cl} \)
(b) \( \text{C}_6\text{H}_5 - \text{NO}_2 \)
(c) \( \text{H}_2\text{C} = \text{CH} - \text{CH}_2\text{Cl} \)
(d) both b & c
Answer: (a) \( \text{H}_2\text{C} = \text{CH} - \text{Cl} \)
In simple words: A positive mesomeric effect happens when a group donates its electrons to a pi-electron system through resonance. Chlorine in vinyl chloride has lone pairs of electrons that it can donate to the double bond, showing a positive mesomeric effect.

🎯 Exam Tip: Groups with lone pairs of electrons directly attached to a conjugated system (like \( \text{Cl} \), \( \text{OH} \), \( \text{NH}_2 \)) usually show a positive mesomeric effect (+M), donating electrons. Groups like \( \text{NO}_2 \) show a negative mesomeric effect (-M), withdrawing electrons.

Question 66. The stability order, in the following carbocations, \( \text{CH}_3\text{CH}_2^+ \) (I), \( \text{(CH}_3\text{)}_2\text{CH}^+ \) (II), \( \text{(CH}_3\text{)}_3\text{C}^+ \) (III), \( \text{CH}_3^+ \) (IV)
(a) I > IV > III > II
(b) I > II > III > IV
(c) III > IV > I > II
(d) III > II > I > IV
Answer: (d) III > II > I > IV
In simple words: Carbocation stability increases as more alkyl groups are attached to the positively charged carbon. So, tertiary carbocations (III) are most stable, followed by secondary (II), then primary (I), and methyl (IV) is the least stable.

🎯 Exam Tip: The inductive effect and hyperconjugation from alkyl groups effectively stabilize carbocations by dispersing the positive charge. Therefore, the order of stability is \( 3^\circ > 2^\circ > 1^\circ > \text{methyl} \).

Question 67. Which is most stable carbocation?
(a) n - propyl cation
(b) iso - propyl cation
(c) Ethyl cation
(d) Triphenylmethyl cation
Answer: (d) Triphenylmethyl cation
In simple words: Triphenylmethyl carbocation is extremely stable because its positive charge can be spread over three benzene rings through resonance. This extensive delocalization of charge makes it far more stable than simple alkyl carbocations.

🎯 Exam Tip: Resonance stabilization, especially by multiple aromatic rings, is a powerful effect that significantly increases the stability of carbocations, often surpassing the stability gained from inductive and hyperconjugative effects alone.

Question 68. Which one of the following carbanions is least stable?
(a) \( \text{CH}_3\text{CH}_2^- \)
(b) \( \text{HC} \equiv \text{C}^- \)
(c) \( \text{(C}_6\text{H}_5\text{)}_3\text{C}^- \)
(d) \( \text{(CH}_3\text{)}_3\text{C}^- \)
Answer: (d) \( \text{(CH}_3\text{)}_3\text{C}^- \)
In simple words: Carbanions are stabilized by electron-withdrawing groups or by resonance. Alkyl groups are electron-donating, so a tertiary carbanion has three electron-donating groups pushing more negative charge onto the already negative carbon, making it the least stable.

🎯 Exam Tip: The stability of carbanions is generally the opposite of carbocations: Primary > Secondary > Tertiary, because electron-donating groups destabilize the negative charge, while electron-withdrawing groups or hybridization (sp > sp2 > sp3) stabilize it.

Question 69. Among the following, the strongest nucleophile is
(a) \( \text{C}_2\text{H}_5\text{SH} \)
(b) \( \text{CH}_3\text{COO}^- \)
(c) \( \text{CH}_3\text{NH}_2 \)
(d) \( \text{NCCH}_2^- \)
Answer: (a) \( \text{C}_2\text{H}_5\text{SH} \)
In simple words: A strong nucleophile readily donates its electron pair to form a new bond. Sulfur-containing compounds are generally stronger nucleophiles than oxygen or nitrogen compounds, especially in polar protic solvents, due to sulfur's larger size and more diffuse electron cloud.

🎯 Exam Tip: Nucleophilicity is influenced by charge (anions are stronger than neutral species), electronegativity (less electronegative atoms are better nucleophiles in the same period), and size/polarizability (larger atoms are often better nucleophiles in the same group, especially in protic solvents).

Question 70. Which of the following is least reactive in a nucleophile substitution reaction?
(a) \( \text{(CH}_3\text{)}_3\text{C} - \text{Cl} \)
(b) \( \text{CH}_2 = \text{CHCl} \)
(c) \( \text{CH}_3\text{CH}_2\text{Cl} \)
(d) \( \text{CH}_2 = \text{CHCH}_2\text{Cl} \)
Answer: (b) \( \text{CH}_2 = \text{CHCl} \)
In simple words: The chlorine in \( \text{CH}_2 = \text{CHCl} \) (vinyl chloride) is directly attached to a double-bonded carbon. The lone pair on chlorine can enter into resonance with the double bond, making the C-Cl bond partially double-bonded and much stronger, which makes it very hard to remove by a nucleophile.

🎯 Exam Tip: Halides where the halogen is directly attached to an \( \text{sp}^2 \) hybridized carbon (e.g., vinyl halides, aryl halides) are significantly less reactive in nucleophilic substitution reactions compared to alkyl halides due to resonance stabilization and the stronger bond.

II. Very Short Question and Answers (2 Marks):

Question 1. What is mechanism of reaction?
Answer: A reaction mechanism describes the detailed, step-by-step sequence of how reactants change into products. These steps often involve temporary molecules or transition states. The slowest step in this sequence controls how fast the entire reaction happens.
In simple words: It's like a recipe for a chemical reaction, showing every tiny step and what happens in between, especially which step is the slowest and sets the pace.

🎯 Exam Tip: When defining reaction mechanism, always include the concepts of "step-by-step sequence," "intermediates/transition states," and "rate-determining step."

Question 2. What is Positive Mesomeric effect? Give example.
Answer: The positive mesomeric effect (or +M effect) occurs when a substituent group donates its electrons to a conjugated system through resonance. This happens if the electron-releasing groups are connected to the conjugated system. In such cases, the group tends to push electrons into the system, stabilizing positive charges or increasing electron density. Common examples of groups showing a +M effect include \( -\text{OH} \), \( -\text{SH} \), \( -\text{OR} \), \( -\text{SR} \), \( -\text{NH}_2 \), and \( -\text{O}^- \).
In simple words: It's when a group connected to a double-bond system gives away its electrons, making that system richer in electrons. Think of it as pushing electrons into the ring or chain.

🎯 Exam Tip: For the positive mesomeric effect, remember that the group must have a lone pair of electrons available for donation and be directly attached to a conjugated system.

Question 3. What is Negative Mesomeric effect? Give example.
Answer: The negative mesomeric effect (or -M effect) happens when a substituent group pulls electrons away from a conjugated system through resonance. This occurs if electron-withdrawing groups are attached to the conjugated system. These groups act by attracting electrons from the system, which can destabilize negative charges or decrease electron density. Examples of groups showing a -M effect include \( -\text{NO}_2 \), \( >\text{C}=\text{O} \), \( -\text{COOH} \), and \( -\text{C}\equiv\text{N} \).
In simple words: This is when a group connected to a double-bond system pulls electrons towards itself, making that system poorer in electrons. Think of it as pulling electrons out of the ring or chain.

🎯 Exam Tip: For the negative mesomeric effect, remember that the group typically has multiple bonds (like \( \text{C}=\text{O} \) or \( \text{C}\equiv\text{N} \)) where the more electronegative atom is connected to the conjugated system, allowing it to withdraw electrons.

Question 4. Explain hyper conjugate effect of vinyl chloride molecule.
Answer: Hyperconjugation is observed when atoms or groups with lone pairs of electrons are attached by a single bond and are in conjugation with a \( \pi \) bond. In vinyl chloride, the lone pair on the chlorine atom can enter into resonance with the adjacent \( \pi \) bond. This delocalization of electrons results in more than one contributing structure, as shown below:
\( \text{H}_2\text{C}=\text{CH}-\ddot{\text{C}}l: \leftrightarrow \overset{\ominus}{\text{H}_2\text{C}}-\text{CH}=\overset{\oplus}{\text{C}}l: \)
In simple words: In vinyl chloride, the electrons that are usually "alone" on the chlorine atom can jump in and help out with the double bond next to it. This makes the electrons spread out more across the molecule, which helps to make it more stable.

🎯 Exam Tip: While hyperconjugation typically refers to \( \sigma \)-\( \pi \) interactions, sometimes similar electron delocalization through resonance involving lone pairs on atoms adjacent to a \( \pi \)-system can be discussed in this context, especially when it stabilizes the molecule.

Question 5. What are the types of substitution reaction?
Answer: A substitution reaction involves replacing an atom or a group of atoms attached to a carbon atom with a new atom or group. These reactions are categorized based on the nature of the attacking reagent. The main types are:

• Nucleophilic substitution
• Electrophilic substitution
• Free radical substitution

🎯 Exam Tip: Clearly differentiate between the three types of substitution reactions by identifying the nature of the attacking species (nucleophile, electrophile, or free radical).

Question 6. Explain free radical substitution reaction with the suitable example.
Answer: A free radical substitution reaction involves the replacement of an atom or group by a free radical. These reactions usually start with homolytic cleavage of a bond to produce free radicals, which then react to form new bonds. A common example is the chlorination of methane, where a hydrogen atom is replaced by a chlorine atom.
\( \text{CH}_4 + \text{Cl}_2 \overset{hv}{\longrightarrow} \text{CH}_3\text{Cl} + \text{HCl} \)
The light energy (\( hv \)) helps to break the \( \text{Cl-Cl} \) bond, forming chlorine free radicals, which then initiate the substitution.
In simple words: In this type of reaction, a very reactive "free radical" (an atom with an unpaired electron) takes the place of another atom in a molecule. For example, light helps chlorine radicals replace a hydrogen in methane.

🎯 Exam Tip: When explaining free radical substitution, mention the initiation step (formation of free radicals, usually by light or heat) and give a simple example like the halogenation of alkanes.

Question 7. Explain Nucleophilic Addition reaction with suitable example.
Answer: A nucleophilic addition reaction occurs when a nucleophile (an electron-rich species) attacks an electron-deficient carbon atom in a multiple bond, such as in a carbonyl compound (\( \text{C}=\text{O} \)). The \( \pi \) bond breaks, and new single bonds are formed. A general representation and an example are shown below:
General reaction:
\( \text{A}=\text{B} + \text{Y}^-\text{W}^+ \rightarrow \text{A}(\text{Y})-\text{B}(\text{W}) \)
Example: Addition of HCN to acetaldehyde
\( \text{H}_3\text{C}-\overset{\text{O}}{\parallel}\text{CH} + \text{H}-\text{C}\equiv\text{N} \rightarrow \text{H}_3\text{C}-\overset{\text{OH}}{\underset{\text{C}\equiv\text{N}}{\mid}}\text{CH} \)
In this reaction, the cyanide ion (\( \text{CN}^- \)) acts as a nucleophile, attacking the carbon atom of the carbonyl group in acetaldehyde.
In simple words: This reaction happens when a "nucleophile" (an electron-rich molecule) adds itself to a double or triple bond, especially in compounds like aldehydes or ketones. The double bond opens up, and the nucleophile attaches to one side.

🎯 Exam Tip: Nucleophilic addition is characteristic of carbonyl compounds (aldehydes and ketones) because the carbonyl carbon is electron-deficient and readily attacked by nucleophiles.

Question 8. Explain free radical addition reaction with suitable example.
Answer: A free radical addition reaction involves free radicals adding across a multiple bond (like a double or triple bond). This type of reaction typically occurs in the presence of peroxides or light, which help generate the initial free radicals. The reaction proceeds through a chain mechanism involving radical intermediates. A general reaction and an example are:
General reaction:
\( \text{A}=\text{B} + \text{Y}\cdot \rightarrow \text{A}(\text{Y})-\text{B}\cdot \)
\( \text{A}(\text{Y})-\text{B}\cdot + \text{Y}-\text{W} \rightarrow \text{A}(\text{Y})-\text{B}(\text{W}) + \text{Y}\cdot \)
Example: Addition of HBr to ethene in the presence of peroxides
\( \text{H}_2\text{C}=\text{CH}_2 + \text{HBr} \overset{\text{Benzoyl Peroxide}}{\longrightarrow} \text{CH}_3-\text{CH}_2-\text{Br} \)
Here, benzoyl peroxide acts as a radical initiator, leading to the anti-Markovnikov addition of HBr to the alkene.
In simple words: This is when very reactive "free radicals" add themselves across a double or triple bond. It often needs a special helper like a peroxide to get started. For example, HBr can add to an alkene in a specific way when peroxides are present.

🎯 Exam Tip: Free radical addition to alkenes is often characterized by anti-Markovnikov regioselectivity (hydrogen adds to the carbon with fewer hydrogens), which is distinct from electrophilic addition (Markovnikov's rule).

Question 9. Write a note on Functional Group inter conversion.
Answer: Functional group interconversion is a key process in organic synthesis, where one functional group in a molecule is chemically changed into another using suitable reagents. This allows chemists to build complex molecules from simpler ones by modifying their reactive parts. For instance, a carboxylic acid group (\( -\text{COOH} \)) can be transformed into various other functional groups, such as an alcohol (\( -\text{CH}_2\text{OH} \)), an amide (\( -\text{CONH}_2 \)), or an acyl chloride (\( -\text{COCl} \)), by reacting it with different chemicals like \( \text{LiAlH}_4 \) (lithium aluminum hydride), \( \text{NH}_3 \) (ammonia), or \( \text{SOCl}_2 \) (thionyl chloride), respectively.
In simple words: It's like changing one part of a chemical molecule into a different part using special chemicals. This helps chemists make new molecules by swapping out the active sections.

🎯 Exam Tip: Be familiar with common reagents and their effects on functional groups (e.g., reducing agents, oxidizing agents, reagents for forming derivatives) as this is fundamental to multi-step organic synthesis.

Question 10. After cutting an apple it turn brown. Why?
Answer: When an apple is cut, its cells are exposed to oxygen in the air. Apples contain an enzyme called polyphenol oxidase (PPO), also known as tyrosinase. This enzyme reacts with phenolic compounds present in the apple in the presence of oxygen. This reaction is called enzymatic browning, and it produces new colored compounds, making the apple turn brown. This browning effect is also seen in other fruits and vegetables like bananas, pears, avocados, and potatoes.
In simple words: Apples turn brown when cut because an enzyme inside them, called PPO, reacts with air (oxygen). This chemical reaction makes brown colored substances form on the cut surface.

🎯 Exam Tip: To prevent enzymatic browning, you can limit the apple's exposure to oxygen (e.g., by submerging it in water), reduce the enzyme's activity (e.g., by refrigeration or adding acid like lemon juice), or use antioxidants.

III. Short Answer Questions (3 Marks):

Question 1. Explain hybridization of carbon in carbocation.
Answer: In a carbocation, the carbon atom that carries the positive charge is \( \text{sp}^2 \) hybridized. This means it uses one s orbital and two p orbitals to form three \( \sigma \) bonds, which are arranged in a trigonal planar geometry. The remaining unhybridized p orbital on the carbocation is empty. Because of its \( \text{sp}^2 \) hybridization, the carbocation has a planar structure.
In simple words: The carbon atom in a carbocation that has a positive charge is shaped like a flat triangle because it uses three special orbitals to make bonds. It also has an empty space above and below it where electrons could go.

🎯 Exam Tip: Always remember that \( \text{sp}^2 \) hybridization leads to a trigonal planar geometry and a bond angle of approximately \( 120^\circ \), which is critical for understanding carbocation reactivity and selectivity.

Question 1. Explain hybridization of carbon in carbocation.
Answer: In a carbocation, the carbon atom with a positive charge is sp² hybridized. This means it has a planar (flat) structure. When a nucleophile (a negatively charged species) approaches, it can attack the carbocation from either side. This allows for diverse reaction pathways. Below are diagrams illustrating the shape of a carbocation, carbanion, and carbon radical, showing their respective orbital structures.

Carbanions usually have a pyramidal shape, with the lone pair of electrons occupying an sp³ hybridized orbital. An alkyl free radical can be either pyramidal or planar, depending on the specific molecule.


In simple words: A carbocation has a central carbon with a positive charge and a flat shape because it uses sp² hybridization. This allows other molecules to attack it from two sides. Carbanions are typically pyramidal, and carbon radicals can be either flat or pyramidal.

🎯 Exam Tip: Remember the geometries for carbocations (planar), carbanions (pyramidal), and free radicals (planar or pyramidal) as this affects how they react in chemical processes.

Question 2. Write three types of electron movement in organic reaction.
Answer: In organic reactions, electrons move in specific ways during bond breaking and forming. There are three main types of electron movement:

• A lone pair of electrons becomes a bonding pair.
• A bonding pair of electrons becomes a lone pair.
• A bonding pair of electrons moves to become another bonding pair (in a different location).

Type: 1 - A lone pair to a bonding pair

Type: 2 - A bonding pair to a lone pair

Type: 3 - A bonding pair to another bonding pair

🎯 Exam Tip: Learning to draw curved arrows accurately is key to showing electron movement in reaction mechanisms. Always start the arrow from the electron pair and point it to where the electrons are moving.

Question 3. Explain hyper conjugate effect of acrylonitrile.
Answer: The hyperconjugation effect is observed when electronegative atoms or groups are linked to a carbon-carbon double bond, allowing them to pull pi (π) electrons from the multiple bond. In the case of carbocations, a greater number of alkyl groups attached to the positively charged carbon leads to more hyperconjugative structures. This increased number of hyperconjugative structures helps stabilize the carbocation. However, when electronegative atoms are in conjugation, the stability of carbocations actually decreases in the order: Tertiary Carbocation > Secondary Carbocation > Primary Carbocation. This means more alkyl groups usually mean more stability due to hyperconjugation. For acrylonitrile, the electron delocalization is shown below:

In simple words: Hyperconjugation helps stabilize carbocations when alkyl groups are attached. In acrylonitrile, the electrons shift around, creating different forms (resonance structures) where positive and negative charges appear on different atoms, which makes the molecule more stable overall.

🎯 Exam Tip: For hyperconjugation, remember that the more alpha-hydrogens (hydrogens on the carbon next to the positively charged carbon), the greater the number of hyperconjugative structures and the more stable the carbocation.

Question 4. What are the different types of organic reactions?
Answer: Organic compounds go through many different types of reactions. Even though there are many, we can group them into six main categories. These categories help us understand how molecules change and form new ones.

1. Substitution reactions
2. Addition reactions
3. Elimination reactions
4. Oxidation and reduction reactions
5. Rearrangement reactions
6. Combination reactions

🎯 Exam Tip: Knowing the basic categories of organic reactions helps you predict products and understand reaction mechanisms, which is fundamental for organic chemistry.

Question 5. Explain nucleophilc substituion reaction with suitable example.
Answer: In a nucleophilic substitution reaction, an atom or a group of atoms connected to a carbon atom is replaced by a new atom or group. The type of reaction depends on the attacking reagent. For example, if a nucleophile (an electron-rich species) attacks, it is a nucleophilic substitution. This reaction can be shown generally as:

Here, \( \text{Y}^- \) is the incoming nucleophile or attacking species, and \( \text{X}^- \) is the leaving group. A common example is the hydrolysis of alkyl halides using aqueous KOH. This reaction follows either an \( \text{S}_N1 \) or \( \text{S}_N2 \) mechanism.
\( \text{CH}_3\text{Br} \xrightarrow{\text{aqueous OH}^-} \text{CH}_3\text{OH} + \text{Br}^- \)
In simple words: In this reaction, one part of a molecule is replaced by another. For example, when an alkyl bromide reacts with water and a base, the bromine is replaced by a hydroxyl group, creating an alcohol.

🎯 Exam Tip: For nucleophilic substitution reactions, always identify the nucleophile, the electrophile, and the leaving group. Consider the solvent and structure to determine if it's an S_N1 or S_N2 pathway.

Question 6. Explain Electrophilic Addition Reaction with suitable example.
Answer: An electrophilic addition reaction happens when an electrophile (an electron-loving species) attacks a multiple bond (like a carbon-carbon double or triple bond). The pi electrons in the multiple bond are attracted to the electrophile, leading to the formation of new sigma bonds. This general reaction can be shown as:

An example of this is the bromination of an alkene to form a dibromo alkane. Here, \( \text{Y}^+ \) is an electrophile.

In simple words: An electrophilic addition reaction happens when a molecule with a double bond reacts with an "electron-loving" particle. The double bond opens up, and two new atoms or groups attach to the carbons that were part of the double bond. For example, when bromine reacts with an alkene, both bromine atoms add across the double bond.

🎯 Exam Tip: In electrophilic addition to unsymmetrical alkenes, Markovnikov's rule helps predict where the electrophile and nucleophile will attach. The positive part usually goes to the carbon with more hydrogens.

Question 7. What are oxidation and reduction reactions? Give suitable example.
Answer: Oxidation and reduction reactions are fundamental processes in chemistry. While they can be complex, for organic compounds, we often define them in terms of oxygen and hydrogen atoms. Oxidation generally means a gain of oxygen atoms or a loss of hydrogen atoms. Conversely, reduction involves a gain of hydrogen atoms or a loss of oxygen atoms. Many organic reactions fall into one of these categories.

**Example:**
Oxidation of Acetaldehyde to Acetic Acid:
\( \text{CH}_3\text{CHO} \xrightarrow{(\text{O}) / \text{Acidic dichromate}} \text{CH}_3\text{COOH} \)

In simple words: Oxidation means adding oxygen or taking away hydrogen, while reduction means taking away oxygen or adding hydrogen. For example, acetaldehyde becomes acetic acid (gains oxygen), and phenol can become p-benzoquinone (loses hydrogen, gains double bonds involving oxygen) during oxidation.

🎯 Exam Tip: Remember these simple definitions: oxidation is "loss of electrons/gain of oxygen/loss of hydrogen," and reduction is "gain of electrons/loss of oxygen/gain of hydrogen." This helps quickly classify reactions.

IV. Long answer questions (5 Marks):

Question 1. Explain different types of Fission of a covalent bond.
Answer: All organic molecules have covalent bonds, which form when atoms share electrons. These bonds can break in two main ways: homolytic cleavage (symmetrical splitting) or heterolytic cleavage (unsymmetrical splitting). How a bond breaks is often influenced by the type of attacking reagent present.

**Homolytic Cleavage:**
This is a symmetrical process where a covalent bond breaks so that each bonded atom gets one electron. This is shown with a half-headed arrow (fish-hook arrow). Homolytic cleavage often happens under high temperatures or with UV light, especially in compounds with non-polar covalent bonds between atoms that have similar electronegativity. This type of bond breaking produces highly reactive species called free radicals. Substances that cause homolytic cleavage, like azobisisobutyronitrile (AIBN) and benzoyl peroxide, are called free radical initiators and are used in polymerization reactions. For example, benzoyl peroxide breaks down as follows:

Free radicals are neutral but unstable because they have an unpaired electron, so they tend to react quickly to become stable. Many organic reactions involve homolytic fission of C-C bonds to form alkyl free radicals. The stability of alkyl free radicals follows this order: \( \text{C(CH}_3\text{)}_3 > \text{CH(CH}_3\text{)}_2 > \text{CH}_2\text{CH}_3 > \text{CH}_3 \).

**Heterolytic Cleavage:**
In heterolytic cleavage, a covalent bond breaks unevenly, meaning one of the bonded atoms keeps both electrons from the shared pair. This results in the formation of a cation (positively charged ion) and an anion (negatively charged ion). The more electronegative atom typically becomes the anion, and the other atom becomes the cation. This cleavage is shown with a curved arrow pointing towards the more electronegative atom.
For example, in tert-butyl bromide, the C-Br bond is polar because bromine is more electronegative than carbon. The bonding electrons are pulled more towards bromine. This causes the C-Br bond to undergo heterolytic cleavage during hydrolysis, forming a tert-butyl cation.

When considering the cleavage of a carbon-hydrogen (C-H) bond in aldehydes or ketones, we know that carbon is more electronegative than hydrogen. Therefore, heterolytic cleavage of C-H bonds leads to the formation of a carbanion (a carbon with a negative charge). For instance, in an aldol condensation, an \( \text{OH}^- \) ion takes an alpha-hydrogen from the aldehyde, forming a carbanion as shown below:

In simple words: Covalent bonds can break in two ways: homolytic, where each atom gets one electron (forming free radicals), or heterolytic, where one atom takes both electrons (forming ions like cations and anions). Free radicals are very reactive, and the stability of different alkyl free radicals varies. The stability of carbocations also varies depending on the number of alkyl groups and other effects.

🎯 Exam Tip: Distinguishing between homolytic and heterolytic cleavage is crucial. Look for fish-hook arrows for homolytic (one electron movement) and full-headed arrows for heterolytic (two electron movement).

Question 2. Explain the acidic nature of carboxylic acid.
Answer: When a halogen atom is connected to the carbon atom closest to a carboxylic acid group, its -I (negative inductive) effect pulls the bonding electrons towards itself. This makes it easier for the hydrogen atom in the -OH group to ionize and leave as a proton, thus increasing the acidity. The acidity of different chloroacetic acids shows this effect: trichloroacetic acid is more acidic than dichloroacetic acid, which is more acidic than chloroacetic acid, and chloroacetic acid is more acidic than acetic acid. This is because the strength of the acid increases as the -I effect of the attached group increases.

Similarly, the following order of acidity in carboxylic acids is due to the +I (positive inductive) effect of alkyl groups. Alkyl groups donate electrons, pushing electron density towards the carboxyl group. This makes it harder for the hydrogen to ionize, thus decreasing acidity. Larger alkyl groups have a stronger +I effect, leading to lower acidity.

In simple words: Carboxylic acids are acidic because the hydrogen atom in their -OH group can leave easily. If electron-pulling atoms like chlorine are nearby, they make the acid stronger. But if electron-donating groups like methyl (CH₃) are nearby, they make the acid weaker because they hold the hydrogen more tightly.

🎯 Exam Tip: To predict the acidity of carboxylic acids, focus on the electron-withdrawing or electron-donating nature of substituents. Electron-withdrawing groups increase acidity, while electron-donating groups decrease it.