Easy Practice — Set 1

Quick Recap — Carbon, Hybridisation & Functional Groups

• Carbon is tetravalent. Hybridisation sets shape: $sp^3$ (tetrahedral, $109.5^\circ$), $sp^2$ (planar, $120^\circ$), $sp$ (linear, $180^\circ$).
• General formulae: alkanes $C_nH_{2n+2}$, alkenes $C_nH_{2n}$, alkynes $C_nH_{2n-2}$.
• A homologous series has members differing by $CH_2$ and sharing a functional group.
• Functional groups: $-OH$ (alcohol), $-CHO$ (aldehyde), $>C=O$ (ketone), $-COOH$ (acid), $-NH_2$ (amine), $-O-$ (ether).

Beyond-NCERT JEE Essentials

1. Degree of unsaturation (DBE). For a formula with C carbons, H hydrogens, N nitrogens and X halogens (oxygen and divalent sulphur are ignored, since they do not change the hydrogen count),

DBE $= \dfrac{2C + 2 + N - H - X}{2}$

Each ring and each pi bond contributes 1; a triple bond counts as 2 (two pi bonds) and a benzene ring counts as 4 (three C=C plus one ring). Read DBE as "rings + pi bonds": a value of 4 usually signals a benzene ring, while 1 means one ring or one double bond. [JEE Tip] Nitrogen ADDS on top and halogen SUBTRACTS (each shifts the count by its extra or missing bond versus C-H); dropping the +N or -X is the commonest slip, and a non-integer answer always means an atom was miscounted.

2. Electronic effects (separate the four).

• Inductive (I): permanent sigma-bond polarisation, transmitted through bonds and dying out within 3-4 bonds. Withdrawing (-I): -NR3+, -NO2, -CN, -COOH, -F, -Cl, -Br, -I, -OR, -OH, -C6H5 (roughly decreasing). Releasing (+I): -O-, -COO- and alkyl groups, with -C(CH3)3 > -CH(CH3)2 > -CH2CH3 > -CH3.
• Resonance / mesomeric (M): permanent delocalisation of pi or lone-pair electrons through a conjugated system, usually STRONGER than induction. +M (lone-pair donors): -O-, -NH2, -NR2, -OH, -OR, -NHCOR, -X. -M (pi-acceptors): -NO2, -CN, -CHO, >C=O, -COOH, -COOR, -CONH2.
• Hyperconjugation: delocalisation of alpha C-H sigma electrons into an adjacent empty p-orbital (cation), pi system (alkene) or half-filled orbital (radical), i.e. "no-bond resonance." The number of hyperconjugative structures equals the number of alpha C-H bonds. [JEE Tip] Count alpha hydrogens, not carbons: the tert-butyl cation has 9 alpha C-H (9 structures), isopropyl 6, ethyl 3, methyl 0.
• Electromeric (E): a temporary, complete shift of a pi pair to one atom that lasts only while a reagent attacks (e.g. H+ adding to C=C) and reverses when the reagent leaves.

3. Stability orders (learn the reasoning, not just the sequence).

• Carbocations (6 electrons, electron-deficient): 3° > 2° > 1° > CH3+, driven by +I and hyperconjugation. Resonance beats both, so allyl and benzyl cations sit at or above tertiary (benzyl > allyl, more resonance structures). An adjacent lone-pair heteroatom is the strongest stabiliser of all: CH3-O-CH2+ becomes the oxocarbenium CH3-O+=CH2 with every atom octet-complete, beating a plain 3° cation. Vinyl and aryl cations (sp carbon) are very unstable. [JEE Tip] Aromaticity flips the sign: the cyclopropenyl cation (2 pi, aromatic) is stabilised, but the cyclopentadienyl cation (4 pi, antiaromatic) is destabilised.
• Carbanions (8 electrons, electron-rich): the alkyl order REVERSES to CH3- > 1° > 2° > 3°, because +I alkyls intensify the negative charge. Stabilisers are -I / -M (EWG) groups and s-character (sp > sp2 > sp3, so the ethynyl anion is unusually stable); resonance onto electronegative atoms dominates, as in the nitronate O2N-CH2- and in enolates.
• Free radicals (7 electrons): same direction as cations, benzyl and allyl > 3° > 2° > 1° > methyl, via hyperconjugation and resonance.
• Alkenes: more alkyl-substituted means more stable (more hyperconjugation), so tetrasubstituted > trisubstituted > disubstituted > monosubstituted > ethene, and trans > cis for a given disubstituted pair. A lower heat of hydrogenation marks the more stable alkene.

4. Aromaticity (Huckel's rule). A species is aromatic only if it is (i) cyclic, (ii) planar, (iii) fully conjugated (an unbroken loop of p-orbitals, every ring atom sp2 or lone-pair-bearing) and (iv) holds $(4n+2)$ pi electrons in the ring ($n = 0, 1, 2, \ldots$, i.e. 2, 6, 10, 14). A planar conjugated ring carrying $4n$ pi electrons (4, 8, …) is antiaromatic and destabilised; break planarity or conjugation (an sp3 centre, a tub shape) and it is merely non-aromatic. Aromatic: benzene (6 pi), cyclopentadienyl anion (4 + lone pair = 6 pi), tropylium cation (6 pi), cyclopropenyl cation (2 pi), pyridine, pyrrole, naphthalene. Antiaromatic: cyclobutadiene and the cyclopentadienyl cation (4 pi). Non-aromatic: cyclooctatetraene (8 pi but tub-shaped). [JEE Tip] For charged rings, count the pi electrons in the ring: an empty p-orbital adds 0, a lone pair in a p-orbital adds 2. That one rule separates the cyclopentadienyl anion (aromatic) from its cation (antiaromatic).

5. Acidity and basicity.

• Acidity ladder: carboxylic acid > carbonic acid > phenol > water > alcohol > terminal alkyne > ammonia > alkene > alkane (pKa roughly 4-5, 6.3, 10, 15.7, 16-18, 25, 38, 44, 50). A carboxylic acid wins because -COO- shares the charge over two equivalent oxygens; phenol beats alcohol because phenoxide delocalises into the ring.
• Substituent effect: -I / -M groups stabilise the anion and RAISE acidity. So Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH (more -Cl), and FCH2COOH > ClCH2COOH (stronger -I of F); the effect fades with distance (2-chlorobutanoic > 3-chloro > 4-chloro > butanoic acid). On phenols, -NO2 at ortho/para (where -M operates) sharply raises acidity: picric acid > p-nitrophenol > phenol > p-cresol; p- and o-nitrophenol > m-nitrophenol.
• Amine basicity: in the gas phase (pure +I) 3° > 2° > 1° > NH3, but in water solvation of the cation scrambles this to roughly 2° >= 1° > 3° > NH3 for methylamines (dimethylamine most basic). Aniline is far weaker than any alkylamine because its lone pair is delocalised into the ring; -NO2 on the ring weakens it further while -CH3 and -OCH3 strengthen it. [JEE Tip] Acid strength and conjugate-base stability always move together, so compare by asking "which anion, or which lone pair, is better stabilised?"
• Keto-enol tautomerism: a carbonyl bearing an alpha hydrogen equilibrates the keto form (>C=O with alpha C-H) with the enol (=C-OH). Keto usually dominates (C=O is stronger than C=C plus O-H), but enol content soars when the enol is stabilised: 1,3-dicarbonyls like pentane-2,4-dione enolise heavily through conjugation and intramolecular H-bonding, and phenol is essentially 100% enol because the keto form would destroy aromaticity. A compound with no alpha H (benzaldehyde) cannot enolise.

Worked Examples — Beyond-NCERT Essentials

Example 1 — Compute the degree of unsaturation. Find DBE for (a) pyridine, C5H5N, and (b) paracetamol, C8H9NO2.

(a) DBE $= \dfrac{2(5) + 2 + 1 - 5}{2} = \dfrac{8}{2} = 4$: one ring plus three pi bonds, exactly the aromatic six-membered ring. (b) DBE $= \dfrac{2(8) + 2 + 1 - 9}{2} = \dfrac{10}{2} = 5$: a benzene ring (4) plus one amide C=O (1). Nitrogen was added and oxygen ignored; a halogen would be subtracted like hydrogen, so benzoyl chloride C7H5ClO gives $\dfrac{14 + 2 - 5 - 1}{2} = 5$, again ring plus C=O.

Example 2 — Rank carbocation stability. Order CH3+, CH3CH2+, (CH3)2CH+, (CH3)3C+ and the benzyl cation C6H5CH2+.

Within the alkyl set, more alkyl groups mean more +I and more hyperconjugation: (CH3)3C+ (3°, 9 alpha C-H) > (CH3)2CH+ (2°, 6) > CH3CH2+ (1°, 3) > CH3+ (0). The benzyl cation spreads its charge into the ring over four resonance structures, and resonance outranks hyperconjugation, so it leads: C6H5CH2+ > (CH3)3C+ > (CH3)2CH+ > CH3CH2+ > CH3+. An alpha-oxygen cation such as CH3OCH2+ would beat even benzyl, since CH3O+=CH2 gives every atom a full octet.

Example 3 — Rank carbanion stability. Order (CH3)3C-, CH3-, the allyl anion CH2=CH-CH2- and the nitromethyl anion O2N-CH2-.

Carbanions are electron-rich, so the alkyl trend REVERSES: +I alkyls pile charge onto the carbon, making (CH3)3C- the least stable and CH3- better. Delocalisation stabilises strongly, since the allyl anion spreads charge over two carbons and the nitronate O2N-CH2- pushes it onto two electronegative oxygens by -M (nitromethane's pKa near 10 proves how stable this anion is). Order: O2N-CH2- > CH2=CH-CH2- > CH3- > (CH3)3C-.

Example 4 — Alkene stability by hyperconjugation. Rank 2-methylbut-2-ene, but-2-ene, but-1-ene and ethene.

Alkene stability grows with alkyl substitution because each alpha C-H hyperconjugates into the pi bond: 2-methylbut-2-ene is trisubstituted, but-2-ene disubstituted, but-1-ene monosubstituted and ethene bare. Stability order (which is the order of INCREASING heat of hydrogenation, because a more stable alkene releases less energy on hydrogenation): 2-methylbut-2-ene > but-2-ene (trans > cis) > but-1-ene > ethene. Equivalently, heat of hydrogenation DECREASES the other way: ethene > but-1-ene > but-2-ene > 2-methylbut-2-ene.

Example 5 — Identify the aromatic species (Huckel). Classify benzene, cyclopentadienyl cation, cyclopentadienyl anion, cyclopropenyl cation, tropylium cation and neutral cyclooctatetraene.

Apply cyclic + planar + fully conjugated + $(4n+2)$ pi. Benzene has 6 pi (n=1), aromatic. The cyclopentadienyl anion has 4 pi from two C=C plus a lone pair in a p-orbital (2), giving 6 pi, aromatic. The cyclopropenyl cation has one C=C (2 pi) with an empty p-orbital, so 2 pi (n=0), aromatic. Tropylium has three C=C = 6 pi, aromatic. The cyclopentadienyl cation has only 4 pi in a planar conjugated ring, antiaromatic. Neutral cyclooctatetraene has 8 pi but puckers into a tub, breaking planarity, so non-aromatic. Aromatic count = 4 (benzene, cyclopentadienyl anion, cyclopropenyl cation, tropylium).

Example 6 — Order acidity of substituted acids. Rank CH3COOH, ClCH2COOH, Cl2CHCOOH, Cl3CCOOH and FCH2COOH.

Acidity rises as -I stabilises the carboxylate anion. More chlorines withdraw more, so Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH; a lone fluorine, being more electronegative, pulls a little harder than a lone chlorine, so FCH2COOH > ClCH2COOH but still less than the dichloro acid. Full order: Cl3CCOOH > Cl2CHCOOH > FCH2COOH > ClCH2COOH > CH3COOH (pKa about 0.7, 1.3, 2.6, 2.9, 4.8).

Example 7 — Order phenol acidity. Rank phenol, p-nitrophenol, p-cresol (p-methylphenol) and picric acid (2,4,6-trinitrophenol).

Electron-withdrawing -NO2 at ortho/para stabilises the phenoxide by -M (charge delocalised onto the nitro oxygens) and raises acidity, whereas electron-releasing -CH3 destabilises the anion and lowers acidity. Three nitro groups make picric acid the strongest: picric acid > p-nitrophenol > phenol > p-cresol. For the nitrophenols alone, para and ortho beat meta (p-/o-nitrophenol > m-nitrophenol) because -M resonance reaches the O- only from the ortho and para positions.

Example 8 — Count hyperconjugative structures. (a) For the tert-butyl cation (CH3)3C+; (b) for propene CH3-CH=CH2.

The number of hyperconjugative (no-bond) structures equals the number of alpha C-H bonds. (a) Three methyls on the cationic carbon, each with 3 C-H, give 9 alpha C-H, hence 9 structures, which is why (CH3)3C+ is so stable. (b) Only the sp3 methyl is alpha to the C=C; its 3 C-H give 3 structures, while the vinylic C-H on the sp2 carbons do not count. This is exactly why propene is more stable than ethene, which has none.