General organic chemistry (GOC) for NEET

General organic chemistry is the grammar of the organic half of the paper. Every named reaction you will learn later, from Markovnikov addition to SN1, is explained by the ideas here: which intermediate is more stable, which atom is electron-rich, which group pulls or pushes electrons. A student who is shaky in GOC ends up memorising organic chemistry; a student who is solid in it can reason most of it out.

NEET tests GOC directly through stability orders (carbocations, free radicals, alkenes), acid strength orders, counting σ and π bonds or hyperconjugating hydrogens, IUPAC names, isomer counts, and "which is an electrophile". It also tests it indirectly in every reaction question in hydrocarbons and later chapters.

Hybridisation, shape and bonds

Carbon is tetravalent. The type of hybridisation depends on how many σ bonds and lone pairs the carbon has.

More s-character means the carbon is more electronegative. That single fact explains why ethyne's hydrogen is acidic and why sp carbons pull electrons more strongly than sp³ carbons.

Counting bonds: a single bond is one σ; a double bond is one σ and one π; a triple bond is one σ and two π.

Classification and IUPAC naming

Organic compounds are classified as acyclic (open chain) or cyclic, and cyclic compounds as alicyclic, aromatic (benzenoid) or heterocyclic. Members of a homologous series share a general formula and functional group and differ by –CH₂–.

IUPAC steps:

1. Find the principal functional group, using the priority order below.
2. Choose the longest chain that contains it (and multiple bonds if possible).
3. Number the chain to give the principal group the lowest locant, then multiple bonds, then substituents.
4. Name the substituents as prefixes in alphabetical order.

Priority order (highest first): –COOH > –SO₃H > –COOR > –COCl > –CONH₂ > –CN > –CHO > >C=O > –OH > –NH₂ > C=C > C≡C. Halogens (–X), –NO₂ and –OR are always named as prefixes.

Isomerism

Isomer counts worth knowing: C₄H₁₀ has 2 structural isomers, C₅H₁₂ has 3 and C₆H₁₄ has 5.

Bond fission and reaction intermediates

• Homolytic fission: each atom keeps one electron of the bond, giving free radicals. It needs heat, light or peroxides and is shown with half-headed (fish-hook) arrows.
• Heterolytic fission: one atom takes both electrons, giving a carbocation and an anion, or a carbanion and a cation.

Stability order:

• Carbocations: 3° > 2° > 1° > CH₃⁺, because alkyl groups donate electrons by the +I effect and by hyperconjugation. Allyl and benzyl cations are further stabilised by resonance.
• Free radicals: the same order, 3° > 2° > 1° > CH₃•, for the same reasons.
• Carbanions: the reverse, CH₃⁻ > 1° > 2° > 3°, because alkyl groups push electrons onto a centre that already has too many.

Electronic effects

Inductive effect:

• −I groups (pull electrons): –NO₂, –CN, –COOH, –F, –Cl, –Br, –I. Among halogens, F > Cl > Br > I.
• +I groups (push electrons): alkyl groups, (CH₃)₃C– > (CH₃)₂CH– > CH₃CH₂– > CH₃–.

Resonance effect:

• +R groups donate a lone pair into the π system: –OH, –OR, –NH₂, –NHR, –X.
• −R groups withdraw π electrons: –NO₂, –CN, –CHO, –COR, –COOH.

Hyperconjugation counts α-hydrogens: hydrogens on carbons directly attached to the positive carbon or the double bond. (CH₃)₃C⁺ has 9, (CH₃)₂CH⁺ has 6, CH₃CH₂⁺ has 3. Among alkenes, (CH₃)₂C=C(CH₃)₂ with 12 α-hydrogens is the most stable.

Acidity, electrophiles and nucleophiles

Anything that stabilises the conjugate base increases acid strength. So electron-withdrawing groups increase acidity and electron-donating groups decrease it.

• Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH (more −I groups).
• FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH (stronger −I).
• 2-Chlorobutanoic acid > 3-chlorobutanoic > 4-chlorobutanoic (−I fades with distance).
• HCOOH > CH₃COOH (the methyl group's +I weakens the acid).
• Phenol > ethanol, because the phenoxide ion is resonance-stabilised.
• HC≡CH > H₂C=CH₂ > H₃C–CH₃ (s-character of the carbon holding the anion).

Electrophiles are electron-seeking: H⁺, NO₂⁺, Br⁺, carbocations, and neutral Lewis acids such as BF₃ and AlCl₃. Nucleophiles are electron-rich: OH⁻, CN⁻, Cl⁻, RO⁻, and neutral molecules with lone pairs such as H₂O, NH₃ and ROH.

Types of organic reactions

• Substitution: one atom or group replaces another. CH₃Cl + OH⁻ → CH₃OH + Cl⁻.
• Addition: a reagent adds across a multiple bond. CH₂=CH₂ + Br₂ → BrCH₂CH₂Br.
• Elimination: atoms are removed from adjacent carbons to form a multiple bond. Alcoholic KOH on bromoethane gives ethene.
• Rearrangement: atoms or groups move within a molecule, often through a carbocation that shifts to a more stable one.

Worked NEET-style questions

Example 1: counting bonds and hybridisation

For CH₂=CH–C≡N, count σ and π bonds and state each carbon's hybridisation.

• σ bonds: 2 C–H + 1 C=C σ + 1 C–H + 1 C–C + 1 C≡N σ = 6.
• π bonds: 1 in C=C + 2 in C≡N = 3.
• Carbons from left: sp², sp², sp.

Example 2: carbocation stability

Arrange (CH₃)₃C⁺, CH₃CH₂⁺, (CH₃)₂CH⁺ and CH₃⁺ by stability.

• α-hydrogens: 9, 3, 6 and 0.
• Order: (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺.

Example 3: acid strength

Arrange CH₃COOH, ClCH₂COOH, Cl₂CHCOOH and FCH₂COOH by acid strength.

• Two chlorines outweigh one fluorine; F has a stronger −I than Cl.
• Order: Cl₂CHCOOH > FCH₂COOH > ClCH₂COOH > CH₃COOH. Their pKa values are about 1.3, 2.6, 2.9 and 4.8, which confirms the order.

Example 4: isomers of C₄H₁₀O

How many structural isomers does C₄H₁₀O have?

• Alcohols: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol = 4.
• Ethers: ethoxyethane, 1-methoxypropane, 2-methoxypropane = 3.
• Total: 7. Butan-2-ol also has a pair of optical isomers, which a stereoisomer count would add.

Practice MCQs

1. The central carbon of CH₂=C=CH₂ is: (a) sp³ (b) sp² (c) sp (d) dsp²
2. The most stable carbocation among these is: (a) CH₃⁺ (b) CH₃CH₂⁺ (c) (CH₃)₂CH⁺ (d) (CH₃)₃C⁺
3. The number of hyperconjugating hydrogens in (CH₃)₂CH⁺ is: (a) 2 (b) 3 (c) 6 (d) 7
4. Which is an electrophile? (a) BF₃ (b) H₂O (c) NH₃ (d) CN⁻
5. The number of structural isomers of C₅H₁₂ is: (a) 2 (b) 3 (c) 4 (d) 5
6. Ethoxyethane and 1-methoxypropane are: (a) chain isomers (b) position isomers (c) functional isomers (d) metamers
7. The most acidic hydrocarbon among these is: (a) ethyne (b) ethene (c) ethane (d) propane
8. Which electronic effect is temporary? (a) inductive (b) hyperconjugation (c) resonance (d) electromeric

Answers

1. (c) It forms two σ bonds and two π bonds.
2. (d) Tertiary, with 9 α-hydrogens and three +I groups.
3. (c) Two methyl groups with three hydrogens each.
4. (a) Boron has an incomplete octet and accepts an electron pair.
5. (b) Pentane, 2-methylbutane and 2,2-dimethylpropane.
6. (d) Different alkyl groups on either side of the oxygen.
7. (a) The sp carbon holds the negative charge best.
8. (d) It operates only while an attacking reagent is present.

What to do next

• Write the four electronic effects on one card with two examples each.
• Practise ten stability orders (carbocations, radicals, carbanions, alkenes) and ten acidity orders, giving the reason for each.
• Name twenty structures by IUPAC rules, starting with the priority order.
• Use these ideas straight away in hydrocarbons and haloalkanes and haloarenes.