Why Carbon is Special

Over 20 million organic compounds are known — dwarfing all inorganic compounds combined.

Hydrocarbon Families

Benzene (C6H6) is the archetypal aromatic. Its 6 pi electrons are delocalized across the ring, creating exceptional stability (aromaticity). The resonance energy of benzene is approximately 150 kJ/mol — it does not undergo the addition reactions typical of alkenes.

IUPAC Nomenclature

Alkane Naming Rules

1. Find the longest continuous carbon chain — this is the parent name.
2. Number the chain from the end nearest the first substituent.
3. Name substituents as prefixes with their position numbers.
4. Use di-, tri-, tetra- for multiple identical substituents.
5. List substituents alphabetically (ignoring di-, tri- prefixes).

Worked Example: Naming a Branched Alkane

Structure: A 6-carbon chain with methyl groups on carbons 2 and 4.

Step 1. Parent chain: hexane (6 carbons). Step 2. Number from the end that gives the lowest locants: positions 2 and 4 (not 3 and 5). Step 3. Two methyl substituents at positions 2 and 4. Name: 2,4-dimethylhexane.

Worked Example: Naming an Alkene

Structure: A 5-carbon chain with a double bond between carbons 1 and 2, and a methyl group on carbon 3.

Step 1. Parent chain includes the double bond: pentene. Step 2. Number from the end nearest the double bond: 1-pentene. Step 3. Methyl at position 3. Name: 3-methyl-1-pentene.

Functional Groups

Functional groups are the reactive sites of organic molecules. They determine chemical behavior.

Priority for naming: carboxylic acid > ester > amide > aldehyde > ketone > alcohol > amine. The highest-priority group becomes the suffix; lower-priority groups are prefixes.

Isomerism

Structural (Constitutional) Isomers

Same molecular formula, different connectivity. C4H10 has two structural isomers: butane (straight chain) and 2-methylpropane (branched).

Geometric (Cis-Trans) Isomers

Arise from restricted rotation around C=C double bonds. Cis: identical groups on the same side. Trans: on opposite sides. For more complex cases, use E/Z notation based on Cahn-Ingold-Prelog priority rules (higher atomic number = higher priority; Z = same side, E = opposite).

Worked example. 2-butene has cis and trans isomers.

cis-2-butene: both CH3 groups on the same side of the double bond. Boiling point: 3.7 C. trans-2-butene: CH3 groups on opposite sides. Boiling point: 0.9 C.

The cis isomer has a higher boiling point because it has a net dipole moment (polar); the trans isomer's dipoles cancel (nonpolar).

Enantiomers (Optical Isomers)

Mirror-image molecules that are non-superimposable. Require a chiral center — typically a carbon bonded to four different groups. Enantiomers have identical physical properties except they rotate plane-polarized light in opposite directions. They can have dramatically different biological activity (thalidomide: one enantiomer treats nausea, the other causes birth defects).

Reaction Mechanisms

Substitution Reactions

SN2 (bimolecular nucleophilic substitution): One step. Nucleophile attacks as leaving group departs. Backside attack causes inversion of configuration. Rate = k[substrate][nucleophile]. Favored by: strong nucleophile, primary substrate, polar aprotic solvent.

SN1 (unimolecular nucleophilic substitution): Two steps. (1) Leaving group departs, forming carbocation. (2) Nucleophile attacks. Rate = k[substrate]. Favored by: tertiary substrate (stable carbocation), polar protic solvent. Produces racemic mixture.

Worked Example: SN2 vs. SN1 Prediction

Problem. Predict the mechanism for: CH3Br + OH- -> CH3OH + Br-

Analysis. Substrate is primary (methyl — no branching). Strong nucleophile (OH-). Backside attack is unhindered. This is textbook SN2.

Problem. Predict the mechanism for: (CH3)3CBr + H2O -> (CH3)3COH + HBr

Analysis. Substrate is tertiary (three methyl groups block backside attack). Weak nucleophile (H2O). The t-butyl carbocation (CH3)3C+ is stable (three hyperconjugating methyls). This is SN1.

Elimination Reactions

E2: One step. Strong base removes a proton while the leaving group departs. Anti-periplanar geometry required. Competes with SN2 (use bulky base to favor E2).

E1: Two steps via carbocation. Competes with SN1. Zaitsev's rule: the more substituted alkene is the major product (more stable).

Addition Reactions

Alkene addition. Electrophilic addition to C=C:

• HX addition: Markovnikov's rule — H adds to the carbon with more hydrogens. Anti-Markovnikov with peroxide (radical mechanism).
• Halogenation (Br2): adds across double bond, anti-addition.
• Hydration (H2O/H+): Markovnikov addition of OH.
• Hydrogenation (H2/Pt): syn-addition, converts alkene to alkane.

Worked Example: Markovnikov Addition

Problem. Predict the major product of HBr addition to propene (CH3CH=CH2).

Step 1. H+ adds to the less-substituted carbon (CH2 end), forming a secondary carbocation on the middle carbon: CH3CH+CH3.

Step 2. Br- attacks the carbocation: CH3CHBrCH3 (2-bromopropane).

Markovnikov's rule in modern terms: the electrophile (H+) adds to form the more stable carbocation intermediate. Secondary > primary carbocation stability explains the regiochemistry.

Polymers

Addition (Chain-Growth) Polymers

Monomers with C=C bonds undergo chain addition. No atoms are lost.

Example: Polyethylene. n CH2=CH2 -> -(CH2-CH2)n-

Other addition polymers: polypropylene (PP), polystyrene (PS), polyvinyl chloride (PVC), polytetrafluoroethylene (PTFE/Teflon).

Condensation (Step-Growth) Polymers

Monomers with two functional groups react, releasing a small molecule (usually H2O) at each step.

Example: Nylon 6,6. Hexamethylenediamine (H2N-(CH2)6-NH2) + adipic acid (HOOC-(CH2)4-COOH) -> amide bonds (-CONH-) + H2O at each linkage.

Example: Polyester (PET). Ethylene glycol + terephthalic acid -> ester bonds (-COO-) + H2O.

Worked Example: Identifying Polymer Type

Problem. Classify the polymerization of styrene (C6H5CH=CH2).

Analysis. Styrene has a C=C double bond. The pi bond opens and monomers chain-link. No small molecule is released. This is addition polymerization, producing polystyrene.

Problem. Classify the formation of a polyamide from a diamine and a diacid.

Analysis. Each linkage forms an amide bond and releases H2O. Two different functional groups react. This is condensation polymerization.

Biochemistry Basics

The four classes of biomolecules are all organic:

Carbohydrates

Monomers: Monosaccharides (glucose C6H12O6, fructose, galactose). Polymers: Disaccharides (sucrose, lactose), polysaccharides (starch, cellulose, glycogen). Bond: Glycosidic linkage (C-O-C between sugar units). Formed by condensation; broken by hydrolysis. Function: Energy storage (starch, glycogen), structural (cellulose in plant cell walls).

Lipids

Not true polymers — diverse hydrophobic molecules. Fats/oils = glycerol + 3 fatty acids (ester bonds). Phospholipids = glycerol + 2 fatty acids + phosphate group (cell membranes). Steroids = four fused rings (cholesterol, hormones). Saturated vs. unsaturated: Saturated fatty acids have no C=C bonds (solid at RT). Unsaturated have one or more C=C (liquid at RT, kinked chains reduce packing).

Proteins

Monomers: 20 amino acids (H2N-CHR-COOH), each with a unique R group. Bond: Peptide bond (amide linkage, -CO-NH-) formed by condensation. Levels of structure: Primary (sequence), secondary (alpha-helix, beta-sheet from H-bonding), tertiary (3D fold from R-group interactions), quaternary (multi-subunit assembly). Function: Enzymes (catalysis), structural (collagen, keratin), transport (hemoglobin), immune (antibodies).

Nucleic Acids

Monomers: Nucleotides (sugar + phosphate + nitrogenous base). DNA: Deoxyribose sugar. Bases: A, T, G, C. Double helix with complementary base pairing (A-T, G-C via hydrogen bonds). RNA: Ribose sugar. Bases: A, U, G, C. Single-stranded. Bond: Phosphodiester linkage (sugar-phosphate backbone). Function: DNA stores genetic information. RNA translates it into proteins (mRNA, tRNA, rRNA).

Worked Example: Hydrolysis of a Dipeptide

Problem. What products form when the dipeptide Gly-Ala is hydrolyzed?

Reaction. The peptide bond (-CO-NH-) between glycine and alanine is broken by adding water.

H2N-CH2-CO-NH-CH(CH3)-COOH + H2O -> H2N-CH2-COOH + H2N-CH(CH3)-COOH

Products: glycine and alanine (the two free amino acids). This is the reverse of condensation — the water molecule that was lost during peptide bond formation is restored.

Common Mistakes

Cross-References

• pauling agent: Molecular structure, bonding, and reaction mechanisms. Primary agent for organic structure and mechanism problems.
• franklin agent: Polymer science and materials applications of organic chemistry.
• chemical-bonding skill: Hybridization, sigma/pi bonds, and VSEPR directly underpin organic molecular geometry.
• reactions-stoichiometry skill: Organic reactions follow the same mass-balance and yield calculations.
• analytical-methods skill: Spectroscopic identification (IR, NMR, mass spec) is the primary tool for organic structure determination.
• materials-chemistry skill: Polymer properties connect to bulk material behavior and green chemistry.

References

• Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry. 2nd edition. Oxford University Press.
• McMurry, J. (2016). Organic Chemistry. 9th edition. Cengage Learning.
• Bruice, P. Y. (2017). Organic Chemistry. 8th edition. Pearson.
• Nelson, D. L. & Cox, M. M. (2017). Lehninger Principles of Biochemistry. 7th edition. W. H. Freeman.
• Vollhardt, K. P. C. & Schore, N. E. (2014). Organic Chemistry: Structure and Function. 7th edition. W. H. Freeman.
• IUPAC. (2013). Recommendations on Organic Nomenclature. International Union of Pure and Applied Chemistry.