Organic Chemistry PDF Download – McMurry (10th Edition, OpenStax) BS Chemistry

Students of BS Chemistry can download “Organic Chemistry: A Tenth Edition” by John E. McMurry free from the official OpenStax site. McMurry is a Professor Emeritus at Cornell University, and he wrote this edition for a two-semester introductory course, keeping the functional-group approach of earlier editions and adding biological examples for biology and pre-med students. The Preface says he received no payment for this edition, which is why it is free to read.

The 31 chapters start with bonding, acids and bases, stereochemistry and reaction mechanisms, then work through alkenes, alkynes, alkyl halides, NMR and IR spectroscopy, aromatic compounds, alcohols, carbonyl chemistry and amines. The last chapters cover carbohydrates, proteins, lipids, nucleic acids, metabolic pathways, pericyclic reactions and polymers. General chemistry is the only prerequisite, and every chapter has worked examples and problems for semester exam practice.

Book Overview

Chapter List

Chapter 1: Structure and Bonding

Difficulty: Easy · Key topics: atomic orbitals, electron configuration, covalent bond, valence bond theory, sp³/sp²/sp hybridization, molecular orbitals, skeletal structures

The chapter reviews the general chemistry that organic chemistry is built on. It starts with atomic structure: the nucleus, atomic and mass numbers, s and p orbitals, and ground-state electron configurations built with the Aufbau principle, Hund’s rule and the Pauli exclusion principle. Valence bond theory then describes a covalent bond as orbital overlap, and sp³, sp² and sp hybrid orbitals explain the shapes of methane, ethane, ethylene and acetylene. Hybridization in nitrogen, oxygen, phosphorus and sulfur, molecular orbital theory, and the drawing of condensed and skeletal structures complete the chapter.

Key Points:

• Organic chemistry is the study of carbon compounds; the old split between organic and inorganic chemistry has no scientific basis
• Atomic number is the number of protons, mass number is protons plus neutrons, and isotopes differ in neutron number
• Electrons fill the lowest-energy orbitals first (Aufbau), two per orbital with opposite spins (Pauli), singly across equal-energy orbitals before pairing (Hund)
• A σ bond forms by head-on orbital overlap and a π bond by sideways overlap of p orbitals; a double bond is one σ plus one π
• Carbon uses sp³ orbitals for tetrahedral single bonds, sp² for planar double bonds and sp for linear triple bonds
• In a skeletal structure every line end and corner is a carbon, and hydrogens are supplied mentally to give four bonds

Memory Tip: To find an atom’s hybridization, count its σ bonds plus lone pairs: four means sp³, three means sp², two means sp.

Common Mistake: Drawing a carbon with five bonds because the hidden hydrogens were not counted. In a skeletal structure, count the bonds you can see and add hydrogens until the carbon has four.

Important Questions:

• What is the difference between a σ bond and a π bond? A σ bond is formed by head-on overlap and is symmetric about the bond axis; a π bond is formed by sideways overlap of p orbitals above and below the axis. A double bond contains one σ and one π bond.
• What is the hybridization and geometry of the carbons in ethylene? Each carbon is sp² hybridized. Its three sp² orbitals form σ bonds in one plane at about 120°, and the leftover p orbital forms the π bond.

Chapter 2: Polar Covalent Bonds; Acids and Bases

Difficulty: Medium · Key topics: electronegativity, dipole moment, formal charge, resonance, pK_a, Brønsted–Lowry and Lewis acids, hydrogen bonding

This chapter supplies the vocabulary used to explain reactivity. It starts with polar covalent bonds, electronegativity, dipole moments and electrostatic potential maps, then formal charges for tracking valence electrons. Resonance follows, with the rules for drawing resonance forms. The second half covers acids and bases: the Brønsted–Lowry definition, K_a and pK_a, predicting acid–base reactions from pK_a values, organic acids and bases, and the Lewis definition. The chapter ends with noncovalent interactions such as hydrogen bonds, dipole–dipole forces and dispersion forces.

Key Points:

• Unequal sharing of electrons makes a bond polar, and a molecule’s overall polarity is measured by its dipole moment μ
• C–H bonds are nearly nonpolar, while C–O bonds are polar because oxygen pulls the shared electrons
• Resonance forms differ only in the position of π and nonbonding electrons; the nuclei do not move, and the real molecule is a resonance hybrid
• pK_a = −log K_a, so the larger the pK_a, the weaker the acid
• A Lewis acid accepts an electron pair into an empty orbital (BF₃, AlCl₃, H⁺); a Lewis base donates one (NH₃, H₂O)
• Hydrogen bonding between an O–H or N–H and a lone pair on another O or N gives proteins and nucleic acids their shapes

Memory Tip: To predict an acid–base reaction, compare pK_a values: the acid with the lower pK_a gives up its proton, and the equilibrium favors the weaker acid.

Common Mistake: Moving atoms when drawing resonance forms. Only π electrons and lone pairs move; if an atom changes position, you have drawn a different compound, not a resonance form.

Important Questions:

• What is the difference between a Brønsted–Lowry acid and a Lewis acid? A Brønsted–Lowry acid donates a proton. A Lewis acid accepts an electron pair, so it also includes species such as BF₃, AlCl₃ and Mg²⁺ that have no proton to give.
• Why can the acetate ion not be shown with one line-bond structure? Its two C–O bonds are equivalent, but any single structure shows one single and one double bond. The real ion is a resonance hybrid of two forms that share the negative charge equally.

Chapter 3: Organic Compounds: Alkanes and Their Stereochemistry

Difficulty: Easy · Key topics: functional groups, alkane isomers, alkyl groups, IUPAC naming, Newman projection, staggered and eclipsed conformations, gauche, steric strain

Alkanes are unreactive, but they carry several ideas that the rest of the book needs. The chapter first introduces functional groups and the compound families they define. It then covers alkanes with the formula C_nH_2n+2, constitutional isomers, alkyl groups, the IUPAC rules for naming alkanes, and their physical properties. The last two sections turn to three dimensions: rotation about C–C single bonds, Newman projections, and the staggered and eclipsed conformations of ethane and larger alkanes, including the anti and gauche forms of butane.

Key Points:

• A functional group is a group of atoms with characteristic reactivity; a molecule’s reactions are largely set by its functional groups
• Alkanes are saturated hydrocarbons with the formula C_nH_2n+2 and no functional group
• Constitutional isomers share a formula but differ in how atoms are connected, as butane and isobutane do
• Staggered ethane is 12 kJ/mol (2.9 kcal/mol) more stable than eclipsed ethane because of torsional strain
• A Newman projection sights down one C–C bond to show the angle between front and back substituents
• In butane the anti conformation is more stable than gauche (by about 3.8 kJ/mol) because of steric strain

Memory Tip: Name an alkane in four moves: find the longest continuous chain, number from the end nearest the first branch, name each branch, then list the branches alphabetically.

Common Mistake: Choosing the longest straight-looking chain on the page instead of the longest continuous chain. Bends in the drawing do not matter; count carbons along any connected path.

Important Questions:

• What is the general formula of alkanes, and what does saturated mean? The formula is C_nH_2n+2. Saturated means the carbons hold the maximum number of hydrogens, with only single bonds in the molecule.
• Why is the staggered conformation of ethane more stable than the eclipsed one? In the staggered form the C–H bonds on neighboring carbons are as far apart as possible. Eclipsing adds torsional strain of about 12 kJ/mol.

Chapter 4: Organic Compounds: Cycloalkanes and Their Stereochemistry

Difficulty: Medium · Key topics: cycloalkane naming, cis–trans isomers, ring strain, chair cyclohexane, axial and equatorial bonds, ring-flip, 1,3-diaxial interaction, polycyclic molecules

Cyclic molecules are common in organic and biological chemistry, so this chapter examines them closely. It covers naming cycloalkanes, cis–trans isomerism in rings, and the three kinds of strain (angle, torsional and steric) that make some rings less stable. Cyclopropane, cyclobutane and cyclopentane are compared, then the chapter spends five sections on cyclohexane: the chair conformation, axial and equatorial positions, ring-flips, and the conformations of monosubstituted and disubstituted cyclohexanes. It closes with polycyclic molecules such as decalin and steroids.

Key Points:

• Cycloalkanes have the formula C_nH_2n; reduced rotation lets disubstituted rings exist as cis and trans stereoisomers
• Ring strain is the sum of angle strain, torsional strain and steric strain
• Cyclopropane (115 kJ/mol) and cyclobutane (110.4 kJ/mol) are strained; cyclopentane is mostly torsional strain
• Chair cyclohexane has bond angles near 109° and all neighboring C–H bonds staggered, so it is essentially strain-free
• Each ring carbon has one axial and one equatorial position, and a ring-flip interconverts them
• Axial substituents suffer 1,3-diaxial interactions, so larger groups prefer the equatorial position

Memory Tip: Axial bonds point straight up or down, parallel to the ring’s axis, and alternate up, down, up, down around the ring; equatorial bonds lie around the ring’s equator.

Common Mistake: Thinking a ring-flip changes cis into trans. A ring-flip only swaps axial and equatorial positions; the cis or trans relationship stays the same.

Important Questions:

• Why is chair cyclohexane free of strain? All its C–C–C angles are close to the tetrahedral 109°, so there is no angle strain, and all neighboring C–H bonds are staggered, so there is no torsional strain.
• Which position does a substituent prefer on a cyclohexane chair, and why? The equatorial position. An axial group is close to the two axial hydrogens on the same side of the ring (a 1,3-diaxial interaction), and this steric strain grows with the size of the group.

Chapter 5: Stereochemistry at Tetrahedral Centers

Difficulty: Hard · Key topics: chirality center, enantiomers, optical activity, R/S configuration, diastereomers, meso compounds, racemate, prochirality

This chapter explains molecular handedness, which matters greatly in biological chemistry. It begins with enantiomers and the tetrahedral carbon, the reason for chirality, optical activity and specific rotation, and Pasteur’s discovery of enantiomers. The Cahn–Ingold–Prelog sequence rules assign R and S configurations. Later sections cover diastereomers, meso compounds, racemic mixtures and their resolution, a review of all types of isomerism, chirality at nitrogen, phosphorus and sulfur, prochirality with Re and Si faces, and chirality in nature and chiral environments.

Key Points:

• A chiral molecule cannot be superimposed on its mirror image; the usual cause is a carbon bonded to four different groups (a chirality center)
• Enantiomers have identical physical properties except that they rotate plane-polarized light in opposite directions
• For R/S, rank the groups by the CIP rules and point the lowest-ranked group away; clockwise 1→2→3 is R and counterclockwise is S
• A compound with n chirality centers has at most 2ⁿ stereoisomers
• Diastereomers are stereoisomers that are not mirror images; epimers differ at only one chirality center
• A meso compound has chirality centers but a plane of symmetry, so it is achiral
• A racemate is a 50:50 mixture of enantiomers; racemates and single diastereomers differ in melting point and solubility

Memory Tip: Hold the lowest-priority group pointing away from you, like a steering column pointing away from the driver. Clockwise through 1, 2, 3 is R (rectus, right) and counterclockwise is S (sinister, left).

Common Mistake: Reading R or S straight from the drawing when the lowest-priority group points toward you. In that case the apparent direction is reversed, so flip your answer.

Important Questions:

• What is the difference between enantiomers and diastereomers? Enantiomers are non-superimposable mirror images and have opposite configuration at every chirality center. Diastereomers are stereoisomers that are not mirror images; they differ at some, but not all, chirality centers.
• What is the maximum number of stereoisomers for a compound with three chirality centers, and why can there be fewer? At most 2³ = 8. There can be fewer when the molecule has internal symmetry, because a meso compound is identical to its own mirror image.

Chapter 6: An Overview of Organic Reactions

Difficulty: Medium · Key topics: addition, elimination, substitution, rearrangement, polar reactions, nucleophile and electrophile, curved arrows, radical reactions, energy diagram, transition state

The chapter asks how and why organic reactions happen. It sorts reactions into four kinds: addition, elimination, substitution and rearrangement. It then contrasts polar and radical mechanisms; in this edition polar reactions come first, and the addition of HBr to ethylene is used to show nucleophiles, electrophiles and curved arrows before radical reactions are introduced. Energy ideas follow: equilibrium and free-energy change, bond dissociation energies, energy diagrams, transition states and intermediates. The last section compares biological reactions, which use enzymes, with laboratory reactions.

Key Points:

• The four reaction types are addition, elimination, substitution and rearrangement
• In a polar reaction a nucleophile (electron-rich) donates an electron pair to an electrophile (electron-poor); the curved arrow runs from nucleophile to electrophile
• Radical reactions involve species with an odd number of electrons, and each partner donates one electron to the new bond
• ΔG = ΔH − TΔS; a negative ΔG is exergonic and gives a favorable equilibrium
• The transition state is the energy maximum, and its height above the reactants is the activation energy ΔG^‡; a higher barrier means a slower reaction
• A reaction intermediate sits in an energy valley between two steps and exists briefly during the reaction

Memory Tip: Curved arrows follow electrons, not atoms: an arrow starts at an electron source (a bond or a lone pair) and points to where the electrons end up.

Common Mistake: Confusing a transition state with an intermediate. An intermediate is a real species in an energy valley; a transition state is the unstable peak between valleys and cannot be isolated.

Important Questions:

• What do ΔG and ΔG^‡ each tell you about a reaction? ΔG tells you how far the reaction goes (the position of equilibrium); ΔG^‡, the activation energy, tells you how fast it goes.
• What are the four general types of organic reactions? Addition (two reactants give one product), elimination (one reactant splits into two products), substitution (two reactants exchange parts) and rearrangement (one reactant reorganizes into an isomer).

Chapter 7: Alkenes: Structure and Reactivity

Difficulty: Medium · Key topics: degree of unsaturation, E,Z designation, alkene stability, electrophilic addition, Markovnikov’s rule, carbocation stability, Hammond postulate, carbocation rearrangement

The chapter opens with the industrial importance of alkenes and the degree of unsaturation, a quick way to find rings and double bonds from a molecular formula. It then covers naming alkenes, cis–trans isomerism, the E,Z designation from the CIP rules, and alkene stability. The rest of the chapter is about electrophilic addition: the two-step mechanism of HX addition through a carbocation, Markovnikov’s rule, carbocation structure and hyperconjugation, the Hammond postulate, and carbocation rearrangements by hydride and alkyl shifts, which are evidence for the mechanism.

Key Points:

• Degree of unsaturation is the number of rings plus π bonds; alkenes have fewer hydrogens than alkanes with the same carbon count
• Z (zusammen) means the higher-priority groups are on the same side of the double bond, and E (entgegen) means opposite sides
• Markovnikov’s rule: in HX addition, H goes to the double-bond carbon with fewer alkyl groups and X goes to the carbon with more
• Carbocation stability runs 3° > 2° > 1° > methyl
• Hammond postulate: the transition state of an endergonic step resembles its product, so a more stable carbocation means a lower-energy transition state
• A carbocation can rearrange by a hydride or alkyl shift to a more stable carbocation

Memory Tip: Markovnikov: the rich get richer. The hydrogen adds to the carbon that already has more hydrogens, so the positive charge lands on the more substituted carbon.

Common Mistake: Forgetting to check for rearrangement. If a carbocation can become more stable by a hydride or alkyl shift, the product you predicted may not be the one actually formed.

Important Questions:

• How do you find the degree of unsaturation of C₆H₁₀? The saturated alkane C₆H₁₄ has four more hydrogens, so there are 2 degrees of unsaturation, for example two double bonds, a ring plus a double bond, or one triple bond.
• State Markovnikov’s rule and explain it. In HX addition to an unsymmetrical alkene, H adds to the carbon with fewer alkyl groups and X to the carbon with more. The reason is that this route passes through the more stable carbocation intermediate.

Chapter 8: Alkenes: Reactions and Synthesis

Difficulty: Hard · Key topics: halogenation, halohydrin, oxymercuration, hydroboration, hydrogenation, epoxidation, ozonolysis, carbene addition, chain-growth polymers

This chapter begins the systematic study of functional groups. Alkenes are prepared by elimination, and most of the chapter covers the reverse: addition. It treats halogenation through a bromonium ion, halohydrin formation, two hydration routes (oxymercuration and hydroboration), catalytic hydrogenation, epoxidation and hydroxylation, and ozonolysis cleavage to carbonyl compounds. It also covers carbene addition to form cyclopropanes, radical addition and chain-growth polymers, biological radical additions, and the stereochemistry of adding water to achiral and chiral alkenes.

Key Points:

• Br₂ and Cl₂ add through a cyclic bromonium or chloronium ion to give anti addition; with water present the product is a halohydrin
• Oxymercuration–demercuration gives Markovnikov hydration; hydroboration–oxidation gives non-Markovnikov, syn hydration
• Catalytic hydrogenation (H₂ with Pt or Pd) reduces an alkene to an alkane
• A peroxyacid makes an epoxide, acid hydrolysis of the epoxide gives a trans-1,2-diol, and OsO₄ gives a cis-1,2-diol
• Ozone followed by zinc cleaves a C=C bond into two carbonyl compounds
• CH₂I₂ with Zn(Cu) converts an alkene into a cyclopropane (Simmons–Smith reaction)
• Chain-reaction polymerization of simple alkenes gives polyethylene, polypropylene and polystyrene

Memory Tip: Choose the hydration route by the alcohol you want: oxymercuration–demercuration gives the Markovnikov alcohol, and hydroboration–oxidation gives the opposite regiochemistry.

Common Mistake: Mixing up syn and anti. Halogen addition is anti (opposite faces), while hydroboration and OsO₄ hydroxylation are syn (same face).

Important Questions:

• Why does bromine add to an alkene with anti stereochemistry? The alkene attacks Br₂ to form a cyclic bromonium ion. Bromide ion can then attack only from the side opposite the bromonium ring, so the two bromines end up on opposite faces.
• Which two methods hydrate an alkene, and how do their products differ? Oxymercuration–demercuration adds H and OH with Markovnikov regiochemistry. Hydroboration–oxidation adds them with non-Markovnikov regiochemistry and syn stereochemistry.

Chapter 9: Alkynes: An Introduction to Organic Synthesis

Difficulty: Medium · Key topics: alkyne hydration, enol and tautomer, Lindlar catalyst, acetylide anion, alkylation, retrosynthesis

Alkynes are less common than alkenes, so this chapter uses them as a vehicle for the strategy of organic synthesis. It covers naming alkynes, preparation by double elimination from dihalides, and addition of HX and X₂, which follows alkene patterns. Hydration goes through an enol that tautomerizes to a ketone, hydroboration gives aldehydes from terminal alkynes, and reduction can stop at a cis alkene (Lindlar catalyst) or a trans alkene (lithium in ammonia). Oxidative cleavage, the acidity of terminal alkynes, alkylation of acetylide anions and an introduction to planning a synthesis finish the chapter.

Key Points:

• Alkyne carbons are sp hybridized, and the triple bond is one sp–sp σ bond plus two p–p π bonds
• Terminal alkynes are weakly acidic; a strong base such as NaNH₂ removes the hydrogen to form an acetylide anion
• An acetylide anion alkylates a primary alkyl halide by an S_N2 reaction, building a longer carbon chain
• Hg²⁺/H₂SO₄ hydration of a terminal alkyne gives a methyl ketone through an enol; hydroboration–oxidation gives an aldehyde
• H₂ over Lindlar catalyst gives a cis alkene, Li in NH₃ gives a trans alkene, and H₂ over Pd gives the alkane
• Retrosynthetic analysis plans a synthesis by working backward from the target molecule

Memory Tip: Hydrogens delivered together on a catalyst surface end up on the same side (cis, Lindlar). Electrons delivered one at a time by a dissolving metal let the groups settle on opposite sides (trans, Li/NH₃).

Common Mistake: Trying to alkylate an acetylide with a secondary or tertiary halide. The strongly basic acetylide eliminates instead of substituting, so use primary alkyl halides.

Important Questions:

• How do you convert an alkyne into a cis alkene and into a trans alkene? Partial hydrogenation over a Lindlar catalyst gives the cis alkene. Reduction with lithium in liquid ammonia gives the trans alkene.
• Why are terminal alkynes more acidic than alkenes or alkanes? The acetylide anion’s negative charge sits in an sp orbital with 50% s character, which holds it closer to the nucleus and stabilizes it more than the sp² or sp³ orbitals of vinylic and alkyl anions.

Chapter 10: Organohalides

Difficulty: Medium · Key topics: alkyl halides, radical halogenation, allylic bromination (NBS), allyl radical, Grignard reagent, Gilman reagent, coupling reactions, oxidation and reduction

This chapter covers how to name and make alkyl halides and their first reactions. It begins with names and structures, then preparation: radical halogenation of alkanes, allylic bromination of alkenes with NBS, and conversion of alcohols with HX, SOCl₂ or PBr₃. The resonance stabilization of the allyl radical explains why allylic bromination works. The chapter then turns alkyl halides into organometallic reagents, Grignard and Gilman reagents, and shows organometallic coupling reactions, including the Suzuki–Miyaura reaction used in drug synthesis. It ends with oxidation and reduction as they apply to organic molecules.

Key Points:

• Radical halogenation of alkanes gives mixtures, and reactivity follows radical stability: 3° > 2° > 1°
• NBS brominates the allylic position through a resonance-stabilized allylic radical
• Tertiary alcohols react well with HX; primary and secondary alcohols are converted to halides with SOCl₂ or PBr₃
• A Grignard reagent RMgX forms from an alkyl halide and Mg in ether and is both nucleophilic and strongly basic
• Gilman reagents (LiR₂Cu) couple with organohalides to form new C–C bonds
• In organic chemistry, oxidation usually means more bonds from carbon to O, N or halogen and fewer to H; reduction is the reverse

Memory Tip: A Grignard reagent behaves like a carbanion (R⁻): it grabs a proton from water or an alcohol instantly, so everything it touches must be dry.

Common Mistake: Making a Grignard reagent from a halide that also carries an OH, NH or C=O group, or in wet ether. Any acidic hydrogen or reactive carbonyl destroys the reagent.

Important Questions:

• Why does NBS brominate at the allylic position? A bromine radical removes an allylic hydrogen to give an allylic radical that is stabilized by resonance. NBS keeps the Br₂ concentration low, so addition to the double bond does not compete.
• How is a Grignard reagent made, and why is it both a nucleophile and a base? It forms when an alkyl halide reacts with magnesium metal in ether. The carbon bonded to magnesium carries a partial negative charge, so it attacks electrophiles and also removes protons from acids.

Chapter 11: Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations

Difficulty: Hard · Key topics: S_N2, S_N1, leaving group, Zaitsev’s rule, E2, E1, E1cB, anti periplanar, kinetics

Substitution and elimination of alkyl halides are treated side by side. The chapter starts with the discovery of nucleophilic substitution, then covers the S_N2 reaction (one step, inversion of configuration, second-order kinetics) and its characteristics, followed by the S_N1 reaction (a carbocation intermediate, first-order kinetics, racemization). After a section on biological substitutions it turns to elimination: Zaitsev’s rule, the E2 reaction with its deuterium isotope effect and anti periplanar geometry, the E1 and E1cB reactions, and biological eliminations. A final section compares all five pathways.

Key Points:

• S_N2 is a one-step, second-order reaction with backside attack and inversion of configuration, favored for primary and secondary substrates
• S_N1 starts with slow ionization to a carbocation, is first-order, gives racemization, and is favored for tertiary substrates
• E2 is a concerted, second-order elimination that needs a strong base and an anti periplanar H and leaving group; it shows a deuterium isotope effect
• Zaitsev’s rule: elimination usually gives the more highly substituted alkene as the major product
• E1 breaks the C–X bond first and shows no isotope effect; E1cB removes H first to give a carbanion and suits a leaving group two carbons from a carbonyl
• In biological substitutions and eliminations, the leaving group is typically a diphosphate ion rather than a halide

Memory Tip: S_N2 is one step with two species in the rate-determining step, and the carbon inverts like an umbrella turning inside out. S_N1 ionizes first, and the flat cation can be attacked from either face, so the product is racemized.

Common Mistake: Using one set of conditions for every substrate. A strong, bulky base with a tertiary halide gives elimination, not substitution; always judge the substrate, the nucleophile or base, and the solvent together.

Important Questions:

• Compare S_N1 and S_N2 reactions. S_N2 is second-order, one step, gives inversion and prefers primary or secondary substrates. S_N1 is first-order, goes through a carbocation, gives racemization and prefers tertiary substrates.
• What is Zaitsev’s rule? In base-induced elimination from an alkyl halide, the more highly substituted (more stable) alkene is usually the major product. An E2 reaction also needs the leaving group and a hydrogen to be anti periplanar.

Chapter 12: Structure Determination: Mass Spectrometry and Infrared Spectroscopy

Difficulty: Medium · Key topics: mass spectrometry, molecular ion, m/z, fragmentation, electromagnetic spectrum, infrared spectroscopy, wavenumber, functional-group absorptions

This chapter introduces structure determination, where spectroscopy tells you what a new molecule is. Mass spectrometry comes first: magnetic-sector instruments ionize molecules with an electron beam and sort the fragments by mass-to-charge ratio, giving the molecular ion and fragmentation patterns for common functional groups, while time-of-flight instruments with electrospray or MALDI ionization handle large biomolecules. The chapter then reviews the electromagnetic spectrum and infrared spectroscopy, explaining how bond stretching and bending absorb IR light, how to interpret a spectrum, and which absorptions identify the common functional groups.

Key Points:

• Mass spectrometry gives the molecular weight and formula; infrared spectroscopy identifies the functional groups present
• Ionized molecules fragment, and the pieces are sorted by mass-to-charge ratio (m/z); the molecular ion M⁺ gives the molecular weight
• Biological mass spectrometry protonates molecules by electrospray ionization (ESI) or MALDI and separates them by time of flight (TOF)
• Higher frequency means shorter wavelength and higher photon energy; IR absorptions are quoted as wavenumbers in cm⁻¹
• IR energy is absorbed when it matches a bond-stretching or bending vibration, so each functional group has a characteristic set of absorptions
• A terminal alkyne ≡C–H absorbs near 3300 cm⁻¹ and an alkene C=C absorbs at 1640–1680 cm⁻¹

Memory Tip: Read an IR spectrum from the left: first look at the region above about 1500 cm⁻¹ for the functional-group peaks (O–H, N–H, C–H, C≡C, C=O), and treat the crowded fingerprint region below it as a match-up tool, not a first-pass guide.

Common Mistake: Taking the tallest peak in a mass spectrum as the molecular ion. The tallest peak is the base peak; the molecular ion is the peak at the highest m/z (apart from isotope peaks) and may be small.

Important Questions:

• What information does mass spectrometry give compared with IR spectroscopy? Mass spectrometry gives the molecular weight and, from fragmentation, structural clues. IR spectroscopy shows which functional groups are present, because each absorbs at characteristic frequencies.
• Where does a terminal alkyne show up in an IR spectrum? The ≡C–H stretch absorbs near 3300 cm⁻¹, a region where few other common functional groups absorb sharply.

Chapter 13: Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Difficulty: Hard · Key topics: NMR absorption, chemical shift, TMS, integration, spin–spin splitting, n+1 rule, ¹³C NMR, DEPT

NMR is presented as the most valuable method of structure determination. The chapter explains how nuclei such as ¹H and ¹³C absorb radiofrequency energy in a strong magnetic field, and how the chemical shift, measured in δ units from the TMS reference, reveals each nucleus’s electronic environment. For ¹H NMR it covers shift ranges, integration to count hydrogens, spin–spin splitting and the n + 1 rule, proton equivalence and more complex splitting patterns. It then turns to ¹³C NMR with signal averaging, FT–NMR, broadband decoupling and DEPT, and ends with the uses of both techniques.

Key Points:

• Magnetic nuclei such as ¹H and ¹³C flip their spins when irradiated with radiofrequency waves in a strong magnetic field
• Chemical shift is reported in δ (ppm); tetramethylsilane (TMS) is the 0 δ reference and appears on the upfield (right) side
• Electrons shield a nucleus from the applied field, so shielded nuclei absorb upfield and deshielded nuclei absorb downfield
• Integration of a ¹H peak gives the relative number of hydrogens that produce it
• n equivalent neighboring hydrogens split a signal into n + 1 peaks, separated by the coupling constant J
• In broadband-decoupled ¹³C NMR each different carbon gives one line, and DEPT shows how many hydrogens are on each carbon

Memory Tip: Ask three questions of every ¹H spectrum: how many signals (kinds of H), how big is each (integration, how many H), and how is each split (the neighbors).

Common Mistake: Expecting equivalent hydrogens to split each other. Chemically equivalent hydrogens do not split one another; only nonequivalent hydrogens on adjacent carbons do.

Important Questions:

• Why is TMS used as the NMR reference, and what shift does it have? TMS absorbs at an unusually high applied field (upfield) in both ¹H and ¹³C spectra, away from most other signals, so it is assigned 0 δ and everything else is measured from it.
• A ¹H signal is a quartet. How many neighboring hydrogens does it have? Three, because n + 1 = 4. A quartet next to a triplet is the classic pattern of an ethyl group, CH₃CH₂–.

Chapter 14: Conjugated Compounds and Ultraviolet Spectroscopy

Difficulty: Medium · Key topics: conjugated diene, allylic carbocation, 1,2- and 1,4-addition, kinetic and thermodynamic control, Diels–Alder reaction, UV spectroscopy, λ_max

The chapter looks at molecules with alternating double and single bonds. It uses molecular orbital theory to explain why conjugated dienes are more stable, then covers electrophilic addition to conjugated dienes through allylic carbocations, which gives both 1,2 and 1,4 products, and the difference between kinetic and thermodynamic control. The Diels–Alder cycloaddition, its characteristics, and diene polymers such as natural and synthetic rubbers follow. The last three sections cover ultraviolet spectroscopy, how conjugation lowers the energy gap and lengthens the wavelength absorbed, and the link between conjugation, color and the chemistry of vision.

Key Points:

• In a conjugated diene four p orbitals combine into four π molecular orbitals; only the two bonding orbitals are filled, and this stabilizes the molecule
• HX adds to a conjugated diene to give both 1,2 and 1,4 products through the same resonance-stabilized allylic carbocation
• The 1,2 adduct is usually formed faster (kinetic control) and the 1,4 adduct is usually more stable (thermodynamic control)
• In the Diels–Alder reaction a diene and an electron-poor alkene (the dienophile) form a cyclohexene in one step, stereospecifically
• The diene must be able to adopt the s-cis conformation for the Diels–Alder reaction to occur
• In UV spectroscopy a π electron moves from the HOMO to the LUMO; 1,3-butadiene absorbs at λ_max = 217 nm, and more conjugation moves the absorption to longer wavelength

Memory Tip: Low temperature favors the fast (kinetic) product and higher temperature lets the mixture settle on the more stable (thermodynamic) one; the 1,2 and 1,4 adducts are the classic pair.

Common Mistake: Forgetting the s-cis requirement. A diene locked in the s-trans shape cannot reach the cyclic transition state, so it does not undergo the Diels–Alder reaction.

Important Questions:

• Why does a conjugated diene give both 1,2 and 1,4 addition products? Protonation gives an allylic carbocation with its positive charge shared between two carbons by resonance. The nucleophile can attack either carbon, giving the 1,2 or the 1,4 product.
• What are the requirements for a Diels–Alder reaction? A conjugated diene that can adopt an s-cis conformation and an alkene partner (the dienophile), which reacts best when it carries electron-withdrawing groups. The product is a cyclohexene formed in a single, stereospecific step.

Chapter 15: Benzene and Aromaticity

Difficulty: Medium · Key topics: aromatic naming, ortho/meta/para, phenyl and benzyl, resonance and MO picture of benzene, Hückel 4n+2 rule, aromatic ions, heterocycles, polycyclic aromatics

This chapter explains why benzene behaves differently from simple alkenes. It begins with naming aromatic compounds, including ortho, meta and para, phenyl and benzyl. It then covers the structure and stability of benzene in both resonance and molecular orbital terms, and the Hückel 4n + 2 rule that defines aromaticity, with antiaromatic compounds as the counterpart. Aromatic ions such as the cyclopentadienyl anion, heterocycles such as pyridine and pyrrole, polycyclic aromatics such as naphthalene, and the NMR and IR signatures of aromatic rings complete the chapter.

Key Points:

• An aromatic compound is cyclic, planar and fully conjugated, with 4n + 2 π electrons delocalized over the ring
• Planar, cyclic, conjugated molecules with other electron counts (4n) are antiaromatic and unusually unstable
• Benzene’s heat of hydrogenation is about 150 kJ/mol less than expected for a cyclic triene, which measures its extra stability
• Aromatic compounds react with electrophiles by substitution, which keeps the ring conjugation, rather than by addition
• The cyclopentadienyl anion and the cycloheptatrienyl cation are aromatic ions
• Pyridine and pyrimidine are six-membered aromatic heterocycles; pyrrole and imidazole are five-membered ones

Memory Tip: Hückel counts: 2, 6, 10 and 14 π electrons are aromatic (n = 0, 1, 2, 3), while 4, 8 and 12 are antiaromatic if the ring is planar.

Common Mistake: Miscounting π electrons in heterocycles. Pyrrole’s nitrogen lone pair is part of the aromatic sextet, so pyrrole is not basic; pyridine’s lone pair lies in the ring plane and is available.

Important Questions:

• State Hückel’s rule and apply it to benzene and cyclobutadiene. A planar, cyclic, conjugated ring is aromatic if it has 4n + 2 π electrons. Benzene has 6 (n = 1) and is aromatic; cyclobutadiene has 4 and is antiaromatic.
• Why does benzene undergo substitution rather than addition? Substitution keeps the ring’s cyclic conjugation and its aromatic stabilization. Addition would destroy the aromatic system, which costs about 150 kJ/mol.

Chapter 16: Chemistry of Benzene: Electrophilic Aromatic Substitution

Difficulty: Hard · Key topics: electrophilic aromatic substitution, Friedel–Crafts reaction, activating and deactivating groups, directing effects, nucleophilic aromatic substitution, benzyne, synthesis of polysubstituted benzenes

Having covered aromaticity, the chapter turns to reactions. Electrophilic aromatic substitution is introduced through bromination, then extended to other substitutions such as nitration and sulfonation and to Friedel–Crafts alkylation and acylation. Substituent effects are explained through resonance and inductive effects, sorting groups into activators and deactivators and into ortho/para and meta directors, with additivity in trisubstituted benzenes. The chapter also covers nucleophilic aromatic substitution, benzyne, oxidation of side chains, reduction of aromatic rings, and planning the synthesis of polysubstituted benzenes.

Key Points:

• Electrophilic aromatic substitution has two steps: E⁺ attacks the ring to form a resonance-stabilized carbocation, then loss of H⁺ restores aromaticity
• Friedel–Crafts alkylation and acylation fail on rings less reactive than a halobenzene, and alkylation can polyalkylate or rearrange the carbocation
• Substituents are ortho/para-directing activators, ortho/para-directing deactivators (the halogens) or meta-directing deactivators
• Resonance effects act through π bonds and inductive effects act through σ bonds
• Nucleophilic aromatic substitution needs a strong electron-withdrawing group ortho or para to the halogen, or it proceeds through a benzyne
• NBS brominates the benzylic position, and aqueous KMnO₄ oxidizes an alkyl side chain to a carboxyl group

Memory Tip: Ask two questions in order: does the group activate or deactivate, and where does it direct? Almost all activators (OH, NH₂, alkyl) direct ortho/para, deactivators direct meta, and the halogens are the exception: they deactivate but direct ortho/para.

Common Mistake: Trying a Friedel–Crafts reaction on a ring that already carries a strong deactivating group such as NO₂. The ring is too unreactive and the reaction fails.

Important Questions:

• Why do halogens deactivate the ring but direct ortho/para? Their electronegativity withdraws electrons through the σ bond (inductive effect), which deactivates the ring. Their lone pairs donate through resonance, which stabilizes the intermediate for ortho and para attack.
• What are the limitations of Friedel–Crafts alkylation? The ring must be at least as reactive as a halobenzene, the product is more reactive than the starting material so polyalkylation occurs, and the carbocation can rearrange before it reacts.

Chapter 17: Alcohols and Phenols

Difficulty: Medium · Key topics: alcohol preparation, reduction with NaBH₄ and LiAlH₄, Grignard addition, oxidation of alcohols, protecting groups, phenols, quinones

Chapter 17 begins the oxygen-containing functional groups. After naming and physical properties, including hydrogen bonding and acidity, it reviews how alcohols are prepared and concentrates on carbonyl routes: reduction with NaBH₄ and LiAlH₄ and addition of Grignard reagents, including a newly added reaction of Grignard reagents with carboxylic acids to give ketones. Reactions of alcohols follow, covering conversion to halides, dehydration and oxidation, then protection as silyl ethers. The chapter ends with phenols, their uses, their greater acidity, oxidation to quinones, and the spectroscopy of alcohols and phenols.

Key Points:

• Alcohols are weak acids (pK_a about 16–18); phenols are stronger (pK_a about 10) because the phenoxide ion is resonance stabilized
• LiAlH₄ reduces aldehydes, esters and carboxylic acids to primary alcohols and ketones to secondary alcohols
• A Grignard reagent gives a primary alcohol with formaldehyde, a secondary alcohol with an aldehyde, and a tertiary alcohol with a ketone or an ester
• Primary alcohols oxidize to aldehydes or carboxylic acids, secondary alcohols to ketones, and tertiary alcohols are not normally oxidized
• Alcohols convert to alkyl halides with PBr₃ or SOCl₂ and can be dehydrated to alkenes with POCl₃
• A trialkylsilyl ether protects an OH group during a reaction that the free alcohol would interfere with
• Phenols oxidize to quinones, and quinones reduce back to hydroquinones

Memory Tip: For a Grignard addition, count the R groups already on the carbonyl carbon: H₂C=O gives a primary alcohol, RCHO a secondary and R₂C=O a tertiary; an ester adds two R groups and also gives a tertiary alcohol.

Common Mistake: Expecting a Grignard reagent to survive in a molecule that also contains an OH group. The acidic hydrogen destroys it, so protect the alcohol first.

Important Questions:

• Why are phenols much more acidic than alcohols? The phenoxide ion’s negative charge is delocalized into the aromatic ring by resonance, which stabilizes it. An alkoxide ion has no such delocalization, so alcohols have pK_a values near 16–18 against about 10 for phenols.
• What products come from oxidizing primary, secondary and tertiary alcohols? Primary alcohols give aldehydes or, with stronger oxidants, carboxylic acids. Secondary alcohols give ketones. Tertiary alcohols are not normally oxidized because they have no hydrogen on the carbon bearing OH.

Chapter 18: Ethers and Epoxides; Thiols and Sulfides

Difficulty: Medium · Key topics: ether synthesis, Williamson synthesis, acidic ether cleavage, epoxide ring-opening, crown ethers, thiols and disulfides, sulfides and sulfonium ions

This chapter finishes the functional groups that have only C–O and C–S single bonds. It covers the names and properties of ethers, their preparation by Williamson ether synthesis and alkoxymercuration, and their cleavage by strong acids such as HI and HBr. Epoxides, strained three-membered cyclic ethers, get two sections on how they form and how acid or base opens the ring, with different regiochemistry in each case. Crown ethers and ionophores, thiols and sulfides, with oxidation to disulfides, sulfoxides and sulfones, and the spectroscopy of ethers complete the chapter.

Key Points:

• The Williamson ether synthesis is an S_N2 reaction of an alkoxide ion with a primary alkyl halide
• Ethers resist most reagents but are cleaved by HI or HBr: S_N2 at the less substituted carbon, or S_N1/E1 if one group is tertiary
• Epoxides are three-membered cyclic ethers whose ring strain lets both acids and bases open them
• Base-catalyzed ring-opening is an S_N2 attack at the less hindered epoxide carbon
• Thiols are made from an alkyl halide and thiourea; mild oxidation gives a disulfide (R–S–S–R) and mild reduction reverses it
• Sulfides are more nucleophilic than ethers, form sulfonium ions with alkyl halides, and oxidize to sulfoxides and sulfones

Memory Tip: Epoxide opening: base attacks the less hindered carbon (a plain S_N2), while acid lets the nucleophile attack the carbon that can better carry a partial positive charge.

Common Mistake: Assuming acid- and base-catalyzed epoxide opening give the same product. With a tertiary epoxide carbon, acid opens at the more substituted carbon and base opens at the less substituted one.

Important Questions:

• How does the Williamson ether synthesis work, and what limits it? An alkoxide ion displaces a halide from a primary alkyl halide by S_N2. Secondary and tertiary halides give mainly elimination, so the halide should be primary or methyl.
• How are thiols and disulfides related? Mild oxidation of two thiols (R–SH) joins them into a disulfide (R–S–S–R), and mild reduction of the disulfide returns the thiols. This is the same chemistry that links cysteine units in proteins.

Chapter 19: Aldehydes and Ketones: Nucleophilic Addition Reactions

Difficulty: Hard · Key topics: nucleophilic addition, hydrates, cyanohydrins, imines and enamines, Wolff–Kishner reaction, acetals, Wittig reaction, conjugate addition

Aldehydes and ketones are introduced through naming, preparation by oxidation of alcohols or partial reduction of esters, and oxidation. The core of the chapter is nucleophilic addition to the polar C=O group, followed by the major nucleophiles in turn: water (hydrates, with new coverage of alpha-keto acid hydrates), HCN (cyanohydrins), hydride and Grignard reagents (alcohols), amines (imines and enamines), hydrazine (Wolff–Kishner), alcohols (acetals) and phosphorus ylides (Wittig). It adds biological reductions, conjugate 1,4-addition to α,β-unsaturated carbonyls, and IR, NMR and mass-spectral identification.

Key Points:

• Nucleophilic addition is the characteristic reaction of C=O: the nucleophile attacks the carbonyl carbon
• NaBH₄ or LiAlH₄ reduces aldehydes and ketones to primary and secondary alcohols; Grignard reagents give secondary and tertiary alcohols
• A primary amine adds to give an imine (Schiff base) and a secondary amine gives an enamine
• Alcohols add under acid catalysis to give acetals, which serve as protecting groups for carbonyls
• The Wittig reaction converts a ketone or aldehyde into an alkene using a phosphorus ylide
• α,β-Unsaturated carbonyls can undergo conjugate (1,4) addition instead of direct 1,2-addition
• IR C=O absorbs at 1660–1770 cm⁻¹, ¹³C NMR shows it at 190–215 δ, and an aldehyde C–H appears near 10 δ in ¹H NMR

Memory Tip: The nucleophile attacks the carbonyl carbon, never the oxygen: carbon is the partially positive end of C=O, so electrons always flow toward it.

Common Mistake: Mixing up 1,2- and 1,4-addition. Hydride and Grignard reagents usually add directly to the C=O (1,2), while amines and diorganocopper reagents add to the β carbon (1,4).

Important Questions:

• What is the difference between the products of a primary and a secondary amine reacting with an aldehyde or ketone? A primary amine gives an imine (C=N, a Schiff base). A secondary amine gives an enamine (C=C–N) because there is no N–H left to form C=N.
• What is an acetal and why is it useful? An acetal forms when a carbonyl compound reacts with two equivalents of an alcohol under acid catalysis. It is stable to bases and nucleophiles, so it protects a carbonyl group during a reaction elsewhere, and acid removes it.

Chapter 20: Carboxylic Acids and Nitriles

Difficulty: Medium · Key topics: carboxylic acid naming, acidity, Henderson–Hasselbalch equation, substituent effects, carboxylation, nitriles, nitrile hydrolysis

Carboxylic acids and their nitrile relatives are treated as building blocks. The chapter covers naming (the -oic acid ending), structure, hydrogen bonding and acidity, the Henderson–Hasselbalch equation for the ionization of biological acids at physiological pH, and how substituents change acidity. Four preparation methods are given, followed by an overview of how carboxylic acids react, then the chemistry of nitriles: preparation, hydrolysis to acids, reduction to amines and Grignard addition to give ketones. IR and NMR then distinguish acids from nitriles.

Key Points:

• Carboxylic acids have pK_a values near 5, far more acidic than alcohols, because the carboxylate ion is resonance stabilized
• Electron-withdrawing substituents raise acidity (lower pK_a); electron-donating substituents lower it
• The Henderson–Hasselbalch equation gives the ionized fraction at a given pH; at the physiological pH of 7.3 acids exist as carboxylate anions
• Carboxylic acids are made by oxidizing alkylbenzenes, oxidizing primary alcohols or aldehydes, reacting Grignard reagents with CO₂, or hydrolyzing nitriles
• Nitriles form by S_N2 with cyanide or by dehydration of an amide, and hydrolyze to acids, reduce to primary amines, or react with Grignards to give ketones
• IR: acid O–H at 2500–3300 cm⁻¹, C=O at 1710–1760 cm⁻¹, nitrile at 2250 cm⁻¹; the acid hydrogen appears near 12 δ in ¹H NMR

Memory Tip: Cyanide adds one carbon: R–Br becomes R–CN and then R–COOH, a one-carbon chain extension in two steps.

Common Mistake: Trying to make a nitrile from a tertiary alkyl halide with cyanide. The basic cyanide ion eliminates instead of substituting; for tertiary halides use the Grignard and CO₂ route to the acid.

Important Questions:

• Why are carboxylic acids more acidic than alcohols? Losing H⁺ from a carboxylic acid gives a carboxylate ion whose charge is shared equally by two oxygens through resonance. An alkoxide has its charge on a single oxygen, so it is much less stable.
• In what form does acetic acid (pK_a 4.76) exist in a cell at pH 7.3? Almost entirely as acetate. The pH is more than two units above the pK_a, so the Henderson–Hasselbalch equation predicts that the carboxylate form dominates.

Chapter 21: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions

Difficulty: Hard · Key topics: acid halides, anhydrides, esters, amides, nucleophilic acyl substitution, tetrahedral intermediate, saponification, thioesters, step-growth polymers

This chapter covers the family of carboxylic acid derivatives and the one mechanism that governs them: nucleophilic acyl substitution, in which a nucleophile adds to C=O to form a tetrahedral intermediate and a leaving group then departs. After naming and the general mechanism, it covers reactions of carboxylic acids, including amide synthesis with EDC (new in this edition), then acid halides, anhydrides, esters, amides, and the biological thioesters and acyl phosphates. Step-growth polymers (nylons and polyesters) and the IR spectra of acid derivatives close the chapter.

Key Points:

• Nucleophilic acyl substitution is addition to C=O giving a tetrahedral intermediate, followed by loss of the leaving group
• Reactivity order: acid halide > acid anhydride > thioester > ester > amide
• Common reactions are hydrolysis (to acid), alcoholysis (to ester), aminolysis (to amide), hydride reduction and Grignard addition (both to alcohols)
• Fischer esterification forms an ester from a carboxylic acid and an alcohol with an acid catalyst
• Saponification is base-promoted ester hydrolysis, the reaction behind soap making
• Nylons (polyamides) form from a diacid and a diamine; polyesters from a diacid and a diol
• Thioesters and acyl phosphates are the acyl-transfer agents of biological chemistry

Memory Tip: Think of a reactivity ladder ranked by leaving-group quality. Chloride (acid chloride) is the best leaving group and NR₂⁻ (amide) the worst, so you can step down the ladder directly, from acid chloride to ester to amide, but not back up.

Common Mistake: Expecting Grignard reagents or LiAlH₄ to stop after one addition to an ester or acid chloride. Both add twice, giving a tertiary alcohol (Grignard) or a primary alcohol (hydride), not a ketone or aldehyde.

Important Questions:

• Rank the carboxylic acid derivatives by reactivity and explain the order. Acid halide > anhydride > thioester > ester > amide. Reactivity follows how good the leaving group is and how much the substituent donates electrons to the carbonyl; the poor leaving group and strong donation in amides make them least reactive.
• What is saponification? It is the hydrolysis of an ester with aqueous base to give an alcohol and a carboxylate salt. The name comes from soap making, where fats (triesters of glycerol) are hydrolyzed to fatty acid salts.

Chapter 22: Carbonyl Alpha-Substitution Reactions

Difficulty: Hard · Key topics: keto–enol tautomerism, alpha halogenation, Hell–Volhard–Zelinskii reaction, enolate ion, LDA, malonic ester synthesis, acetoacetic ester synthesis

This chapter covers the first of two reaction types that build on the acidity of the alpha carbon, substitution at the position next to a C=O. It is one of the four fundamental reaction types of carbonyl chemistry. It starts with keto–enol tautomerism and the reactivity of enols with electrophiles, giving alpha halogenation of aldehydes and ketones and the Hell–Volhard–Zelinskii bromination of carboxylic acids. It then turns to the acidity of alpha hydrogens and the formation of enolate ions with bases such as LDA, their reactivity, and the alkylation of enolates, including the malonic ester and acetoacetic ester syntheses.

Key Points:

• A carbonyl compound and its enol are tautomers in equilibrium; the enol is the minor form but has a nucleophilic double bond
• Alpha halogenation of ketones in acid goes through the enol; the HVZ reaction brominates a carboxylic acid with Br₂ and PBr₃
• Alpha hydrogens are weakly acidic, and a strong base such as LDA converts a carbonyl compound to its enolate ion
• Enolate ions react with primary alkyl halides by S_N2 alkylation
• The malonic ester synthesis converts RX into RCH₂CO₂H; the acetoacetic ester synthesis converts RX into RCH₂COCH₃
• Base-induced E2 elimination of an alpha-halo carbonyl gives an α,β-unsaturated carbonyl compound

Memory Tip: Count the carbons you add: the malonic ester synthesis adds CH₂CO₂H (two carbons, giving an acid) and the acetoacetic ester synthesis adds CH₂COCH₃ (three carbons, giving a methyl ketone).

Common Mistake: Mixing up tautomers and resonance forms. Keto and enol are different compounds with a hydrogen in different places; resonance forms differ only in the position of electrons.

Important Questions:

• What is keto–enol tautomerism and why does an enol react with electrophiles? A carbonyl compound with an alpha hydrogen exists in equilibrium with an enol isomer. The enol’s C=C bond is electron-rich, so it attacks electrophiles at the alpha carbon, which is why alpha halogenation occurs.
• How can you convert an alkyl halide into a carboxylic acid with two more carbons? Use the malonic ester synthesis: deprotonate diethyl malonate, alkylate the enolate with the alkyl halide, then hydrolyze and decarboxylate to give RCH₂CO₂H.

Chapter 23: Carbonyl Condensation Reactions

Difficulty: Hard · Key topics: aldol reaction, mixed aldol, intramolecular aldol, Claisen condensation, Dieckmann cyclization, Michael reaction, Stork enamine reaction, Robinson annulation

The last of the four carbonyl reaction types combines enolate chemistry with nucleophilic addition. The chapter begins with the aldol reaction, how condensation differs from alpha substitution, and dehydration of aldol products to enones, then covers mixed and intramolecular aldol reactions, which build five- and six-membered rings. The ester versions follow: the Claisen condensation, mixed Claisen reactions and the Dieckmann cyclization. The Michael reaction, the Stork enamine reaction and the Robinson annulation extend the toolkit, and the chapter ends with biological carbonyl condensations.

Key Points:

• In a carbonyl condensation one partner becomes an enolate (alpha substitution) and adds to the C=O of the other partner (nucleophilic addition)
• The aldol reaction joins two aldehyde or ketone molecules to give a β-hydroxy carbonyl, which can dehydrate to an α,β-unsaturated product
• A mixed aldol works when one partner is an unusually good donor or can only act as an acceptor, as formaldehyde and benzaldehyde do
• The Claisen condensation joins two esters to give a β-keto ester; the Dieckmann cyclization does this intramolecularly to make 5- and 6-membered rings
• The Michael reaction is conjugate addition of a stabilized donor, such as a β-keto ester, to an α,β-unsaturated acceptor
• The Robinson annulation is a Michael addition followed by an intramolecular aldol cyclization, forming a substituted cyclohexenone

Memory Tip: The word aldol is aldehyde plus alcohol: the first-step product contains both groups. Remember that the aldol makes β-hydroxy carbonyls and the Claisen makes β-keto esters.

Common Mistake: Running a mixed aldol between two partners that can both form enolates and expecting one product. You can get up to four, so use one partner that cannot enolize or a very acidic donor.

Important Questions:

• What is the difference between an aldol reaction and a Claisen condensation? The aldol joins two aldehydes or ketones and gives a β-hydroxy carbonyl compound (or its dehydrated enone). The Claisen joins two esters and gives a β-keto ester.
• How does the Robinson annulation build a ring? A Michael addition first joins a donor (a β-diketone or β-keto ester) to an α,β-unsaturated ketone. An intramolecular aldol cyclization then closes the ring, giving a substituted cyclohexenone.

Chapter 24: Amines and Heterocycles

Difficulty: Medium · Key topics: amine naming, basicity of amines and arylamines, reductive amination, Gabriel synthesis, Hofmann and Curtius rearrangements, diazotization, Sandmeyer reaction, heterocyclic amines

Amines are among the most abundant and chemically rich functional groups, found in proteins, nucleic acids, coenzymes and most drugs. This chapter covers naming, structure, and the basicity of alkylamines and arylamines, including ionization at physiological pH through the Henderson–Hasselbalch equation. Synthesis methods include S_N2 alkylation, the Gabriel synthesis, reductions, reductive amination, and the Hofmann and Curtius rearrangements. Reactions of amines (Hofmann elimination) and arylamines (diazotization and the Sandmeyer reaction) follow, then heterocyclic amines such as pyrrole, pyridine, imidazole, quinoline, indole and purine, and the spectroscopy of amines.

Key Points:

• Amines are both basic and nucleophilic because of the lone pair on nitrogen
• Arylamines are weaker bases than alkylamines because the lone pair is delocalized into the ring; electron-withdrawing ring substituents weaken them further
• At the physiological pH of 7.3 alkylamines exist almost entirely in their protonated form
• Reductive amination converts an aldehyde or ketone plus an amine into a more substituted amine with NaBH₄ or a similar reducing agent
• The Hofmann and Curtius rearrangements give an amine with one fewer carbon than the starting carboxylic acid derivative
• Diazotization of an arylamine with nitrous acid gives an arenediazonium salt, and the Sandmeyer reaction replaces the diazonio group with Cl, Br, I or CN
• Pyrrole is nonbasic because its nitrogen lone pair is part of the aromatic π system

Memory Tip: Basicity ladder: alkylamines are stronger bases than ammonia, ammonia is stronger than arylamines, and an amide nitrogen is essentially not basic because its lone pair is tied up with the carbonyl.

Common Mistake: Expecting an amine alkylation with an alkyl halide to stop after one step. The product is more nucleophilic than the starting amine, so you get a mixture, which is why the Gabriel synthesis and reductive amination are preferred.

Important Questions:

• Why is aniline a weaker base than methylamine? In aniline the nitrogen lone pair is delocalized into the aromatic ring, so it is less available to accept a proton. In methylamine the lone pair is localized on nitrogen and the alkyl group donates electrons.
• How does the Hofmann rearrangement change the carbon count? It converts a primary amide into a primary amine with one fewer carbon, because the carbonyl carbon is lost as the R group migrates to nitrogen. The Curtius rearrangement has the same net effect.

Chapter 25: Biomolecules: Carbohydrates

Difficulty: Hard · Key topics: aldoses and ketoses, Fischer projection, D and L sugars, furanose and pyranose, anomers, mutarotation, glycosides, disaccharides, polysaccharides

This chapter opens the biomolecule section with carbohydrates. Sugars are classified by carbon count and carbonyl type, drawn as Fischer projections, and assigned D or L configuration. The chapter covers the configurations of the aldoses, the cyclic hemiacetal forms (furanose and pyranose) with α and β anomers and mutarotation, and the reactions of monosaccharides: reduction, oxidation, glycoside formation, Kiliani–Fischer chain lengthening and Wohl degradation. It then describes the eight essential monosaccharides, disaccharides, polysaccharides such as cellulose, starch and glycogen, and other important carbohydrates.

Key Points:

• Carbohydrates are polyhydroxy aldehydes and ketones; glucose is an aldohexose
• A sugar is D or L according to the configuration of the chirality center farthest from the carbonyl, usually shown in a Fischer projection
• Monosaccharides exist mainly as cyclic hemiacetals: five-membered furanoses and six-membered pyranoses
• Cyclization creates a new chirality center, the anomeric carbon, giving α and β anomers that interconvert (mutarotation)
• NaBH₄ reduces a sugar to an alditol, aqueous Br₂ oxidizes it to an aldonic acid, and HNO₃ to an aldaric acid
• A glycoside forms when the anomeric OH is replaced by an OR group from an alcohol under acid catalysis
• Cellulose is a structural polysaccharide, while starch and glycogen store energy

Memory Tip: In a Fischer projection, horizontal bonds come toward you and vertical bonds go away. For a D sugar the OH on the lowest chirality center points to the right (D for dextro, right).

Common Mistake: Mixing up α and β anomers. In the Haworth drawing of a D-sugar, α has the anomeric OH pointing down (trans to the CH₂OH group) and β has it pointing up (cis).

Important Questions:

• What are anomers, and what is mutarotation? Anomers are two cyclic sugar forms that differ only in configuration at the new chirality center formed on ring closure (the anomeric carbon): α and β. Mutarotation is their interconversion in solution through the open-chain form.
• Why do cellulose and starch behave so differently when both are glucose polymers? Starch has α glycosidic links and cellulose has β links. The α links give a helical, digestible energy-storage polymer, while the β links give straight, rigid chains suited to structural fibers.

Chapter 26: Biomolecules: Amino Acids, Peptides, and Proteins

Difficulty: Medium · Key topics: amino acids, zwitterion, isoelectric point, peptide bond, Edman degradation, peptide synthesis, Merrifield method, protein structure, enzymes

The chapter follows amino acids from structure to protein function. It describes the structures of the 20 common amino acids, their zwitterionic forms and isoelectric points from the Henderson–Hasselbalch equation, and laboratory synthesis. It then treats peptides and proteins, amino acid analysis, Edman degradation sequencing, peptide synthesis with protecting groups (Boc, Fmoc, DCC or EDC) and the Merrifield solid-phase method, with details updated in this edition. Protein structure at four levels, then enzymes, coenzymes and a close look at how citrate synthase works, complete the chapter.

Key Points:

• Twenty α-amino acids occur in proteins; all except glycine have L-type stereochemistry, and in neutral solution they exist as zwitterions
• The isoelectric point is the pH at which an amino acid carries no net charge
• Peptides and proteins are chains of amino acid residues linked by amide (peptide) bonds
• Amino acid analysis tells which amino acids are present; Edman degradation with PITC removes the N-terminal residue one at a time for sequencing
• Peptide synthesis couples an N-protected amino acid (Boc or Fmoc) using DCC or EDC; the Merrifield method does this on insoluble polymer beads
• Proteins have four levels of structure: primary (sequence), secondary (α helix, β sheet), tertiary (overall fold) and quaternary (assembly of chains)
• Enzymes are protein catalysts in six classes, and many need cofactors such as metal ions or coenzymes

Memory Tip: The four levels of protein structure: primary is the sequence (the spelling), secondary is the local repeating shape (helix or sheet), tertiary is the full fold of one chain, and quaternary is several chains together.

Common Mistake: Treating amino acid analysis as a sequencing method. It shows which amino acids are present and in what proportion, but not their order; Edman degradation gives the order.

Important Questions:

• What is a zwitterion, and what is the isoelectric point? A zwitterion has a positive NH₃⁺ and a negative CO₂⁻ in the same molecule, the usual form of an amino acid in neutral solution. The isoelectric point is the pH at which its net charge is zero.
• How does the Merrifield solid-phase method work? The first N-protected amino acid is anchored to insoluble polymer beads. Each cycle removes the protecting group and couples the next protected residue, and excess reagents are simply washed away. The finished peptide is then cleaved from the beads.

Chapter 27: Biomolecules: Lipids

Difficulty: Medium · Key topics: fats and oils, triacylglycerols, saponification and soap, phospholipids, eicosanoids, terpenoids, steroids, steroid biosynthesis

Lipids are defined by how they are isolated, extraction with nonpolar solvents, rather than by one structure. The chapter covers waxes, fats and oils (triacylglycerols), soap and saponification, phospholipids in cell membranes, eicosanoids such as prostaglandins, terpenoids built from isopentenyl diphosphate, and steroids with their characteristic four-ring skeleton. The last section traces steroid biosynthesis: acetate feeds the mevalonate pathway to isopentenyl diphosphate, squalene cyclizes cationically, and lanosterol is formed. A closing feature looks at saturated fats, cholesterol and heart disease.

Key Points:

• Lipids are isolated from plants and animals by extraction with a nonpolar organic solvent and have very diverse structures
• Fats and oils are triacylglycerols, triesters of glycerol with long-chain fatty acids; animal fats are mostly saturated and vegetable oils mostly unsaturated
• Saponification (base hydrolysis of a fat) gives glycerol and soap, the salts of fatty acids
• Phospholipids, glycerophospholipids and sphingomyelins, are major components of cell membranes
• Eicosanoids, including prostaglandins, are made from arachidonic acid and are found in all body tissues
• Terpenoids are built from the five-carbon unit isopentenyl diphosphate (IPP); steroids share a tetracyclic skeleton and arise from lanosterol, itself made from squalene

Memory Tip: Count terpenoids in five-carbon units: 10 carbons is two units, 15 is three, 20 is four and 30 is six.

Common Mistake: Thinking soap works by chemically reacting with grease. It forms micelles instead: the hydrocarbon tails dissolve in the oil and the charged heads face the water.

Important Questions:

• Why are vegetable oils usually liquid while animal fats are solid? Vegetable oils contain mostly unsaturated fatty acid residues, whose cis double bonds put kinks in the chains and keep them from packing tightly. Animal fats are mostly saturated, so the chains pack closely and the fat is solid.
• What is the biosynthetic origin of steroids? Steroids come from lanosterol, which forms by cationic cyclization of the hydrocarbon squalene. Squalene itself is built from isopentenyl diphosphate, which the mevalonate pathway makes from three acetate units.

Chapter 28: Biomolecules: Nucleic Acids

Difficulty: Medium · Key topics: nucleotides, DNA double helix, base pairing, replication, transcription, translation, codons and tRNA, Sanger sequencing, PCR

DNA and RNA are the chemical carriers of genetic information, and this chapter describes them as organic molecules. It covers nucleotides and nucleic acids, the base pairing that holds the DNA double helix together, and the three processes that decipher genetic information: replication, transcription and translation, in which mRNA codons are read by tRNA to build proteins. The last three sections turn to laboratory methods: DNA sequencing by the Sanger dideoxy method, automated DNA synthesis, and the polymerase chain reaction, which can amplify small amounts of DNA by a factor of about 10⁶.

Key Points:

• A nucleoside is a purine or pyrimidine base bonded to a sugar; a nucleotide adds a phosphate; the sugar is ribose in RNA and 2-deoxyribose in DNA
• Nucleotides are linked by phosphate bridges from the 5′ phosphate of one to the 3′ hydroxyl of the next
• In DNA, adenine pairs with thymine and cytosine pairs with guanine by hydrogen bonds, and the two strands coil into a double helix
• Replication copies DNA, transcription makes mRNA from a DNA segment, and translation uses mRNA codons and tRNA anticodons to assemble a protein
• The Sanger dideoxy method sequences DNA, and automated instruments can synthesize short DNA segments
• PCR amplifies small amounts of DNA by a factor of about 10⁶

Memory Tip: Transcription keeps the same language (DNA text is copied into RNA text), while translation changes the language (nucleotide codons are read out as amino acids).

Common Mistake: Treating the two DNA strands as interchangeable. The mRNA sequence matches the sense (coding) strand, with U in place of T, and is complementary to the antisense (template) strand that is actually read.

Important Questions:

• Which bases pair in DNA, and what holds the pairs together? Adenine pairs with thymine and cytosine pairs with guanine. Hydrogen bonds between the heterocyclic bases on opposite strands hold the double helix together.
• What are the roles of mRNA and tRNA in translation? mRNA carries the genetic message copied from DNA, divided into three-base codons. Each tRNA carries one amino acid and recognizes a codon through its anticodon, delivering amino acids in the order the mRNA specifies.

Chapter 29: The Organic Chemistry of Metabolic Pathways

Difficulty: Hard · Key topics: catabolism and anabolism, β-oxidation, glycolysis, citric acid cycle, gluconeogenesis, transamination, ATP

Metabolism, the sum of all body reactions, is presented here as ordinary organic chemistry. The chapter opens with an overview of metabolism and biochemical energy, then follows fat catabolism (glycerol and β-oxidation), fatty acid biosynthesis, and carbohydrate catabolism by glycolysis. It covers the conversion of pyruvate to acetyl CoA, the citric acid cycle, gluconeogenesis, and protein catabolism by deamination. Each pathway is broken into steps that use reactions met earlier, such as hydrolysis, oxidation, aldol and Claisen condensations, and eliminations. A final section draws general conclusions about biological chemistry.

Key Points:

• Catabolism breaks large molecules into smaller ones; anabolism builds large molecules from small pieces; both follow the normal rules of organic reactivity
• β-Oxidation is a four-step pathway that removes two carbons at a time from a fatty acid as acetyl CoA
• Glycolysis is a ten-step pathway that converts glucose to pyruvate, which is then converted to acetyl CoA
• The eight-step citric acid cycle degrades acetyl CoA to CO₂, and oxaloacetate from the last step is the reactant of the first
• Most amino acids lose nitrogen by transamination, in which the NH₂ group trades places with the keto group of an α-keto acid such as α-ketoglutarate to give glutamate
• Energy from catabolism is used in the electron-transport chain to make ATP, which drives unfavorable reactions
• Anabolic pathways are not the reverse of catabolic ones: fatty acids are built from acetate in 8 steps and glucose is made from pyruvate by 11-step gluconeogenesis

Memory Tip: Pin each pathway to a number (glycolysis 10 steps, citric acid cycle 8, gluconeogenesis 11, β-oxidation a repeating 4-step cycle), then name the organic reaction type behind every step.

Common Mistake: Assuming a biosynthetic pathway is the reverse of the matching breakdown pathway. The chapter stresses that anabolism and catabolism use different routes; gluconeogenesis has 11 steps while glycolysis has 10.

Important Questions:

• Define catabolism and anabolism. Catabolism is the set of reactions that break large molecules into smaller fragments, releasing energy. Anabolism is the set that builds large molecules from small pieces, using energy.
• Why is the citric acid cycle called a cycle? It is a closed loop: oxaloacetate, the product of the final step, is a reactant in the first step, where it combines with acetyl CoA. Each turn degrades acetyl CoA to CO₂ and regenerates oxaloacetate.

Chapter 30: Orbitals and Organic Chemistry: Pericyclic Reactions

Difficulty: Hard · Key topics: pericyclic reactions, conjugated π systems, electrocyclic reactions, conrotatory and disrotatory, cycloaddition, suprafacial and antarafacial, sigmatropic rearrangement

This chapter uses orbital symmetry to explain reactions that occur in one step through a cyclic transition state, with no intermediates. It starts with the molecular orbitals of conjugated π systems, then covers electrocyclic reactions with conrotatory and disrotatory ring closure, the stereochemistry of thermal and photochemical electrocyclic reactions, cycloadditions with suprafacial and antarafacial pathways, and sigmatropic rearrangements such as the Claisen and Cope rearrangements, with expanded Cope and new oxy-Cope coverage in this edition. A final section summarizes the rules in one mnemonic.

Key Points:

• A pericyclic reaction is a single step through a cyclic transition state with no intermediates; the three classes are electrocyclic, cycloaddition and sigmatropic
• Electrocyclic ring closure is conrotatory (the lobes turn in the same direction) or disrotatory (opposite directions), depending on the symmetry of the HOMO
• A Diels–Alder reaction joins a diene (4 π electrons) and a dienophile (2 π electrons) to give a cyclohexene
• Cycloadditions and sigmatropic shifts can be suprafacial (same face) or antarafacial (opposite faces)
• A sigmatropic rearrangement moves a σ-bonded group across a π system; the Claisen and Cope rearrangements are examples
• Photochemical (excited-state) reactions follow exactly the opposite rules from thermal (ground-state) reactions

Memory Tip: Use the book’s mnemonic, The Electrons Circle Around: a Thermal reaction with an Even number of electron pairs is Conrotatory or Antarafacial, and photochemical reactions reverse it.

Common Mistake: Counting electrons instead of electron pairs when applying the rule. A thermal Diels–Alder reaction involves three pairs (an odd number), so it is suprafacial, not antarafacial.

Important Questions:

• What are the three classes of pericyclic reactions? Electrocyclic reactions (ring closure or opening of conjugated polyenes), cycloaddition reactions (two unsaturated molecules joining to form a ring) and sigmatropic rearrangements (a σ-bonded group migrating across a π system).
• What is the difference between conrotatory and disrotatory ring closure? In a conrotatory closure both terminal lobes rotate in the same direction; in a disrotatory closure they rotate in opposite directions. Orbital symmetry decides which path a reaction takes.

Chapter 31: Synthetic Polymers

Difficulty: Easy · Key topics: chain-growth polymers, Ziegler–Natta catalyst, tacticity, copolymers, step-growth polymers, olefin metathesis, thermoplastics and elastomers

The final chapter links organic reactions to materials. It covers chain-growth polymers made from vinyl monomers by radical, cationic or anionic initiation, then the stereochemistry of polymerization with Ziegler–Natta catalysts, which give isotactic or syndiotactic chains. Copolymers (graft and block), step-growth polymers such as polycarbonates and polyurethanes, olefin metathesis polymerization and, new in this edition, intramolecular olefin metathesis for ring closing follow. The chapter ends with how structure, crystallinity and cross-linking set the properties of thermoplastics, fibers, elastomers and thermosetting resins.

Key Points:

• The two main classes are chain-growth polymers (made from alkenes) and step-growth polymers (made from difunctional molecules)
• Monomers with electron-donating groups, such as 2-methylpropene, polymerize cationically; those with electron-withdrawing groups, such as methyl α-cyanoacrylate, polymerize anionically
• Ziegler–Natta catalysts give stereoregular chains: isotactic (substituents on one side) or syndiotactic (alternating sides), instead of atactic (random)
• Copolymers join two monomers; graft and block copolymers are examples whose properties differ from either homopolymer
• Polycarbonates form from a diester and a diol, and polyurethanes from a diisocyanate and a diol
• Polymers fall into four property classes: thermoplastics, fibers, elastomers and thermosetting resins

Memory Tip: Tacticity in one line: isotactic means all substituents on the same side, syndiotactic means they alternate, and atactic means there is no pattern.

Common Mistake: Assuming two polymers made from the same monomer behave alike. Isotactic polypropylene is strong and crystalline, while atactic polypropylene is soft and rubbery; branching, tacticity and crystallinity change the properties.

Important Questions:

• What is the difference between chain-growth and step-growth polymers? Chain-growth polymers form by a chain reaction that adds monomer units one at a time to an active end, usually from alkenes. Step-growth polymers form by independent bond-forming reactions between difunctional molecules, as in nylons, polyesters and polyurethanes.
• What do isotactic, syndiotactic and atactic mean, and how does a Ziegler–Natta catalyst help? They describe the arrangement of substituents along the chain: same side, alternating sides, or random. A Ziegler–Natta catalyst controls the polymerization so the chains have little branching and a regular (isotactic or syndiotactic) stereochemistry.

Appendices: Nomenclature, Acidity Constants, Glossary and Periodic Table

Appendix · Difficulty: Easy · Key topics: polyfunctional nomenclature, functional-group precedence, acidity constants, glossary, periodic table

Four appendices follow the 31 chapters. The first extends the IUPAC naming rules from single functional groups to polyfunctional compounds, in four steps: choose the suffix by functional-group precedence, choose the parent chain or ring, add the prefixes and locants, and write out the name. The second is a table of acidity constants (pK_a values) for about a hundred organic compounds. The third is a glossary of roughly 690 defined terms, each pointing back to the section that introduces it. The last is a periodic table.

Key Points:

• Naming a polyfunctional compound starts with the highest-precedence functional group, which sets the suffix
• The parent is the main chain or ring that carries the principal group, then prefixes and locants are added and the name is written out
• The acidity table lists pK_a values for roughly one hundred organic compounds, from methanesulfonic acid (−1.8) down to methane (about 60)
• The glossary defines about 690 terms and links each to the section where it is introduced
• The periodic table appendix is a quick reference for atomic numbers, symbols, names and atomic masses

Memory Tip: Keep the pK_a table open when you work the acid–base problems in Chapter 2: find the two acids, compare their pK_a values, and the equilibrium favors the weaker acid.

Important Questions:

• Where can I find a quick way to compare how acidic two organic compounds are? In the acidity constants appendix. It lists pK_a values for about a hundred compounds, and the lower the pK_a, the stronger the acid.
• How do I name a compound that has more than one functional group? Use the nomenclature appendix: the group with the highest precedence becomes the suffix, the main chain or ring carrying it is the parent, and the other groups are named as prefixes with locants.

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How to Study This Book

Organic Chemistry: A Tenth Edition has 31 chapters and four appendices, and the book does not group its chapters into parts. The Preface says it is meant for a two-semester introductory course and that it keeps the functional-group approach and the table of contents of earlier editions. Read in order, the chapters fall into natural groups. Chapters 1 to 6 are the foundations: bonding, acids and bases, alkanes, rings, stereochemistry and the language of mechanisms. Chapters 7 to 11 cover alkenes, alkynes, organohalides, and substitution and elimination. Chapters 12 and 13 are spectroscopy, Chapters 14 to 16 cover conjugation and aromatic chemistry, and Chapters 17 and 18 the oxygen functional groups. Chapters 19 to 23 are the carbonyl family, Chapter 24 is amines, Chapters 25 to 29 are biomolecules and metabolism, and Chapters 30 and 31 are pericyclic reactions and polymers. This grouping is only a study aid, not a division made by the author.

Chapters 1 and 2 mostly review general chemistry, so skim them if hybridization and pK_a are fresh, but do not skip resonance and the acid–base discussion in Chapter 2, because every later chapter uses them. Chapter 6 is the hinge of the book: it introduces nucleophiles, electrophiles and curved arrows, and in this edition it teaches polar reactions before radical reactions, so you learn arrow-pushing first. Chapter 5 on stereochemistry comes back in Chapters 8, 11, 13, 19, 25 and 26, so make R and S configuration and meso compounds solid early.

Chapters 7 to 11 form one chain. Chapter 7 gives electrophilic addition and carbocation stability, Chapter 8 applies it to the main alkene reactions, Chapter 9 uses alkynes to introduce planning a synthesis, and Chapter 10 prepares the alkyl halides and Grignard reagents that Chapter 11 then pushes through S_N2, S_N1, E2, E1 and E1cB. Chapter 11 is the one most courses test hardest. Its closing section summarizes when each pathway wins; turn it into a one-page table of substrate, nucleophile or base, solvent and result.

Spectroscopy (Chapters 12 and 13) sits after Chapter 11, but many courses teach it earlier so students can read spectra in the lab. Chapter 13 uses ideas from Chapter 5 (enantiotopic and diastereotopic hydrogens), so you can read Chapters 12 and 13 right after Chapter 5 if your course needs them early. Every later functional-group chapter ends with its own spectroscopy section (Chapters 15, 17, 18, 19, 20, 21 and 24), so you will return to the IR and NMR chapters often. Chapter 14 adds UV spectroscopy alongside conjugated dienes and the Diels–Alder reaction.

Chapters 19 to 23 are best studied as a set. Chapter 22 calls alpha substitution one of the four fundamental reaction types in carbonyl chemistry and Chapter 23 calls condensation the fourth and last, and the other two are nucleophilic addition (Chapter 19) and nucleophilic acyl substitution (Chapter 21). Chapter 20 on carboxylic acids and nitriles is the bridge to Chapter 21. Learn the alcohol, hydride and Grignard chemistry of Chapter 17 first, because the same reagents return in Chapters 19 and 21. A four-column table, one column per reaction type, is the best revision tool for these five chapters.

Chapters 24 to 29 turn to biological chemistry, and the book keeps stressing that biological reactions follow the same rules. Chapter 25 reuses stereochemistry from Chapter 5 and hemiacetal and acetal chemistry from Chapter 19, and Chapters 26 and 27 lean on the amide and ester chemistry of Chapter 21. The Henderson–Hasselbalch equation appears in Chapters 20, 24 and 26, so learn it once properly. Chapter 29 reuses reactions from nearly every earlier chapter, so it works as a final review. Courses differ in how much of Chapters 25 to 29 they cover, so check your syllabus before you plan your time.

Chapters 30 and 31 are the most likely to be left out of a two-semester course. Chapter 30 on pericyclic reactions extends the Diels–Alder reaction of Chapter 14 and the molecular orbital ideas of Chapters 1 and 14, and its closing mnemonic is worth learning if your syllabus includes it. Chapter 31 on polymers draws together alkene polymerization from Chapter 8, step-growth polymers from Chapter 21 and olefin metathesis, and it is largely descriptive, so you can read it on your own.

The Preface counts more than 1,800 in-text and end-of-chapter problems. Each chapter opens with a short “Why This Chapter?” section, includes worked examples with a strategy and a detailed solution, and ends with a summary, a key-terms list and problems; the Preface also names Visualizing Chemistry problems and mechanism problems at the end of each chapter. Sixteen chapters (8 to 11, 14 and 16 to 26) also end with a Summary of Reactions that lists the key reactions in one place, which is ideal for revision. Learn mechanisms by drawing them with curved arrows on paper. The book teaches no laboratory technique, so you will need a lab manual alongside it. The Study Guide and Student Solutions Manual by Susan McMurry are separate books, and the Preface says both are openly licensed and free for students in digital formats. The four appendices are a guide to naming polyfunctional compounds, a table of acidity constants, a glossary of about 690 terms and a periodic table.

Used In These Programs

This book is used for the Organic Chemistry course in BS Chemistry and in biology and pre-medical programs that require organic chemistry. Browse all Organic Chemistry books or the full Chemistry category.

Who Should Read This

This book is written for students taking a two-semester introductory organic chemistry course after general chemistry. It is the natural text for BS Chemistry students and works equally well for biology and pre-medical students, because the author brings in biological examples throughout. Chapters 1 and 2 review atoms, bonds and acids, and the book needs algebra and logarithms but not calculus. It is a full textbook, not a short summary, so if your syllabus covers only part of it, use the Chapter List above to find the chapters you need. It teaches no laboratory technique and it is not an advanced synthesis text. If you want a school-level run-up, the Chemistry 2nd Year (FSc) book covers aromatic hydrocarbons, halogenoalkanes, hydroxy and carbonyl compounds, organic nitrogen compounds and spectroscopy at an introductory level, and the Chemistry 1st Year (FSc) book covers bonding, acid–base ideas and hydrocarbons. This page covers the 10th edition, so if your department uses a different edition, the chapter order and numbering may differ a little.

Applicable Universities

This book is useful for students at Pakistani universities offering BS Chemistry and biology or pre-medical programs, including Punjab University, COMSATS, UET, NUST and other HEC-recognized institutions, wherever a two-semester organic chemistry course is taught.

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Related Books

• Chemistry 2nd Year FSC Book (Punjab Board) – the school-level preview of aromatic hydrocarbons, halogenoalkanes, hydroxy and carbonyl compounds, and spectroscopy
• Chemistry 1st Year Book (Punjab Board FSC) – bonding, acid–base ideas and hydrocarbons at college level
• Chemistry Class 10 English Medium (Punjab Board PCTB) – a first look at organic chemistry, hydrocarbons and biochemistry
• Chemistry Class 9 English Medium (Punjab Board PCTB) – quick revision if atoms and bonding feel unfamiliar

Organic Chemistry: A Tenth Edition gives BS Chemistry students a free, complete two-semester course in McMurry’s functional-group approach. Browse more Chemistry books to find the rest of your semester.

Organic Chemistry: A Tenth Edition, by John E. McMurry. OpenStax, Rice University. Free under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International licence (CC BY-NC-SA 4.0). Access for free at https://openstax.org/details/books/organic-chemistry