Slides and Notes For Midterm 1
Slides and Notes For Midterm 1 Chem 241
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th Organic Chemistry, 6 Edition L. G. Wade, Jr. Chapter 1 Introduction and Review Jo Blackburn Richland College, Dallas, TX Dallas C© 2006, Prentice Hallge District Definitions • Old: ▯derived from living organisms▯ • New: ▯chemistry of carbon compounds▯ • From inorganic to organic, Wöhler, 1828 O + - heat NH 4 OCN H2N C NH 2 urea => Chapter 1 2 Atomic Structure • Atoms: protons, neutrons, and electrons. • The number of protons determines the identity of the element. • Some atoms of the same element have a different number of neutrons. These are called isotopes. 12 13 14 • Example: C, C, and C. => Chapter 1 3 Electronic Structure • Electrons: outside the nucleus, in orbitals. • Electrons have wave properties. • Electron density is the probability of finding the electron in a particular part of an orbital. • Orbitals are grouped into ▯shells,▯ at different distances from the nucleus. => Chapter 1 4 First Electron Shell The 1s orbital holds two electrons. Chapter 1 5 Second Electron Shell 2s orbital (spherical) => 2p orbital Chapter 1 Three p orbitals 6 Electronic Configurations • Aufbau Principle: Place electrons in lowest energy ↑ ↑ orbital first. • Hund▯s Rule: ↑↓ Equal energy orbitals are half- filled, then filled. ↑↓ • 6C: 1s 2s 2p 2 => Chapter 1 7 Electronic Configurations => Chapter 1 8 Bond Formation • Ionic bonding: electrons are transferred. • Covalent bonding: electron pair is shared. => Chapter 1 9 Lewis Structures • Bonding electrons • Nonbonding electrons or lone pairs H H H C O H Satisfy the octet rule! => Chapter 1 10 Multiple Bonding => Chapter 1 11 Dipole Moment • Amount of electrical charge x bond length. • Charge separation shown by electrostatic potential map (EPM). • Red indicates a partially negative region and blue indicates a partially positive region. => Chapter 1 12 Electronegativity and Bond Polarity Greater ΔEN means greater polarity => Chapter 1 13 Calculating Formal Charge • For each atom in a valid Lewis structure: • Count the number of valence electrons • Subtract all its nonbonding electrons • Subtract half of its bonding electrons 3 H O O H C C O O P O H O => Chapter 1 14 Ionic Structures H H + - H C N H Cl H H _ Na O CH 3 or Na+ O CH 3 X => Chapter 1 15 Resonance • Only electrons can be moved (usually lone pairs or pi electrons). • Nuclei positions and bond angles remain the same. • The number of unpaired electrons remains the same. • Resonance causes a delocalization of electrical charge. Chapter 1 Example=>16 Resonance Example _ _ _ O O O N N N O O O O O O • The real structure is a resonance hybrid. • All the bond lengths are the same. • Each oxygen has a -1/3 electrical charge. => Chapter 1 17 Major Resonance Form • Has as many octets as possible. • Has as many bonds as possible. • Has the negative charge on the most electronegative atom. • Has as little charge separation as possible. Example=> Chapter 1 18 Resonance Hybrid minor contributor, major carbon does contributor not have octet => Chapter 1 19 Chemical Formulas H O • Full structural formula H C C O H (no lone pairs shown) H • Line-angle formula • Condensed structural formula • CH COOH 3 • Molecular formula • 2 H4O2 • Empirical formula • CH2O => Chapter 1 20 Calculating Empirical Formulas • Given % composition for each element, assume 100 grams. • Convert the grams of each element to moles. • Divide by the smallest moles to get ratio. • Molecular formula may be a multiple of the empirical formula. => Chapter 1 21 Sample Problem • An unknown compound has the following composition: 40.0% C, 6.67% H, and 53.3% O. Find the empirical formula. 40.0gC =3.33molC = 1 12.0gC/molC 3.33 6.67gH 1.01gC/molH =6.60molH = 1.98 = 2 3.33 53.5gO =3.33molO = 1 CH 2 16.0gO/molO 3.33 Chapter 1 22 Arrhenius Acids and Bases + • Acids dissociate in water to give H 3 ions. • Bases dissociate in water to give OH ions. • Kw= [H 3 ][OH ] = 1.0 x 10 -14at 24°C + • pH = -log [H3O ] • Strong acids and bases are 100% dissociated. + - HCl + H 2 H 3 + Cl 1 M 1 M => Chapter 1 23 Br Ø nsted-Lowry Acids and Bases • Acids can donate a proton. • Bases can accept a proton. • Conjugate acid-base pairs. O O - + CH3 C OH + CH3 NH2 CH3 C O + CH3 NH3 acid base conjugate conjugate base acid Chapter 1 => 24 Acid and Base Strength • Acid dissociation constant, K a • Base dissociation constant, K b • For conjugate pairs, (K )(K ) a K b w • Spontaneous acid-base reactions proceed from stronger to weaker. O O - + CH 3 C OH + CH3 NH2 CH3 C O + CH3 NH3 pK a.74 pK b.36 pKb9.26 pK a0.64 Chapter 1 =>25 Structural Effects on Acidity • Electronegativity • Size • Resonance stabilization of conjugate base => Chapter 1 26 Electronegativity As the bond to H becomes more polarized, H becomes more positive and the bond is easier to break. => Chapter 1 27 Size • As size increases, the H is more loosely held and the bond is easier to break. • A larger size also stabilizes the anion. => Chapter 1 28 Resonance • Delocalization of the negative charge on the conjugate base will stabilize the anion, so the substance is a stronger acid. • More resonance structures usually mean greater stabilization. O O CH3CH2OH < CH 3 OH < CH 3 S OH O => Chapter 1 29 Lewis Acids and Bases • Acids accept electron pairs = electrophile • Bases donate electron pairs = nucleophile => Chapter 1 30 End of Chapter 1 Chapter 1 31 th Organic Chemistry, 6 Edition L. G. Wade, Jr. Chapter 5 Stereochemistry Jo Blackburn Richland College, Dallas, TX Dallas © 2006, Prentice Hallege District Stereoisomers • Same bonding sequence. • Different arrangement in space. • Example: HOOC-CH=CH-COOH has two geometric (cis-trans) isomers: O O O H C OH HO C C OH C C C C HO C H H H O => fumaric acid, 87 2 maleic acid, 38 1C essential metabolite toxic irritant Chapter 5 2 Chirality • ▯Handedness▯: right glove doesn▯t fit the left hand. • Mirror-image object is different from the original object. => Chapter 5 3 Chirality in Molecules • The cis isomer is achiral. • The trans isomer is chiral. • Enantiomers: nonsuperimposable mirror images, different molecules. => Chapter 5 4 Stereocenters • Any atom at which the exchange of two groups yields a stereoisomer. • Examples: • Asymmetric carbons • Double-bonded carbons in cis-trans isomers => Chapter 5 5 Chiral Carbons • Tetrahedral carbons with 4 different attached groups are chiral. • If there▯s only one chiral carbon in a molecule, its mirror image will be a different compound (enantiomer). => Chapter 5 6 Mirror Planes of Symmetry • If two groups are the same, carbon is achiral. (animation) • A molecule with an internal mirror plane cannot be chiral.* Caution! If there is no plane of symmetry, molecule may be chiral or achiral. See if mirror image can be superimposed. => Chapter 5 7 (R), (S) Nomenclature • Different molecules (enantiomers) must have different names. • Usually only one enantiomer will be biologically active. • Configuration around the chiral carbon is specified with (R) and (S). Chapter 5 => 8 Cahn-Ingold-Prelog Rules • Assign a priority number to each group attached to the chiral carbon. • Atom with highest atomic number assigned the highest priority #1. • In case of ties, look at the next atoms along the chain. • Double and triple bonds are treated like bonds to duplicate atoms. => Chapter 5 9 Assign Priorities O 2 C OH Cl 4H 3 C 4 3 H 3 H 1H 2 2 * natural alanine H Cl 1 3 H CH2 C C C CH * 4 expands to H C 2 H C C CH(CH 3 2 CH2OH H C *C CH(CH3 2 1O 2 O CH2OH => C O Chapter 5 10 Assign (R) or (S) • Working in 3D, rotate molecule so that lowest priority group is in back. • Draw an arrow from highest to lowest priority group. • Clockwise = (R), Counterclockwise = (S) => Chapter 5 11 Properties of Enantiomers • Same boiling point, melting point, density • Same refractive index • Different direction of rotation in polarimeter • Different interaction with other chiral molecules – Enzymes – Taste buds, scent => Chapter 5 12 Plane-Polarized Light • Polarizing filter – calcite crystals or plastic sheet. • When two filters are used, the amount of light transmitted depends on the angle of the axes. => Chapter 5 13 Polarimetry • Use monochromatic light, usually sodium D • Movable polarizing filter to measure angle • Clockwise = dextrorotatory = d or (+) • Counterclockwise = levorotatory = l or (-) • Not related to (R) and (S) => Chapter 5 14 Specific Rotation Observed rotation depends on the length of the cell and concentration, as well as the strength of optical activity, temperature, and wavelength of light. α (observed) [α] = c • l c is concentration in g/mL l is length of path in decimeters. => Chapter 5 15 Calculate [α] D • A 1.00-g sample is dissolved in 20.0 mL ethanol. 5.00 mL of this solution is placed in a 20.0-cm polarimeter tube at 25°C. The observed rotation is 1.25° counterclockwise. => Chapter 5 16 Biological Discrimination => Chapter 5 17 Racemic Mixtures • Equal quantities of d- and l- enantiomers. • Notation: (d,l) or (±) • No optical activity. • The mixture may have different b.p. and m.p. from the enantiomers! => Chapter 5 18 Racemic Products If optically inactive reagents combine to form a chiral molecule, a racemic mixture of enantiomers is formed. => Chapter 5 19 Optical Purity • Also called enantiomeric excess. • Amount of pure enantiomer in excess of the racemic mixture. • If o.p. = 50%, then the observed rotation will be only 50% of the rotation of the pure enantiomer. • Mixture composition would be 75-25. => Chapter 5 20 Calculate % Composition The specific rotation of (S)-2-iodobutane is +15.90°. Determine the % composition of a mixture of (R)- and (S)-2-iodobutane if the specific rotation of the mixture is -3.18°. => Chapter 5 21 Chirality of Conformers • If equilibrium exists between two chiral conformers, molecule is not chiral. • Judge chirality by looking at the most symmetrical conformer. • Cyclohexane can be considered to be planar, on average. => Chapter 5 22 Mobile Conformers H H H H Br Br H H Br Br Br Br Nonsuperimposable mirror images, Use planar but equal energy and interconvertible. approximation. => Chapter 5 23 Nonmobile Conformers If the conformer is sterically hindered, it may exist as enantiomers. => Chapter 5 24 Allenes • Chiral compounds with no chiral carbon. • Contains sp hybridized carbon with adjacent double bonds: -C=C=C-. • End carbons must have different groups. Allene is achiral. => Chapter 5 25 Fischer Projections • Flat drawing that represents a 3D molecule. • A chiral carbon is at the intersection of horizontal and vertical lines. • Horizontal lines are forward, out-of-plane. • Vertical lines are behind the plane. Chapter 5 26 Fischer Rules • Carbon chain is on the vertical line. • Highest oxidized carbon at top. • Rotation of 180° in plane doesn▯t change molecule. • Do not rotate 90°! • Do not turn over out of plane! => Chapter 5 27 Fischer Mirror Images • Easy to draw, easy to find enantiomers, easy to find internal mirror planes. • Examples: CH 3 CH 3 CH 3 H Cl Cl H H Cl Cl H H Cl H Cl CH => 3 CH 3 CH 3 Chapter 5 28 Fischer (R) and (S) • Lowest priority (usually H) comes forward, so assignment rules are backwards! • Clockwise 1-2-3 is (S) and counterclockwise 1-2-3 is (R). • Example: (S) CH 3 H Cl Cl H CH 3 (S) => Chapter 5 29 Diastereomers • Stereoisomers that are not mirror images. • Geometric isomers (cis-trans). • Molecules with 2 or more chiral carbons. => Chapter 5 30 Alkenes Cis-trans isomers are not mirror images, so these are diastereomers. H H H CH 3 C C C C H C CH H C H => 3 3 3 cis-2-butene trans-2-butene Chapter 5 31 Ring Compounds • Cis-trans isomers possible. • May also have enantiomers. • Example: trans-1,2-dimethylcyclopentane => Chapter 5 32 Summary of Isomers => Chapter 5 33 Two or More Chiral Carbons • Enantiomer? Diastereomer? Meso? Assign (R) or (S) to each chiral carbon. • Enantiomers have opposite configurations at each corresponding chiral carbon. • Diastereomers have some matching, some opposite configurations. • Meso compounds have internal mirror plane. • Maximum number is 2 , where n = the number of chiral carbons. => Chapter 5 34 Examples COOH COOH H OH HO H HO H H OH COOH COOH (2R,3R)-tartaric acid,3S)-tartaric acid COOH H OH H OH COOH (2R,3S)-tartaric acid => Chapter 5 35 Fischer-Rosanoff Convention • Before 1951, only relative configurations could be known. • Sugars and amino acids with same relative configuration as (+)-glyceraldehyde were assigned D and same as (-)-glyceraldehyde were assigned L. • With X-ray crystallography, now know absolute configurations: D is (R) and L is (S). • No relationship to dextro- or levorotatory. => Chapter 5 36 D and L Assignments CHO H OH CHO * CH2OH H OH D-(+)-glyceraldehyde HO H COOH H OH H2N H => * H * OH CH2CH2COOH L-(+)-glutamic acid CH 2H D-(+)-glucose Chapter 5 37 Properties of Diastereomers • Diastereomers have different physical properties: m.p., b.p. • They can be separated easily. • Enantiomers differ only in reaction with other chiral molecules and the direction in which polarized light is rotated. • Enantiomers are difficult to separate. => Chapter 5 38 Resolution of Enantiomers React a racemic mixture with a chiral compound to form diastereomers, which can be separated. => Chapter 5 39 Chromatographic Resolution of Enantiomers => Chapter 5 40 End of Chapter 5 Chapter 5 41 Organic Chemistry, 6 Edition L. G. Wade, Jr. Chapter 2 Structure and Properties of Organic Molecules Jo Blackburn Richland College, Dallas, TX Dallas © 2006, Prentice Hallege District Wave Properties of Electrons • Standing wave vibrates in fixed location. • Wave function, ψ, mathematical description of size, shape, orientation. • Amplitude may be positive or negative. • Node: amplitude is zero. Chapter 2 =2 Wave Interactions • Linear combination of atomic orbitals ! on different atoms produce molecular orbitals ! on the same atom give hybrid orbitals. • Conservation of orbitals. • Waves that are in phase add together. Amplitude increases. • Waves that are out of phase cancel out. => Chapter 2 3 Bonding Region • Electrons are close to both nuclei. => Chapter 2 4 Sigma Bonding • Electron density lies between the nuclei. • A bond may be formed by s-s, p-p, s-p, or hybridized orbital overlaps. • The bonding MO is lower in energy than the original atomic orbitals. • The antibonding MO is higher in energy than the atomic orbitals. => Chapter 2 5 Bonding Molecular Orbital Two hydrogens, 1s constructive overlap => Chapter 2 6 Anti-Bonding Molecular Orbital Two hydrogens, destructive overlap. => Chapter 2 7 H : s-s overlap 2 => Chapter 2 8 Cl : p-p overlap 2 Constructive overlap along the same axis forms a sigma bond. => Chapter 2 9 HCl: s-p overlap Question: What is the predicted shape for the bonding MO and the antibonding MO of the HCl molecule? => Chapter 2 10 Pi Bonding • Pi bonds form after sigma bonds. • Sideways overlap of parallel p orbitals. => Chapter 2 11 Multiple Bonds • A double bond (2 pairs of shared electrons) consists of a sigma bond and a pi bond. • A triple bond (3 pairs of shared electrons) consists of a sigma bond and two pi bonds. => Chapter 2 12 Molecular Shapes • Bond angles cannot be explained with simple s and p orbitals. Use VSEPR theory. • Hybridized orbitals are lower in energy because electron pairs are farther apart. • Hybridization is LCAO within one atom, just prior to bonding. => Chapter 2 13 sp Hybrid Orbitals • 2 VSEPR pairs • Linear electron pair geometry • 180° bond angle Chapter 2 14 sp Hybrid Orbitals • 3 VSEPR pairs - • Trigonal planar e pair geometry • 120° bond angle => Chapter 2 15 sp Hybrid Orbitals • 4 VSEPR pairs - • Tetrahedral e pair geometry • 109.5° bond angle => Chapter 2 16 Sample Problems • Predict the hybridization, geometry, and bond angle for each atom in the following molecules: • Caution! You must start with a good Lewis structure! • NH2NH 2 • CH -C≡C-CHO 3 O _ CH3 C CH 2 => Chapter 2 17 Rotation around Bonds • Single bonds freely rotate. • Double bonds cannot rotate unless the bond is broken. Chapter 2 =18 Isomerism • Same molecular formula, but different arrangement of atoms: isomers. • Constitutional (or structural) isomers differ in their bonding sequence. • Stereoisomers differ only in the arrangement of the atoms in space. => Chapter 2 19 Structural Isomers CH O CH and CH 3 CH 2 OH 3 3 CH 3 and CH3 => Chapter 2 20 Stereoisomers Br Br Br CH 3 C C and C C H C CH H C Br 3 3 3 Cis - same side Trans - across Cis-trans isomers are also called geometric isomers. There must be two different groups on the sp carbon. H H C C H3C No cis-trans isomers possible H Chapter 2 => 21 Bond Dipole Moments • are due to differences in electronegativity. • depend on the amount of charge and distance of separation. • In debyes, µ = 4.8 x δ (electron charge) x d(angstroms) => Chapter 2 22 Molecular Dipole Moments • Depend on bond polarity and bond angles. • Vector sum of the bond dipole moments. => Chapter 2 23 Effect of Lone Pairs Lone pairs of electrons contribute to the dipole moment. => Chapter 2 24 Intermolecular Forces • Strength of attractions between molecules influence m.p., b.p., and solubility, esp. for solids and liquids. • Classification depends on structure. ! Dipole-dipole interactions ! London dispersions ! Hydrogen bonding => Chapter 2 25 Dipole-Dipole Forces • Between polar molecules. • Positive end of one molecule aligns with negative end of another molecule. • Lower energy than repulsions, so net force is attractive. • Larger dipoles cause higher boiling points and higher heats of vaporization. => Chapter 2 26 Dipole-Dipole => Chapter 2 27 London Dispersions • Between nonpolar molecules • Temporary dipole-dipole interactions • Larger atoms are more polarizable. • Branching lowers b.p. because of decreased surface contact between molecules. Chapter 2 2=> Dispersions => Chapter 2 29 Hydrogen Bonding • Strong dipole-dipole attraction. • Organic molecule must have N-H or O-H. • The hydrogen from one molecule is strongly attracted to a lone pair of electrons on the other molecule. • O-H more polar than N-H, so stronger hydrogen bonding. => Chapter 2 30 H Bonds => Chapter 2 31 Boiling Points and Intermolecular Forces CH 3 CH 2 OH CH 3 O CH 3 ethanol,b.p.= 78°C dimethylether,b.p.= -25°C H3C N CH3 CH 3H2 N CH3 CH 3H2CH2 N H CH3 H H propylamine, b.p. 49°C trimethylamine, b.p. 3.5°Cthylamine, b.p. 37°C CH CH OH CH CH NH 3 2 3 2 2 ethanol, b.p. = 78° C ethyl amine, b.p. = 17 ° C Chapter 2 32 Solubility • Like dissolves like. • Polar solutes dissolve in polar solvents. • Nonpolar solutes dissolve in nonpolar solvents. • Molecules with similar intermolecular forces will mix freely. => Chapter 2 33 Ionic Solute with Polar Solvent Hydration releases energy. Entropy increases. => Chapter 2 34 Ionic Solute with Nonpolar Solvent => Chapter 2 35 Nonpolar Solute with Nonpolar Solvent => Chapter 2 36 Nonpolar Solute with Polar Solvent => Chapter 2 37 Classes of Compounds • Classification based on functional group. • Three broad classes !Hydrocarbons !Compounds containing oxygen !Compounds containing nitrogen. => Chapter 2 38 Hydrocarbons 3 • Alkane: single bonds, sp carbons • Cycloalkane: carbons form a ring • Alkene: double bond, sp carbons • Cycloalkene: double bond in ring • Alkyne: triple bond, sp carbons • Aromatic: contains a benzene ring => Chapter 2 39 Compounds Containing Oxygen • Alcohol: R-OH • Ether: R-O-R' O • Aldehyde: RCHO CH 3H 2 C H O • Ketone: RCOR' CH3 C CH3 => Chapter 2 40 Carboxylic Acids and Their Derivatives O • Carboxylic Acid: RCOOH C OH • Acid Chloride: RCOCl O • Ester: RCOOR' • Amide: RCONH C Cl 2 O O C NH 2 C OCH 3 => Chapter 2 41 Compounds Containing Nitrogen • Amines: RNH2, RNHR', or R3N • Amides: RCONH , RCONHR, RCONR 2 2 • Nitrile: RCN CH 3 C N O N CH 3 => Chapter 2 42 End of Chapter 2 Chapter 2 43 Organic Chemistry, 6 Edition L. G. Wade, Jr. Chapter 3 Structure and Stereochemistry of Alkanes Jo Blackburn Richland College, Dallas, TX Dallas © 2006, Prentice Hallege District Classification Review Chapter 3 2 Alkane Formulas • All C-C single bonds • Saturated with hydrogens • Ratio: C n 2n+2 • Alkane homologs: CH (CH 3 CH 2 n 3 • Same ratio for branched alkanes H H H H H H C H H C C C C H H H H H H H H C C C H H H H Butane, 4H10 Isobutane4 10 => Chapter 3 3 Common Names • Isobutane, ▯isomer of butane▯ • Isopentane, isohexane, etc., methyl branch on next-to-last carbon in chain. • Neopentane, most highly branched • Five possible isomers of hexane, 18 isomers of octane and 75 for decane! => Chapter 3 4 Alkane Examples => Chapter 3 5 IUPAC Names • Find the longest continuous carbon chain. • Number the carbons, starting closest to the first branch. • Name the groups attached to the chain, using the carbon number as the locator. • Alphabetize substituents. • Use di-, tri-, etc., for multiples of same substituent. =>" Chapter 3 6 Longest Chain • The number of carbons in the longest chain determines the base name: ethane, hexane. (Listed in Table 3.2, page 82.) • If there are two possible chains with the same number of carbons, use the chain with the most substituents. H3C CH CH 2 CH3 CH 3 H3C CH 2 C CH CH 2 CH 2 CH 3 CH3 => Chapter 3 7 Number the Carbons • Start at the end closest to the first attached group. • If two substituents are equidistant, look for the next closest group. 1 CH 3 CH3 3 4 5 H3C CH CH CH 2 CH 2 CH CH 3 2 CH CH 6 7 2 3 => Chapter 3 8 Name Alkyl Groups • CH 3, methyl CH • CH 3H -,2ethyl 3 CH3 CH CH 2 • CH 3H CH2-, n2propyl isobutyl • CH CH CH CH -, n-butyl 3 2 2 2 CH 3 CH 3 CH CH 3 CH 3 CH CH 2 CH 3 H 3 C CH3 sec-butyl tert-butyl isopropyl => Chapter 3 9 Propyl Groups H H H H H H H C C C H H C C C H H H H H H H n-propyl isopropyl Aprimary carbon Asecondary carbon => Chapter 3 10 Butyl Groups H H H H H H H H H C C C C H H C C C C H H H H H H H H H n-butyl sec-butyl Asecondary carbon Aprimary carbon => Chapter 3 11 Isobutyl Groups H H H H C H C H H H H H H C C C H H C C C H H H H H H H isobutyl tert-butyl Aprimary carbon Atertiary carbon => Chapter 3 12 Alphabetize • Alphabetize substituents by name. • Ignore di-, tri-, etc. for alphabetizing. CH 3 CH 3 H C CH CH CH CH CH CH 3 2 2 3 CH CH 2 3 3-ethyl-2,6-dimethylheptane => Chapter 3 13 Complex Substituents • If the branch has a branch, number the carbons from the point of attachment. • Name the branch off the branch using a locator number. • Parentheses are used around the complex branch name. 1 2 3 1-methyl-3-(1,2-dimethylpropyl)cyclohex=>e Chapter 3 14 Physical Properties • Solubility: hydrophobic • Density: less than 1 g/mL • Boiling points increase with increasing carbons (little less for branched chains). • Melting points increase with increasing carbons (less for odd- number of carbons). Chapter 3 15 Boiling Points of Alkanes Branched alkanes have less surface area contact, so weaker intermolecular forces. => Chapter 3 16 Melting Points of Alkanes Branched alkanes pack more efficiently into a crystalline structure, so have higher m.p. => Chapter 3 17 Branched Alkanes • Lower b.p. with increased branching • Higher m.p. with increased branching • Examples: CH 3 CH C CH CH CH 3 CH CH 3 2 3 CH CH C2 CH2 3 3 CH CH 3 CH 3 CH3 CH3 CH 3 bp 60°C bp 58°C bp 50°C mp -154°C mp -135°C mp -98°C Chapter 3 18 => Major Uses of Alkanes • C -C : gases (natural gas) 1 2 • C -C : liquified petroleum (LPG) 3 4 • C5-C 8 gasoline • C9-C 16diesel, kerosene, jet fuel • C17up: lubricating oils, heating oil • Origin: petroleum refining => Chapter 3 19 Reactions of Alkanes • Combustion heat 2 CH 3H 2H C2 3 + 13 O 2 8 CO 2 + 10H 2 • Cracking and hydrocracking (industrial) catalyst long-chainalkanes shorter-chainalkanes • Halogenation heat or light CH 4 + Cl2 CH3Cl + CH 2l 2 + CHCl 3 + CCl4 => Chapter 3 20 Conformers of Alkanes • Structures resulting from the free rotation of a C-C single bond • May differ in energy. The lowest-energy conformer is most prevalent. • Molecules constantly rotate through all the possible conformations. => Chapter 3 21 Ethane Conformers • Staggered conformer has lowest energy. • Dihedral angle = 60 degrees H H H H H H => Newman sawhorse model projection Chapter 3 22 Ethane Conformers (2) • Eclipsed conformer has highest energy • Dihedral angle = 0 degrees => Chapter 3 23 Conformational Analysis • Torsional strain: resistance to rotation. • For ethane, only 12.6 kJ/mol => Chapter 3 24 Propane Conformers Note slight increase in torsional strain due to the more bulky methyl group. => Chapter 3 25 Butane Conformers C2-C3 • Highest energy has methyl groups eclipsed. • Steric hindrance • Dihedral angle = 0 degrees totally eclipsed => Chapter 3 26 Butane Conformers (2) • Lowest energy has methyl groups anti. • Dihedral angle = 180 degrees anti => Chapter 3 27 Butane Conformers (3) • Methyl groups eclipsed with hydrogens • Higher energy than staggered conformer • Dihedral angle = 120 degrees => eclipsed Chapter 3 28 Butane Conformers (4) • Gauche, staggered conformer • Methyls closer than in anti conformer • Dihedral angle = 60 degrees => gauche Chapter 3 29 Conformational Analysis => Chapter 3 30 Higher Alkanes • Anti conformation is lowest in energy. • ▯Straight chain▯ actually is zigzag. CH CH CH CH CH 3 2 2 2 3 H H H H H H C C C C C H H H H H H => Chapter 3 31 Cycloalkanes • Rings of carbon atoms (-CH - groups) 2 • Formula: C n 2n • Nonpolar, insoluble in water • Compact shape • Melting and boiling points similar to branched alkanes with same number of carbons => Chapter 3 32 Naming Cycloalkanes • Cycloalkane usually base compound • Number carbons in ring if >1 substituent. • First in alphabet gets lowest number. • May be cycloalkyl attachment to chain. CH2CH3 CH2CH3 CH3 => Chapter 3 33 Cis-Trans Isomerism • Cis: like groups on same side of ring • Trans: like groups on opposite sides of ring => Chapter 3 34 Cycloalkane Stability • 5- and 6-membered rings most stable • Bond angle closest to 109.5° • Angle (Baeyer) strain • Measured by heats of combustion per -CH -2 => Chapter 3 35 Heats of Combustion/CH 2 Alkane + O → CO + H O 2 2 2 697.1 686.1 658.6 kJ 664.0658.6 662.4 663.6 kJ/mol Long-chain => Chapter 3 36 Cyclopropane • Large ring strain due to angle compression • Very reactive, weak bonds => Chapter 3 37 Cyclopropane (2) Torsional strain because of eclipsed hydrogens => Chapter 3 38 Cyclobutane • Angle strain due to compression • Torsional strain partially relieved by ring-puckering => Chapter 3 39 Cyclopentane • If planar, angles would be 108 °, but all hydrogens would be eclipsed. • Puckered conformer reduces torsional strain. => Chapter 3 40 Cyclohexane • Combustion data shows it▯s unstrained. • Angles would be 120 °, if planar. • The chair conformer has 109.5 ° bond angles and all hydrogens are staggered. • No angle strain and no torsional strain. => Chapter 3 41 Chair Conformer => Chapter 3 42 Boat Conformer => Chapter 3 43 Conformational Energy => Chapter 3 44 Axial and Equatorial Positions => Chapter 3 45 Monosubstituted Cyclohexanes => Chapter 3 46 1,3-Diaxial Interactions => Chapter 3 47 Disubstituted Cyclohexanes => Chapter 3 48 Cis-Trans Isomers Bonds that are cis, alternate axial- equatorial around the ring. CH 3 CH 3 One axial, one equatorial => Chapter 3 49 Bulky Groups • Groups like t-butyl cause a large energy difference between the axial and equatorial conformer. • Most stable conformer puts t-butyl equatorial regardless of other substituents. => Chapter 3 50 Bicyclic Alkanes • Fused rings share two adjacent carbons. • Bridged rings share two nonadjacent C▯s. bbicyclo[3.1.0]hexane bicyclo[2.2.1]heptane => Chapter 3 51 Cis- and Trans-Decalin • Fused cyclohexane chair conformers • Bridgehead H▯s cis, structure more flexible • Bridgehead H▯s trans, no ring flip possible. H H H => H cis-decalin trans-decalin Chapter 3 52 Bicyclo[4.4.0]decane => Chapter 3 53 End of Chapter 3 Chapter 3 54
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