CHEM 230 Lecture Notes (S09)

  1. *Introduction
    1. *Study strategies
      1. Web Page - CHEM 130 A copy of the text is in HT 306/307
      2. Reading - Recognize review material vs. new material
      3. Be sure to check answers to homework.
      4. Be sure to DO THE HOMEWORK ON TIME
      5. Don't get behind - More theoretical and abstract than Organic I
    2. *Energy in chemistry and biology
      1. *Energy flow in muscle contraction
      2. *Energy of substances in glycolysis
      3. *Rates of reaction also - enzymes
        1.  Glucose (green) in active site of hexokinase (Hexokin_glu.pdb)
          Hexokinase and glucose.hin
    3. *Glucose
      1. *Representations of β-D-glucose (b-glucose.pdb)
        1.   bglucose.hin      

          Need to know usual number of bonds and bond angles
            
      2. Rotation of CH2OH group - conformers  (glu_rot30.xyz)
      3. *Equilibrium of β- and α-forms
        1.  
    4. *For Friday Review Structure and Nomenclature material from Organic I (Structures on Sheet)
       QUIZ  Be able to:
      1. *Use the periodic table to predict the usual number of bonds for C, N, P, O, S, H and halogens.
        Click to expand outline and review.
        1. Number of bonds only applies to (covalent) bonds in molecules (uncharged), not ions (charged)
        2. Related to position in periodic table - click here to review
        3. Halogens are F, Cl, Br and I
      2. *Use formulas to determine bonding pattern (which pairs of atoms are bonded) and number of hydrogen atoms in line (skeletal) formulas.
        Click to expand outline and see example.
        1. Draw formula showing all bonds for the molecule below
        2. Click here to see answer.
      3. *1) Use octet rule to locate unshared pairs of electrons,
          2) Use
        VSEPR theory to predict bond angles and
          3) Use dashed-line and wedge formulas.
        Click to expand outline and see an example.
        1. Predict the bond angles around all atoms other than H in the molecule below:
            See answer
        2. *Click to review VSEPR
          1. Determine the bonding pattern (and H atoms) using the usual number of bonds.
          2. Use the octet rule to determine location of unshared pairs.
             See unshared pairs in structure above.
          3. Use the number of groups of electrons to determine the bond angles:
            One Group = single bond, double bond, triple bond or unshared pair
            Two Groups => 180o   Three Groups => 120o   Four Groups => 109.5o
      4.  *Name hydrocarbons (alkanes, alkenes and alkynes, including rings and cis/trans) and
             halogenated (F, Cl, Br, I) compounds. (Demonstrate practice ex.)
        Click to expand outline and review.
        1. Draw trans-9,10-dibromo-4-ethyl-3,7,7-trimethyl-3-decene.
          Do
          Practice Exercise to check your answer.
        2. Name the compound below:
          *Click to see name.
          1. 2,5,7-tribromo-7-methyl-5-ethyl-3-decyne
        3. *Procedure for IUPAC names
          1. Find longest chain or largest ring.
          2. Identify functional groups (if any) - for now, just carbon-carbon multiple bonds.
          3. Number the chain or ring so that highest priority group has lowest number.
          4. Use numbers to indicate positions of functional groups and alkyl groups or other substituents, such as halogen atoms.
          5. Use di-, tri-, tetra, etc. to indicate number of identical substituents, even if they are attached to different carbon atoms.
          6. Use meth-, eth-, prop-, but-, pent-, etc. indicate number of carbon atoms in a chain, ring or alkyl group.
          7. Use bromo, chloro, iodo to indicate halogen atoms.
          8. Ending on name indicates highest priority group in molecule - for now, -ene indicates a carbon-carbon double bond, -yne indicates a C-C triple bond, -ane no C-C multiple bond.
          9. *Cis/trans at beginning of name for rings or double bonds - *Click to see summary
            1. Cis = parts of longest chain on same side of line through carbons in double bond, trans opposite
            2. Cis = substituents on same side of plane of ring, trans opposite
        4. *Some substances have common names as well.
          1. Halogenated compounds - alkyl halides - name of alkyl group followed by bromide, chloride or iodide.
          2. Draw propyl bromide  *Click to see structure.
      5. Recognize and name alcohols, ethers and amines - Naming Simple Functional Groups Click link to review
  2. *Structures and Names of Organic Molecules
    1. For Wednesday Review Stereochemistry (Demonstrate Chart) as well as
                                            Formal Charge and Resonance - Click a link to review
    2. *New nomenclature
      1. *IUPAC vs. common names
        1. *IUPAC
          1. Root name based on longest chain, other atoms (except hydrogen) named as substituents - substitutive -  e.g., 1-chloro-2-methypropane
            (remember - hydrocarbon groups that are not part of longest chain are substituents, too)
          2. Name is generally one word (numbers separated by commas, letters and numbers separated by hyphens)
        2. *Common
          1. "Pieces" of molecule named separately - additive - like inorganic,
            e.g. ethyl chloride.
          2. Name is generally two words
        3. Practice - Give both types of name for
          *
          Click to see names.
          1. IUPAC: 2-bromopropane
          2. Common: isopropyl bromide
        4. Note: IUPAC root name based on number of carbons just in longest chain.
          Common names based on total number of carbons in molecule or group.
          Example: isobutyl chloride vs. 2-methyl-1-chloropropane
      2. *Chains with more than 10 carbons - Know up to 20.
        1. 10   decane
          11   undecane
          12   dodecane
          13   tridecane
          14   tetradecane
          .....   (Greek prefixes followed by "decane")
          19   nonadecane
          20   eicosane
      3. *Complex alkyl groups
        1. *IUPAC
          1. Longest chain in the group starting at point of attachment
          2. Carbon in group attached to rest of molecule is highest priority so is always number 1.
          3. Rest of name follows pattern of other names; i.e., find longest chain in group, number the carbons in the chain - Substituents on a substituent
          4. Use parentheses to enclose names of complex groups
          5.    *Click to see name.
            1. 3-(1-ethyl-3-methyl-3-butenyl)cycloheptene
          6. Note: Rings can be substituents; e.g., 4-(2-cyclopentenyl)undecane
        2. *Common names - know butyl groups and "iso" groups
          1. Secondary and tertiary alkyl groups - based on type of carbon attached to rest of chain.
          2. Primary, secondary and tertiary carbon based on number of other carbons attached to it (H's and others based on type of C it's attached to)
          3. "Normal" alkyl groups are unbranched (use primary carbon to attach other atom or group)
          4. Note - Can be applied to atoms attached to a carbon, also.
          5. Draw isohexyl chloride *Click to see answer
        3. Note - Common names based on total number of carbon atoms in group; IUPAC based on number in longest chain (starting at point of attachment)
      4. *Examples of nomeclature involving complex alkyl groups: Alcohols, Ethers and Amines
        1. Draw sec-butyl alcohol *Click to see answer
        2. Give an appropriate name for the following compound:
            *
          Click to see answer
          1. 1,3-dimethylbutyl 3-methyl-4-chloro-5-ethylphenyl ether
        3. Give an appropriate name for the following compound:
          *
          Click to see answer
          1. 3-bromo-3-cyclopentenyl di(2-methylpentyl) amine
    3. *Stereoisomers
      1. *Stereoisomers of alkenes (E/Z assignments)
        1. Cis/trans isomers of alkenes determined by "pieces" of longest chain:
          is trans
        2. But what about
            vs. 
          Use position of higher priority atom (based on atomic number) on each carbon in double bond
          Br >  Cl, F> H; Two higher priority atoms (one on each double bond carbon) on same side = Z (opposite side = E)
          Molecule on left above would be E.
        3. *Assigning priorities to groups of atoms
          1. First, use atoms attached directly to double bond carbons (all the same here)
          2. Next - compare highest priority atom attached to those
            HOCH2- > CH3CH2- but Cl in both cases on right
          3. If highest priority atom is same element, compare the number of high priority atoms:
            -CHCl2 > -CH2Cl ; more Cl's attached to C
          4. Double bond treated as two single bonds to two atoms

            Triple bond dealt with similarly
          5. Molecule above would be Z.
        4. Assign E or Z to the molecule below:
          *Click to see answer
          1. E
          2. BrCH2 > CF3
          3. CH=CHCH3 > CH2CH2CH2CHBr2
      2. *Absolute configuration around asymmetric centers (R/S assignments)
        1. Assign priorities to groups as for E/Z determination
        2. Point lowest priority group away from viewer
          2-fluorobutane.hin   R_2_fluorobutane.pdb
        3. Note direction as proceed from highest priority down; clockwise = R; counterclockwise = S
        4. *Strategies when lowest priority group is not pointing away
          1. Lowest priority group pointing toward viewer reverses direction of R and S - With H pointing toward you, other groups arranged counterclockwise, so molecule is R.
          2. OR  Exchanging two groups gives opposite configuration
            This would be S, so original molecule is R
          3.  
          4. OR Use hand as molecular model
        5. Assign R or S to the Fischer projection below:
          *Click to see answer
          1. CH2Br > CF3 > CH=CHCH3 > CH2CH2CH2CHBr2
          2. Clockwise, so this is R.
    4. *Resonance
      1. *Resonance in peptide bonds (proteins)
        1. All of the atoms in the structure above lie in the same plane.
        2. All of the bond angles are 120o.
        3. Rotation around the bond indicated by arrow is restricted; that is, rotation occurs, but not as readily as usual.
        4. All of the above affect protein folding - shapes of enzymes.
        5. Explanation - resonance  
          *Click to see resonance structures.
      2. Need new approach to bonding that involves orbitals to understand resonance and explain why rotation occurs around single bonds but not double.
    5. *Orbital models of bonding
      1. *Review orbitals and electronic configurations - Reading assignment
        1. s,p,d,f orbitals - Shapes of s orbitals; one p orbital and set of p orbitals
        2. Energy levels
        3. Order of filling and electronic configuration from periodic table - Chart
      2. *Valence bond theory - covalent bonds are pairs of electrons in overlapping orbitals (NOT IN TEXT) 
        1. H2 - Pairs of electrons are formed from unpaired electrons on atoms 
        2. Cl2 - Orbitals with unpaired electrons?
        3. HF - Orbitals with unpaired electrons?
        4. H2O - Note bond angle predicted to be 90o 
        5. Expect CH2 to form rather than CH4  
      3. *Hybridization approach: methane
        1. How does carbon form 4 bonds when there are only two unpaired electrons?
        2. 4 unpaired electrons in electron dot formula
        3. Suggests "promotion" of electron from 2s to 2p, but would expect different kinds of bonds and all bonds are equivalent in methane
        4. Mix orbitals together and form average (hybrid orbitals) - sp3 
        5. *Arranged tetrahedrally, overlap with 1s on hydrogen to form bonds - Picture
      4. *Hybridization: Carbon-carbon single bonds
        1. Ethane - overlap two sp3 hybrids, one on each carbon - Picture
        2. Same strategy for bonds in other alkanes
      5. *Double bonds: sp2 Hybridization
        1. Bonds up to this point have been σ bonds: head-to head overlap along bond axis (a line connecting the two atoms)  
        2. p-orbitals can also overlap side-to-side with overlap above and below bond axis to form a π bond  
        3. Double bond - one σ and one π bond
        4. p-orbitals in π bonds must be unhybridized, so only two p-orbitals can mix with s - sp2 hybrid orbitals - Picture
        5. Picture of hybrid orbitals in ethene
      6. *Triple bonds: sp Hybridization
        1. Triple bond - one σ and two π bonds
        2. Only one p-orbital can mix with s - sp hybrid orbitals - Picture
        3. Picture of hybrid orbitals in ethyne
      7. *Representations of molecules and ions - electrostatic potential maps
        1. Charge distribution in ionic, covalent and polar bonds - See map
        2. Methane, Ammonia and Water - See map
        3. Ethane
        4. Ethene
        5. Ethyne
      8. *Hybridization: Atoms other than carbon - unshared pairs in hybrid orbitals; Ions treated in same way as molecules
        1. Ammonia
        2. Ammonium ion
      9. *Determining hybridization of an atom in a molecule
        1. Keep track of unhybridized p-orbitals needed for π bonds.  Other p-orbitals mix with s to form hybrids
                                  OR
        2. Predict bond angle using VSEPR, hybridization is type that has that angle
        3. Examples p. 27 in text
      10. Orbitals for single bonds between atoms having different hybridization

        NOTE: Single bond does not mean sp3 !!!!!!!
      11. *Rotation around double bonds does not occur because o bond would break.
        1. No overlap when p-orbitals are perpendicular - Ethene
        2. Rotation around σ bond maintains overlap - Ethane
      12. *Orbitals explain resonance
        1. Benzene: p-orbitals overlap on both sides, each pair of π electrons is delocalized completely around the ring
        2. Peptides: Nitrogen is hybridized sp2 to allow delocalization of unshared pair of electrons (in unhybridized p-orbital)
                         
        3. *What about aniline? (PreLecture)
          1. In order for unshared pair to be delocalized all the time, need unhybridized p orbital on N all the time.
          2. N is hybridized sp2, planar, 120o bond angles
        4. *Pyridine
    6. *Spectroscopy - Sample Spectra
      1. *Mass Spectroscopy - Not in text
        1. Schematic diagram and operation - Picture
        2. *Examples
          1. Krypton
            Mass (amu)  80  82 83 84 86
            % of Total    2  12 12 57 17
            Measuring mass of individual particles (vs. average)
          2. Chlorine
            Mass (amu)  35  37 70 72 74
            %   75  25 56 38 6
              Atoms Molecular ions
            Observe peaks from molecules and fragments -
                 Note isotopic distribution

        3. *Accounting for pattern in molecular ion region = M, M+1, M+2, etc.
          1. For most elements, most common isotope has lowest mass, so M peak is generally largest of molecular ion peaks.
            Chart of distribution of isotopes
          2. H, N and O have only one isotope
            Carbon is 1.1% C-13
          3. *What combination of isotopes account for the for the M+5 peak in the mass spectrum of a substance containing C, H and Cl? (Isotopes of Cl are 35 and 37 amu.)  *Click to see answer
            1. M has C-12 and Cl-35
            2. Additional mass comes from C-13 and Cl-37, so two Cl-37 and one C-13
            3. Very low probability of two C-13
          4. *Predict the peaks in the molecular ion region (M, M+1, M+2, etc.) for C5H11Br. (Isotopes of Br are 79 and 81 amu.) *Click to see answer
            1. M has C-12 and Br-79
            2. M+1 is one C-13 in place of C-12
            3. M+2 is Br-81 in place of Br-79 (two C-13 very low probability)
            4. M+3 is  Br-81 and C-13
        4. Dibromobenzene mass spectrum
        5. Determining number of atoms of elements from molecular ion region
        6. Nitrogen Rule
        7. Determining identity of unknown; e.g., Cocaine
              Fragmentation pattern = Fingerprint
        8. Determining molecular formula by peak matching
        9. Electrospray Mass Spectrometry
      2. *Infrared Spectroscopy
        1. Electromagnetic Spectrum p. 366 in text
        2. CD Introduction
        3. *Applications
          1. Functional Group Region (> 1500 cm-1) vs. Fingerprint Region  (< 1500 cm-1Typical spectrum
          2. Absorptions to know - Chart
      3. *Nuclear Magnetic Resonance - Proton NMR
        1. Number of absorptions = number of sets of equivalent hydrogens
          (unless accidental degeneracy - see below)
        2. Position of absorption = chemical shift, δ scale in ppm
        3. Equivalent H's have same chemical shift - *Click to see procedure for determining equivalency
          1. Substitute a different atom (e.g., Br) in place of H's, if get identical molecule, H's are equivalent.
          2. If can rotate molecule so that H's exchange positions, but molecule is still exactly the same as it was, H's are equivalent.
          3. How many sets of equivalent H's do we expect in toluene - ?
             *Click to see answer.
            1. Four - the CH3, ortho (2H's), meta (2H's) and para (1H).
        4. But non-equivalent H's can have same chemical shift (accidental degeneracy) - See toluene, above
        5. *Simple coupling
          1. NMR of 2-pentanone (below) - *Click to see larger image.
          2. Four absorptions - four sets of equivalent hydrogens
          3. Triplet at δ1.0, 6 peaks at 1.6, singlet at 2.2 and triplet at 2.5
          4. *Number of peaks in absorption = number of adjacent hydrogens + 1
             # of peaks = N + 1
            1. Absorption at δ1.0 is triplet because of adjacent CH2 (2 H's at 1.6)
            2. Absorption at δ2.2 is singlet because there are no adjacent H's
            3. Absorption at δ2.6 is triplet because of adjacent CH2 (2 H's at 1.6)
            4. Absorption at δ1.6 is sextet (6) because of 5 adjacent H's (2 H's at 2.5 and three H's at 1.0)
        6. H's on sp3 carbons in hydrocarbons absorb around δ 1-2 ppm
        7. H's on sp2 and sp hybridized carbons absorb at higher δ values
        8. Nearby electronegative atoms shift absorption to higher δ values
        9. H's attached to carbons in aromatic (benzene) rings absorb around δ 7-8
        10. Only H's in aldehydes and COOH absorb around δ 9 - 10
        11. Total areas of absorptions are proportional to number of H's in absorbing group (total area is proportional to total number of H's
        12. Splitting of absorptions into multiple peaks caused by coupling to nearby hydrogens (generally on adjacent carbons)
        13. Distance between adjacent peaks in split absorption is coupling constant (J).
        14. Absorptions of coupled H's have same coupling constant.
        15. All H's on adjacent carbons (with same J) cause absorption to split into N + 1 peaks
        16. Coupling to groups with different J's cause more complex coupling patterns Equal J's
        17. H's with same chemical shift do not couple (even if not equivalent)
        18. H's in OH groups do not couple (don't cause splitting, are not split)
          * Click to see spectrum of ethanol

          1. Peak at δ 2.6 is OH - isn't split, doesn't cause CH2 to be split into more than a quartet (due to CH3)
        19. Sometimes non-adjacent H's can cause splitting (long range coupling)
      4. *Other NMR
        1. *Deuterium
          1. Does not couple (doesn't cause splitting)
          2. Absorbs in different region - doesn't appear in proton NMR spectrum
        2. *Carbon-13
          1. 1% natural abundance - computer adds up multiple scans to get good spectrum
          2. Number of absorptions (signals) = number of sets of equivalent carbons
          3. Area proportional to number of carbons, but not as precise as hydrogens
          4. *Couples in usual way - N + 1
            1. Don't see coupling between 13C's - very low probability of adjacent 13C's
            2. Do see coupling between a 13C and H's attached directly to it.
            3. Can remove H coupling from 13C spectrum by double irradiation
        3. Phosphorus-31 also
      5. *Index of Hydrogen Deficiency (IHD) - for drawing possible isomers
        1. Non-ring alkane = CnH2n+2 , even if branched - Degree of Unsaturation = 0
        2. Ring or double bond causes molecule to have 2 H's fewer - IHD = 1
        3. Triple bond has same effect as two double bonds (4 fewer H's) - IHD = 2
        4. IHD = # of "missing" H's (compared to saturated)/2
        5. What are possibilities for C7H10? *Click to see answer
          1. Index of hydrogen deficiency = (16-10)/2 = 3
          2. 3 double bonds or rings (or combination) OR
            Triple bond and either ring or double bond
        6. *Strategy for heteratoms (other than C and H)
          1. Work with molecular formula
          2. Remove heteroatoms without changing degree of unsaturation, then calculate degree of unsaturation for resulting hydrocarbon.
          3. Oxygen has no effect - just remove
          4. Halogens take place of H in molecule - replace with H
          5. To remove nitrogen, have to remove a hydrogen as well.
          6. Calculate degree of unsaturation for C17H30N3O4Cl5  *Click to see answer
            1. Same degree of unsaturation as C17H32
            2. (36-32)/2 = 4
      6. *Structure determination from NMR, IR, Mass Spec
        1. Mass spec (if needed) to get molecular weight and molecular formula
        2. IHD to get possible combinations of rings, multiple bonds
        3. IR to determine multiple bonded groups (e.g., C=O), OH, etc.
        4. *NMR to determine arrangement of hydrogens
          1. Number of absorptions to determine number of equivalent sets of H's
          2. Unique chemical shifts to determine presence of aromatic rings (δ 7-8), aldehyde and carboxylic acid (both δ 9 - 10)
          3. Integration to determine relative numbers of H's
          4. Coupling to determine adjacent groups of H's and numbers of H's in groups
        5. Example
      7. *Theory
        1. Experiment - imagine proton with no electrons in field of magnet
        2. Effect of electrons - shielding
        3. Effect of electronegative atoms - less electron density, less shielding
        4. Mechanism of coupling - rapid exchange explains why OH generally doesn't couple and sometimes causes very broad peaks for hydrogens attached to N or O.
  3. *Energies of molecules and ions; Equilibrium and Rates of Reaction
    1. *Introduction
      1. *Equilibrium - Axial and Equatorial substituted cyclohexanes as example
        1. *From Organic I
          1. Cyclohexane can exist in two forms with bond angles of 109.5o - chair and boat
            Nearly all of cyclohexane is in chair.
          2. Boat and chair interconvert - They are conformers
          3. Two different positions for substituents in chair form - axial and equatorial
          4. Axial and equatorial interconvert - They are conformers
          5. Since they interconvert, they automatically come to equilibrium.
            M
            ore equatorial than axial because equatorial is lower energy.
            Axial vs. equatorial methylcyclohexane
        2. *Amounts (at equilibrium) determined by relative energy of the two forms
          1. Energy diagrams - methylcyclohexane vs. t-butylcyclohexane
          2. Axial t-Butyl "bumps into" other hydrogen atoms more than methyl - See molecular models
          3. Effect of energy difference on amounts at equilibrium - Demo
          4. For any reaction, ΔGo = energy difference between substances on right of equation and those on left;
            i.e., for
            Axial Equatorial,  ΔGo = energy of Equatorial - energy of Axial
          5. Energy determines size of Keq: ΔGo = -RTlnKeq or  Keq = e -ΔG/RT 
        3. Ax-Eq energies for substituents - Chart
        4. Note - The direction we write the reaction doesn't matter.
        5. *Using Keq to predict reactions
          1. For Axial Equatorial, 
          2. For methylcyclohexane, Keq is 18
          3. Does equilibrium lie mainly to the right, to the left or roughly in the middle?
          4. In which direction will reaction go if [Ax] = 0.1 M and [Eq] = 0.9 M?
          5. In which direction will reaction go if [Ax] = 0.01 M and [Eq] = 1.0 M?
      2. *Why do reactions come to equilibrium?
        Why aren't all of the higher energy molecules reacted at equilibrium?
        Rates of reactions
        1. Rate - related to number of molecules (or ions) that react per second not how long it takes one molecule to react
        2. Regulation of enzymes is rate, not equilibrium, phenomenon.
        3. *t-Butylcyclohexane - All axial at start - Axial Equatorial
          1. *Where is the rate fastest? 
              Is the rate increasing or decreasing?
              Why is the rate changing?
            1. Rate is fastest at the beginning, then slows down.
            2. Eventually, no change - Equilibrium
            3. Rate proportional to concentration of axial: rate = k [axial] Demo
               k is called the rate constant for the reaction (different for different reactions)
            4. In general, rate = k [X]a [Y]b [Z]c where k is the rate constant for this reaction, X, Y and Z are reactants and exponents are not necessarily the coefficients in the balanced equation (determined by doing experiments).
        4. *Chlorocyclohexane - All axial at start- Net reaction is Axial  Equatorial
          1. Observed rate initially proportional to concentration of axial, but becomes zero before axial is zero.
          2. Observed rate slows down faster than for t-butyl
          3. Once some equatorial has formed, the reverse reaction (Ax Eq) begins to occur at the same time as the forward reaction.
          4. Observed rate is the net change (difference between the two processes) Iteration Demo
          5. At equilibrium, the two opposing reactions occur at the same rate (but haven't stopped) so observe no net change in amounts of axial and equatorial.
          6. Rate of individual processes is always proportional to concentration of reactants in process:
            For Ax Eq   rateR = kR [Ax] (to right)
            For Ax Eq   rateL = kL [Eq] (to left)
          7. For this reaction, k can be thought of as the fraction of the molecules that react per second (or some other time period).
          8. Observed rate decreases with time because:
          9. If a reaction is not at equilibrium, the species whose concentration is "too high" reacts faster than those for which it is "too low" and its concentration goes down until equilibrium is reached (equal rates).
            (And vice versa)
          10. Rates account for LeChatelier's principle Iteration Demo
          11. *For processes like this (one step), we can derive Keq by setting the net rate equal to zero.
            1. Net Rate = rateR - rateL = kR [Ax] - kL [Eq]
            2. At Equilibrium,
              1) Concentrations are equilibrium concentrations (
              [  ]eq) and
              2) Net rate is zero (no observed change)
                     Net Rate
              = kR [Ax]eq - kL [Eq]eq = 0
                  so rates of opposing processes are equal
                    
              kR [Ax]eq = kL [Eq]eq
              Rearranging gives
          12. *What determines magnitude of rate constants?  Activation Energy
            1. Transition state (TS) is arrangement of atoms at the peak (maximum energy)
            2. Activation energy in each reaction is difference in energy between TS and reactants in that reaction
            3. Larger activation energy causes k to be smaller, so reaction is slower.
            4. Blue arrow is activation energy for reaction to the right
                (smaller activation energy, larger k, faster reaction )
            5. Red arrow is activation energy for reaction to the left
                (larger activation energy, smaller k, slower reaction)
        5. Kinetic vs. Thermodynamic stability
    2. *Some of the factors that affect energies of molecules and ions
      1. *Review reactions of hydrocarbons with oxygen and hydrogen (Organic I)
        1. All hydrocarbons react with excess oxygen in the presence of a flame to produce carbon dioxide and water.
                                           flame
          CxHy  +  O2 (excess)   CO2   +  H2O
        2. All unsaturated hydrocarbons (alkenes, alkynes, aromatic) react with hydrogen in the presence of catalyst (usually Pt) to add H2 across the multiple bond. (Alkynes require two moles of H2 - excess - to react completely to form a single bond.)
      2. *Methods for measuring energy differences
        1. Measure Keq and use ΔGo = -RTlnKeq
        2. *Heat of combustion of β- (-650.8 kcal/mole) and α-D-glucose (-670.8)
          1. Both are C6H12O6 so form same amount of CO2  and H2O
          2. Can compare energy of each (+ O2) to CO2 and H2O
          3. β-D-glucose is 20.0 kcal/mole more stable than α-D-glucose
        3. *Heat of hydrogenation of cis- (-28.6 kcal/mole) and trans-2-butene (-27.6)
      3. *Nonbonded interactions
        1. *Axial/Equatorial substituents in cyclohexane (chair form) - Can add ΔGo for more than one substituent.
          1. Ax-Eq energies for substituents - Chart
          2. Compare energy of each substituent in one of the molecules to the energy of that substituent in the other molecule.
          3.        
          4.                                      Axial i-propyl = +2.2 kcal/mole (higher than eq)
                                                 Equatorial CH3 = -1.7
            kcal/mole (lower than ax)
          5. ΔGo = 0.5 kcal/mole; K<1 (i-Pr bigger than methyl)
        2. *Conformations of open chain compounds
          1. Staggered always lower energy than eclipsed - Ethane and Newman projections
          2. Butane
          3. chair and boat cyclohexane
      4. *Angle strain from heat of combustion
        1. Expect strain: cyclobutane > cyclopentane
        2. Heat of combustion: cyclobutane = -650.2 kcal/mole, cyclopentane = -786.6
        3. Can't compare directly because have different number of atoms reacting:
          C4H8 + 6 O2 4 CO2 + 4 H2O         ΔH =  -650.2 kcal/mole
          C5H10 + 7.5 O2 5 CO2 + 5 H2O    ΔH =  -786.6 kcal/mole
        4. But C4H8 = (CH2)4 and C5H10 = (CH2)5 so
          1/4 C4H8 + 1.5 O2   CO2 +  H2O      ΔH =  -162.6 kcal/mole
          1/5 C5H10 + 1.5 O2   CO2 +  H2O    ΔH =  -157.3 kcal/mole
          gives energy change per CH2 and allows us to compare the same number of atoms -  cyclopentane has less strain
        5. Strain in other ring compounds - Chart 
      5. *Review Lowrey-Bronstad acids and bases and formal charge.
        1. *Conjugate acids and bases
          1. When an acid reacts (loses a proton) it forms its conjugate base.
            A- is the conjugate base of HA in the following reaction.
          2. When an base reacts (gains a proton) it forms its conjugate acid.
            HB+ is the conjugate acid of B in the reaction above.
        2. *Ka and Kb and relationship to pK's
          1. Ka is the equilibrium constant for a reaction of an acid with water, except that water is not included in the equilibrium constant.
            For HA (as an acid),
          2. Kb is the equilibrium constant for a reaction of an base with water, except that water is not included in the equilibrium constant.
          3. For any equilibrium constant, pK = -log K
            Note that as K gets larger, pK gets smaller and vice versa.
        3. Ka (for conjugate acid) x Kb (for conjugate base) = Kw = 10-14 
          This allows us to calculate the K for any acid or base from the K for the conjugate.
        4. *Relationship of Ka, Kb and pK's to strength of acid or base
          1. Since larger K's indicate that the reaction lies farther to the right at equilibrium and stronger acids or bases react more, the larger the Ka or Kb, the stronger the acid or base is.
          2. Since pK gets smaller as K gets larger, a stronger acid or base will have a smaller Ka or Kb and vice versa.
        5. In any reaction, the stronger acid and base react more, so the weaker acid and base are favored at equilibrium.
        6. Effect of pH on amounts of conjugate acid and base
          1. At lower pH (more acidic) there will be larger amounts of conjugate acid in the solution (for any conjugate acid-base pair).
          2. And vice versa.
      6. *Effect of electronegativity of atom bearing formal charge
        1. *What is position of equilibrium in
          ROH  +  RNH3+    RNH +  ROH2+ ?
          1. Positive charge is more stable on less electronegative atom. (Negative charge more stable on more electronegative atom.)
        2. *Which equilibrium lies farther to the right?
          ROH  +  H2  OH-  +  ROH2+   OR
          RNH +  H2  OH-  +  RNH3+ 
          1. Position of positive charge makes conjugate acid of amine more stable (relative to amine) than conjugate acid of alcohol (relative to alcohol).
        3. *When CH3OH acts as an acid, is the hydrogen atom lost from the oxygen or the carbon atom?
          1. From oxygen
          2. Negative charge on oxygen is more stable than on carbon
      7. *Electronic effects of nearby atoms (inductive effect)
        1. *As a Base: CH3NH2  > ClCH2NH2  Why?
          1. Consider either
            1) Rxn 1

                                                        OR

            2)     Rxn a
                                                    VS.
                   Rxn b
          2. Cl is more electronegative than H and attracts electrons more, destabilizes positive charge (would stabilize negative charge)
          3. Ion on left (ClCH2NH3+) in Rxn 1 is less stable than ion on right(CH3NH3+), so Rxn 1 lies mainly to right indicating that CH3NH2 reacts more and therefore is the stronger base.
          4.  ClCH2NH3+ is less stable compared to it's conjugate base than CH3NH3+ is compared to it's conjugate base, so Rxn a lies farther to right than Rxn b indicating that CH3NH2 reacts more and therefore is the stronger base.
          5. Strength of acid or base depends on difference in energy between conjugate acid and conjugate base in pair.
          6. Stronger base reacts more with any acid and vice versa.
          7. So we can just look at the energies of the conjugate acid base pairs.
        2. *Which is a stronger acid, CH3OH or ClCH2OH?
          1. ClCH2OH
          2. Negative charge in conjugate base is stabilized by inductive effect of chlorine.
        3. *Which is a stronger acid, CH3CHClOH or ClCH2CH2OH?
          1. CH3CHClOH
          2. Cl is closer to negative charge in conjugate base and does a better job of stabilizing it.
        4. *Which is a stronger acid, ClCH2OH or Cl2CHOH?
          1. Cl2CHOH
          2. More chlorines stabilize negative charge better.
        5. *CH3NH2 is a weaker base than (CH3)2NH. Do alkyl groups stabilize positive charge or destabilize it (compared to hydrogen atoms)?
          1. Alkyl groups stabilize positive charge.
          2. Since dimethyl amine reacts more, its conjugate acid must be more stable.
      8. *Resonance
        1. *The conjugate base of a carboxylic acid exhibits resonance as shown below:

          Explain how that accounts for the fact that carboxylic acids are stronger acids than alcohols.
          1. Resonance stabilizes anion, shifts equilibrium to right, acid reacts more, stronger acid
          2. But also resonance in conjugate acid!!

            Need to look at resonance more carefully
        2. *Resonance hybrid (the species which actually does exist) is more stable if it has 1) More resonance structures or 2) better resonance structures.
          *What would make a resonance structure "better"?
          1. "Better" means more stable IF it actually existed.
          2. Atoms satisfy octet rule OR don't satisfy octet rule?
          3. More covalent bonds OR Fewer covalent bonds?
          4. Positive charge on more electronegative atom OR Less electronegative atom?
          5. Negative charge on more electronegative atom OR Less electronegative atom?
          6. Opposite charges closer OR Farther apart?
          7. Like (same sign) charges closer OR Farther apart?
        3. *Are the structures below resonance structures? (Trick question!)
          In what type of orbital is the unshared pair of electrons in the structure on the left?
                 
          1. Not resonance structures - orbital containing unshared pair (sp2) does not overlap with p-orbital on adjacent carbon
          2. Electrons can't be shared, bond can't form
        4. Chart of Resonance Generalizations
        5. Better resonance structures make greater contribution to the resonance hybrid - amides, for example
        6. *Resonance in carboxylic acids
          1. Same number of resonance structures in both, but separation of charge in conjugate acid (not as good).
        7. Aniline exhibits resonance as shown below:

          *Which will be the stronger base, aniline or cyclohexyl amine?
          1. No resonance in conjugate acid or base of amine.
          2. Resonance in both conjugate acid and base of aniline.
          3. Resonance Chart
          4. More resonance structures in aniline (conjugate base) than its conjugate acid, making it more stable, reacts less, weaker base
        8. *Which, if either, of the following amines is a stronger base? Once again, we need to be careful.
          1. They are approximately the same!
          2. No resonance in molecule on right or its conjugate acid.
          3. But same amount of resonance in both molecule on left and its conjugate acid, so ΔGo is the same for both reactions.
        9. *Explain why phenol is a stronger acid than an alcohol.
          1. Resonance Structures
          2. Resonance in both phenol and its conjugate base (none in alcohol)
          3. Same number of resonance structures in both conjugate acid and base of phenol.
          4. Better resonance structures in conjugate base.
  4. *Addition Reactions of Unsaturated Compounds
    1. Examples
    2. *Addition of HCl, HBr or HI (gases bubbled into nonaqueous solution of alkene)
      1. *Observations (same for all three reagents)
        1. *
           
          1. Cis reacts faster than trans
          2. Equal amounts of R and S are formed (Review stereochemistry, if necessary.)
        2. *
           
          1. Molecules with methyl groups both cis and trans are formed.
          2. Suggests that H and Br attach at different times.
        3. *
          1. Keep track of atoms in rearranged product
          2. Most of product is rearranged.
          3. Rearranged product suggests that H and X do not add at the same time.
        4. *
          1. Conjugate addition product (on right) - not expected - keep track of atoms and electrons (bonds)
          2. Conjugate addition suggests that H and Br do not add at the same time.
        5. *In presence of other ions
          1. Again suggests that H and "other atom" attach at different times
        6. *
          1. Both cis and trans are formed
          2. More trans than cis
        7. *
          1. Markownikov's Rule
          2. "Hydrogen attaches to carbon (in double bond) that already has the most hydrogens."
        8. *
          1. Markownikov's Rule again
        9. *

          Does this follow Markownikov's Rule?
          1.  Nope.
      2. *Explanation - the Reaction Mechanism
        1. Most reactions occur in steps - the sequence of steps is called the mechanism of the reaction
        2. *Observations that suggest that addition of HX occurs in at least two steps:
          1. Molecules with methyl groups both cis and trans are formed in the following reaction:
          2. Rearranged products form in some reactions:
          3. Conjugate addition can occur in some reactions:
          4. Other ions can be incorporated in the product:
        3. HCl, HBr and HI are all strong acids which suggests that the first step involves donation of H+ to alkene:
        4. In second step, bromide ion attaches to positively charged carbon to complete the reaction:
        5. *Keep track of atoms, electrons and charge
          1. CLICK ON THE QUESTION TO SEE THE ANSWER!
          2. *Where is the hydrogen from HBr (after the first step)?
            1. On the second carbon from the left
          3. *What electrons are used to form the new C-H bond?
            1. The π-electrons of the double bond
          4. *What happens to the electrons in the bond between H and Br (in the first step)?
            1. They become an unshared pair on the Br.
          5. *How do you know that there will be a positive charge in the organic species (after the first step)?
            1. With a negative charge on Br there has to be a positive charge to keep the charge the same on both sides of the equation.
          6. *How can you tell which atom has the positive charge?
            1. Calculate formal charge - C has 3 valence electrons if shared equally, would have 4 if uncharged. Note - Usual number of bonds strategy doesn't work. C doesn't satisfy the octet rule.
          7. *What electrons are used to form the new bond between C and Br?
            1. One of the unshared pairs on Br -
        6. *Terminology
          1. Species (ions, molecules, etc.) formed during steps of mechanism that are not products or reactants are called "intermediates."
            *What are the intermediates in the addition of HBr to 2-butene?
            1. Bromide ion and the organic cation
          2. Ion with positive charge on carbon is called a "carbocation," classified as primary, secondary or tertiary based on number of alkyl groups attached to positively charged carbon.
            is a secondary carbocation
          3. *When a new covalent bond is formed, the species that provides the pair of electrons is called a "nucleophile" and the species that accepts the pair of electrons is called an "electrophile."
            What are the nucleophile and electrophile in each of the steps in the mechanism above? Click to see answer.
            1. First step - Electrophile is H+ from HBr, nucleophile is alkene.
            2. Second step - Electrophile is cation, nucleophile is bromide ion.
      3. *Why does cis-2-butene react faster than trans?
        1. *What determines the rate of a multi-step reaction?
          1. Energy diagram for entire reaction of addition of HBr to alkene
             
          2. Each step has a transition state and each step (both directions) has a rate constant determined by the activation energy for that step.
          3. Overall, rate = k [alkene] [HBr]
          4. Activation energy for multistep reaction is difference in energy between reactants and highest energy transition state in the reaction (called the transition state for the overall reaction):
          5. Rate determining step is the step in which the overall transition state occurs
          6. Note how activation energy and rate determining step changes if second transition state is higher energy:
        2. The energy diagram for the addition of HBr to trans-2-butene is: 
          *
          Draw energy diagram for addition of HBr to cis vs. trans-2-butene.
          1. Transition states in rate determining step for the two reactions nearly equal in energy (forming same intermediate)
          2. Cis alkene is higher in energy, needs less energy to reach transition state (lower activation energy)
          3. Arrangement of atoms in transition state is in between reactants and intermediates, but "looks more like" intermediates.

            Causes energy of two transition states to be nearly equal - both look like the same intermediates.
      4. *Markownikov's Rule - Predict product of following reaction:
        1.  
        2. *Draw the possible carbocations and classify them as 1o, 2o or 3o:
          1. Attack at "a" would yield - a tertiary carbocation
              
          2. Attack at "b" would yield - a primary carbocation
        3. Note that only one product can be formed from each carbocation.  The formation of the carbocation is the product determining step.
          (It is also the rate determining step in most of these addition reactions.)
        4. In most organic reactions, proportions of products are determined by rates of the steps in the reaction - kinetic control.
        5. *Reaction that proceeds through tertiary carbocation goes faster - Why? Hint: Why is (CH3)2NH a stronger base than CH3NH2?
          1. Alkyl groups stabilize positive charge, so tertiary carbocation is more stable than primary
          2. Transition state has partial positive charge which ought to be stabilized by alkyl groups
          3. Expect transition state leading to tertiary to be lower energy than leading to primary:
          4. Assuming first step is rate determining, activation energy for reaction through tertiary carbocation is lower than for primary, so reaction goes faster.
          5. Would come to same conclusion if second step were rate determining.
      5. *Reaction of HX with 4-methylcyclopentene - Rate determining vs. Product determining step
        1. *Why are both isomers (cis and trans) formed?
          1. Carbocation is planar (flat) so bromide can attach to either side.
          2. Attack from above yields cis, attack from below yields trans.
        2. *Why is more trans formed than cis?
          1. Proportions of cis and trans products determined by rates of second steps (product determining steps) in the reaction. (NOT EQUILIBRIUM!!!!!)
          2. Cis product higher in energy than trans - nonbonded interactions
          3. Transition states parallel energies of products (they come from the same intermediate)
          4. Transition state (2nd step) leading to trans is lower in energy than one leading to cis.
          5. Trans has lower activation energy, faster rate, greater amount of product
          6. Note: First step is still rate determining - determines rate at which reactant reacts (or the rate of formation of total products).
      6. *Why do we get equal amounts of R and S from 2-butene (either isomer)?
        1. R vs. S proportions determined by rates of second steps - Product Determining Step (as opposed to rate determining)
        2. Transition states in seconds steps are enantiomers so they have the same energy
        3. Same activation energy, same rate, equal amounts of R and S
      7. *Why does Anti-Markownikov addition occur sometimes?
        1. Different type of stabilization of carbocation
        2. In this case, resonance stabilization by benzene ring:
      8. *Addition to alkynes - Two moles of HX add in normal manner (Markownikov addition)  
        1. Addition of first HX follows Markownikov addition for usual reason (alkyl groups stabilize positive charge)
        2. *But halogen atom is more electronegative and might destabilize positive charge in addition of second molecule of HX:
          1. Carbocation in this step is stabilized by resonance:
      9.  *How does conjugate addition occur?
                                                                   
        1. *Where does H+ attach? Why?
          1. Addition to either end of conjugated system produces resonance stabilized ion:
          2. Ion produced when H+ attaches to a middle carbon is not stabilized by resonance:
             
        2. *Why two products?
          1. Carbocation formed in reaction of 1,3-butadiene with HX exhibits resonance:
          2. Halide ion can attach at either of two carbon atoms, yielding the two products.
                           
        3. *Why different products at different temperatures?
          1. Energy diagram
          2. Second step is product determining.
          3. *At low temperatures, there is no reversal of second step, proportions of products determined by rates of formation - Kinetic Control
            1. Rates determined by energies of transition states in second step.
            2. As bond to halide ion begins to form in transition state, partial positive charge is concentrated on carbon to which it is attaching:
            3. Partial positive charge is more stable on secondary carbon (on left) than primary, so normal product is formed faster.
          4. *At higher temperatures, second step comes to equilibrium so more stable product predominates.
            1. Alkyl groups stabilize double bonds - Heats of Hydrogenation
            2. Conjugate addition product is more stable, favored at equilibrium.
          5. Note - Transition state "looks like" both carbocation and product (alkene), but TS with most stable double bond has least stable positive charge and vice versa.  Position of positive charge is more important than double bond in transition state. 
      10.  *How does rearrangement take place?
        1.  In a carbocation, alkyl groups or hydrogen atoms can move to an adjacent one:
        2. Why does the positive charge end up where it does?  - Keep track of electrons.
        3. Why does rearrangement take place?  Think about stability.
    3. *Other electrophilic additions Nucleophile and Electrophile
      1. *Addition of sulfuric acid
        1. Just like addition of HX (Markownikov, rearrangement, etc.), same mechanism, just different nucleophile in second step.
      2. *Addition of water or alcohols (acid catalyzed) to alkenes
        1. *      
          1. Follows Markownikov's rule.
          2. Acid catalyst generates carbocation in usual way.
          3. Water acts as nucleophile reacting with carbocation.
          4. Oxonium ion loses H+ to conjugate base of acid, regenerating catalyst.
        2. Only differences are 1) acid catalyst used in first step, 2) water acting as nucleophile and 3) additional acid-base step to form product and regenerate catalyst.
        3. Be sure to keep track of electrons, bonds and charge.
        4. Alcohols react exactly the same as water:
      3. *Addition of water to alkynes - forms ketones!
        1.  Same reaction (and mechanism) as addition of water to alkene - (follows Markownikov's Rule).
        2. But product after addition of water molecule is an enol which rearranges to a ketone:
           
        3. Terminal alkynes need HgSO4 along with H2SO4 as catalyst.
      4. *Addition of Br2 or Cl2  Not in text
        1.  
        2. Conjugate addition also occurs (along with normal product), suggesting two step mechanism again:
        3. Obtain additional products in presence of other nucleophiles:

          Suggests two steps (again).
        4. Generally trans addition:
        5. Rearrangement generally does not occur, suggesting not a typical carbocation.
        6. *Explanation - Mechanism of the Reaction
          1. Electrophilic Addition - again
          2. Br+ is electrophile.
          3. Cyclic bromonium ion is formed.
          4. Br- has to attack from side opposite bromonium, giving trans addition:
        7. *But why does conjugate addition occur? (Bromonium ion would yield only normal addition.)
          1. Resonance stabilization makes noncyclic ion more stable than cyclic bromonium ion (which would have no resonance:
          2. Note - won't get trans addition from ion on left.
        8. *Addition of Br2 or Cl2 in presence of other nucleophiles:
          1.             
          2. Still trans addition where appropriate.
        9. Alkynes react in same way as alkenes (one mole adds trans), but can react with two molecules of X2 if they are available.
    4. *Summary of electrophilic addition
      1. *Electrophile attacks multiple bond
        1. *H+
          1. From reagent if it is an acid
          2. From catalyst if reagent is not an acid (water, alcohols)
        2. X+  from Br2 or Cl2.
      2. *Forms most stable cation
        1. Tertiary > secondary > primary
        2. Bridged halonium ion
        3. Resonance stabilized
      3. *Nucleophile attaches to carbon in cation
        1. Conjugate base of acid reagent
        2. Halide ion (or other negatively charged ion)
        3. Other nucleophile in solution (water, alcohol, ion other than conjugate base or halide from reagent)
      4. If nuclophile is uncharged, there will be a positive charge after attack, H+ will be lost to form final product.
      5. Enols rearrange to carbonyl compounds.
    5. *Addition of Hydrogen
      1.  
      2. *Generally cis addition:
        1. Explanation
  5. *Reactions of Aromatic Compounds
    1. *Observations
      1. *Reaction with hydrogen
        1. Benzene reacts with three H2 molecules to form cyclohexane.
        2. But it requires more vigorous conditions (high temperature and pressure) than typical alkenes or alkynes.
      2. *Reactions with halogens
        1. Benzene and most of its derivatives do not react with Br2 or Cl2 unless FeX3 (or other catalyst) is present and substitution occurs (rather than addition)
          *Click to see equation.
        2. *Some substituted benzenes will react without catalyst, but only some of the possible isomers are formed.
          1.  
          2. Only ortho and para in this case
        3. If a substituent is present on the benzene ring, the products are either predominately meta or predominately ortho and para.
          *Click to see examples.
        4. Some substituents cause the reaction to go faster than on benzene, some cause it to go slower.
        5. Generally, the substituents which cause the reaction to go faster than with benzene produce mainly ortho and para products (ortho-para directors) and vice versa.
      3. HBr and HCl don't react with benzene.
      4. Sulfuric acid does react, but requires a catalyst and does substitution instead of addition (again).
        *Click to see equation.
        1. Note that group that attaches is different from the nucleophile in the addition of sulfuric acid to alkenes.
      5. Nitric acid also reacts by substitution and needs sulfuric acid as catalyst.
        *Click to see equation.
      6. Alkyl and acyl halides react in the presence of catalyst to substitute the alkyl or acyl group in place of hydrogen.
        *Click to see equations.
    2. *Why is benzene so unreactive?
      1. Benzene is exceptionally stable. Heats of hydrogenation
      2. *Why is benzene so stable? Aromaticity
        1. *Resonance is only part of the answer.
          1. *Cyclobutadiene is very unstable in spite of resonance.
          2. *The cyclopropenium ion is surprisingly stable.
          3. *The cycloheptatrienylium ion is very stable but the corresponding anion is very unstable.
            1.  
          4. *The cyclopentadienide ion is very stable, but the corresponding cation is very unstable.
            1.  
        2. *An odd number of delocalized pairs of electrons (4n+2 π electrons) in a continuous ring of atoms with an unhybridized p orbital on each atom is exceptionally stable - Aromatic.
          1. *Delocalized electrons can include unshared pairs.
            1. Pyrrole is aromatic.
          2. Continuous ring can include carbocation or carbanion.
          3. *There can be more than one ring involved as long as there is an unhybridized p orbital on each atom.
            1. Naphthalene is aromatic
            2. is not aromatic.
            3. is aromatic.
        3. *Need to be sure that orbitals are available for delocalization of unshared pairs of electrons.
          1. The unshared pair on nitrogen is in an sp2 orbital in the plane of the ring which does not overlap with the p orbital on the adjacent carbons.
          2. Only 6 electrons are delocalized (3 pairs) and the molecule is aromatic.
        4. Molecules and ions which satisfy all the criteria for aromaticity but have an even number of delocalized pairs (4n delocalized electrons) are exceptionally UNSTABLE and are antiaromatic. (All others are neither aromatic nor antiaromatic.)
    3. *The mechanism of electrophilic aromatic substitution on benzene
      1. Need to account for:
        a) Lack of reactivity of benzene
        b) Purpose of catalysts
        c) Substitution rather than addition
      2. *Reaction with Br2 or Cl2 - Halogenation
        1. From reactions with unsaturated compounds, we expect:
          1) Attack by Br+ (or Cl+) to form cation.
          2) Since a resonance stabilized cation can form, the cyclic bromonium ion will not form.
        2. The intermediate is what we expect - 
          (*Click to see resonance structures.)
          1.  
        3. Note that the intermediate is resonance stabilized but not aromatic.  Since benzene is aromatic, the activation energy is large and benzene tends not to react.
        4. *The purpose of the catalyst is to generate the electrophile.
          1.  
          2. Catalyst is acting as a Lewis acid - accepts a pair of electrons (similar to electrophile)
        5. *The cation loses H+ which combine with one of the bromine atoms attached to the iron atom forming HBr and completing the substitution.
        6. *Substitution occurs because the product is aromatic.  Addition would yield cyclohexadiene which is less stable than benzene.
        7. Energy diagram for reaction
      3. *Summary of the mechanism - Same for other reactions
        1. Generation of the electrophile, usually involving the catalyst - this is the only step that will be different for different reactions.
        2. Attack of the electrophile to form a cation (resonance stabilized)
        3. Loss of H+ to regenerate the aromatic ring
      4. *Reaction with alkyl and acyl halides - Friedel-Crafts alkylation and acylation
        1. *Alkyl halides often yield rearranged products. (Suggests carbocation intermediate.)
        2. PROPOSE MECHANISM FOR ALKYLATION
        3. *Catalyst acts as Lewis acid and generates a carbocation.
        4. *Carbocation acts as electrophile and attaches to benzene forming resonance stabilized ion.
        5. *Loss of H+ to form HX and regenerate aromatic ring 
          1. Intermediate from attack of acylium ion reacts in same way.
        6. How does rearranged product form?
      5. *Reaction with nitric acid - Nitration
        1. *Reaction of nitric acid with sulfuric acid followed by loss of water generates nitronium ion (electrophile)
          1. Note - 2 steps to generate electrophile.
          2. Also, nitric acid is acting as a base in the first step.
        2. *Nitronium ion attaches to benzene to form cation.
        3. Cation loses H+ to HSO4- to complete the substitution and regenerate the catalyst.
      6. *Reaction with sulfuric acid -Sulfonation
        1. *Basically the same mechanism as nitration except sulfuric acid is the base.
          1. Electrophile can also be generated by reaction of catalyst (SO3) with sulfuric acid:
        2. *Electrophile attacks benzene ring and cation loses H+ as before.
    4. *Reactions of substituted benzenes - Directive and Activation Effects
      1. *Observations
        1. Chart of nitration of benzene derivatives
        2. *Activating vs. Deactivating Groups - Rates relative to same reaction with benzene
          1. Some substituted benzenes react faster than benzene (substituent is an activating group, ring is an activated ring)
          2. Some react slower than benzene (substituent is a deactivating group, ring is a deactivated ring)
        3. *Ortho-para vs. Meta Directing Groups - Relative rates of formation of isomers
          1. Some substituted benzenes undergo substitution mainly at the ortho or para position (substituent is an ortho-para director)
          2. Some substituted benzenes undergo substitution mainly at the meta position (substituent is a meta director)
        4. Activating groups always are ortho-para directors.
        5. Deactivating groups are generally meta directors.
        6. Halogens are exceptions: They are deactivators but they are ortho-para directors.
        7. Note: Directive effects listed above depend only on the group already present in the ring.  The results are the same for all electrophiles.
      2. *Example of approach - Nitration of toluene
        1. It is observed that:
          1) Nitration of toluene goes faster than nitration of benzene - Methyl group activates the ring.
          2) Nitration of toluene yields mainly o- and p-toluene
        2. *Explanation - Surprise!!! - Mechanism of the reaction
          1. Energy Diagram
          2. *Products are mainly o- and p-nitrotoluene.
            1. Intermediate from either ortho or para attack is more stable than from attack at the meta position.
            2. Reaction at ortho or para positions occurs faster than at meta leading to mainly ortho and para products.
            3. The methyl group is an ortho-para director.
          3. *Toluene reacts faster than benzene (in same reaction).
            1. Both intermediates from toluene are more stable than the intermediate from benzene.
            2. The methyl group is an activating group (activator).
      3. *Nitration of trimethylanilinium ion
        1. *Reaction goes slower than nitration of benzene.
          1. Because of charge in substituent, the positive charge in the ring in the intermediate is less stable than the positive charge in the intermediate from benzene.
          2. The trimethylanilinium is a deactivating group.
        2. *Product is mainly meta
          1. Intermediates from ortho or para attack are destabilized more than meta because positive charge in ring is partially on carbon with positively charged group.

            *Click to see resonance structures
          2. Meta intermediate avoids having charge on same atom as substituent.

            *Click to see resonance structures.
          3. Trimethylanilinium group is a meta director.
        3. Energy diagram 
      4. *Reactions of methoxybenzene
        1. Observe - o/p director, activator (E+ is electrophile)
        2. *Important intermediate is one from para attack (ortho is similar) which is stabilized by resonance, but destabilized by inductive effect of oxygen atom.
          1. Electron withdrawing effect of oxygen destabilizes positive charge.
          2. Extra resonance structure stabilizes ion. Structure on right is not possible in intermediate from meta attack or from attack on benzene.  (Is possible with ortho attack.)
        3. Methoxy group is o/p director because of resonance in o/p intermediate that is not possible in meta (all have inductive effect).
        4. Methoxy group is activator because resonance stabilization is more important than inductive destabilization.
        5. Energy diagram
      5. *Directive and activation effects of halogens
        1. Observe - o/p director, but deactivator (E+ is electrophile)
        2. *Important intermediate is one from para attack (ortho is similar) which is stabilized by resonance, but destabilized by inductive effect of halogen atom.
          1. Electron withdrawing effect of halogen destabilizes positive charge.
          2. Extra resonance structure stabilizes ion. Structure on right is not possible in intermediate from meta attack or from attack on benzene.  (Is possible with ortho attack.)
        3. Halogen atom is o/p director because of resonance in o/p intermediate that is not possible in meta (all have inductive effect).
        4. Halogen atom is a deactivator because resonance stabilization is less important than inductive destabilization.
        5. Energy diagram
  6. *Reactions of Carbonyl Compounds
    1. *Many biologically important reactions occur a carbonyl groups
      1. *Ring forms of sugars (from Organic I)
      2. *Glycolysis
        1. Bisphosphoglycerate from glyceraldehyde-3-phosphate
        2. 3-Phosphoglycerate from bisphospoglycerate
    2. Reactions above are examples of a large category of reactions in organic chemistry that includes addition across the double bond and substitution at the carbonyl carbon as well as others.
      1. Hydrolysis of amides (this is one of the reactions that occurs when proteins are digested)
    3. *Mechanisms of addition to carbonyl groups
      1. Most reactions at carbonyl groups either are addition to the double bond or begin with addition followed by other reactions, such as substitution.
      2. *Addition of HCN to aldehydes and ketones as example (From Organic I)
        1. General reaction
        2. Reaction is catalyzed by cyanide ion (CN-), but can occur without catalyst.
        3. *Mechanism in absence of catalyst is essentially the same as the addition of  acids across a carbon-carbon double bond - attack by H+ followed by attachment of the nucleophile.
          1. The hydrogen attaches to the oxygen atom because the resulting ion is resonance stabilized.
          2. If the H attached to carbon the ion would not have resonance, would have a positive charge on oxygen and the oxygen would not satisfy the octet rule: 
          3. Note: Protonation followed by attack by nucleophile
        4. *In the catalyzed mechanism, the nucleophile attaches in the first step and HCN protonates the resulting ion in the second step.
          1. Cyanide attaches to carbon because that puts the negative charge on oxygen.
          2. Note: Nucleophiles generally don't attach first in addition to C=C because that would put negative charge on carbon.
  7.  
    1. Aromaticity - heat of hydrogenation
    2. *Distinguish among Conformers, Other Isomers and Identical Molecules.
      Click to expand outline and review.
      1. Conformers interconvert by rotation around a single bond - See Example
      2. Identical molecules differ only in their orientation in space
      3. All others require bonds to be broken to interconvert and are isomers other than conformers.
      4. Do exercise: Practice Same vs. Different 
    3. *Determine when nonidentical molecules are the same compound (and, therefore, have the same name) vs. different compounds (different names).
      Click to expand outline and review.
      1. Molecules of different compounds can be separated.  Molecules that can not be separated are the same compound (and have the same name) even if they are not identical.
      2. Molecules which interconvert at room temperature can not be separated. Conformers are constantly interconverting, so they are the same compound and have the same name.
    4. *Alcohols, ethers and amines
      1. *Alcohols (IUPAC)
        1. "-ol" ending, takes precedence over carbon-carbon double or triple bonds
        2. Give the IUPAC name for the following compound:
          *
          Click to see answer
          1. 3-ethyl-4-methyl-3-cyclohexenol
        3. "Hydroxy" when not highest priority functional group.
        4. Give the IUPAC name for lactic acid:
          *
          Click to see answer
          1. 2-hydroxypropanoic acid
      2. *Alcohols (common names)
        1. Name of alkyl group followed by "alcohol," e.g., ethyl alcohol
        2. Alcohols classified as 1o, 2o, 3o based on type of C to which OH is attached.
        3. Give the common name for *Click to see answer
          1. sec-butyl alcohol
      3. *Ethers
        1. Previous naming system was like common naming strategy, but is included in IUPAC
        2. Name the following compound
          *
          Click to see answer
          1. propyl phenyl ether
        3. "Alkoxy" when not highest priority (ethoxy, for example), but can also be used if one of the groups is very complex and the other is quite simple;
          e.g., 4-ethoxy-6,6,8,9-tetramethyltridecane.
        4. Name the following compound
          *
          Click to see answer
          1. 4-propoxyoctanoic acid
      4. *Amines
        1. Name alkyl groups attached to N followed by "amine" - like common name, but part of IUPAC
        2. Name the following compound:
          *
          Click to see answer
          1. dimethyl cyclopentyl amine
        3. "Amino" or "alkyl amino" when not highest priority (dimethylamino, for example)
        4. Name the following compound:
          *
          Click to see answer
          1. 4-(N,N-diethylamino)octanoic acid
        5. *New IUPAC - alkanamine - note: one word
          1. Amine group takes precedence over carbon-carbon multiple bonds
          2. Parent compound is N atom and longest chain or largest ring attached to it.
          3. Parent name is name of longest chain or largest ring followed by
             "-amine" (all one word) with number to indicate position of N
          4. Substituents attached to nitrogen indicated by "N" instead of a number.
          5. Example - Draw N-ethyl-4-methyl-2-heptanamine  *Click to see answer
            1. Parent compound in box: 2-heptanamine
          6. Name the compound below as many ways as you can: *Click to see answers
            1. 1-ethyl-4-methylhexyl diisopropyl amine
            2. 1-ethyl-4-methylhexyl di(1-methylethyl) amine
            3. N,N-diisopropyl-6-methyl-3-octanamine
            4. N,N-di(1-methylethyl)-6-methyl-3-octanamine
        6. Amines are classified as 1o, 2o and 3o based on number of alkyl groups (carbons) attached to N (Note difference from alcohols, for example.)
    5. *Carboxylic acid
      1. -OH attached to carbonyl group
      2. -oic acid ending, no number - has to be #1
      3. Draw 3-methylpentanoic acid *Click to see structure.

        1.  
      4. Name *Click to see name.
        1. 5-Bromohexanoic acid
    6. -ol ending, number to indicate position of OH
    7. Draw 3-methyl-2-pentanol *Click to see structure.

      1.  
    8. Name *Click to see name.
      1. 5-bromo-3-heptanol
    9. Chiral objects - different from mirror image
    10. Chiral substances rotate plane polarized light (optically active)
    11. Planes of symmetry -  if present, molecule is not chiral
                     
    12. Meso compounds - molecules that are not chiral although they do contain asymmetric centers - *Click to see example
      1.   Fischer Projection
      2. Note plane of symmetry
    13. Isomers *Click to see definition
      1. Different molecules with same molecular formula
    14. Stereoisomers *Click to see definition
      1. Isomers with the same bonding pattern
    15. Constitutional (Structural) Isomers *Click to see definition
      1. Isomers with different bonding patterns
    16. Cis-trans isomers  *Click to see definition
      1. in rings - same side vs. opposite sides of ring - Example
      2. in alkenes - longest chain on same side vs. opposite side of double bond- *Click to see example
        1. is trans
    17. Isomers that contain asymmetric centers
    18. *Molecular orbital theory - new orbitals encompassing more than one atom formed by mixing orbitals together
      1. *H2 - Bonding molecular orbital 
                 Both molecular orbitals
      2. *F2 - Molecular orbitals