OrganicConjSystems

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    on ugate ystems,,

    Ultraviolet S ectrosco

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    Wave Properties of Electrons

    Standing wave vibrates in fixed location. Wave function, , is a mathematical description of

    s ze, s ape, an or en a on. Amplitude may be positive or negative. Nod : Am litude is zero.

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    Wave Interactions

    Linear combination of atomic orbitals:

    between different atoms is bondformation.

    on the same atom is hybridization.

    n rv i n f r i l

    Waves that are in phase add together.

    . Waves that are out of phase cancel out.

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    Sigma Bonding

    A bond may be formed by s-s, p-p, s-p, or

    The bonding molecular orbital (MO) is

    orbitals.

    than the atomic orbitals.

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    Bonding MO

    Formation of a -bonding MO: When the 1sorbitals of two hydrogenatoms overlap in phase with each other, they interact constructively to

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    orm a on ng .

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    Antibonding MO

    Formation of a * antibondin MO: When two 1sorbitals overla out of

    Chapter 2 6

    phase, they interact destructively to form an antibonding MO.

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    H2: s-sOverlap

    Chapter 2 7

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    Cl2: p-pOverlap

    When two porbitals overlap along the line betweenthe nuclei, a bonding orbital and an antibondingorbital result.

    Most of the electron densit is centered alon the linebetween the nuclei.

    This linear overlap is another type of sigma bonding

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    .

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    Carbon: sp3 hybridization

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    Methane

    109.5o

    Each of the four C-H bonds results from head-on s overlaof a singly occupied carbon sp3 hybrid orbital with a singlyoccupied hydrogen 1s orbital. Sigma bonds are formed by

    - -

    singly occupied sp3

    hybrid orbital of carbon.

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    Carbon: sp2 hybridization

    - three hybrid sp2 orbitals that lie on the same plane at 120o

    - a third p orbital that is lying perpendicular to the plane

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    Carbon: sp hybridization

    The carbon atoms forms: two hybrid sp orbitals and two porbitals.

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    Conjugated double bondsare separated by one single

    bond. Example: 1,3-pentadiene.

    Isolated double bonds are

    separated by two or moresin le bonds. Exam le: 1 4-pentadiene.

    Cumulated double bonds areC HH

    on adjacent carbons.Example: 1,2-pentadiene. H CH2CH3

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    Stabilities of Dienes

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    Structure of 1,3-Butadiene

    Most stable conformation is planar. Single bond is shorter than 1.54 . Electrons are delocalized over molecule.

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    Constructing Molecular Orbitals

    mo ecu ar or ta s are t e s eways over ap o porbitals.

    porbitals have 2 lobes. Plus (+) and minus (-),

    not electrical charge.

    When lobes overlap constructively, (+ and +, or -

    - .

    - ,

    and a node forms; antibonding MO.

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    Ethylene MOs

    The combination oftwo porbitals mustg ve wo mo ecu arorbitals.

    is a bonding MO.

    is an anti-bondingM .

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    1 MO for 1,3-Butadiene

    Lowest energy. All bonding

    interactions. Electrons aredelocalized over

    .

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    2 MO for 1,3-Butadiene

    2 bondinginteractions. 1 antibonding

    interaction. A bonding MO.

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    * MO for 1 3-Butadiene

    Antibonding MO. Empty at ground

    state. Two nodes.

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    4* MO for 1,3-Butadiene

    All antibonding.

    Highest energy.

    state.

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    MO Energy Diagram

    The averageenergy of

    electrons islower in theconjugated

    compound.

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    Conformations of 1,3-Butadiene

    s-trans conformer is more stable than the s-. .

    .

    H

    HH

    HH

    HH

    H- H

    H H

    -

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    Allylic Cations

    Carbon adjacent to C=C is allylic. .

    Stability of 1 allylic 2 carbocation. a y o a y c car oca on.

    H+

    H+

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    1,2- and 1,4-Addition to Conjugated

    enes

    u

    produces the most stable intermediate.

    For conjugated dienes, the intermediate is a.

    ,of which have the delocalized positive charge.

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    Addition of HBr

    _Br Br

    _

    H H H H

    H3C C C CH2

    Br

    H3C C C CH2

    Br

    1,2-addition product 1,4-addition product

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    Kinetic vs. Thermodynamic Control

    Major product

    at 40C

    a or pro uctat -80C

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    Allylic Radicals

    Stabilized by resonance.

    Radical stabilities: 1 < 2 < 3 < 1 allylic.

    Substitution at the allylic position competes witha on o ou e on .

    ,of reagent with light, heat, or peroxides to initiatefree radical formation.

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    Bromination Using NBS

    N-Bromosuccinimide (NBS) provides a low,2.

    -produce Br2 and prevent HBr addition.

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    MOs for the Allylic System

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    SN2 Reactions of Allylic Halides

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    Diels-Alder Reaction

    Otto Diels, Kurt Alder; Nobel prize, 1950

    Produces cyclohexene ring

    Diene + alkene or alkyne with electron-withdrawing group (dienophile)

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    Examples of Diels-Alder Reactions

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    Stereochemical Requirements

    .

    Dienes C1 and C4 orbitals must overlawith dienophiles porbitals to form newsigma bonds.

    Both sigma bonds are on same face of thediene: synstereochemistry.

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    s-cis Conformation of the Diene

    When the diene is in s-trans conformation, the endorbitals are too far a art to overla with the

    orbitals of the dienophile

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    Cyclopentadiene: Diels Alder Reaction

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    Cyclopentadiene: Diels-Alder Reaction

    Cyclopentadiene

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    Concerted Mechanism

    4 + 2 cycloaddition concerted cyclic movement of 6 electrons

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    the 2 end p orbitals of the diene with those of the dienophile

    syn Stereochemistry

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    synStereochemistry

    -diene and dienophile

    The dienophile adds to one face of the diene, and the diene

    adds to one face of the dienophile

    Substituents that are on the same side of the diene ordienophile will be cis in the newly formed ring

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    E d R l

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    Endo Rule

    -on the dienophile have a secondary overlap with.

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    R i ifi it

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    Regiospecificity

    reaction will have electron-donating and electron-withdrawin rou s1 2 or 1 4 but not 1 3.

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    Pericyclic Reactions

    Diels-Alder reaction is example.

    Woodward and Hoffmann predicted reaction

    orbital symmetry.

    MOs must overlap constructively to stabilizethe transition state.

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    Symmetry-Allowed Reaction

    Diene contributeselectrons from itshighest energyoccup e or a(HOMO).

    Dienophile receives

    lowest energyunoccu ied orbital

    43(LUMO).

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    Forbidden Cycloaddition

    [2 + 2] cycloaddition

    form cyclobutene has

    -of HOMO and LUMO.

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    Photochemical Induction

    Absorption of correct energy photon willpromote an electron to an energy level thatwas previously unoccupied.

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    [2 + 2] Cycloaddition

    allowed, but

    thermall forbidden.

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    Ul i l S

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    Ultraviolet Spectroscopy

    200-400 nm photons excite electrons from a * .

    in energy.

    A compound that has a longer chain ofconjugated double bonds absorbs light at a

    longer wavelength.

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    * Transition for Ethylene and Butadiene

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    Transition for Ethylene and Butadiene

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    Obtaining a UV Spectrum

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    Obtaining a UV Spectrum

    of a reference beam through solvent only

    r

    solution of the sample (Is).

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    sor ance s t e og o t e rat o r s

    Th UV S t

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    The UV Spectrum

    Usually shows broad peaks.

    Read max from the graph.

    Absorbance, A, follows Beers Law:=

    ,sample concentration in moles per liter, and li h l n h f h li h h in n im r .

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    Sample UV Absorptions

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    Sample UV Absorptions

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    Woodward-Fieser Rules

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    Woodward Fieser Rules

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    Problem on

    Diels-Alder Cycloaddition

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    SOLVED PROBLEM 15-1

    Use the endo rule to predict the product of the following cycloaddition

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    Use the endo rule to predict the product of the following cycloaddition.

    Solution

    Imagine this diene to be a substituted cyclopentadiene; the endo product will be formed.

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    SOLVED PROBLEM 15-1 (continued)

    Solution (continued)In the imaginary reaction, we replaced the two inside hydrogens with the rest of the cyclopentadiene ring.

    Now we put them back and have the actual product.

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    SOLVED PROBLEM 15-2

    Predict the products of the following proposed DielsAlder reactions.

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    ed ct t e p oducts o t e o ow g p oposed e s de eact o s.

    o ut on

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    Solution (continued)

    SOLVED PROBLEM 15-2 (continued)

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    Solution (continued)

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    Provide the structure of the major organic product

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    Provide the structure of the major organic product

    n e o ow ng reac on.

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