Ch. 8 Alkene Reactions & Synthesis

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27 Terms

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Alkene addition reaction types

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  1. Halogen —> 1,2-dihalide

  2. Hypohalous acid (HOX) —> halohydrin

  3. H2O —> alkane

  4. 1 oxygen —> epoxide

  5. 2 hydroxyl group —> 1,2-diol

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If the electrophile is a small atom (like H)

a carbocation intermediate will form

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If the electrophile is a larger atom (like a halogen)

a 3-membered ring intermediate is formed

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Four ways to prepare alkenes via elimination reactions

dehydrohalogenation, dehydration, dehydrogenation, dehalogenation

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The preparation of alkenes is dominated by

elimination reactions (Ex. the loss of HBr from an alkyl halide to form an alkene, or the loss of H2O from an alcohol to form an alkene)

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Dehydrohalogenation

“Lose-hydrogen-& halogen”

Removes hydrogen and adjacent halogen to form alkene using 1, 2, or 3 alkyl halides as a reactant and a strong base or strong bulky base

<p>“Lose-hydrogen-&amp; halogen”</p><p>Removes hydrogen and adjacent halogen to form alkene using 1, 2, or 3 alkyl halides as a reactant and a strong base or strong bulky base</p>
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Dehydration

“Lose-water”

Loss of water from an alcohol to form an alkene using a strong acid (H2SO4), tertiary reactants best and carbocation intermediate formed so REARRANGEMENTS POSSIBLE!!!

<p>“Lose-water”</p><p>Loss of water from an alcohol to form an alkene  using a strong acid (H2SO4), tertiary reactants best and carbocation intermediate formed so REARRANGEMENTS POSSIBLE!!!</p>
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Dehydrogenation

“Lose-hydrogen”

Removal of H2 from alkane using heat and a catalyst (Pd or Pt) to form alkene, results in multiple products

<p>“Lose-hydrogen”</p><p>Removal of H2 from alkane using heat and a catalyst (Pd or Pt) to form alkene, results in multiple products</p>
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Dehalogenation of Vicinal Dibromides

“Lose halogen”

Both bromides removed by KI or ZN in CH3COOH (acetic acid), halogens are ANTI to each other…stereochemistry of reactants = stereochemistry of products (if reactant cis then product cis)

<p>“Lose halogen”</p><p>Both bromides removed by KI or ZN in CH3COOH (acetic acid), halogens are ANTI to each other…stereochemistry of reactants = stereochemistry of products (if reactant cis then product cis)</p>
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Halogenation of Alkenes

“Add halogen”

Addition of HX (Br or Cl) to form a 1,2-dihalide. Bromonium or chloronium ion intermediate forms. Anti-stereochemistry reaction: halogens attack from opposite faces (“backside attack”). No regiochemistry/Markovnikov product b/c adding same atom (H2)

<p>“Add halogen”</p><p>Addition of HX (Br or Cl) to form a 1,2-dihalide. Bromonium or chloronium ion intermediate forms. Anti-stereochemistry reaction: halogens attack from opposite faces (“backside attack”). No regiochemistry/Markovnikov product b/c adding same atom (H2)</p>
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Halohydrin

“Add halogen and hydroxide”

Addition of HO-X to alkene using halogen (Br or Cl) and water to form a Markovnikov halohydrin. Bromonium or chloronium ion intermediate forms. Bromohydrin in DMSO solvent and using NBS.

<p>“Add halogen and hydroxide”</p><p>Addition of HO-X to alkene using halogen (Br or Cl) and water to form a Markovnikov halohydrin. Bromonium or chloronium ion intermediate forms. Bromohydrin in DMSO solvent and using NBS.</p>
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Acid-Catalyzed Hydration of Alkenes 

Addition of H and OH to an alkene using water and a strong acid (H2SO4 or H3PO4) to form Markovnikov alcohol, can also just be H3O+. Carbocation intermediates possible.

<p>Addition of H and OH to an alkene using water and a strong acid (H2SO4 or H3PO4) to form Markovnikov alcohol, can also just be H3O+. Carbocation intermediates possible.</p>
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Oxymercuration-Demercuration Hydration of Alkenes

Addition of H and OH to alkene using mercuric acetate [Hg(OAc)2] with water (1st step) and sodium borohydride (NaBH4) (2nd step) to form Markovnikov alcohol. Forms mercurinium ion, but NOT carbocation intermediate!!

<p>Addition of H and OH to alkene using mercuric acetate [Hg(OAc)2] with water (1st step) and sodium borohydride (NaBH4) (2nd step) to form Markovnikov alcohol.&nbsp;Forms mercurinium ion, but NOT carbocation intermediate!!</p>
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Hydroboration-Oxidation Hydration of Alkenes

“Adds hydrogen and boron”

ANTI-MARKOVNIKOV addition of H and OH to alkene using boran (BH3) and THF (1st step) and peroxide (H2O2) under basic conditions (NaOH) (2nd step). Transition state and triborane intermediate formed. 

<p>“Adds hydrogen and boron”</p><p>ANTI-MARKOVNIKOV addition of H and OH to alkene using&nbsp;boran (BH3) and THF (1st step) and peroxide (H2O2) under basic conditions (NaOH) (2nd step). Transition state and triborane intermediate formed.&nbsp;</p>
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Catalytic-Hydrogenation Reduction of Alkenes

Forms alkane from alkene using H2 & a catalyst (Pd, Pt, Ni). No regioselectivity b/c H2 = no markovnikov product. Stereoselectivity b/c syn addition of H2.

<p>Forms alkane from alkene using H2 &amp; a catalyst (Pd, Pt, Ni). No regioselectivity b/c H2 = no markovnikov product. Stereoselectivity b/c syn addition of H2.</p>
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Epoxidation Oxidation of Alkenes

Forms epoxide (cyclic ether R—O—R in 3-membered ring) from oxidizing alkene with peroxyacid (RCO3H…Ex. mCPBA). Oxygen adds with syn-stereochemistry

<p>Forms epoxide (cyclic ether R—O—R in 3-membered ring) from oxidizing alkene with peroxyacid (RCO3H…Ex. mCPBA). Oxygen adds with syn-stereochemistry</p>
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Hydroxylation Oxidation of Alkenes

Forms 1,2-diol by using water (hydrolysis) and catalytic acid (H3O) on epoxide. Two OH groups added with anti-stereochemistry b/c being added to ring.

<p>Forms 1,2-diol by using water (hydrolysis) and catalytic acid (H3O) on epoxide. Two OH groups added with anti-stereochemistry b/c being added to ring.</p>
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Hydroxylation using OsO4 Oxidation of Alkenes

Directly forms 1,2-diol without an epoxide intermediate by treating alkene with osmium tetraoxide (OsO4). Two OH groups added with syn-stereochemistry. 

<p>Directly forms 1,2-diol without an epoxide intermediate by treating alkene with osmium tetraoxide (OsO4). Two OH groups added with syn-stereochemistry.&nbsp;</p>
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ketone

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carboxylic acid

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aldehyde

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Cleavage to Carbonyl Compounds Oxidation of Alkenes

Forms 2 carbonyl containing fragments from alkene C=C bonds cleaved by a strong oxidizing agent: ozone (O3), potassium permanganate (KMnO4)…can be mild, and periodic acid (HIO4)…ozonolysis. Oxygen double-bonds to each cleaved carbon. 

tetrasubstituted = 2 ketones

trisubstituted = 1 ketone, 1 aldehyde

disubstituted = 2 aldehydes

Under oxidizing conditions aldehydes oxidize further into carboxylic acids.

<p>Forms 2 carbonyl containing fragments from alkene C=C bonds cleaved by a strong oxidizing agent: ozone (O3), potassium permanganate (KMnO4)…can be mild, and periodic acid (HIO4)…ozonolysis. Oxygen double-bonds to each cleaved carbon.&nbsp;</p><p>tetrasubstituted = 2 ketones</p><p>trisubstituted = 1 ketone, 1 aldehyde</p><p>disubstituted = 2 aldehydes</p><p>Under oxidizing conditions aldehydes oxidize further into carboxylic acids.</p>
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Mild Oxidation of Alkenes

KMnO4, NaOH (aq), cold = CIS 1,2-diol

<p>KMnO4, NaOH (aq), cold = CIS 1,2-diol</p>
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Oxidative Cleavage of Alkenes

KMnO4, H2SO4 (aq), heat = ketones, carboxylic acids, and sometimes CO2

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Carbenes

neutral molecule thats only ever a reaction intermediate, electron deficient, highly reactive, behaves as electrophile when reacting with alkenes, stereospecific

Dichlorocarbene formed by reaction of chloroform (CHCl3) with strong base (Ex. KOH). When dichlorocarbene reacts with alkene, dichlorocyclopropane forms. 

Simmons-Smith Rxn uses a carbenoid instead.

<p>neutral molecule thats only ever a reaction intermediate, electron deficient, highly reactive, behaves as electrophile when reacting with alkenes, stereospecific</p><p>Dichlorocarbene formed by reaction of chloroform (CHCl3) with strong base (Ex. KOH). When dichlorocarbene reacts with alkene, dichlorocyclopropane forms.&nbsp;</p><p>Simmons-Smith Rxn uses a carbenoid instead.</p>
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Addition of H2O to Achiral Alkene

R & S enantiomers result, cation has plane of symmetry, transition states are mirror images and equally likely to occur

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Addition of H2O to Chiral Alkene

# of chiral centers found by 2^n, if 2 chiral centers then 4 possible stereoisomers, unequal amounts of products