Chapter 8 handouts

Chapter 8 handouts - Chapter 8 Reactions of Alkenes...

Info iconThis preview shows pages 1–18. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 8 Reactions of Alkenes
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 2 Reactivity of C=C • Electrons in _______ are loosely held. • ____________ are attracted to the pi electrons. • ____________ intermediate forms. • ____________ adds to the carbocation. • Net result is addition to the double bond.
Background image of page 2
Chapter 8 3 Electrophilic Addition • Step 1: Pi electrons attack the electrophile. • Step 2: Nucleophile attacks the carbocation.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 4 Types of Additions
Background image of page 4
Chapter 8 5 Addition of HX (1) Protonation of double bond yields the most stable __________. Positive charge goes to the carbon that was not protonated.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 6 Addition of HX (2)
Background image of page 6
Chapter 8 7 Regiospecificity • Markovnikov’s Rule: The proton of an acid adds to the carbon in the double bond that ___________________. “Rich get richer.” • More general Markovnikov’s Rule: In an electrophilic addition to an alkene, the ____________ adds in such a way as to form the ____________ intermediate. • HCl, HBr, and HI add to alkenes to form Markovnikov products.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 8 Free-Radical Addition of HBr • In the presence of ________, HBr adds to an alkene to form the “anti- Markovnikov” product. • Only HBr has the right ___________. • ______ bond is too strong. • HI bond always tends to break ______________ to form ions.
Background image of page 8
Chapter 8 9 Free Radical Initiation • Peroxide O-O bond breaks easily to form free radicals. • Hydrogen is abstracted from HBr.
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 10 Propagation Steps • Bromine adds to the double bond. • Hydrogen is abstracted from HBr.
Background image of page 10
Chapter 8 11 Anti-Markovnikov ?? • Tertiary radical is more stable, so that intermediate is formed faster. • (Karasch – peroxide effect, 1933.) CH 3 C 3 CH CH 3 Br + 3 C 3 3 3 C 3 3 X
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 12 Question • So, here’s a question. • Why don’t we see allylic bromination (i.e. Br removing allylic H to generate allyl radical) in this case?
Background image of page 12
Chapter 8 13 Hydration of Alkenes • Reverse of __________ of alcohol • Use very dilute solutions of H 2 SO 4 or H 3 PO 4 to drive ___________ toward hydration (have excess of H 2 O). CC + H 2 O H + C H C OH alkene alcohol
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 14 Mechanism for Hydration + C H C + H 2 O C H C OH H + + H 2 O C H C H + C H C O H H 3 O + + CC O HH H + + + H 2 O C H C +
Background image of page 14
Chapter 8 15 Orientation for Hydration • Markovnikov product is formed. + CH 3 C 3 CH CH 3 O HH H + + H 2 O + H 3 3 C 3 H 2 O 3 C 3 3 H O H H + H 2 O 3 C 3 3 H O H
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 8 16 Indirect Hydration Two methods: • __________________________ ¾ Markovnikov product formed ¾ Anti addition of H-OH ¾ No rearrangements • _______________ ¾ Anti-Markovnikov product formed ¾ Syn addition of H-OH
Background image of page 16
Chapter 8 17 Oxymercuration (1) • Reagent is mercury(II) acetate which dissociates slightly to form + Hg(OAc).
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 18
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/08/2008 for the course CHEM 223 taught by Professor Pearson during the Fall '07 term at Case Western.

Page1 / 66

Chapter 8 handouts - Chapter 8 Reactions of Alkenes...

This preview shows document pages 1 - 18. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online