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Organic Chemistry Week 7 Notes

by: Jernelle John

Organic Chemistry Week 7 Notes CHEM 231

Marketplace > University of Maryland > Chemistry > CHEM 231 > Organic Chemistry Week 7 Notes
Jernelle John
GPA 3.5

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These are notes from week 7 of Organic Chemistry 231.
Organic Chemistry 1
Dr. Monique Koppel
Class Notes
Organic Chemistry, Organic Chem, orgo, organicchemistry231
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This 6 page Class Notes was uploaded by Jernelle John on Sunday October 16, 2016. The Class Notes belongs to CHEM 231 at University of Maryland taught by Dr. Monique Koppel in Fall 2016. Since its upload, it has received 4 views. For similar materials see Organic Chemistry 1 in Chemistry at University of Maryland.


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Date Created: 10/16/16
Saturday, October 15, 2016 Organic Chemistry Subject - Free Radical Halogenation homolytic cleavage- breakage of covalent bond to produce to new radical • compounds where each product gets one of the shared electrons • we use halogens (Br and Cl) • Mechanism (initiation, propagation, termination) - initiation- breaking the weakest bond to form two radical atoms using heat or light (hv) - propagation- using one of the radical atoms formed in step one and an alkane to form an alkyl radical • the alkyl radical will than react with another Cl-Cl (or Br-Br) bond to form another Cl radical - propagation chain reaction- the Cl radical can than react with another alkane and so on and so forth - termination- reacting two radicals together to get covalent bonds and destroy all radicals • can react a halogen (Cl or Br) radical and alkyl radical • can react a halogen radical with another halogen radical • can react two methyl radicals together - when breaking a bond with more than two carbons, the + charge should always be placed on the more stable position 1 Saturday, October 15, 2016 - tertiary positions of an alkyl radical are more stable than secondary positions which are more stable than primary positions due to the inductive effect • treat radical carbons as sp2 - Hyperconjugation- donation through space of parallel bonds - Bond dissociation energy (BDE)- energy it takes to break a bond • energy gained as a bond breaks and energy loss as a bond forms • exothermic- negative • endothermic- positive - Chlorine vs. Bromine Radical Halogenation bond dissociation energy of a Br-Br bond is 192kJ • bond dissociation energy of a Cl-Cl bond is 247kJ • • Cl is much more exothermic than Br due to bond dissociation energy BDE: 104 BDE:103 delta H1 (change in enthalpy)=104-103 = +1 kcal/mol Cl-Cl BDE: 84 BDE: 58 delta H2 (change in enthalpy)= 58-84= -26 kcal/mol delta H = +T- 26 = -25 BDE: 104 BDE: 87 delta H1= 104-87 = +17 2 Saturday, October 15, 2016 BDE; 46 BDE: 70 delta H2 = 46-70 = -24 delta H = +T7 - 24 = -7 - Hammond’s Postulate- species that are close in structure are close in energy • Br is closer to the transition state - Br will always go to the tertiary position, but Cl will not • Cl is closer to the starting material - Alkenes cis/ trans isomerism cis is less stable than trans due to steric strain • trans-2-butene cis-2-butene - Alkenes E/Z naming system • E/Z naming is used in place of cis/trans naming when there is more than two substituents surrounding the double bond - E: substituents with the highest priority are on different sides of the double bond - Z: substituents with the highest priority are on the same side of the double bond - Stabilize a molecule • tertiary state • resonance 3 Saturday, October 15, 2016 • hyperconjugation • inductive effect - allylic compounds have lower BDE than tertiary positions • they are more stable due to resonance - the more negative delta H is, the more stable the compound - Addition of HBr anti-maokovnikov addiction Anti-Markovnikov addition for HBr refers to the addition of Br to the less substituted • (less stable) position using peroxides (ROOR) - HCl and HI cannot participate in anti-Markovnikov addition, only HBr • mechanism - initiation- breakage of the weakest bond using light or heat to form two radicals • the O-O bond of the peroxide bond is the weakest one of the alkyl radicals from peroxide than reacts with HBr to form a radical • bromine - propagation • the radical bromine than reacts with an alkene to form a radical on the more stable position 4 Saturday, October 15, 2016 • the new radical compound than reacts with another HBr molecule - giving the final product with Br on the less substituted/ less stable position and H on the more substituted position - termination- destruction of radicals by forming covalent bonds - Allylic Carbon- carbon next to carbon=carbon double bond - Rearrangement- the movement of of a single bond of a compound to another position to form a more stable positive charge rearranges to - hydration of alkene - the H+ (from a compound like H O+, H SO -, HNO +) bonds to the compound getting 3 2 4 3 rid of the double bond - H2O is than added to the compound to get rid of the positive charge 5 Saturday, October 15, 2016 6


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