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CH 3
CH 3 CH 3
Br
CH 3
Br
H
CH 3
H
CH 3
H
Br
CH 3
CH 3 CH 3 CH 3
X
OH
CH 3 CH 3
BH 2
CH 3
OH
H
H BH 2
H
CH 3 CH 3 CH 3
Hg(OAc)
OR
CH 3
H
CH 3
H
OH
X+
CH 3 CH 3 CH 3
X
X
X+
Hg(OAc)
β-Mercury-
substituted
carbocation
H 2 SO 4 or H 3 PO 4 or NaHSO 4
Which upon
reaction
with:
Reagent Yields
Halonium Ion
"X+"
Carbocation
Halonium Ion
What adds?
HOOH/NaOH Hg(OAc)+ "X+"
To give the more
stable:
Radical
Substrate
(NaBH 4 reduction
replaces Hg(OAc) with
H)
Radical Hydrohalogenation: X 2 /H 2 O
Organoborane
Halogenation: Hydration: Halohydrin Formation: Hydroboration/Oxidation: Oxymercuration: Summary of Alkene Reactions Hydrohalogenation: HBr/peroxides
Carbocation
B 2 H 6 Br• Hg(OAc) 2 ROH X– H+^ Br– H+ HBr HBr X 2 H 2 O H 2 O
ORGANIC CHEMISTRY SYSTEMATIC NOMENCLATURE OF BICYCLIC COMPOUNDS Bicyclic compounds require the breaking of two carbon-carbon bonds to convert them to open chained compounds--containing no rings. Bicyclic compounds are named by prefixing bicyclo- to the name of the parent hydrocarbon. The name of the parent hydrocarbon is obtained by counting the total number of carbon atoms in all of the rings of the compound. Thus, because norbornane (below) contains seven carbon atoms in the two rings of the molecule it is a bicycloheptane. Carbon atoms shared by both rings are referred to as bridgehead carbons. 1 2 3 4 5 6 7 1 2 3 4 5 6 7 norbornane bicyclo[2.2.1]heptane The numbers of carbon atoms between bridgehead carbons in the molecule are specified by counting from the bridgehead carbon and listing each of the numbers in brackets in decreasing order prior to the name of the hydrocarbon. Thus, norbornane is bicyclo[2.2.1]heptane. The following are additional examples of bicycloheptanes: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 bicyclo[3.1.1]heptane bicyclo[3.2.0]heptane
H O O C H
The Curved Arrow Formalism
Curved arrows are used by chemists to indicate the flow of
electrons in reactions.*
For each electron pair (either a bonding pair or lone pair) that changes position in a reaction, one arrow is required.
The tail of the arrow starts at the initial position of the electron pair (at an atom for a lone pair; at a bond for a bonding pair).
The head of the arrow points to the new position of the electron pair (to an atom for formation of a lone pair; midway between atoms for formation of a bond.)
CURVED ARROWS DO NOT REPRESENT THE MOVEMENT OF NUCLEI!!!
H O CH 3 H H O H O CH 3
Ex. 1 H
H H H C H O O H
Ex. 2
Ex. 3
O H H C CH 3 H 3 C CH 3 O H H C CH 3 H 3 C CH 3 O H H N otohteer setlriecct (^) trboanlasn acned o af (^) tcohmaircg pe (^) oasnitdio tnhsa tr (^) eomnlayi (^) nre tdh (^) ee lseacmtroen).s aKreee mp ocavrinegfu (l atrllack of formal charges.
- nToht ecyo (^) nafrues ael sthoe s otwmoe utismaegse (^) su—seind (^) tteor (^) cinodnivceatrsei (^) othne o ifn rteesrcoonnavnecres iofonr (^) mosf r iess noonta an creea fcotrimons!.! (^) !Do
Guide to Displacement Reactions Reactivity Type of carbon (^) S N^1 (carbocation intermediate) E (carbocation intermediate) SN 2 (concerted) E (concerted) 1 ° X X
- good LG
- good Nu:
- polar aprotic solvent - strong, bulky bases 2 °
- good LG
- poor Nu:
- polar protic solvent
- competes with SN 1 when base is present
- good LG
- good Nu: (weaker base than HO–)
- polar aprotic solvent
- competes with SN 2
- favored by bulky bases stronger than HO– 3 °
- good LG
- any Nu:
- polar protic solvent
- competes with SN 1 when base is present X
- strong base Nucleophilicity Classification Nucleophiles Rel. React. Excellent I–, HS–, RS–^ >10^5 Good Br–, HO–, RO–, CN–, N 3 –^104 Fair (^) NH 3 , Cl–, F–, RCO 2 –^103 Weak H 2 O, ROH 1 Very Weak RCO 2 H 10 – Leaving Groups Classification Leaving Groups Excellent ROSO 2 – , H 2 O, ROH, N 2 Good (^) I–, Br–, Cl–, NR 3 , RCO 2 – Poor (^) HO–, RO–, F–, CN–, NH 2
CH 3 O CHH 3 CH 3 CH^3 Cl **Give all possible products, including stereochemistry! Cl 2 H 2 O
- Hg(OAc) 2
- NaBH 4 /OH–
- BH 3
- H 2 O 2 /OH–
- OsO 4
- H 2 O 2 /OH–**
+ enantiomer
CH 3 OH OH OH OH
+ enantiomer + enantiomer
OH Cl
+ enantiomer
OH H H OH
+ enantiomer
+ enantiomer
+ enantiomer
H OH
+ enantiomer
HO OH OH Cl
+ enantiomer
OH H H OH
+ enantiomer
+ enantiomer
+ enantiomer
H OH HO OH Cl OH HO Cl OH OH OH OH OH HO OH Reagent Substrate
Analysis of Isomerism Two molecules with the same molecular formula Molecules are Identical Molecules are Constitutional Isomers Mirror Images? Same Connectivity? Superimposable? YES YES Molecules are Diastereomers Molecules are Enantiomers YES NO NO NO
LEWIS STRUCTURES OFTEN DO NOT ADEQUATELY REPRESENT ACTUAL STRUCTURES:
Cl N
O
O
Remember...
I Emxppleierism uennetqaullya,l tbhoenyd ainreg othf (^) eN s taom twe ole onxgythg!ens! Linus Pauling (Feb. 28 , 1901 - Aug. 19 , 1994 ) introduced the concept of... R wehseorena thnec a e: r (^) ra W n h g e e n m a e m nt o o le f c a u t l o e m ca s n i (^) s b e th r e e p sa re m se e n , t b e u d (^) t b e y l (^) e (^2) c o tr r o m ns or a e (^) r L e e s w h is if s te tr d uctures R heesadonedan).c eD (^) oar nroowt (^) u(nddoeurb laeny c ainrc euqmusiltibanricuems acorrnofwus!e with T phuesh sinmga"l la crurorwvesd. aTrhroewy sr (^) eaprere "seelentc ttrhoen f Tighuisra isti vfeor m boovoekmkeeenpt (^) inofg apnu erpleocsteros no (^) npalyi!r. Cl N
O
O
2 1 ...RE HSYOBNRAINDCE 2 Cl N O
O
2 2 1 THE ACTUAL MOLECULE IS MORE STABLE THAN ANY INDIVIDUAL RESONANCE F SOPRECMI (^) EIMSP, TLIHEES .M IONR FEA CSTT,A TBHLEE MITO ISR.E RESONANCE FORMS ONE CAN DRAW OF A O INZSOTNAEB,I (^) LOI (^3) T,Y A..P.PEARS TO HAVE A SEPARATION OF CHARGE, IMPLYING SEVERE 1 1 N ineteitrhceorn Lveewrtiisng s.t ruRcatuthreer b, tyh iets aeclft (^) ureapl rsetsruecntutsr (^) ere isal iat ym. (^) ixTh oef tshtreu ctwtuor e(osr a mreor n e o ) t r reasopindalynce forms and is called a...
RESONANCE THEORY
...T HHYEB RREIDS OSHNOANWCSE THAT
B SOHTAHR ET ETRHMEI NNEAGL AOTXIVYEGENS
C GHRAERAGTEER, R SETSAUBLILTIITNYG IN
RULES OF RESONANCE THEORY
Cl N
O
O
3. R BEUSTOEN EAQNUCAEL LFYO RTOM ST HFEO RO VAE GRIVAELNL SMTORLUECCTUULREE N (ETEHED RNOESTO CNOANNTCREI-
HYBRID) SINCE THEY MAY DIFFER IN ENERGY
AND
1. ATOMIC POSITIONS REMAIN THE SAME
2. FOLLOW THE OCTET RULE
are not resonance forms!!! ...is not! are resonance forms!!! are valid resonance forms but... O
O
O
Separation of charge indicates a higher energy resonance form... co^ mntroirbeu^ stetas bmleore colnetsrsib^ usttaesb lleess O
O
O
more stable ( on O) less stable ( on C) F eonermrgayl nreesgoantiavnec ceh faorrgme, oans ewleecllt arosp poossitiivtivee e clehmaregnet so inn deliecacttreosn ae (^) ghaigtihveer elements... O O O ( aknindd y oefl (^) lloikwe mbliuxieng to make green) 4. (^) TINHSEU SRAEM TEH NAUT (^) MABLELR R EOSFO ENLAENCCTRE (^) OFNOSR,M ASN HD ATVHEE T SHAEM SEA NMUEM NBEETR COHFARGE, UNPAIRED ELECTRONS
N C O C N O
N C O N C O
N
O
H C C
H H H
H N
O
H C C
H H H
H
N
O
H C C
H H H
H
C C
H
H H
H
C C
H
H H
H
N
O
H C C
H H H
H N
O
H C C
H H H
H
A
B
A
B
Orbital Hybridization Linear Carbon (Atom)
- Only two other atoms or
- One atom and one lone pair surrounding carbon (atom)
{one s + one p}
two sp orbitals
ORIENT THEM: 180 ° apart
Two p orbitals left over
σ bonds are formed by the
sp hybrids and π bonds by
the p orbitals
Example Trigonal Planar Carbon (Atom)
- Only three other atoms or
- Two atoms and one lone pair or
- One atom and two lone pairs (rare) surrounding carbon (atom)
H C N
{one s + two p}
three sp^2 orbitals
ORIENT THEM: 120 ° apart Tetrahedral Carbon (Atom)
- Four other atoms or - Three atoms and one lone pair or
- Two atoms and one lone pair surrounding carbon (atom)
{one s + three p}
four sp^3 orbitals
ORIENT THEM: 109. 5 ° apart Example Hydrogen cyanide
No p orbitals left over
σ bonds are formed by the
sp^3 hybrids—no π bonds
C C
Methane
(side view)
H
H
H
H
(top view)
Ethylene
H
H
C
H^ H
One p orbital left over
σ bonds are formed by the
sp^2 hybrids and a π bond by
the p orbital
Example
C C
H
H
H
H
