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CHEM:341 Organic Chemistry I Week 5 Notes

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CHEM:341 Organic Chemistry I Week 5 Notes CHEM341

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These note cover lectures for week 5. Bond hybridization, mechanisms, etc.
Organic Chemistry
Anna Elizabeth Allen
Class Notes
Organic Chemistry, Chemistry
25 ?




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This 9 page Class Notes was uploaded by Raffasarru on Friday September 23, 2016. The Class Notes belongs to CHEM341 at Colorado State University taught by Anna Elizabeth Allen in Fall 2016. Since its upload, it has received 4 views.


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Date Created: 09/23/16
CHEM:341 Organic Chemistry I Week 5 Notes  Hybridization Consequences: Bond Length and Strength o ???? − ???? bonds  Single: long, weak  Double: medium, medium  Triple: short, strong o ???? − ???? bonds: same deal as ???? − ???? bonds but slightly stronger and shorter bonds o Can be explained by quantity called percent s-character 3 1????  ???????? : ???? + ???? + ???? + ???? 4 ???????????????????????????????? 0.25 = 25% s-character (longer and thinner orbitals) 2 1????  ???????? : ???? + ???? + ???? 3 ???????????????????????????????? 0.33 = 33% s-character  ????????:???? + ???? 1???? = 0.50 = 50% s-character (closest orbital to nucleus 2 ???????????????????????????????? because it’s more s (shorter/fatter) than the other hybridized orbitals) o Acidity of ???? − ???? bonds  Methane is an extremely weak acid  Acidity:  Least Acidic → alkanes > alkenes > alkynes ⟵ Most Acidic  Difference is where the lone pair is held 3 2  In alkanes, lone pair is held in ???????? ; alkenes, ???????? ; alkynes, ????????  Most stable: ???????? because lone pair (negative charge) is held closer to nucleus (positive charge), in other words, the lone pair is held in an orbital with higher s-character (lower energy) o Alkenes: Nomenclature  Steps: 2  1) find longest carbon chain that contains both ???????? carbons of the alkene and change ending to “ene”  2) number carbon chain to give alkene the lowest numbers o If same, consider other groups o If other groups are the same/no other groups, alphabetize  3) name substituents on carbon chain  4) use prefixes as needed (may need to use cis/trans) o  Note: Two double bonds are -dienes, three double bonds are -trienes, etc.  Example: cis-2-trans-4-hexadiene   Example: 2,5-dimethyl-2-hexene or 2,5-dimethylhex-2-ene   When Cis and Trans are not enough  Divide molecule by drawing line down middle of double bond  Number substituents by priority (higher number on periodic table)  Are first priority groups on same or opposite side? If opposite: E, if same: Z  Priorities: one atom at a time out, higher atomic number o Meaning, if say, an ???? is on a longer chain and a ???????? is bonded to a shorter chain to the double bond, the chain with ???????? would be first priority because of ???????? vs ???? (you don’t even get to the ????) o If an atom has a higher atomic number it trumps multiples of those with lower atomic number  Example: cis-2-trans-6-decadiene or 2-Z-6-E-decadiene   Example: E-3-chloro-3-hexene  o Alkynes: Nomenclature  Steps  1) find longest carbon chain that contains both ???????? carbons of alkyne and change ending to “yne”  2) number carbon chain to give alkyne lowest numbers o If same, look at substituents o If other substituents same/no substituents, use alphabet  3) name substituents on carbon chain  4) use prefixes as needed  Notes o Two triple bonds are diynes, three triple bonds are triynes, etc. o Compounds with both double and triple bonds are enynes: 2 numbering gives first ???????? OR ???????? carbon lowest number possible  Example: oct-2-en-6-yne   Molecular Formulas and Degrees of Unsaturation o Alkenes and alkynes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon when compared with alkanes o General Formulas:  Alkanes: ???? ???? 2????+2  Alkenes AND Rings: ???? ???? ???? 2????  Alkynes: ???? ???? 2????−2 o Rings also reduce number of hydrogens in a molecule o Each ????-bond or ring removes two hydrogen atoms from a carbon chain and introduces a degree of unsaturation o How to calculate number of degrees of unsaturation for formula ???? ???? ???? ???? 2????+2 −????  Degrees of Unsaturation (# of rings/# ????-bonds) = 2 where x is number of hydrogens and n is number of carbons o Example: degrees of unsaturation in ???? ???? 8 14  (2 8 +2 −14= 2 2 o This can help us propose structures from molecular formulas  Introduction to Alkene Reactivity: Hydrohalogenation o Alkenes contain a reactive, electron rich ????-bond and participate in addition reactions with electrophiles (Lewis Acids) o Hydrohalogenation is the addition of hydrogen halides (acids: ????????) to alkenes to form alkyl halides    Dipole between carbons with double bond, electrons form a bond to ???? and one on ???? goes to ????????  ???????? bonds to the positively charged 2 carbon atom  Hydrohalogenation Mechanism: Orbital interactions o Step 1.1: use ????-bonding orbital (on nucleophile) and empty orbital for electrophile o Step 1.2: break bond between ???? and ???????? so use ???? -bonding orbital o Step 2: Bromide bonds to the positively charged carbon atom using the carbon atoms empty p orbital and the non-bonding lone pair on chloride  Energetics of Halogenation: An Energy Diagram o Energy diagram is schematic representation of a reaction that shows the energy change as starting material turns into the product o o First step: endothermic; second step: exothermic o Parts of Diagram:  Transition state: highest point in energy for each step of reaction, occurs when bonds are in process of breaking/forming  Partial ????-bond shows breaking/forming in transition state  Partial ????-bond shows forming in transition state  Use partial charges to show movement of charges: partial negative charges on halogen, partial positive on opposite middle carbon in this case  Intermediate: molecule that is formed and then consumed in a reaction, usually higher energy than reactants and products  Activation Energy (∆????): energy barrier that must be overcome for a reaction to occur; typically, highest point  Enthalpy (∆????): energy difference between reactants and products  Carbocation Stability: Hyperconjugation o As number of carbon groups around a carbocation increases, stability of carbocation increases, stability of carbocation increases o 3° Carbocation > 2° Carbocation > 1° Carbocations > Methyl Carbocations o The first two are the only two that are stable enough to make o Hydroconjugation: stabilizing interaction of electrons in a ????-bond with an adjacent empty or partially filled ???? orbital or ????-orbital   Pi bonds are at 90° in a methyl carbocation so none of the hydrogens line up (???? orbital in the way)  In a 3° carbocation, the ????-bonds can align with partially filled ???? orbital to make the molecule more stable o Larger number of alkyl groups around a carbocation, the greater the possibility for hyperconjugation and the larger the stabilization  Hydrohalogenation of Alkenes: Regioselectivity o o o One way will result in a more stable carbocation, which is where the halogen bonds o Markovnikov’s Rule: In addition of ???????? to an alkene, the proton adds to the carbon that forms the most stable carbocation  Catalyzed Addition to Alkenes: Hydration o Hydration of an alkene is an overall addition of water across the ????-bond o Reaction works same way as adding ???????? to an alkene, however, water is not reactive enough to protonate an alkene, so a strong acid is also required as a catalyst o o  Synthesis in Organic Chemistry: A Beginning o Creating a target molecule using a set of reactions is called synthesis o Requires us to think about reactions starting with product o Retrosynthesis: thinking backward o Chapter 4: Classification of Isomers  Isomers are different compounds with same molecular formula  Two major types: o Constitutional Isomers: same formula but different atom connectivity o Stereoisomers: same formula and same atom connectivity but different orientation in space (3D arrangement of atoms)  Trans (E), Cis (Z), also rings  Chiral and Achiral Molecules   Chiral: o no plane of symmetry o enantiomers (special class of stereoisomers): mirror images of each other that are not superimposable, also cannot be their own mirror image o Racemic mixture: 50:50 mixture of two enantiomers  Have two mirror images together o Stereogenic Carbon: ???????? hybridized carbon with four different groups around it  Does not have mirror plane o If a molecule has one stereogenic carbon, it must be chiral (have a non-superimposable mirror image) o If a molecule has two or more stereogenic carbons, it may or may not be chiral o Must be clear about where groups are: tetrahedral with plane through it (with top and right groups in plane)


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