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OU CHEM 3153 - Exam 1 Study Guide

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CHEM 3153 1nd EditionExam # 1 Study Guide Lectures: 1 - 9Lecture 1 (January 12)Summary of Orbitals- Electrons possess characteristics of both a particle and a wave- Orbital: region of space that can be occupied by an e-o Orbitals have different phases (positive and negative) This is due to the wave-like nature of e-‘so e- density is equally likely to be found in both + and – phaseso Orbitals have nodes e- density is never found at nodesSummary of Molecular Orbitals (MO’s)- Molecular orbitals are a result of in-phase and out-of-phase addition of atomic orbitals - In-Phase = constructive interferenceo A bonding MO is createdo A bond is formed when e-‘s are in this orbital- Out-of-Phase = destructive interferenceo An antibonding MO is created These MO’s have nodeso A bond is broken when e-‘s are put into this orbital The node is located between the nucleiExample of a carbon double bonded to another carbon:Summary of Hybridizationsp sp2sp32 “things” attached 3 “things” attached 4 “things” attachedThings = atoms or lone pairsIt’s very important that you be able to identify the hybridization of an atomLecture 2 (January 14) Summary of Mechanisms- You need to be able to know how to draw mechanisms- You’ll have to identify major components of the reaction in order to determine how it will proceed.1. Look for areas of high and low electron densities that are likely to react under the given conditions.High e - density- Atoms with lone pairs, except neutral halogens (Cl, Br, I)o Ex:- Initiators: N, O, P, S, and charged halogens (Cl, Br, I)- Pi bonds (π) between carbon atomso Ex:- Sigma bonds (σ)o Ex: o The attack is initiated by the more electronegative atom in the bond. In these examples, they are carbon and hydrogenLow e - density- The less electronegative (EN) atom in a polar σ or π bond.o Ex: o Usually, we attack the hydrogen because they are less EN- Atoms lacking an octeto Ex: o Carbocations have low e- density which allows e- density to flow to it2. Look at the conditions the reaction is under.- Acidic- Neutral- Basic- Temperature- Order of additionStrong acids usually protonate first (proton transfer) and strong bases usually deprotonate first.- Do not generate strong acids under basic conditionso Ex: o This won’t exist in strong basic conditions (such as NaOH)- Do not generate strong bases under acidic conditionso Ex: o This won’t exist in strong acidic conditions (such as HCl and H2SO4)3. Draw arrows from high to low e- density (ed)If the atom has the high ed- Start the arrow at lone pairs and point to low ed atom on the other moleculeo Ex: o *IMPORTANT*o We want a neutral atom, so oftentimes we will end the reaction with a proton transfero Remember to keep track of formal charges (FC) during the mechanismo Do not exceed octetsIf the bond has the high ed- Start the arrow in the middle of the bond and point to the low ed atomo Ex:o Make sure to draw the bond breaking between hydrogen and bromine becausehydrogen doesn’t have room for 2 bondso Bromine is more EN than hydrogen, so hydrogen is the atom with lower edIntegrated Example:A) Indicate the area of high ed.Answer: The π bond. B) Indicate the area of low ed.Answer: The hydrogen. C) Indicate e- flow involved in the 1st step.Answer: π bond to H (bond forming) and bond between H and Br to Br (bond breaking)D) Indicate formal charges.Answer: The carbon molecule gives electrons to the hydrogen. So the carbocation will receive apositive charge on the more substituted carbon because it is more stable in that position. The bromine has received electrons from the hydrogen so it will receive a negative charge.E) Indicate e- flow involved in 2nd stepAnswer: Br to positive end of carbocation. Remember that arrows gofrom high to low ed!!Lecture 3 (January 16)Dienes: 2 double bonds in a molecule3 Types of Dienes1) Isolated Dienes- Ex: 2) Conjugated Dienes- Ex:3) Cumulated Diense- Ex:Focus on Conjugated Dienes!4) Conjugated Dienes- Ex:- Double bonds are on carbons next to one another- P orbitals can interact with each other- They are more stable than isolated dieneso π bonds interacting with each other spreads out e- density- the single bond is shorter than a typical C—C bondo Ex:o The single bond is between 2 sp2 hybridized carbons sp2 orbitals have more s character than sp3 s orbitals are closer to the nucleus than p orbitals More s character = e- held closer to the nucleus for each sp2 carbono Bond between 2 sp2 orbitals will be shorter than a bond between 2 sp3 orbitals- Simplest conjugated diene is 1,3 – butadieneo Ex: o It has two important conformers at room temp Conformers: shapes that a molecule can assume by only rotating single bondso The switch back and forth between these 2 conformers is fast at room temp s-trans: double bonds on opposite sides of single bond s-cis: double bonds on the same side of single bondo Do not confuse with trans and cis!Electrophilic Addition to Conjugated Dienes- Ex: o The more substituted carbon holds the formal chargeAnother product is attainable in the above reaction.A) Indicate e- flow involved to obtain the other productAnswer: π bond to σ bond (resonance) then Br to positively charged C (bond forming) Lecture 4 (January 21)- The above products have distinct nameso The names are based on the positions of H and Bro 1,2 – adduct and 1,4 – adducto 2 carbons of separation and 4 carbons of separation1,2 – addition vs 1,4 – addition- The ratio of products can be controlled by tempo Ex: - This is possible due to an energy difference in the products and the transition states leading to the productKNOW HOW TO ANSWER THE FOLLOWING QUESTIONS:A) Which product is more stable and why?Answer: The 1,4 adduct (the second product) because the double bond is more substitutedB) Which transition state (TS) is more stable and why?Answer: The 1,2 adduct (the first TS) because the partial positive charge is on the more substituted carbonC) Which product will be the major product and why?Answer:- At low temps (0°C), the 1,2 – adduct forms faster because there isn’t enough energy to overcome the TS barrier to form the 1,4 – adducto 1,2 – adduct will be the major product Its TS is lower in energy- At high temps (40°C), 1,4 – adduct forms predominantlyo 1,4 – adduct will be the major producto There is enough energy to overcome the energy barrier to make the 1,4 – adducto Once the 1,2 –


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