Chem 241a-Ist Edition Lecture 9 Outline of Last Lecture I. Naming Chiral Centers – R+S formalismII. Determining Number of StereoisomersIII. Properties of StereoisomersOutline of Current Lecture I. Optical ActivityII. Chapter 6-Organic Reactionsa. Equilibrium vs. Rateb. Energy diagramsIII. Chapter 7-Alkylhalides and Substitute Reactionsa. Alkylhalidesb. Reactivity of R-XCurrent LectureI. Optical Activity - Polarimetera. One of the lab techniques that we can use to identify how much of an R or S molecule there is in a solutionb. Enantiomers interact with plane polarized light (PPL) because it is chirali. Whichever direction the R-enantiomer rotates, the S-enantiomer will rotate in the opposite direction1. If it rotates to the right, (+ rotation)- it is dexrotatory - d2. If it rotates to the left, (- rotation)- it is levarotatory – lii. The magnitude of α (the angle of rotation) is the same for each enantiomer, the direction is opposite (the sign switches)1. Ex: If the R is +45°, then the S is -45°These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.2. NOTE: R and S are not related to +/-, there are multiple solutions where R-enantiomers are both + and – and vice versaiii. Any substance that rotates PPL is said to be optically active (often synonymous with chiral) iv. 1:1 mixture of enantiomers is a racemic mixture; the R and S enantiomerscancel each other out and no α is observedv. An unequal mix of enantiomers rotate PPL in the direction of the enantiomer in excess (EE)vi. The %EE=%one enantiomer - %of the other enantiomer1. Ex: 90% EE means that 90% is one enantiomer, and 10% is racemicmix. Therefore, half of the racemic mix is the R enantiomer, and half is the S enantiomer. Thus, 95% of the mix would be one of theenantiomers, and 5% is the other.vii. We can also determine EE based on observed α knowing the [α]D of pure compound (which would be given):Therefore, EE= α observed/ α pure x 100Ex: The observed α for a sample of R and S 2-iodobutane = +7.95. The R-[α]D= +15.9; S-[α]D=-15.9. What is the %EE and how much (%) of the enantiomer is in excess?Answer: EE=+7.95/+15.9 x 100 = 50%. (NOTE: we used the standard R value because the observed α was positive, indicating that the majority is the R enantiomer.) Half of 50% is 25%, so we know that 25% of the racemic mix is R. Therefore, 75% of the mix is R-enantiomer, and 25% is S-enantiomer.========================CHAPTER 6=========================II. Organic Reactionsa. Equilibrium i. How far a reaction goes in one direction (the completeness of a reaction)ii. Governed by laws of thermodynamicsb. Ratei. How fast (speed) the reaction goesii. Governed by the laws of kineticsc. Both the equilibrium vary independently, meaning each reaction is different; these can be described using energy diagramsd. For a reaction A→BIn these situations, the rate=k=[concentration of species present in rate determining step]e. For reaction A→B→C (B is intermediate)Remember that the slow step is the rate determining step, therefore:k=[R] (left); k=[I]====================CHAPTER 7=======================III. Alkylhalides and Substitution Reactionsa. Alkyl carbon compounds bounded to halogens b. This is usually denoted generically by R-X, where R represents a carbon group and X represents some halogenEx: propylchloride (common); or 1-chloropropane (IUPAC name) We can classify these X as 1°, 2°, 3°Ex: Not all of these carbon’s are sp3 hybridized:Primary is propylchloride, Secondary: Tertiary:c. Some of the molecules have special names attached:i. Vinyl bromide (X directly bonded to a carbon with a double bond):ii. If the double bond is a benzene ring, then it is a aryl Chloride:iii. If it is once removed, one away from carbon, it is allyliciv. Once removed on a benzene ring is a benzyllicIV. Reactivity of R-Xa. R-X is a polar compound towards the halides:b. Alkylhalides undergo substitution reactions:R-X + Nu: - → R-Nu + X: - ; where R-X is the substrate (has to be sp3), Nu: is the nucleophile and has an available pair of electrons, R-Nu is the product, and X: is the leaving group (started on the substrate and left) REMEMBER: the net charge on each side of the stays the same ((-) in this