Chemistry 30A Review Sheet, Lecture notes of Physical Chemistry

Download Chemistry 30A Review Sheet and more Lecture notes Physical Chemistry in PDF only on Docsity! Chemistry 30A Review Sheet Tau Beta Pi – Boelter 6266 Contents 1. General Chemistry Topics 3 1.1 Electronegativity…………………………………………………………………………………………………………3 1.2 VESPR and Hybrid Orbitals……………………………………………………………………………………….3 1.3 Resonance………………………………………………………………………………………………………………...3 2. Isomerism and Chirality 3 2.1 Terminology……………………………………………………………………………………………………………..3 2.2 Differentiating between diastereomers and enantiomers using the R.S. Convention………….4 3. Acids and Bases 4 3.1 Terminology……………………………………………………………………………………………………………..4 3.2 ARIO……………………………………………………………………………………………………………………….5 4. Alkane Nomenclature 5 5. Alkane Conformations and Strain 7 5.1 Conformations…………………………………………………………………………………………………………..7 5.2 Types of Strain…………………………………………………………………………………………………………..8 6. Alkenes 8 6.1 Physical Properties and Conformations………………………………………………………………………..8 6.2 Nomenclature……………………………………………………………………………………………………………8 6.3 Reactions…………………………………………………………………………………………………………………..8 7. Alkynes 10 7.1 Physical and Chemical Properties……………………………………………………………………………….10 7.2 Nomenclature…………………………………………………………………………………………………………..10 7.3 Reactions………………………………………………………………………………………………………………….10 8. Haloalkanes 13 8.1 Terminology……………………………………………………………………………………………………………..13 8.2 Reactions………………………………………………………………………………………………………………….13 9. SN1, SN2, E1, E2 15 9.1 SN2 Nucleophilic Substitution……………………………………………………………………………………..15 9.2 SN1 Nucleophilic Substitution……………………………………………………………………………………..15 9.3 E2 Elimination………………………………………………………………………………………………………….15 Tau Beta Pi – UCLA Chemistry 30A Page 2 9.4 E1 Elimination…………………………………………………………………………………………………………16 9.5 Which mechanism?................................................................................................................16 Tau Beta Pi – UCLA Chemistry 30A Page 5 3.2 A.R.I.O Method of Evaluating Acid/Base Stability ‘A’ stands for ‘Atom’ Consider the electronegativity and size of the atom. The greater the electronegativity of the atom that will bear the negative charge following proton donation, the more strongly the electrons are held. The anion is more stable, and the acid is stronger. Used to compare atoms across a period. The larger the atom bearing the negative charge, the more stable the anion is (because larger atoms have larger electron clouds and are more polarizable) and the stronger the acid. ‘R’ stands for ‘Resonance’ Resonance delocalizes negative charge in the base, making the base more stable and its conjugate acid more acetic. ‘I’ stands for ‘Inductive Effect’ The inductive effect is the polarization of electron density around one covalent bond by the electronegativity of a neighboring covalent bond. Electron donating groups (like methyl groups) induce a partial negative charge on whatever atom they are bound to. Electron withdrawing groups (all electronegative atoms) induce partial positive charges on whatever atoms they are bound to. These partial charges can stabilize or destabilize anions The inductive effect is weak, and falls of rapidly with increasing distance. ‘O’ stands for ‘Orbital Hybridization’ For anions differing only in the hybridization of the charged atom, the greater the percent ‘s’ character in the hybridization (i.e. the more bonds) the more stable the anion (see diagram below). Therefore, triple bonds are more stable than double bonds and single bonds. 4. Alkane Nomenclature First, name the parent chain (in cycloalkanes, this is the number of carbons in the ring). Start with the prefix. The prefix indicates the length of the longest carbon chain in the atom, or the parent chain (see table below). Number of atoms in parent chain Prefix 1 Meth 2 Eth 3 Prop 4 But Tau Beta Pi – UCLA Chemistry 30A Page 6 5 Pent 6 Hex 7 Hept 8 Oct Next, the infix indicates the type of bond. In this case, since the molecule is an alkane, it only has single bonds, and the infix is –an-. Finally, the suffix is used to denote what type of compound the alkane is (see table below). Type Suffix Hydrocarbon -e Alkane -ane Alcohol -ol Aldehyde -al Amine -amine Ketone -one Carboxylic Acid -oic acid Next, number the parent chain, giving the first substituent the lowest number possible. This will allow you to locate the position of all substituents along the parent chain. For cycloalkanes, start numbering from the substituent that comes first alphabetically. Locate and identify all substituents. Place the names of the substituents, along with its numerical positions on the parent chains, before the name of the parent chain alphabetically. Separate the substituents with hyphens. Below is a list of common substituent groups, and their names. Tau Beta Pi – UCLA Chemistry 30A Page 7 5. Alkane Conformations and Strain 5.1 Conformations Staggered conformation - conformation a that has the atoms or groups on one carbon as far apart as possible from the atoms or groups in an adjacent carbon. There are two types: a) Anti conformation - staggered conformation where groups lie at a dihedral angle of 180 degrees. b) Gauche conformation - staggered conformation about a carbon-carbon single bond in which groups lie at a dihedral angle of 30 degrees. Eclipsed conformation - conformation where atoms or groups on one carbon are as close as possible to the atoms or groups on an adjacent carbon. That is, groups lie at dihedral angles of 0 degrees. Cyclohexane chair conformation – This conformation is the most stable conformation of a cyclohexane ring, and is generally considered to have no strain. Around the cyclohexane ring, there are six axial and six equatorial substituents, as shown below. In general, it is better if larger groups are equatorial. The chair has two equivalent conformations, which switch in a process called a “chair flip”. When a chair flips, substituents pointing up stay pointing up and vise versa. Axial substituents become equatorial and equatorial substituents become axial. All substituents are moved over one position. An example chair flip is shown below. Tau Beta Pi – UCLA Chemistry 30A Page 10 Oxidation with O3 (O3, Me2S): The double bond is cleaved into two carboxylic acids. Hydrogenation/Reduction (H2Pd or H2Pt or H2Ni): The double bond is reduced to a single bond via syn addition of two hydrogens across the double bond. 7. Alkynes 7.1 Physical and Chemical Properties Like alkenes, the only intermolecular forces acting on alkynes are London Dispersion forces. Alkynes are more acidic than alkanes and alkenes (the pH of a terminal alkyne is approximately 25). Terminal alkynes can be deprotonated by NaNH2, NaH or LDA. 7.2 Nomenclature Use the infix –yn- to show the presence of a carbon-carbon triple bond. The format of the name is as follows: (position of substituent)-(substituent)-…-(position of triple bond)-(longest carbon chain) 7.3 Reactions Addition of Br2: Bromine is added across the triple bond to form an alkene with Bromine on either side of the double bond. If Br2 is present in excess, the reaction will repeat to form an alkane bonded to four bromine atoms. Tau Beta Pi – UCLA Chemistry 30A Page 11 Addition of HBr: Hydrogen and bromine are added across the triple bond to form an alkene connected to one hydrogen and one bromine. The reaction follows Markovnikov’s rule, and the hydrogen will add to the less substituted side of the triple bond. If HBr is present in excess, the reaction will repeat to form an alkane bonded to two Bromine molecules. Hydroboration: a) Internal alkyne (BH3, H2O2, NaOH): Addition of H and OH across the triple bond. b) Terminal alkyne [(sia)2BH, H2O2, NaOH)]: Addition of H and OH across the triple bond. The addition does not follow Markovnikov’s rule; the hydrogen is added to the more substituted side of the triple bond. Hydroboration-Proteolysis [internal alkyne, BH3/THF, (CH3COO)3]: The internal alkyne is reduced to an internal alkene via addition of H2 across the double bond. Tau Beta Pi – UCLA Chemistry 30A Page 12 Dissolving Metal Reduction [internal alkyne, Na°, NH3 (l)]: Reduction of the internal alkyne to an trans- alkene. Hydration (H2O, H2SO4, HgSO4): Terminal alkynes are transformed into ketones. Reduction of Alkynes: a) H2Pd reduces the alkyne to an alkane. b) H2Pd and Lindlar’s Catalyst reduce the alkyne to a cis alkene Tau Beta Pi – UCLA Chemistry 30A Page 15 9. SN1, SN2, E1, E2 9.1 SN2 Nucleophilic Substitution Mechanism: one step, both reactants are involved in the transition state of the rate determining step. The nucleophile attacks from the back. Stereochemistry: for SN2 reactions at a chiral center the result is an inversion of the chiral center because the nucleophile attacks from the back and pushes all other substituents to the other side. Note: depending on the nucleophile you may or may not have an inversion of ‘R’ and ‘S’ at the chiral center. 9.2 SN1 Nucleophilic Substitution Mechanism: two steps, the bond between the carbon and the leaving group is entirely broken before a new bond is formed with the nucleophile. Stereochemistry: R and S enantiomers are both formed; the product is racemic. 9.3 E2 Elimination Mechanism: breaking of the carbon-leaving group bond and the formation of the carbon-hydrogen bond occurs at the same time. Note that this type of elimination requires the presence of a strong base. Regioselectivity: if a strong base is used the more substituted alkene will be the major product. If a strong, sterically hindered base is used, the major product will the less substituted alkene because the base will deprotonate the most accessible hydrogen. Stereochemistry: this type of elimination is most favorable when the hydrogen and leaving group are oriented antiperiplanar (or 180 degrees apart) Tau Beta Pi – UCLA Chemistry 30A Page 16 9.4 E1 Elimination Mechanism: the bond between the carbon and the leaving group breaks before the base deprotonates the alkane, so a carbocation is formed. Regioselectivity: the major product is the most stable alkene, because the base is able to choose which proton to deprotonate because the reaction occurs in two steps. Stereochemistry: the alkene can be either cis or trans. Since the reaction occurs in two steps, the base will deprotonate whichever hydrogen allows for formation of the most stable isomer. 9.5 Which mechanism? If you have a primary alkyl group: a) SN1 and E1 are NOT POSSIBLE because primary carbocations are unstable and will not form b) Strong sterically hindered base -> E2 c) Strong unhindered base that is an ammide anion will result in E2. If your base is a good nucleophile, (OH-, acetylide, methoxide, RS-, CN-, I-, N3 -) you will have SN2. If your base is a moderate nucleophile you will have SN2 If you have a secondary alkyl group: a) Strong bases (sterically hindered or unhindered) will result in E2 b) Weak bases that are good/moderate nucleophiles will result in SN2 c) Weak bases that are bad nucleophile and in a polar protic solvent will result in E1 and SN1 If you have a tertiary alkyl group: a) SN2 is not possible because a tertiary carbocation is too sterically hindered b) Strong bases will result in E2 c) Polar Protic solvents will facilitate SN1 and E1