Organic Chemistry 1 - Week 1
Organic Chemistry 1 - Week 1 CHM 2210
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This 8 page Class Notes was uploaded by Madison Williams on Tuesday August 16, 2016. The Class Notes belongs to CHM 2210 at University of North Florida taught by Corey Causey in Fall 2015. Since its upload, it has received 12 views. For similar materials see Organic Chemisty 1 in Chemistry at University of North Florida.
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Date Created: 08/16/16
Week #1 o organic chemistry: the study of carbon based molecules and a few other elements (primarily nitrogen, oxygen, and hydrogen) Electron Configuration o Electrons are held in orbitals. o Electron configurations tell us where electrons are located. o example #1: 2 2 2 Carbon: 6 electrons, 1s 2s 2p o example #2: Hydrogen: 1 electron, 1s1 o example #3: 2 2 4 Oxygen: 8 electrons, 1s 2s 2p Lewis Dot Structures o Lewis dot structures help us determine how bonds form between atoms. o These structures deal with valence electrons (the outermost electrons) because these are the electrons involved in bonding. o Atoms are most happy (stable) when they have 8 valence electrons. o The Lewis model of bonding tells us that atoms are most stable when the valence shell is filled (“octet rule”). o example #1: Carbon o example #2: Hydrogen o example #3: Oxygen Octet and Bonding o Atoms can gain or lose electrons in order to become “filled”. - This changes the charge of the atom. This creates ions that form ionic bonds. o Atoms can share electrons with one of more atoms in order to complete their valence shell. - This is called a covalent bond. o When electrons are unequally shared between atoms, they create a polar covalent bond. - This occurs when the atoms (elements) involved are of different electronegativities. - This type of bond contains some ionic and some covalent characteristics - primarily due to the different electronegativities involved. - As you move up and to the right of the periodic table, the elements increase in electronegativity. (Fluorine is the most electronegative element.) o Ionic bonds exist if the difference in electronegativity is greater than 1.9. If a polar covalent bond exists, the difference is between 0.5 and 1.9. If a nonpolar covalent bond exists, the difference is between 0 and .5. Lewis Structures for Polyatomics 1. Determine the number of valence electrons. (The electrons in the core are not involved with bonding.) - These come from two places: the electrons on each atom, and the charge of the molecule. - example #1: CH 4 Carbon: 4e x 1 = 4e - Hydrogen: 1e x 4 = 4e - - Total: 8e 2. Determine the connectivity of the atoms. - example #1: CH 4 - example #2: C4H 10 Carbon: 4e x 4 = 16e - - - Hydrogen: 1e x 10 = 10e Total: 26e - note: Hydrogen can only form one bond so it must be on the outside of the skeleton. *both of these are accurate structures they are structural isomers (empirical formulas are not as important as how they are formed) 3. Connect atoms with single bonds and add extra electrons to fill octets. - example #1: C 2 O6 - - Carbon: 4e x 2 = 8e Hydrogen: 1e x 6 = 6e - - - Oxygen: 6e x 1 = 6e Total: 20e - 4. Consider multiple bonds (double/triple). - example #1: C 2 4 Carbon: 4e x 2 = 8e - - - Hydrogen: 1e x 4 = 4e Total: 12e - 5. Consider the formal charge: (1) draw Lewis structure, (2) determine how many electrons “belong” to each atom (bond counts as one, lone pairs count as two), (3) compare the number of electrons to that of a neutral atom, (4) sum of all formal charges must equal the charge on the molecule - extra electron = -1 charge - missing electron = +1 charge - example #1: H 3 + Hydrogen: 1e x 3 = 3e - Oxygen: 6e x 1 = 6e - “ ”: -1e - - Total: 8e formal charges: Hydrogen (same for all three): owns 1e , needs 1e 0 - - - Oxygen: owns 5e , needs 6e +1 - example #2: HCO 3- Hydrogen: 1e x 1 = 1e - Carbon: 4e x 1 = 4e - - - Oxygen: 6e x 3 = 18e “ “: +1e - Total: 24e - formal charges: Hydrogen: owns 1e , needs 1e 0 - Carbon: owns 4e , needs 4e 0 - Oxygen: owns 7e , needs 6e -1 - - example #3: HNO 3 Hydrogen: 1e x 1 = 1e - Nitrogen: 5e x 1 = 5e - - - Oxygen: 6e x 3 = 18e Total: 24e - formal charges: Hydrogen: owns 1e , needs 1e 0 - Oxygen (leftmost/top): owns 6e , needs 6e 0 - - - Oxygen (rightmost): owns 7e , needs 6e -1 Nitrogen: owns 4e , needs 5e +1 - check: -1 + (+1) = 0 o structural isomers: contain all the same atoms but the skeletons are different (such as C H 4ro10the previous example) o formal charge: - The overall charge may be zero but each individual atom may have a charge that gets canceled out by another. Generally Speaking: o Carbon: usually has 4 bonds and 0 lone pairs o Oxygen: usually has 2 bonds and 2 lone pairs o Nitrogen: usually has 3 bonds and 1 lone pair o Hydrogen: usually has 1 bond and 0 lone pairs Exceptions to the “Octet Rule”: o Boron and Aluminum are neutral with six electrons. - The formal charge on a boron (or aluminum) atom surrounded by 4 hydrogen atoms (8 electrons), is -1. - The formal charge on a boron (or aluminum) atoms surrounded by 3 hydrogen atoms (6 electrons), is 0. Bond Angles and Molecular Shapes o Valence Shell Electron Pair Repulsion (VSEPR) Theory: explains why molecules take the three dimensional shape that they take o Regions of Electron Density: - includes single bonds and lone pairs RoED Shape Bond Angle Hybridization 4 tetrahedral 109.5° sp3 2 3 trigonal planar 120° sp 2 linear 180° sp - example #1: CH 4 This structure looks planar in the drawing, but in reality it is tetrahedral because the bonded electrons want to be as far away from each other as possible (VSEPR Theory). - example #2: CH 2 - example #3: C 2 2 Valence Bond Theory o hybridization: two different things coming together to create a unified piece similar to the individual parts o example #1: sp hy3ridization note: You must conserve the number of orbitals, but for this hybridization they will all look the same. This is how the tetrahedral shape comes about. o example #2: sp hybridization 2 note: Because this is an sp2hybridization, you only combine one “s” orbital with 2 of the “p” orbitals. The fourth orbital is unhybridized and, therefore, remains the same. In addition, the single π bond will prevent the molecule from twisting. o example #3: On each carbon atom, there are 3 bond angles, so each side of the molecule is trigonal planar with a sp 2ybridization. Using hybridization, the following structure can be created. - The blue overlaps represent the 5 σ bonds. - The red “p” orbitals are positioned 90° to the plane of the rest of the molecule. (Think you this as then sticking straight up from the paper.) - The π bonds, represented by the dotted line connected the “p” orbitals, keep the two sides of the molecule from twisting. o example #4: If two addition hydrogen atoms were added to the molecule above, each carbon would then possess 4 regions of electron density. It would therefore contain an sp hybridization, and would be able to twist freely because of the absence of π bonds and the presence of σ bonds. o example #5: C H 2 2 On each carbon atom, there are 2 bond angles, so each side of the molecule is linear with a sp hybridization. Using this information, we can create this structure. - The blue overlaps represent the 3 σ bonds. - The red and purple “p” orbitals are positioned 90° to the linear molecule. - The π bonds, represented by the dotted line connected the red “p” orbitals together, and the purple “p” orbitals together keep the two carbon atoms from twisting. Polar vs. Non-Polar o To be polar, the molecule must contain polar bonds. (However, the presence of polar bonds doesn’t always lead to a polar molecule.) o Polarity is a scale that can be measured quantitatively. o Polarity is derived from the orientation and types of atoms involved in a molecule. o example #1: This molecule only contains nonpolar bonds, and its therefore a nonpolar molecule. o example #2: By replacing one of the hydrogen atoms from example #1 with a chlorine atom, there is now a polar bond between the chlorine atom and the carbon atom because of the difference in electronegativity. This is a polar molecule. o example #3: This molecule is nonpolar because all of the dipoles cancel each other out. Resonance o Some molecules are not accurately represented by a single Lewis structure. o Resonance allows us to take several structure to describe a single structure. o The best Lewis structure contributes the most to the “average” of the resonance structures (weighted average). o Resonance structures do not actually exist and they are not in equilibrium with one another. o example #1: CO 3-2 - The Double bonds are shorter than single bonds. However, if you look at CO 3 2, all of the bonds appear to be the same length. The actual structure (below) looks somewhat like each of the Lewis structures. In this model, the electrons move between atoms and each oxygen has a -2 formal charge of about / . 3 Rules for Drawing Resonance Structures 1. All structures that you draw must have the same number of electrons. 2. You must not violate the “octet rule” for 1 and 2 row. (You can have less than 8, but not more. 3. The connectivity (skeleton) cannot change. You cannot break sigma bonds. Pi bonds and lone pair electrons are the only parts that can move. 4. You must have the same number of unpaired electrons.
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