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Organic Chemistry Exam #1 Study Guide

by: Madison Williams

Organic Chemistry Exam #1 Study Guide CHM 2210

Marketplace > University of North Florida > Chemistry > CHM 2210 > Organic Chemistry Exam 1 Study Guide
Madison Williams
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This contains all of the previous class notes, as well as additional notes from the textbook from chapters 1 and 2.
Organic Chemisty 1
Corey Causey
Study Guide
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This 24 page Study Guide was uploaded by Madison Williams on Saturday September 10, 2016. The Study Guide belongs to CHM 2210 at University of North Florida taught by Corey Causey in Fall 2015. Since its upload, it has received 10 views. For similar materials see Organic Chemisty 1 in Chemistry at University of North Florida.


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Date Created: 09/10/16
Exam #1 – Chapters 1 &2 Chapter 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 Ground-state electron configurations are the electron configurations of the lowest energy. o Electron configurations tell us where electrons are located. o example #1: Carbon: 6 electrons, 1s 2s 2p2 o example #2: 1 Hydrogen: 1 electron, 1s o example #3: Oxygen: 8 electrons, 1s 2s 2p4 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 as a result of the attraction between oppositely charged ions. 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. (Electronegativity is the measure of the force of an atom’s attraction for electrons.) - 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. o If a polar covalent bond exists, the difference is between 0.5 and 1.9. o 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 o 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 the molecules are formed.) 3. Connect atoms with single bonds and add extra electrons to fill octets. - example #1: C2H 6 - - 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: C2H 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 - formal charge = # of valence electrons in neutral/unbounded atom – (unshared electrons + half of shared electrons) - example #1: H O + 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 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: a method for predicting bond angles based on the idea that electron pairs repel each other and keep as far apart as possible (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 o 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 o example #2: CH2O o example #3: C2H2 Valence Bond Theory: a model of bonding that places electron pairs between adjacent atoms to create bonds o hybridization: the combination of atomic orbitals of different types o example #1: sp hyb3idization 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 hyb2idization Because this is a 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 sp2hybridization. Using hybridization, the following structure can be created. Notes about this structure  - 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 in example #3, each carbon would then possess 4 regions of electron density. It would 3 therefore contain a 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. Notes about 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 bond dipole moment: a measure of polarity of a covalent bond o molecular dipole moment: the vector sum of individual bond dipoles o The more electronegative atom will have a slightly negative charge; while the less electronegative atom will have a slightly positive charge. o example #1: This molecule only contains nonpolar bonds, and it’s 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: a theory that many molecules are best described as a hybrid of several Lewis structures 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 contributing structures: representations of a molecule or ion that differ only in the distribution of valence electrons o resonance hybrid: a molecule or ion described as a composite of a number of contributing structures o You use arrows to show the movement of electrons. o Major contributors contribute more to the average than the minor contributors. Rules for Drawing Resonance Structures o All structures that you draw must have the same number of electrons. o You must not violate the “octet rule” for 1 and 2 row. (You can have less than 8, but not more. o The connectivity (skeleton) cannot change. You cannot break sigma bonds. Pi bonds and lone pair electrons are the only parts that can move. Contributing structures must only differ in the distribution of valence electrons. o You must have the same number of paired and unpaired electrons. o example #1: CO 3-2 The double bonds are shorter than single bonds. However, if you look at CO 3- , 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 formal charge of about / . 3 o example #2: Follow up notes  - The first three structures are all major contributors because they are essentially all the same. - The fourth structure is not as good because the carbon atom does not have a full octet. Similarly, the fifth structure is not as good because the oxygen atom on the lower right does not have a full octet. - All of these structures are valid, but none of them really exist. o The point of creating resonance structures is to predict reactivity. Contributions for Resonance Structures (most important  least important) 1. filled valence shells - example #1: The first structure is the major contributor because all of the valence shells have a full octet. 2. maximum number of covalent bonds - In the example from above, the first structure is still better because it has 5 covalent bonds, whereas the second structure only has 4. 3. minimize separation of charge - If charges are present in the resonance structure, it is best if the opposite charges are close together (one bond length apart is better than two bond lengths apart). - example #1: The first structure is the best, the second is good, and the third is the worst. 4. charges on most appropriate atom - If you have a negative charge, it is better for it to be on an electronegative atom. If you have a positive charge, it is better for it to be on a less electronegative atom. Nitrogen Hybridization o Recall that CH h4s sp hy2ridization. o NH al3o has sp hyb3idization, but in some cases will adopt an sp 2 hybridization. o If the atom beside nitrogen has a p-orbital, then nitrogen will also “want” a p- orbital. o example #1: o example #2: The nitrogen atom adopts a sp hy2ridization rather than have a sp 3 hybridization because of its resonance structures. o In order for nitrogen to form this resonance structure, it must have a p- orbital (delocalization of electrons). o delocalization: the spreading of charge and/or electron density over a large volume of space o For an atom to be involved in π bonding, it must have a 2p orbital – which sp 2 atoms have. (page 51) Functional Groups Alcohols: a compound containing a hydroxyl group (-OH) bonded to a tetrahedral carbon (a carbon having single bonds to four other atoms) o example #1: CH OH 3methanol) o primary (1°): a compound containing a function group bonded to a carbon atom which is bonded to only one other carbon atom and two hydrogen atoms - structure: R – CH –2OH  where R represents an alkyl (carbon) group - example #1: CH – 3H – O2 (ethanol) o secondary (2°): a compound containing a function group bonded to a carbon atom which is bonded to two other carbon atoms and two hydrogen atoms - structure: - example #1: o tertiary (3°): a compound containing a functional group bonded to a carbon atom bonded to three other carbon atoms - structure: - example #1: o Although these are all similar, they have very different physical and chemical properties. Amines o The functional group of an amine is an amino group. o amino group: compound containing a nitrogen atom bonded to one (1°), two (2°), or three (3°) carbon atoms by a single bond o They are based off of NH3(Ammonia). o primary (1°): nitrogen is bonded to one carbon atom and two hydrogen atoms - structure: - example #1: o secondary (2°): nitrogen is bonded to two carbon atoms and one hydrogen - structure: - example #1: o tertiary (3°): nitrogen is bonded to three carbon atoms - structure: - example #1: o quaternary (4°): nitrogen is bonded to four carbon atoms (this structure changes the formal charge of the nitrogen atom) - structure: - example #1: ammonium Aldehydes and Ketones o The function group of both of these is a carbonyl group (C-O double bond). o Because they have structural similarities, their chemistry is very similar. o Aldehydes: The carbonyl group must be bonded to at least one hydrogen (- CHO group). - structure: - example #1: o Ketones: The carbonyl group is bonded to two carbons. - structure: If one of these R-groups was a hydrogen atom, the molecule would be an Aldehyde. - example #1: acetone Carboxylic Acids o The functional group of these is a carboxyl group (-COOH or –C2 H) o Carboxyl groups are essentially a hydroxyl bonded to a carbonyl. o structure: o example #1: o example #2: Carboxylic Ester o The hydrogen of the carboxylic acid structure is replaced with a –OR group (carbon-containing group). o structure: o example #1: o example #2: Carboxylic Amide o The hydroxyl group of the carboxylic acid structure is replaced by an amine. o structure: o example #1: o example #2: Amino Acids o Amino acids, called the “building blocks of life”, make up proteins. o Amino acids are held together by peptide bonds, also called amide bonds. o example #1: glycine (the simplest amino acid) This contains both an amine and a carboxylic acid. o example #2: alanine o example #3: serine Chapter 2 Hydrocarbons: molecules composed of carbon and hydrogen Alkanes: saturated molecules (contains only carbon-carbon single bonds and the maximum amount of hydrogen atoms possible ) o All alkanes have this composition.  C H n 2n+2 o Oxidation is the primary reaction of Alkanes. o Nomenclature - suffix: “-ane” - prefix: indicates the number of carbons in the molecule - 1: CH =4methane (“meth” refers to 1 carbon) - 2: CH -3H = 3thane - 3: CH -3H -C2 = p3opane - 4: CH -3H -C2 -CH2= b3tane Alkenes: unsaturated molecules that contain at least one C-C double bond o example #1: Alkynes: unsaturated molecules that contain a C-C triple bond o example #1: Arenes (Aromatics) *we won’t be dealing with these much* o degrees of unsaturation: For every pi bond you lose, you need two hydrogen atoms. Line-Angle Formulas o The “ends” of lines and the breaks/intersections/bends imply carbons. o Hydrogen atoms on carbons are assumed. o methane (1 carbon): CH 4 o ethane (2 carbons): C H 2 6 o propane (3 carbons): C H 3 8 o butane (4 carbons): C H 4 10 The bottom left end of this structure is assumed to be CH , t3e top left bend is assumed to be CH , 2he bottom right bend is assumed to be CH , and2the top right end is assumed to be CH . 3 o pentane (5 carbons): C H 5 12 o hexane (6 carbons): C H 6 14 o heptane (7 carbons): C H 7 16 o octane (8 carbons): C H 8 18 o nonane (9 carbons): C H 9 20 o decane (10 carbons): C H 10 22 Constitutional Isomers: compounds with the same molecular formula but a different connectivity of their atoms o These have different structural formulas, meaning they differ in the kinds of bonds they have and/or the connectivity of their atoms. o example #1: C H 6 14 All of these have the same formula, but they have different structures. Hexane is normal and unbranched, whereas the other ones contain branches (points where they diverge) off of the main chain. Nomenclature of Alkanes o The prefix indicates the number of carbons. - 1: meth- - 2: eth- - 3: prop- - 4: but - 5: pent- - 6: hex- - 7: hept - 8: oct- - 9: non- - 10: dec- o The suffix “-ane” indicates that the molecule is an alkane o Unbranched alkanes are named by the prefix + “-ane” suffix. For Branched Alkanes 1. Select the longest continuous chain of carbons – known as the parent/root. Then, name this chain as a normal, unbranched alkane. 2. Give each substituent (any carbon not apart of the main chain) a name and a locant (location of the main chain). 3. Use the same prefix. The “-ane” suffix is converted to a “-yl”. - example #1: CH - “3ethyl” - example #2: CH -CH3- “e2hyl” 4. Number the parent chain to give the lowest number to the substituents. - example #1: 2-methyl pentane You number this one from right to left so that the substituent is at 2. (If you numbered it the other way, the substituent would be at 4.) 5. If there are multiple identical substituents, use “di-“, “tri-“, “tetra-“, “penta-“, “hexa-“, etc. - example #1: 2, 3 – dimethyl butane For this one, it does not matter which direction you number the parent chain because either way there will be a methyl at 2 and 3. 6. For non-identical substituents, list their names in alphabetical order. (If they have prefixes, do not take them into consideration when putting them in alphabetical order.) - example #1: 4-ethyl-2, 2-dimethyl hexane You number this parent chain from left to right because it is better to have substituents at 2 and 4, than at 3 and 5. The substituent at 2 is two methyls because they are separated, unlike the bonded ethyl on 4. 7. If two parent possibilities have the same length, then choose the chain that is more branched. - example #1: 3-ethyl-2-methyl hexane We choose the second one because it has two branches/two substituents. (The first one only has one branch/one substituent.) Common Sub. Names 1. propyl: isopropyl: an isomer of propyl 2. butyl: bonded to parent chain at carbon 1 isobutyl: 3. secbutyl: bonded to chain at carbon 2 4. tert-butyl (t-butyl) Other Nomenclature o Infix: - “-an-“: all single bonds (alkane) - “-en-“: one of more double bonds (alkene) - “-yn-“: one or more triple bonds (alkyne) o Suffix - “-e”: hydrocarbon - “-ol”: alcohol - “-al”: aldehyde - “-amine”: amine - “-one”: ketone - “-oic acid”: carboxylic acid Cycloalkane: a saturated hydrocarbon that contains carbons joined to form a ring o The formula for cycloalkanes is C n .2n The Shapes Nomenclature o Name these by adding the prefix “cyclo-“. o For substituents on the ring  1. If there is only one substituent, then no locant is needed. - example #1: methyl cyclopentane 2. If there are two substituents, you give the lower number to the substituent that is first alphabetically. - example #1: 1, 2-dimethyl cyclohexane - example #2: 1-ethyl-2-methyl cyclohexane Conformations of Alkanes o conformation: any three-dimensional arrangement of atoms in a molecule that results from rotation about a single bond Newman Projections: a way to view a molecule by looking along a carbon-carbon single bond to help evaluate the relative orientations of attached groups o eclipsed: a conformation about a carbon-carbon single bond in which the atoms or groups on one carbon are as close as possible to the atoms or groups on an adjacent carbon o staggered: a conformation about a carbon-carbon single bond in which the atoms or groups on one carbon are as far apart as possible from atoms or groups on an adjacent carbon o There will always be three eclipsed (high energy), and three staggered (low energy). o dihedral angle: the angle created by two intersecting planes o example #1: ethane - O° (eclipsed) The back hydrogen atoms are the three that are not connected to the middle carbon. (Only the front three hydrogen atoms are drawn in this way.) - 60° (staggered) - 120° (eclipsed) - 180° (staggered) - 240° (eclipsed) - 300° (staggered) - If you graph these, the eclipsed structures all exhibit the same energy. o example #2: butane – We are viewing this from the C -C 2on3. - O° (eclipsed): This position exhibits the highest energy. - 60° (staggered): This position is called gauche (staggered position in which the large groups are 60° apart). - 120° (eclipsed): This does not exhibit energy as high as 0°. - 180° (staggered): This exhibits the lowest energy and is the best case scenario. It is called the anti conformation. - 240° (eclipsed): This conformation is no better than that of 120°. - 300° (staggered): This conformation is also gauche and is no better than 60°. - If you graph these, the eclipsed structures all exhibit the same energy. Cycloalkanes o Cyclopropane is much higher in energy than propane because of torsional strain and angle strain. - torsional strain: strain that arises when nonbonded atoms separated by three bonds are forced from a staggered conformation to an eclipsed conformation This molecule must be planar and as a result the carbon atoms cannot twist like they can in ethane and butane. In the structure above, all six hydrogen atoms are eclipsing each other. This causes a significant amount of torsional strain. note: This was drawn in class as just a triangle with the hydrogen atoms (carbon was not shown). - angle strain/ring strain: the strain that arises when a bond angle is either compressed or expanded to its optimal value The hybridization of each carbon atom should be sp mea3ing each angle would be 109.5°; however, in actuality each angle is only 60°. This compression from the optimal bond angle introduces a lot of angle strain. o Cyclobutane does not exhibit as much torsional strain as cyclopropane, but it does possess some angle strain. - The image on the right is called a puckered conformation and it decreased the torsional strain associated with eclipsed interactions, and increases angle strain caused by the compression of the C-C-C bonds. - The puckered conformation is more stable because the decrease in torsional strain is greater than the increase in angle strain. - Puckered conformations are favored in all cycloalkanes larger than cyclopropane. o Cyclopentane does not exhibit much angle strain because it’s angles are 108° and as an sp 3ybridization they want to be 109.5°. However, torsional strain is present because there are 10 pairs of fully eclipsed C-H bonds. The image on the right is known as an envelope conformation - which relieves some of the torsional strain. - Four of the carbon atoms are in a plane, and the fifth bends out of the plane. - Cyclopentane exists as a dynamic equilibrium of five equivalent envelope conformations in which each carbon atom alternates as the out-of-plane atom. o Cyclohexane - Cyclohexane adopts a number of puckered conformations – the most stable of which is the chair conformation (all bond angles are close to 109.5° and all bonds on adjacent carbons are staggered). - Six C-H bonds are called axial bonds (roughly perpendicular to the equator of the ring). - 3 of the axial bonds point straight up; while the other 3 point down. - Six of the C-H bonds are called equatorial bonds (lies rought along the equator of the ring). - The equatorial bonds alternate, first slightly up and then slightly down as you move from one carbon to the next. - If the axial bond on a carbon points upward, the equatorial bond on that carbon points slightly downward, and vice versa.


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