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Introduction to Organic Chemistry

by: Jordy Rice

Introduction to Organic Chemistry CHM 122

Marketplace > Wake Forest University > Chemistry > CHM 122 > Introduction to Organic Chemistry
Jordy Rice
GPA 3.97

Paul Jones

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Paul Jones
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This 6 page Class Notes was uploaded by Jordy Rice on Wednesday October 28, 2015. The Class Notes belongs to CHM 122 at Wake Forest University taught by Paul Jones in Fall. Since its upload, it has received 18 views. For similar materials see /class/230750/chm-122-wake-forest-university in Chemistry at Wake Forest University.

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Date Created: 10/28/15
Chemistry 122 Review of General Chemistry Some skills you should have at the beginning of Chm 122 1 Given a molecular formula and connectivity data draw a Lewis Structure 2 Recognize when a single Lewis Structure does not adequately represent a molecule and be able to then draw appropriate resonance structures and to judge their relative contributions to the overall structure 3 Calculate formal charge 4 Determine approximate bond angles and molecular shape from Lewis Structure 5 Determine bonding and molecular dipoles 6 Determine the hybridization of carbons in molecules These topics will be reviewed quickly at the beginning of the semester 1 Lewis Structure see Carey 11 14 A Lewis Structure conveys the following information connectivity of atoms type of bonding single double triple and location of nonbonding lone pairs of electrons Below is an example of how to deduce the Lewis Structure of acetone a chemical you will encounter often in class and lab as well in fingernail polish remover if you do that sort of thing from the condensed formula 9 HCCCH H 39139 39139 H simply ll the remaining six valences on the two carbons with the six hydrogens from the condensed formula 0 II gt c gt acetone oxygen will make C C 2 Valem bonqs so begin with the while carbon Will CO group the carbon make four and 0 gen must be double bonded in the condensed formula the oxygen is only bonded to the carbon The carbon t en has two additional covalent bonds these must be to the other two carbons if they were to hydrogen there would be no more available bonds to make to the other carbons CH32CO I H H H H count electrons oxygen as six valence electrons and is only using two in bonding therefore it must have four in non bonding lone pairs other atoms have the proper number of electrons from bonding pairs Learn to recognize patterns in bonding in organic molecules Most atoms involved in stable organic molecules will have similar or identical types of bonding For e e an oxygen atom participating in two covalent bonds eithertwo single bonds or one double bond will have two lone pairs An oxygen with only one covalent bond will almost always have three lone pairs and a formal charge of1 Carbon will have four covalent bonds and no lone pairs in a stable molecule Hydrogen will never have more than one covalent bond Nitrogen that has three covalent bonds will have one lone pair and a formal charge of zero Nitrogen that has four covalent bonds will have no lone pairs and a formal charge of 1 The halogens will generally have one covalent bond and three lone pairs 2 Resonance structures see Carey 19 Molecules with multiple bonds such as acetone above generally can not be accurately represented by a single Lewis Structure The simplest explanation for this is that the electrons in a multiple bond can be divided unevenly between the two atoms involved without breaking the connection between the atoms Resonance structures must have the same connectivity and the same overall molecular charge Neither atoms nor electrons can be added or eliminated atoms can not be moved the connectivity can not change the structures are meant to represent a single compound and a single compound can not have varying amounts of atoms or electrons or different connectivity of the atoms Each resonance structure is a correct Lewis structure for that particular connectivity of atoms See below for the resonance structures of acetone 6 o o o H 39339 H 399 lt gt HI3 IH HE Q EH H Four electrons are shared by the carbon and oxygen Without breaking the connection between the two atoms two of the electrons can be moved to the electronegative oxygen The quotcurved arrowquot in the first structure represents the quotmovementquot of these electrons During the course man reactions will involve the movement of electrons and the use of curved arrows will be used to keep track of this The and signs in the structure on the right represent the formal charges of the atoms the sign is nearest The two structures shown are resonance structures of the compound acetone Neither structure alone is adequate to explain all of the properties and behaviors of the compound The true structure is a hybrid of the two resonance structures with a greater contribution from the more stable structure In this case that is the structure on the left All atoms are neutral and have full octets in contrast to the structure on the right in which two atoms are charged and one the carbon does not have a full octet 3 Formal Charge see Carey 16 Molecules can be neutral or electrically charged The overall charge of any molecule is equal to the sum of the formal charges of the atoms in the molecule Formal charges are generally indicated in organic structures by a single or sign next to the charged atom An atom without a sign is assumed to be neutral It is very unusual for an organic molecule to possess an atom with a formal charge of more than 1 or less than 1 Using the resonance structures for acetone shown above the formal charges for two atoms in the CO double bond are calculated The formula for formal charge is group valence electrons number of bonds number of unshared electrons So for the structure on the left the more stable resonance structure Note that for main group elements the group number is the number of FC carbon 4 4 6 2 valence electrons 0 0 FC oxygen 4 0 For the structure on the right the less stable structure FC carbon 4 3 0 1 FC oxygen6 l 6l In both structures the sum of the formal charges is zero 4 Molecular shape see Carey 110 A useful tool for predicting molecular shape is the VSEPR theory Part of this theory is that electrons same charge will repel one another and that molecular bonds and lone pairs are simply electrons Therefore molecular shape should conform to whichever conformation maximizes the separation of the bonds and lone pairs Despite serious aws with the overall theory it does an excellent job in approximating bond angles in molecules For predicting molecular shape VSEPR theory considers only how many quotthingsquot are attached to an atom For this purpose lone pairs are considered the same as atoms In the case of most organic molecules there are three possibilities of concern Four substituents SP3 hybridized atom H 3 10950 H6 C quot H H H methane ammon1a Carbon having four single bonds is called quottetrahedral39 because the ammonia also has four molecular shape can be accurately substituents Therefore represented by a tetrahedron The the HNH bond angle will be four quotsubstituentsquot atoms to which close to 1095 degrees the carbon is bonded occupy the comers of a tetrahedron with the carbon at the center The bond angles for any two substituents and the carbon are therefore 1095 degrees This is of course atheoretical value but experimentally most tetrahedral carbons are found to have bond angles quite close to the theoretical number Counting the lone pair Three substituents spz hybridized atom O H 3 120 C H H Formaldehyde Carbon having two single bonds and a double bond has only three substituents the double bond is counted only once in VSEPR Therefore the maximum separation of the bonding electrons will be if the three substituents occupy the same plane The bond angles must then be 120 degrees The carbon is trigonal planar 5 Molecular Dipoles see Carey 15 111 Two substituents Sp hybridized atom H C N hydrogen cyanide Carbon having one single bond and one triple bond or two double bonds has only two substituents Obviously these two substituents will lie on opposite sides of the carbon giving the molecule a linear shape and a bond angle of 180 degrees Note that the nitrogen has only two substituents as well the carbon and the lone pair Therefore the carbon and the lone pair must be linear with the nitrogen Electrons are not distributed evenly throughout molecules Some atoms are more electronegative than others you should know which and therefore attract electron density toward themselves more than other atoms This is particularly apparent when two atoms of widely disparate electronegativities are bonded to one another Consider the carbonchlorine bond Chlorine is extremely electronegative while carbon is not The two electrons that make up this bond will be quotpulledquot closer to the chlorine than the carbon The bond will then be quotpolarizedquot so that the chlorine end has some negative character and the carbon end has some positive character Each bond not between two identical atoms will be polarized to some extent The sum of these bond dipoles will confer a molecular dipole These dipoles are important in understanding molecular reactivity A useful piece of information to keep in mind is that no matter how many polarized bonds a molecule may contain if the molecule is completely symmetric it will not have a molecular dipole Consider carbon tetrachloride four highly polar carbonchlorine bonds are present in the molecule arranged in the familiar tetrahedral pattern about the carbon The resultant from adding the four dipole vectors is a zero dipole The four dipoles cancel one another and the molecule is nonpolar 6 Orbital Hybridization see 115 118 A useful way of thinking about orbitals and bonding in organic chemistry is orbital hybridization Carbon has four valence orbitals one s atomic orbital and three p atomic orbitals In the orbital hybridization model one imagines that carbon mixes various atomic orbitals to make bonding orbitals There are three ways the four orbitals can be mixed and be consistent with observed bonding They are shown below J i u Ill The carbon that is bonded to three The carbon that iS bonded to four substituents one of which is a double different substituents is said to be bond is said to be sp2 hybridized The carbon that is bonded to only 8133 hybridized That is the four There are three equal bonding orbitals two substituents is said to be sp orbitals involved in bonding are sp2 orbitals that are formed by the hybridized One bond to each equivalent orbitals formed by mixing mixing of one s orbital and two p substituent is formed with an sp one s and three p orbitals Remember orbitals This leaves one p orbital hybridized orbital formed by the the number of orbitals mixed must that is unmixed It is this p orbital mixing of one s and one p atomic equal the number of orbitals formed that participates in the pi bond orbital This leaves two p atomic So an sp3 hybridized carbon as is the the double bond is made of one orbitals unmixed These orbitals one above will be a tetrahedral carbon sigma bond and one pi bond participate in pi bonds Suggested Review Problems To be sure you understand these concepts you should work the following problems 16171819110116117118119123124125126129130132133134 142 145


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