Exam One Study Guide
Exam One Study Guide CHM 1240
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This 4 page Study Guide was uploaded by Maggie Bruce on Saturday September 24, 2016. The Study Guide belongs to CHM 1240 at Wayne State University taught by Stanislav Groysman in Fall 2016. Since its upload, it has received 290 views. For similar materials see Organic Chemistry 1 in Chemistry at Wayne State University.
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Date Created: 09/24/16
CHM 1240 Exam One Study Guide Chapter 1 o Section 4 – Formal charge The charge on atoms and molecules Formal charge = number of normal valence electrons – minus number of bonds – minus number of lone electrons o Section 5 – Resonance forms and the curved arrow formalism More than one Lewis structure may be possible, called resonance Each resonance form is needed to completely describe the molecule Curved arrows are used to move electron pairs to get new resonance forms Double headed arrows are used between resonance forms o Section 6 – Molecular orbitals Atomic orbitals combine to become molecular orbitals When figuring out the molecular orbitals, only count lone pairs and single bonds 2 is sp 3 is sp2 4 is sp3 o Section 7 – Bond strength The energy that holds molecules together can be quantified Enthalpy is the difference in total bond energy between products and reactants Negative enthalpy means the product is more stable Positive enthalpy means the reactant is more stable Bond dissociation energy is the amount of energy needed for homolytic bond cleavage, where atoms split electrons evenly Fish hook arrows are used to draw this o Section 8 – Acids and bases A Lewis acid has an empty orbital and will accept two electrons (electrophile) A Lewis base has two electrons that it will give (nucleophile) Chapter 2 – Alkanes o Section 1 - Preview C n 2n+2 Methane is the simplest, CH 4 o Section 2 – Hybrid orbitals Methane has sp3 hybridization and is tetrahedral o Section 3 – the methyl group and methyl compounds Methyl group is CH 3 A methyl compound is CH X 3 o Section 5 – Ethane, ethyl compounds and Newman projections Ethane is C 2 6 Newman projection allow us to see the three dimensional structure when there is free rotation around the single bond Look down the C-C bond The hydrogens will be staggered (you can see the hydrogens on the front and back carbons), or eclipsed (you can only see the front hydrogens) Eclipsed has a higher energy Ethyl compounds C H X 2 5 o Section 6 – structure drawings Three dimensional structures are drawn with lines, wedges, and dashes Lines mean it is in line with the plane Wedges mean it is coming out of the plane towards you Dashes mean it is going into the plane away from you o Section 7 – Propane and Propyl compounds Propane is C H3 8 For newman projections, look down the bond between carbon one and two, there will be a methyl group coming off of the second carbon Methylene is the CH gr2up in the middle Replacing one of the hydrogens give an isopropyl compound Methyl is the two outside CH gr3ups Replacing one of the hydrogens gives a propyl compound o Section 8 – Butane and Butyl compounds Butane is C H4 10 There are four possible newman projections Two staggered – one with the two methyl groups opposite each other and one with them close Two eclipsed – one with the methyl groups right next to each other and one with them further apart In linear butane there are two butyl compounds Butyl – replacing one of the outside hydrogen Sec-butyl – replacing one of the inside hydrogens Isobutane is when there is a central carbon atom attached to three carbons and a hydrogen Tert-butyl – replacing the hydrogen on the central carbon Isobutyl – replacing one of the other carbons o Section 9 – Pentanes and Pentyl compounds Pentane is C H5 12 Isopentane has a methyl group coming off of a secondary carbon Neopentane has a central carbon attached to four other carbons o Section 10 – Naming alkanes The IUPAC system 1. Find the longest chain of carbons, use the root word that matches that number and put –ane at the end 2. Number the carbons so that the substituents are coming off the lowest number possible use the number and the substituent name as a prefix 3. Order the substituents alphabetically o Section 12 – Cycloalkanes C n 2n All the carbons connect in a circle Count the carbons and use that root word, add ane at the end and cyclo at the beginning Cis and trans may be needed because there is no free rotation around the bond o Section 15 – Acid and base reactions Any reaction between an empty orbital and a pair of electrons The basic electrons always move from negative to neutral to positive When the electrons move, the atom is bonded to the new molecule Chapter 3 – Alkenes and Alkynes o Section 2 – Alkenes: structure and bonding Alkenes contain a double bond (they have a pi bond) o Section 3 – Derivatives and isomers of alkenes Ethene – C 2 4 Replacing one hydrogen forms vinyl Replacing one hydrogen with a methyl group forms propene Alkenes cannot rotate around the double bond Changing each hydrogen results in a different molecule Alkenes can be cis (z) or trans (e) Cis means the two most significant groups are on the same side Trans means the two most significant groups are on opposite sides o Section 4 – Nomenclature The double bond must be a part of the main chain The double bonds must be given the lowest number possible when numbering the carbons Double bonds are more significant that substituents The suffix is –ene If there is more than one double bond you need to put the number of the bonds and use diene, triene, etc. as the suffix o Section 5 – The Cahn-Ingold-Prelog priority system Used when deciding which groups off of a double bond are more significant First look at the atomic number of the atoms attached to the carbons, the higher ones are more significant If there is a tie, add up the atomic number of the atoms attached to the first atom, the one that is higher is more significant o Section 7 – Double bonds in rings Any size ring can have a double bond To name use correct root word and add cyclo- and –ene o Section 9 – Alkynes: structure and bonding Contain a triple bond These compounds are linear o Section 11 – Derivatives and isomers of alkynes Ethyne – C 2 2 Replacing one hydrogen forms an ethynyl compound Replacing one hydrogen with a methyl group forms propyne o Section 12 – triple bonds in rings Because the triple bond makes a linear compound, only large rings can have a triple bond To name use correct root word and add cyclo- and –ene o Section 13 – acidity of alkynes Hydrocarbons are usually very weak acids Alkynes are reasonably strong bronsted acids It is much easier to remove a hydrogen from one of the ends of an alkyne than it is to remove one from an alkane or alkene o Section 15 – Degrees of unsaturation Degrees of unsaturation are the number of pi bonds and rings in a molecule So a ring would have one degree of unsaturation A ring with a double bond would have two degrees of unsaturation o Section 16 – addition reactions of alkenes and alkynes Alkenes and alkynes react with bronsted acids The bronsted acid breaks up and attaches where the double or triple bond was o Section 17 – Mechanism of the addition of hydrogen halides to alkenes First the double bond breaks and attaches to the hydrogen, the electrons from the bond in the acid go to the halide Then a pair of electrons on the halide bond with the now positive carbon
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