Chapter 2: Polar Bonds and Their Consequences Coverage: 1. Polar Covalent and Dipole Moment 2. Intermolecular Forces 3. Resonance 4. Acids and Bases – Bronsted-Lowry and Lewis 5. Prediction of Acid/Base Reactions 6. Drawing Organic Structures Goals: 1. Be able to calculate the charge separation on two bonded atoms given the bond length and the measured dipole moment. 2. Be able to calculate formal charges on atoms. 3. Be able to draw resonance forms for simple organic molecules. 4. Know the definitions of Bronsted-Lowry and Lewis acids and bases. 5. Understanding the meaning of pKa and use it to predict the direction of an acid-base reaction. 6. Know how to used curved arrows to indicate flow of electrons in a chemical reaction. 7. Know the structures of common organic bases and acids. 8. Know the nature of intermolecular forces, including hydrogen bonding, dipole-dipole interactions, and London dispersion forces. 9. Know how to draw condensed and skeletal structures for simple hydrocarbons. 2-1 Bond Dipoles A bond is polar if there is a difference in electronegativities of the two atoms. If the difference, DEN, is greater than or equal to 0.9 (Pauling scale), the bond is ionic. DEN > 2.0 DEN 2.0 and 0.5 DEN < 0.5 Ionic Bond Polar Bond Nonpolar Bond Calculation of Bond Dipole Moment = Q x r x 4.8 where Q is magnitude of the charge separation (electrons) r is internuclear distance (Angstroms) is the bond dipole moment (debyes, D) 2-2 2-3 Example: The carbonyl group in a ketone O Carbonyl Group of Ketone CH3CCH3 The dipole moment of the group is measured at 2.4 D. The bond distance, r, is 1.21 Angstroms. Calculate the charge separation Q = r - 0.413 = _____2.4_____ = 0.413 electrons O 1.21 x 10-10 CH3CCH3 The symbol (delta) indicates a partial charge +0.413 on the atom, either negative or positive. 2-4 Some typical Bond Dipoles: Bond Debyes C-N C-O C-F C-Cl C-Br C-I H-C H-N H-O C=O C=N - 0.22 0.86 1.51 1.56 1.48 1.29 0.3 1.31 1.53 2.4 3.6 Q: Why is the C-F bond less polar than the C-Cl, since F is more electronegative than Cl? 2-5 Molecular Dipole Moments Molecular Dipole Moments are the vector sum of the individual bond dipole moments. They depend on the magnitude and direction of the bond dipoles. NH3 H2O H H N H H : = 1.5 D H CO2 : O : 1.9 D .. .. :O=C=O: 0.0 D CH3Cl H H H C Cl 1.87 D 2-6 2-7 Intermolecular Forces Intermolecular Forces (IMFs) determine the physical properties (e.g. melting point, boiling point) of organic molecules. There are three major types of IMFs; hydrogen bonding, dipole – dipole and London dispersion forces 1. Hydrogen Bonding. This force is dominant if present. It is not a true bond. Organic compounds must possess an O-H or N-H group to hydrogen bond to itself. Example: Methanol : H3C O : O CH3 - H H + Hydrogen Bond 2-8 Hydrogen Bonding raises the boiling point of organic molecules. Isomers of C3H9N .. H3C N CH3 .. N CH3 CH3CH2 CH3 Boiling Point No. of NH Groups CH3CH2CH2 H 3.50C 0 370C 1 .. H N H 490C 2 2-9 2. London Dispersion Forces - the next most dominant force. It arises from temporary dipoles induced by nearby molecules. This force depends on the surface area and, hence, molecular weight. Nonpolar Nonpolar + - + - Coordinated Temporary Dipole caused by distortion of electron density around molecule. When these molecules diffuse away from each other, the coordinated dipole disappears. 2-10 London Dispersion forces are proportional to surface area, and therefore molecular weight. Which hydrocarbon among these straight-chain alkanes has the highest boiling point? CH3CH2CH2CH3 CH4 Lowest CH3CH2CH2CH2CH2CH3 Highest If hydrocarbons have the same molecular weight, the molecule with the least branching will possess the highest boiling point. Consider isomers of C6H14 Boiling Point 60oC 69oC 58oC 2-11 3. Dipole- Dipole Interaction. These forces result from permanent dipole moments in polar molecules. + - + - Rank these molecules, which have similar molecular weights, according to boiling point. OH O Lowest Highest Summary 1. 2. 3. Hydrogen bonding raises boiling point. Increased molecular weight raises boiling point Polar molecules possess higher boiling point than nonpolar molecules if same molecular weight. 2-12 Geckos climb vertical and even inverted surfaces with ease using millions of micronscale adhesive foot-hairs on each toe (setae). Each foot-hair splits into hundreds of tips only 200 nanometers in diameter, permitting intimate contact with rough and smooth surfaces alike. All of this occurs due to van der Waal forces (sum total of intermolecular forces), not sticky substances. 2-13 Resonance Structures Some structures are not adequately expressed by a single structure. If two or more valence bond, structures are possible, the molecule shows characteristics of both – termed resonance structures. H + C H H H H C N N H H H + Resonance forms differ in the placement of p and nonbonding valence electrons. The atoms themselves occupy the same positions in both resonance forms. The molecule is not changing, it is just represented by two or more different Lewis structures. .. :O: :O: H3C C - H3C C :O: .. - :O: The charged is shared by the two oxygen atoms. 2-14 Rules for Resonance 1. The real structure is a composite of the resonance forms. The structure does not change, but is a resonance hybrid of the individual structures. :O: H H C C H .. :O: C C H :O: .. - :O: -1/2 :O: H C C H :O: -1/2 2. Resonance forms differ only in the placement of electrons, not the placement of atoms. 3. Different resonance forms are not necessarily equivalent. 4. Resonance forms must be valid Lewis structures and obey normal rules of valency. Usually the octet rule must be obeyed, but in some cases a valid structure shows only six electrons around an atom. (See previous example). 2-15 To become proficient in drawing resonance structures, we need to learn to recognize five patterns. I will illustrate all of these on the chalkboard. 1. Allylic lone pair (two curved arrows): 2. Allylic positive charge (one curved arrow): 3. Lone pair adjacent to positive charge (one or two curved arrows): 4. A π bond between two atoms of differing electronegativity. (one arrow): 5. Conjugated π bonds enclosed in a ring: (three curved arrows): 2-16 Assessing Resonance Structures; Three Rules 1. Minimize charges. Resonance forms with no charges are more significant. Example on board. 2. Electronegative atoms, such as N, O, and Cl, can bear a positive charge, but only if they possess an octet of electrons. Example on board 3. Avoid drawing resonance structures in which two carbon atoms bear opposite charges. Example on board. 2-17 Acids and Bases Bronsted-Lowry Acid – substance that donates a hydrogen ion (proton). Bronsted-Lowry Base – substance that accepts a hydrogen ion (proton). HA + H2 O Ka = [H2O] [A-] [HA] [H2O] A- + H3O+ pKa = -log Ka The lower the pKa, the stronger the acid Acid-Base Reaction Conjugate Acids Conjugate Bases Ka pKa HBr + H2O H3O(+) + Br(-) HBr H3O(+) Br(-) H2O 105 -5 CH3CO2H + H2O H3O(+) + CH3CO2(-) CH3CO2H H3O(+) CH3CO2(-) H2O 1.77*10-5 4.75 C2H5OH + H2O H3O(+) + C2H5O(-) C2H5OH H3O(+) C2H5O(-) H2O 10-16 16 NH3 + H2O H3O(+) + NH2(-) NH3 H3O(+) NH2(-) H2O 10-34 34 2-18 Common Organic and Inorganic Acids Weakest Strongest Acid ethane ammonia ethanol water methylammonium ion bicarbonate ion phenol ammonium ion carbonic acid acetic acid benzoic acid hydrogen fluoride phosphoric acid hydronium ion oxonium ion sulfuric acid hydrogen chloride hydrogen bromide hydrogen iodide Formula CH3CH3 NH3 CH3CH2OH H2 O CH3NH3+ HCO3– C6H5OH NH4+ H2CO3 CH3CO2H C6H5CO2H HF H3PO4 H3 O+ R2OH+ H2SO4 HCl HBr HI pKa 48 33 15.9 15.7 10.64 10.33 9.95 9.24 6.26 4.76 4.19 3.5 2.1 – 1.74 – 1.75 – 5.2 –7 –8 –9 Conjugate Base CH3CH2– NH2– CH3CH2O– OH – CH3NH2 CO32– C6H5O– NH3 HCO3– CH3CO2– C6H5CO2– F– H2PO4– H2O R2O HSO4– Cl – Br – I– 2-19 Predicting Acid –Base Reactions In general, an acid with a lower pKa will react with a conjugate base of an acid with a higher pKa. O O CH3COH Acid pKa + Na+OHBase + CH3CO- Na Conjugate Base 4.76 + H2O Conjugate Acid 15.7 Keq ≈ Ka reactant/ Ka product ≈ 10-5/10-16 = 1011 Very Large. This means that the above reaction is quantitative. In general, if there is at least 5 pKa units difference, then the reaction will be quantitative, that is, the reaction will be nearly complete.* 2-20 Will the following acid and base reaction be quantitative, i.e go to completion? CH3CH2O Acid pKa H + Na+ OH- Base + CH3CH2O- Na+ Conjugate Base H2O Conjugate Acid 15.9 15.7 Answer: the reaction will take place, but it will not be quantitative. Acetone, a ketone, possesses an acidic hydrogen on the methyl group. Will acetone quantitatively react with sodium amide (NaNH2)? Answer: Yes H O H H C C C H H + Na+NH2- H Acid pKa H O 19 -.. H C C C H H Base Na + + NH3 H Conjugate Base Conjugate Acid 33 2-21 Irving Langmuir His interest in fundamentals involved him in the theory of chemical bonding in terms of electrons, and he elaborated on ideas first expressed by Gilbert Lewis. Langmuir proposed that octets could be filled by sharing pairs between two atoms—the "covalent" bond. His studies of surface chemistry—the study of chemical forces at the contact surfaces (interfaces) between different substances, where so many biologically and technologically important reactions occur—earned him the Nobel Prize in chemistry in 1932. Gilbert Newton Lewis 1875-1946. An American chemist at the University of California, Berkeley who revolutionized concepts of bonding. The Lewis octet rule is named after him. He also generalized the concept of acids and bases and we now think of Lewis acids and Lewis bases. He had a summer home in Marin County, Calif. 2-22 Lewis Acids and Bases Lewis Acid – substance that accepts an electron pair Lewis Base – substance that donates an electron pair B : Base – electron rich Nucleophile + A B : A Acid – electron poor Electrophile Curved Arrow Formalism - a curved arrow always indicates the direction of electron flow. The electron flow is from the Lewis base (electron rich) to the Lewis acid (electron poor) or from the nucleophile to the electrophile. 2-23 F3 B + :NH3 F3B---NH3 Boron trifluoride : Acid Acid-Base Adduct Base 2-24 Drawing Structures Focus on Hydrocarbons Alkanes 1. Complete Lewis Structures H H H H H C H H H H H C C C C C C C H H H H H H H H CH3 2. Condensed Formula Structure 3. Skeletel or Line-angle Formulas CH3CH2CHCH2CH2CH2CH3 2-25 Alkenes H H H H H C H H H C C C C C C C H H H H H H H H CH3 CH3CH2CHCH=CHCH2CH3 2-26