EM 3013 RGANI EMISTRY I LETURE NTES 1 APTER 2 1. ormal harge The Lewis structures we have drawn thus far have all been neutral covalent molecules. owever, some covalently bonded molecules may contain charged atomic components. urthermore, many ionic species also contain covalent bonds. To manage the electronic bookkeeping we assign a formal charge to each of the atoms of a structure. The formula, ormal harge = Valence Electrons - 1/2( Bonding Electrons) - N on- Bonding Electrons will be used for this purpose. + or Nitrogen or ydrogen N = VE-(1/2BE + NBE) = VE-(1/2BE + NBE) = 5 - ( 8/2 + 0) = 1 = 1 - (2/2 + 0) = 0 Nitrogen bears the positive charge alculation of ormal harge 2. Polar ovalent Bonds Like most things in nature there are "gray" areas between the ionic and covalent bond. A polar covalent bond is a covalent bond between two atoms of differing electronegativity. Electronegativity increases as you move to the right and the top of the periodic table. Electronegative elements need more electron density near them. We can predict the nature of a chemical bond by using the difference in Pauling Electronegativities, ( E.Neg.). or example, acetonitrile is a compound containing a triple bond where the electronegativity of nitrogen is greater than that of carbon. Thus the electron density on the nitrogen is greater than on the carbon. δ+ δ 3 N Pauling 2.5 3.0 Electroneg. E.NEG = 0.5 PLAR VALENT BND 3. Dipole Moments The polarization of a covalent bond (separation of charge densities) gives rise to a dipole along the internuclear axis. This dipole is a measurable quantity which is exerted in a localized direction ( a vector). The dipole moment, µ is a product of the charge separation in electrostatic units, q,( esu), the bond length, r, (A) and a constant term (10 10 debye/esu. A ).
2 µ = 2.76 D No Net Dipole µ = q x r x 10 10 debye/esu A X Y A vector can be resolved into it's three component directions. A combination of dipole moments may result in a net cancellation. Z Bond Dipoles 4. Resonance Structures Resonance structures are groups of Lewis-type representations which differ only in t he position of localized electron density. A resonance structure is a formula which can be written which involves the movements of bonds, unshared electron pairs, single electrons or charges. It is important to understand that atoms stay fixed in space in resonance structures, only electrons are shared over one atom. Structures which involve the movement of atoms are not valid resonance forms. The standard rules for writing valid Lewis structures (octet rule, no pentavalent carbons, etc.) apply for resonance forms. Two possible Lewis Structures for the acetate ion: 3 a b or xygen a: = 0 = -1 or xygen b: = -1 = 0 RESNANE STRUTURES D NT USE TW SINGLE ARRWS, TIS DEPITS AN EQUILIBRIUM a b - 1/2-1/2 Each xygen has a -1/2 ormal harge. The one double bond is shared. RESNANE YBRID Use of Double-eaded Arrows to indicate Resonance Structures The formalism used to depict resonance structures is the double-headed arrow. This must be well-understood to be distinct from the pair of single headed arrows used to describe an equilibrium reaction. It must be remembered that while we may draw several separate structural formulae to depict the various resonance contributors, these are not i ndividual molecules in equilibrium with one another, but a composite depiction of the
chemical species being discussed. Resonance and equilibrium ARE NT TE SAME TING. 3 A molecule is a weighted average of all its contributing resonance structures. The structural depiction of the weighted average is called the Resonance ybrid. The best real world analogy for a resonance hybrid would be to consider a biological hybrid formed when two related species form offspring. A good example of this is the mule, a hybrid formed in the crossing of a horse and a donkey. A mule is always a mule, it show some characteristics of horses and some of donkeys, but it is always a mule. It never spends part of its time as a horse and part of its time as a donkey ( this would be like equilibrium).\ While a molecule is an average of its contributing resonance structures, it need not be true that all resonance forms contribute equally. To assess the weighting of the contributions we examine the various structures and consider their stability as if they were real individual molecules ( which, of course, they are not). The forms that have the greatest stability are the forms which have the greatest contribution to the resonance hybrid. 1. Identical structures contribute equally. 2 2 2 2 allyl anion 2. Structures with the greater number of bonds are more important 2 2 BEST 2 2 2 2 3. Structures that delocalize charge or unshared electrons are important 2 3 2 3 4. When considering structures with separated charges, atoms with appropriate electronegativity will best accomodate the charges. 3 3 3 3 BEST Rules for Writing Resonance Structures In General, the more important resonance structures one can write for a compound, the more stable will be that compound. Resonance structures imply additional orbital overlap, and hence, stability.
4 5. Bronstead Acidity and Basicity A general acid A, in aqueous solution, can dissociate into its component parts, + (the proton) and A - (the conjugate base of A). The larger the equilibrium constant for this ionization reaction, the stronger the acid will be. Because of the great range of organic acids, chemists usually discuss this subject using a log scale, where pka is defined as the negative log of the equilibrium constant Ka. Most compounds fall within the pka range of 55 (exceedingly weak organic acids like methane) to -7 (strong inorganic acids like 2S4). hemical reactivity and reaction mechanisms are often intimately related to acid strength. actors related to acid strength will be of critical importance throughout this course. Ka A + + A - 1/Ka Ka = + A - A pka = - log Ka + + Ka = 10-16 pka = 16 Equilibrium lies to the left, water is only slightly ionized. Water is a relatively weak acid. Smaller, more negative pka...stronger acid Larger, more positive pka...weaker acid Acids, Bases and pka 6. Acid, Base Reactions Being able to predict whether an acid-base reaction will proceed or not is very important. A reaction will proceed to the right (towards products) only if the conjugate acid is a weaker acid.
5 A + B :- -B + A: - acid base conjugate conjugate acid base N 3 + : - + :N 2 - pka = 35 pka = 16 Equilibrium favors reactants, little products formed. Keq =? 7. Lewis Acids and Bases To solve : combine the Ka for each of the individual species. Keq = Ka(N 3 ) x 1/Ka() = 10-35 /10-16 = 10-19 The reaction is extremely unfavorable! alculation of acid-base reaction using pka data. A Lewis acid is defined as an electron acceptor. They are species which are at least one electron pair short of a filled outer shell configuration(either octet or duet), because of this they are very reactive toward electron sources. Another term for Lewis acids is electrophiles (electron loving species).typically, a Lewis acid is a cation, (e.g. +, Li +, (3)3 + ) or a metal atom in a salt.a Bronstead acid (proton source) is also a Lewis acid A Lewis base is defined as an electron donor. They are species that react with Lewis acids by supplying an electron pair. Another name for Lewis bases is nucleophiles (nucleus loving species). A typical Lewis base is an anion (e.g. -) or a neutral heteroatom with at least one pair of non-bonding electrons (e.g. :N3). A Lewis acid must have available a low energy empty orbital to accept an electron pair from the Lewis base. empty orbital + B B louride anion Lewis base Boron triflouride Lewis acid Tetrafluoroborate anion Reaction of Lewis Acid with a Lewis Base 7. ombustion Analysis onverting organic compounds to carbon dioxide and water by combustion provides an analytical means for determination of the % composition of the molecule in question. This allows an organic chemist to assign an empirical formula to the compound in question. The equations:
Weight in sample = weight 2 x (2.016g / 18.016 g 2) 6 Weight in Sample = weight 2 x (12.01 g / 44.01 g 2) % in Sample = Weight in sample / sample weight % in Sample = Weight in Sample / sample weight