Name: Unit 1 Packet: Background, Lewis Structures, Resonance, Formal Charge, VSEPR, Hybridization, Isomerism, Acids and Bases

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1 Name: Unit 1 Packet: Background, Lewis Structures, Resonance, Formal Charge, VSEPR, Hybridization, Isomerism, Acids and Bases

2 General Organic Chemistry Orbital Key Terms For Unit 1 Orbital Notation Isomerism Structural Isomerism Geometric Isomerism Bonding Anion Aufbau Principle Bond Angle Bond Length Cation Covalent Bond Double Bond Electronegativity Formal Charge Full Octet Hindered Rotation Hund s Rule Hybridization Ionic Bond Line Angle Non-polar covalent bond Octet Rule Pauli exclusion Principle Polar covalent bond Pi bond (π) Resonance hybrid Sigma bond (σ) sp hybrid orbital sp 2 hybrid orbital sp 3 hybrid orbital Tetrahedron Trigonal Planar Valence Electrons Valence Shell Valence Shell Electron- Pair Repulsion Nomenclature Alcohol Aldehyde Carbonyl group Carboxyl group Carboxylic Acid Ether Functional Group Ketone Acids and Bases Ka pka ph Strong Acid Weak Acid Conjugate Acid/Base Pair Bronsted-Lowry Acid Bronsted-Lowry Base Lewis Acid Lewis Base Nucleophile Electrophile 2

3 What is Organic Chemistry? Organic Chemistry: What makes carbon so special? What other elements are commonly involved? Important Dicsovery: Friedrich Wohler (1828) Accidental synthesis of Urea ( ) from ammonium cyanate ( ). Atomic Structure Review Any atom is made up of three subatomic particles:, and. Two atoms of the same element with different numbers of are called. An atom of an element with different numbers of and are called. Anions are charged ions. Cations are charged ions. Ex: Ex: In Chemistry, the subatomic particle that we are most concerned with is the, and particularly, the in the outermost, or shell. These are the particles that allow different atoms to with one another. Electrons are located in Principle Energy Levels: Electrons Sublevels: Orbitals: Electrons can be represented visually using orbital notation 3

4 Ground State Orbital Notation Orbital Notation is a visual method to represent the,, and of each electron in an atom or ion. The ground state of an atom represents its lowest energy, preferred configuration. The placing of electrons into an orbital notation is governed by three rules: Aufbau Principle: Hund s Rule: Pauli Exclusion Principle: To practice using orbital notation, consider the ground state configurations of carbon and bromine: Carbon, Z = 6 Bromine, Z = 35 Lewis Dot Diagrams Since we are primarily concerned with the in an atom, a scientist named G. N. Lewis devised another method of representing an atom that is commonly used called a Lewis Dot Diagram. Let s consider how these are drawn for several elements. H C P Xe I **We can piece these diagrams together to make Lewis structures for molecules. Ex: H 2 O, water** 4

5 Lewis Structures History: Purpose: Process: 1. Predict the Arrangement of the atoms: CH 4 COCl 2 NO 3-2. Count up the valence electrons 3. Connect the surrounding atoms to the central atom with single bond (one shared pair) 4. Determine how many electron pairs you have left 5. Place lone pairs around terminal atoms to satisfy octet (stick any leftovers on central atom) 6. If the central atom has not achieved an octet, form multiple bonds to do so 5

6 Lewis Structures Additional Practice CO CO 3 2- NH 3 SOCl 2 CH 2 O C 2 H 4 Resonance Structures Often, it is possible to draw more that one legal Lewis structure for a molecule. e.g. NO 3 - Each of these structures fulfills the requirements of a, therefore, they are each valid resonance structures. No single one of these resonance structures gives a picture of what the ion looks like. A resonance structure is a Lewis structure that contributes to the overall resonance of a molecule or polyatomic ion. Another example of resonance, the sulfate ion (SO 4 2- ), involves a concept called expanded octet. e.g. SO 4 2-6

7 Resonance Hybrids What does a molecule really look like? Similar to restaurant Diet Coke mixtures, not all of the resonance structures that can be drawn for a sulfate ion are created equal. How do we decide which ones contribute most to the actual picture of what a molecule looks like? To do this, we have to minimize the formal charge of each atom. e. g. CO 2 By minimizing the formal charge on each atom, we can isolate the resonance structure that best represents what a molecule or ion looks like. Try some others: C 2 O 4 2- SO 3 2- Evidence for Resonance Benzene A commonly cited example of evidence for the resonance theory is the molecule benzene (C 6 H 6 ). In a Lewis diagram of benzene, the carbons can be connected by either a single or double covalent bond. The known length of a C=C double bond is 133 pm, the known length of a C-C single bond is 154 pm, but the bond length of every carbon-carbon bond in benzene is 139 pm, a value in between that of a single and double. General Guidelines for Resonance Structures 1. Try to draw structures that are as low in energy as possible 2. The best structures tend to have the maximum number of bonds and the most octets 3. When structures are equivalent in terms of bonds and octets, minimize formal charge to find the more stable structure 4. All structures must be valid. Only electrons may be moved to change between structures, bonding sequence of atoms must remain the same. 5. Use curved arrows to show the movement of electrons. Only move lone pairs and multiple bonds. 6. Separate resonance structures by a double headed arrow. 7. Resonance stabilization is very important when it delocalizes or spreads a charge over two or more atoms 8. Negative formal charges are more stable on atoms with higher electronegativities. 7

8 Worksheet Resonance and Formal Charge 1. For each of the following compounds, draw the important resonance structures. Indicate which structures are major and minor contributors or whether they have the same energy.. (-) H :O: H.. (-) a. H C = N O:.. (+) e. H C C N:.. (-) :O:.. b. H C NH 2 :O:.... f. H 2 C C O CH 3.. (-).. :O H c. H C H (+) :O: :O:.. g. H C C C H H (-) H.... d. H C N = N: (+) (-) 2. Draw the important resonance structures for the following ions (+) (-) a. H 2 C = CH CH 2 b. H 2 C = CH CH 2 8

9 Summary of Bonding Patterns Atom Valence Electrons Positively Charged Neutral Negatively Charged B C N O Halogen Important Note: The bonds shown do not always have to be single bonds Ex: 9

10 Line-Angle Structures Often times when we are drawing organic molecules, a large number of atoms are involved and drawing standard Lewis structures can become somewhat hairy. e.g. 2-methylpropane (aka isobutane) A remedy for this is to represent molecules in what is called line-angle format. There are several important rules to remember when drawing line-angle structures: 1. Every angle and every end of a line that does not have an element symbol represents a carbon atom. 2. All non-carbon atoms (except hydrogens attached to a carbon) are shown with their chemical symbol 3. Hydrogen atoms attached to carbon atoms are disregarded. 4. When four lines are drawn from an intersection, an attempt should be made to show 3-D perspective using dashes and wedges. e.g. 5,5-dimethylhex-3-en-1-ol becomes 2,4,6-trinitrotoluene becomes 4-bromo-3-ethylpent-1-yne becomes 10

11 Worksheet Line-Angle Structures For the following structures, if the standard structure is shown, draw the line-angle equivalent. If the line angle structure is shown, draw the standard equivalent

12 VSEPR Theory VSEPR: VSEPR theory is used to predict the 3-D geometry of the terminal atoms around the central atom of a molecule. VSEPR theory tells us the shape of the molecule and the bond angles to expect. How to determine the 3-D geometry of a molecule from a Lewis Structure 1. Draw a Lewis structure for a molecule or polyatomic ion 2. Count the regions of electron density around the central atom a. Any bond (single, double, or triple) counts as one region b. A lone pair counts as a region c. The number or regions tells you the parent shape or electron arrangement of the structure 3. If there are any lone pairs around the central atom, the actual shape will be based on the parent shape, but differ from it in practice. Parent Shape Regions of Electron Density Example Molecule w/ Lewis Structure Actual Shape Bond Angles Drawing Hybridization of Central Atom CO 2 BF 3 CH 4 NH 3 H 2 O 12

13 Parent Shape Linear Trigonal Planar Tetrahedron Trigonal Bipyramid Octahedron Pentagonal Bipyramid 13

14 Hybridization Consider the molecule methane, CH 4. One carbon atom bonded to four different hydrogen atoms. Their orbital notations are shown below. C: H: 1s 2s 2p 1s Hund s Rule: Pauli Exclusion Principle: We know that single bonds are formed by the overlap of two orbitals that each have one unpaired electron to be able to form 4 bonds to carbon, one electron will have to be promoted from the 2s to the empty 2p orbital. C: 1s 2s 2p At present, of the four bonds that carbon is making to hydrogen, there are three 1s overlapping 2p bonds and one 1s overlapping 2s bond. From your knowledge of s and p orbitals, draw a diagram of what this molecule would look like. There are a couple of problems with the diagram above. First, experimental evidence shows us that the bond angles for the above molecule are incorrect all 4 hydrogen atoms should be equivalently spaced from one another. Second, all experimental evidence tells us that the four bonds in methane are equal in energy. To take into consideration these two problems, a bonding theory called hybridization was devised Hybridization: 14

15 Hybridization Continued sp hybrids: produced from one s and one p orbital. Two p orbitals remain unchanged and available for π bonding. Ex: Beryllium in BeCl 2 two sp hybrid orbitals form two σ bonds. Drawing 2p Electron 2p Hybridization 2p Promotion 2s 2s sp Ex: Carbon in C 2 H 2 two sp hybrid orbitals form two σ bonds. Two p orbitals remain available for π bonding. Drawing 2p Electron 2p Hybridization 2p Promotion 2s 2s sp sp 2 hybrids: produced from one s and two p orbitals. One p orbital remains unchanged and available for π bonding. Ex: Boron in BF 3 three sp 2 hybrid orbitals form three σ bonds. Drawing 2p Electron 2p Hybridization 2p Promotion 2s 2s sp 2 Ex: Carbon in C 2 H 4 three sp 2 hybrid orbitals form two σ bonds. One p orbital remains available for π bonding. Drawing 2p Electron 2p Hybridization 2p Promotion 2s 2s sp 2 sp 3 hybrids: produced from one s and three p orbitals. Ex: Carbon in CH 4 four sp 3 hybrid orbitals form four σ bonds. Drawing 2p Electron 2p Hybridization Promotion 2s 2s sp 3 Ex: Oxygen in H 2 O two sp 3 hybrid orbitals with unpaired electrons will form two σ bonds. The two sets of paired electrons exist as lone pairs. Drawing There is no electron 2p Electron promotion because Hybridization Promotion all orbitals are occupied 2s sp 3 15

16 Orbital Diagrams for Unhybridized s, p and d Orbitals 16

17 Representing Molecules with 3-D Perspective: The Orbital Diagram When we first learned to draw Lewis structures, we typically drew bond angles that were evenly spaced in 2-D. e.g. CH 4 Upon the introduction of VSEPR theory, it became clear that the original manner of representation did not give an accurate reflection of the 3-D shape of most molecules, so we began drawing molecules with perspective. e.g. CH 4 As we continue to draw more and more complex molecules in 3-D, such as those that involve pi bonds, their drawings become more and more intricate. Let s consider a few. Ethane, C 2 H 6 Ethene, C 2 H 4 Ethyne, C 2 H 2 **Challenge** Allene, H 2 C=C=CH 2 17

18 Worksheet: Hybridization and Lewis Structures 1. For each of the following, predict the hybridization and bond angles for the indicated atoms. a) :O:.... H O C O - H b) H H.. H C C O : (-) H H c) CH 3 (+) CH 3 N CH 3 CH 3 d) H H H C C (+) H H e) H.... :Cl C = C C = N H H H f) H C C H a. Hybridization: Bond Angle: b. Hybridization: Bond Angle: c. Hybridization: Bond Angle: d. Hybridization: Bond Angle: e. i. Carbon - Hybridization: Bond Angle: ii. Nitrogen - Hybridization: Bond Angle: f. Hybridization: Bond Angle: 2. From question #1, which structures are ions? 3. What type of orbitals are overlapping between the atoms in the following? a) b) c) H.. H C = O: H H H C = C = C H H.. H C O: H H 18

19 4. Draw Lewis Structures for the following. Remember, hydrogen and terminal halogens do not make double bonds. a. CO 2 d. N 2 H 4 g. NO 2 + b. CCl 4 e. NH 2 - h. BF 3 c. NH 4 + f. CO State the hybridization of each central atom from Question #4 a. b. c. d. e. f. g. h Use wedges and dashes for σ bonds to draw an orbital diagram of CH 3 C O: Draw any p orbitals, label bond angles. :O: H 19

20 Worksheet Bonding and Hybridization 1. For each of the following compounds: a. Give the hybridization for each atom except hydrogen b. Give the approximate bond angles for each atom except hydrogen c. Draw an orbital diagram using lines, wedges and dashed lines for sigma bonds. Draw the p orbital interaction for pi bonds.. I. H 3 O + III. CH 3 C = N H H II. (CH 3 ) 4 N + IV. CH 2 O 2. For each of the following: a. Draw the Lewis structure b. Indicate what type of orbitals are overlapping to form each bond c. Give approximate bond angles for each atom except hydrogen.. I. CH 3 C C C = O H III. (CH 3 ) 2 NH II... H 2 N CH 2 CN: IV. CH 3 CH = C(CH 3 ) 2 20

21 3. Predict the hybridization and geometry of each carbon atom in the following anion: :O: (-).. CH 3 C CH 2 4. Draw orbital diagrams of the pi bonding in the following compounds. Use lines, dashes and wedges to show sigma bonds a. CH 3 C CH 3 :O:.. b. CH 3 C C C = O H c. CH 3 CH = CH CH 2 CH 3 (you may realize there are two ways to do this one, circle the six coplanar atoms in this molecule) 21

22 Isomerism In the simplest terms, isomers are compounds with the same but a different arrangement of atoms. We will investigate this topic in greater detail in the future, but for now, it will suffice to introduce a few categories of isomerism and give examples of each. 1. Structural Isomers isomers that have different - - bonding Example: C 5 H Cis-Trans Isomers (aka Geometric) involves either a double bond or a ring; isomers have the same - - bonding, but a different spatial arrangement about the double bond or ring (because of rotation) Example: but-2-ene, CH 3 -CH=CH-CH 3 Extra Practice: Draw all of the isomers of C 6 H 14 22

23 Worksheet Bonding Part pentyne has the formula CH 3 C C CH 2 CH 3. Use dashed lines and wedges to draw a 3-D diagram of this molecule. Draw p orbitals as clouds. Circle the four atoms that are in a straight line. 2. Which of the following show cis-trans isomerism? Draw the cis and trans isomers of the ones that do. a. CH 2 =C(CH 3 ) 2 b. CH 3 CH=CHCH 3 c. CH 3 C CCH 3 d. e. CH 3 CH = C CH 2 CH 3 CH 2 CH 3 23

24 Worksheet Bonding Part 2 (cont.) 3. State the relationships between the following pairs of structures. Your choices are: identical compound, cis-trans isomers, structural isomers, totally different molecules. a. CH 3 CH 2 CH 2 CH 3 and (CH 3 ) 3 CH b. CH 2 = CH CH 2 Cl and CHCl = CHCH 3 c. CH 3 CH 3 CH 3 \ / and \ CH = CH CH = CH \ CH 3 d. CH 3 CH 3 CH 2 \ / and CH = CH CH 3 C CH 3 e. and f. and g. and 24

25 Bronsted-Lowry Acid: Bronsted-Lowry Base: Acids and Bases Bronsted-Lowry Definition Example reaction: H :O: H H :O: (-) H C C O H + H N H H N H + H C C O: (+) H H H H Bronsted-Lowry Bronsted-Lowry Conjugate Conjugate Acid Base Acid Base The reaction of Bronsted-Lowry Acid and a Bronsted-Lowry Base produces a new acid and a new base. To distinguish the products from the reactants, they are referred to as the conjugate acid and conjugate base. Lewis Acid: Lewis Base: Acids and Bases Lewis Definition Example reaction: F H F H (-) (+) F B + :N H F B N H F H F H Lewis Acid Lewis Base The electron pair involved in a Lewis Acid-Base reaction is shared to form a new covalent bond. Ex: Using curved arrows, show how acetaldehyde, CH 3 CHO, can act as a Lewis base. :O: + H A CH 3 C H 25

26 Equilibrium Any acid and base reaction is represented by an equilibrium. An equilibrium is a representation of a reaction that takes into account both the forward and reverse reactions present in a system. Both forward and backward arrows are written to show this ebb and flow. Let s take a look at the reaction of HCl with water: H 2 O + HCl H 3 O + + Cl - An important question to ask here is? That answer hangs on the following considerations: 1. Which of the two acids (HCl and H 3 O + ) are stronger? 2. Which of the two bases (H 2 O and Cl - ) are stronger? Having a clear answer to any one of these questions can lead you to know which direction the equilibrium favors. Equilibrium always drives in the direction of the. Since Cl - is the weaker base and H 3 O + is the weaker acid, the equilibrium will move in the direction of the products. H 2 O + HCl H 3 O + + Cl - Acid Strength K a and pk a Acids vary in terms of their strength, or acidity. Strong acids (such as HCl) will react almost completely with water, while weak acids (such as acetic acid) react only slightly. The exact measure of the strength of an acid is given by its acid constant, K a. For the reaction of any acid (shown as HA) with water, the acidity constant, K a, is HA + H 2 O A - + H 3 O + K a = [H 3 O + ][ A - ] [HA] [ ] Concentration in mol/l The stronger the acid in the above equation, the more the reaction favors the. This results in a large value (greater than 1) for the K a. K a values for acids can vary anywhere from the order of for the strongest acids to for the weakest. These numbers are unwieldy, so we use the concept of pk a to make the numbers more workable. pk a = - log(k a ) K a = 10 -pka 26

27 Acid Strength K a and pk a - Continued For the convenience of the numbers, pk a values are usually listed for us to determine the relative strength of an acid or base. The lower the pk a value, the stronger the acid. The lower the pk a value of the conjugate acid, the weaker its conjugate base. Thus, a very strong acid will have a very conjugate base. This makes logical sense with what we learned above. The Relationship Between [H ], ph K a and pk a Chances are that you have heard of the concept of ph before. ph is a measure of the or of a water solution, with a low ph corresponding to an acidic solution and a high ph corresponding to a basic solution. Although the values for ph and pka of a strong acid would both tend to be small, and the values for ph and pka of a strong base would both be large, the concepts of ph and pka are inherently different. ph = p[h + ] = p[h 3 O + ] = - log[h 3 O + ] While the pka takes into account all of the components of an acid base reaction, ph only focuses on the concentration of the (H 3 O + ). Thus, the ph of a solution change with depending on how much acid or base is added to the solution (molarity), while the pka. Example Problem: A sample of 6.00 g of acetic acid (CH 3 COOH) is diluted with enough water to make 1 L of solution. After waiting for equilibrium to become established, the ph is measured to be What is the pka of acetic acid? Additional Relationship: (Henderson-Hasselbalch Equation) pka [ conjugatebase ] ph log [ conjugateacid ] 27

28 Acid Strength What makes one stronger than the other? There are three factors that influence the strength of any particular acid. They all have to do with an acid s conjugate base. 1. Electronegativity The more electronegative the atom, the more likely it is to be willing to accommodate a charge. For the elements in the 2 nd period, this trend is easy to see. Since F is the most element, its anion F - should be the easiest to make. HF > H 2 O > NH 3 > CH 4 Acid Strength 2. Size The relative size of an atom or species, the easier it can accommodate a negative charge. Atoms or species that are larger allow for charge to be more spread out, which tends to make ions considerably happier. For the elements in column 7A, this trend is clearly apparent. As you increase in principle, the diameter of the atoms increases, outer electrons are held less tightly and the atom becomes more. HI > HBr > HCl > HF Acid Strength 3. Resonance A species ability to spread charge around between atoms contributes highly to the stability of the. Consider HSO - 4, the conjugate base of sulfuric acid (pka = -10). Three additional resonance forms that are all in energy exist for the bisulfate ion. This is the major contribution to the stability of the conjugate base, and in turn, the strength of the acid. 28

29 Acids and Bases Worksheet 1. Identify the conjugate acid and conjugate base of each of the following. Write none if there is none. Conjugate Acid Conjugate Base CH 4 NH 3 CH 3 OCH 3 CCl 4 2. Using your pka table, predict the products of the following reaction. If there is no reaction, write NR. a. CH 3 COOH + NH 3 b. + NaNH 2 c. + HCl 3. In which direction do the following equilibriums lie? Explain your choice. + -OH + H 2 O (+) CH 3 CH 2 NH 2 + H 2 S CH 3 CH 2 NH 3 + HS - 4. Use your pka table to suggest an appropriate base to accomplish the following reaction: R C C H R C C (-) 29

30 Acids and Bases Worksheet (Cont.) 5. For the following list of compounds: a. Circle the strongest acid b. Put a box around the strongest base c. Show the conjugate acid and base of each compound listed. Write none if there is none. d. Is the first compound a strong enough acid to protonate the third? Explain your reasoning. Conjugate Acid Conjugate Base CH 3 OH NH 3 H 2 S SH - 6. Explain why phenol (see below) is more acidic than methanol, CH 3 OH ml of an aqueous solution containing 13.0 g of a mystery acid (FW = 142 g/mol) is found to have a hydronium ion concentration of mol/l. What is the pka of this mystery acid? 8. Show the product of a reaction between NH 3 and the following secondary carbocation (see below). Show any formal charges that are required. NH

31 Material Covered on the Unit 1 Test 1. Be able to define and give examples of the all of the key terms on page 2 of this packet 2. Know the structure and names of all functional groups mentioned in page 2 of this packet 3. Be able to draw valid Lewis structures for polyatomic ions and organic molecules 4. Be able to assign formal charges to atoms in Lewis structures and also be able to determine which structures are the major and minor contributors to the resonance hybrid 5. Be able to predict bond angles and molecular geometry (shape) from Lewis structures 6. Be able to determine the hybridization that carbon, nitrogen, oxygen and sulfur atoms have undergone in a molecule or ion 7. Memorize the relative order of electronegativities of the following elements: F, O, Cl, N, Br, S, I, C, P, H. 8. Be able to assign polarity to bonds in Lewis structures 9. Be able to draw Lewis structures and line-angle diagrams for structural isomers if given a molecular formula 10. Be proficient in the use of curved arrows to show electron movement when drawing contributing resonance structures 11. Be able to write and interpret condensed structural formulas 12. Draw orbital diagrams of sigma and pi bonding 13. Be able to draw representations of 3-D molecules by using wedges and dashed lines 14. Be able to do calculations between ph, pka and Ka. 15. Be able to describe the effects of electronegativity, size and resonance on acid strength be able to illustrate these with examples. 16. Use curved arrows in acid base reactions 17. Use pka table to predict products of acid-base reactions and label conjugates in Bronsted-Lowry reactions 18. Predict directions of equilibria using pka tables 31

What Is Organic Chemistry?

What Is Organic Chemistry? EQ: What is Organic Chemistry? Read: pages 1-3 Answer the questions in your packet Basics of Organic Chem 1 Chapter 1: Structure and Bonding Key terms Organic Chemistry Inorganic