Stereochemistry of Alkanes and Cycloalkanes the Shapes of Molecules

CHEM 212 Exam 2 Part 1

Professor Kelly Boebinger Chapter 4: Stereochemistry of Alkanes and Cycloalkanes The Shapes of Molecules

 Stereochemistry: is concerned with the three- dimensional shapes of molecules resulting from many forces  Conformations: different arrangement of atoms that result from the rotation around a single bond.  Conformers: A specific conformation. Cannot be isolated since they move too rapidly.  Conformers interconvert rapidly and a structure is an average of conformers

3 Representing Conformations

 Sawhorse representations show molecules at an angle, showing a molecular model  C-C bonds are at an angle to the edge of the page and all C-H bonds are shown  Newman projections show how the C-C bond would project end-on onto the paper  Bonds to front carbon are lines going to the center  Bonds to rear carbon are lines going to the edge of the circle

4 Ethane’s Conformations

 The most stable conformation of ethane has all six C– H bonds away from each other (staggered)  The least stable conformation has all six C–H bonds as close as possible (eclipsed) in a Newman projection – energy due to torsional strain

5 Conformations of Propane

 Propane (C3H8) torsional barrier around the carbon– carbon bonds 14 kJ/mol  Eclipsed conformer of propane has two 6 kJ/mol ethane-type H–H interactions and an interaction between

C–H and C–C bond 4 kJ/mol

6 Total 14 kJ/mol Conformations of Butane

 anti conformation has two methyl groups 180° away from each other  Rotation around the C2–C3 gives eclipsed conformation  Staggered conformation with methyl groups 60° apart is gauche conformation Energy: Lowest 2nd Lowest Highest

The most stable conformation of any alkane has the carbon- carbon bonds staggered, with the large substituents anti to one another 7 Stability of Cycloalkanes: The Baeyer Strain Theory

 Baeyer (1885): since carbon prefers to have bond angles of approximately 109°, ring sizes other than five and six may be too strained to exist

 Rings from 3 to 30 C’s do exist but are strained due to bond bending distortions and steric interactions

8 The Nature of Ring Strain

 Rings larger than 3 atoms are not flat and assume nonplanar conformations to minimize angle strain and torsional strain by ring-puckering  Larger rings have many more possible conformations than smaller rings and are more difficult to analyze

 Angle strain - expansion or compression of bond angles away from most stable  Torsional strain - eclipsing of bonds on neighboring atoms  Steric strain - repulsive interactions between

nonbonded atoms in close proximity 9 Cyclopropane: An Orbital View

 3-membered ring must have planar structure  Symmetrical with C–C–C bond angles of 60°  Requires that sp3 based bonds are bent (and weakened)  All C-H bonds are eclipsed

10 Conformations of Cyclobutane

 Cyclobutane has less angle strain than cyclopropane but more torsional strain because of its larger number of ring hydrogens  Cyclobutane is slightly bent out of plane - one carbon atom is about 25° above  The bend increases angle strain but decreases torsional strain until a minimum energy balance is achieved

11 Conformations of Cyclopentane

 Planar cyclopentane would have no angle strain but very high torsional strain  Actual conformations of cyclopentane are nonplanar, reducing torsional strain  Four carbon atoms are in a plane  The fifth carbon atom is above or below the plane – looks like an envelope

12 Conformations of Cyclohexane

 Substituted cyclohexanes occur widely in nature  The cyclohexane ring is free of angle strain and torsional strain  The conformation is has alternating atoms in a common plane and tetrahedral angles between all carbons  This is called a chair conformation

13 How to Draw Chair Cyclohexane

14 Cyclohexane Video Link

Here is the link if the video doesn’t work Cyclohexane Drawing Video: https://youtu.be/JV0d-Qqelto

More information in following slides 15 Axial and Equatorial Bonds in Chair Cyclohexane

 The chair conformation has two kinds of positions for substituents on the ring: axial positions and equatorial positions  Chair cyclohexane has six axial hydrogens perpendicular to the ring (parallel to the ring axis) and six equatorial hydrogens near the plane of the ring

16 Drawing the Axial and Equatorial Hydrogens

17 Conformational Mobility of Cyclohexane  Chair conformations readily interconvert, resulting in the exchange of axial and equatorial positions by a ring-flip

18 19 Bromocyclohexane

 When bromocyclohexane ring-flips the bromine’s position goes from equatorial to axial and so on  At room temperature the ring-flip is very fast and the structure is seen as the weighted average

20 Conformations of Monosubstituted Cyclohexanes

 The two conformers of a monosubstituted cyclohexane are not equal in energy  The equatorial conformer of methyl cyclohexane is more stable than the axial In general, equatorial positions give more stable isomer

21 4.12 Conformational Analysis of Disubstituted Cyclohexanes

 In disubstituted cyclohexanes the steric effects of both substituents must be taken into account in both conformations  There are two isomers of 1,2-dimethylcyclohexane. cis and trans  In cis-1,2, both conformations are equal in energy

22 Trans-1,2-Dimethylcyclohexane

 One trans conformation has both methyl groups equatorial and only a gauche butane interaction between methyls (3.8 kJ/mol) and no 1,3- diaxial interactions  The ring-flipped conformation has both methyl groups axial with four 1,3-diaxial interactions  Steric strain of 4  3.8 kJ/mol = 15.2 kJ/mol makes the diaxial conformation 11.4 kJ/mol less favorable than the diequatorial conformation  trans-1,2-dimethylcyclohexane will exist almost exclusively (>99%) in the diequatorial conformation 23 24 Boat Cyclohexane

 Cyclohexane in a boat conformation

25 Conformations of Polycyclic Molecules  Two or more rings are fused together along a common bond.  Decalin consists of two cyclohexane rings joined to share two carbon atoms (the bridgehead carbons, C1 and C6) and a common bond  Two isomeric forms of decalin: trans fused or cis fused  Flips and rotations do not interconvert cis and trans

26 CHEM 212 Exam 2 Part 2

Professor Kelly Boebinger Chapter 5 An Overview of Organic Reactions

H H h H H H C C H + Cl Cl H C C Cl + HCl chlorine H H H H ethane chloroethane Kinds of Organic Reactions Four General Types

 Addition reactions – two reactants add together to form a single new product with no left over atoms: A + B  C

H H H H C C + HBr H C C Br H H H H ethylene bromoethane

 Elimination reactions – single reactant splits into two products: A  B + C

H H H H base H C C Br C C + HBr H H H H bromoethane ethylene 29 Kinds of Organic Reactions

 Substitution –two reactants exchange parts to give 2 new products: AB + CD  AD + CB

H H hv H H H C C H + Cl Cl H C C Cl + HCl chlorine H H H H ethane chloroethane

 Rearrangement reactions – a single reactant undergoes reorganization of bonds and atoms to yield an isometric product: A  B

acid catalyst

CH2 CH3 H3C H3C 1-butene 2-butene

30 Identify the following reactions (addition, elimination, substitution, or rearrangement)

a) H3CCH=CH2 + H2  CH3CH2CH3

b) CH3CH2X + KOH  CH3CH2OH + KX

c) CH3CH2CH2OH  CH3CH=CH2 + H2O

d) acid catalyst

a) addition b) substitution c) elimination d) rearrangement 31 Reactions

 Bond breaking and bond making occurs in all chemical reactions.

 Bonds are broken in the reactants

 Bonds are made in the products

32 Radicals

 Alkyl groups are abbreviated “R” for radical

 Example: Methyl iodide = CH3I, Ethyl iodide = CH3CH2I, Alkyl iodides (in general) = RI

 A “free radical” is an “R” group on its own:

 CH3 is a “free radical” or simply “radical” .  Has a single unpaired electron, shown as: CH3

33 Radical Reactions and How They Occur  Note: Polar reactions are more common  Radicals are highly reactive because of the odd number of valence electrons (usually 7) and want to complete electron octet of valence shell  A radical can add to an alkene to give a new radical, causing an addition reaction

34 Radical Reactions and How They Occur  A radical can break a bond in another molecule and abstract a partner with an electron, giving substitution in the original molecule, and leaving behind a new radical.

35 Radical Substitution Requires 3 Steps

 Three types of steps  Initiation – homolytic formation of two reactive species with unpaired electrons  Propagation – reaction with molecule to generate radical  Termination – combination of two radicals to form a stable product, can be more than one termination step.

36 Radical Substitution Process is a Chain Reaction

 Step 1: Initiation – Produces a small number of reactive radicals.

 Example – formation of Cl atoms from Cl2 and light

37 Radical Substitution Process is a Chain Reaction

 Step 2: Propagation – reactive chlorine radical collides with methane molecule

 Example - reaction of chlorine atom with . methane to give HCl and CH3

38 Radical Substitution Process is a Chain Reaction

 Step 3: Termination – When the cycle is broken and the chain is ended. There can be more than one possible termination step. These occur infrequently.

39 Written together as an overall Radical Reaction

Step 1

Step 2

Step 3

40 Next we will look at Polar Reactions and How They Occur  Polar reactions occur because opposite charges attract and the electron-rich sites in one molecules reacts with the electron poor sites of another molecule.  Bond polarity (Table 5.1) occurs in polar functional groups that contain O,N,F,Cl, and Br (elements higher electronegativity than C) and C becomes partially (+)  When C is bonded to a less electronegative element, (Li, Mg) it becomes partially (-) charged.  The more electronegative atom has the greater electron density

41 Generalized Polar Reaction

 Polar reactions occur because opposite charges attract.  The combination is indicated with a curved arrow from nucleophile to electrophile

Donates e- for bond

42 Polar bonds also result from interactions of functional groups with Lewis acids or bases

 An electrophile “electron loving” is a Lewis acid, an electron-poor, accepts electrons, can be neutral or positively charged.  Examples include; acids, alkyl halides, and carbonyl compounds

43 Polar bonds also result from interactions of functional groups with Lewis acids or bases

 A nucleophile “nucleus loving” is a Lewis base,electron rich, donates electrons, can be neutral or negatively charged. (Look for negative charge and electron pairs) - -  Examples include; NH4, H2O, OH , Br

44 Polar bonds also result from interactions of functional groups with Lewis acids or bases

 Some species can act as either, depending on what else is present. Example of water.

45 Add curved arrows, then identify and label the nucleophile and electrophile in the reactions below.

O O + H - + H Cl + Cl H+ Cl- nucleophile electrophile

OH- H+ _ + H O H + H3C MgBr CH4 + HO MgBr

eelectrophilelectrophile nnucleophileucleophile

46 Describing How Organic Reactions Occur: Mechanisms

 An overall description of reactions occur. It describes in detail exactly what takes place at each stage of a chemical transformation.  It describes what bonds are broken and in what order.  Relative rates of steps.  Accounts for all reactants used, products formed, and amounts.

47 Steps in Mechanisms

 We classify the types of steps in a sequence  A step involves either the formation or breaking of a covalent bond  Steps can occur in individually or in combination with other steps  When several steps occur at the same time they are said to be concerted

48 Rules for Using Curved Arrows

 The arrow goes from the nucleophilic reaction site to the electrophilic reaction site  Curved arrows are a way to keep track of changes in bonding in polar reaction  The arrows track “electron movement” Full arrow 2e- moving, single head (fish hook) 1e- moving. -  Charges change during the 2 e reaction

 One curved arrow corresponds to - one step in a reaction mechanism 1 e

49 Rules for Using Curved Arrows

 Octet rule must be followed

50 Rules for Using Curved Arrows

 Octet rule must be followed

51 Mechanism of Addition of HBr to Ethylene

intermediate  HBr electrophile is attacked by  electrons of ethylene (nucleophile) to form a carbocation intermediate and bromide ion  Bromide adds to the positive center of the carbocation, which is an electrophile, forming a C-Br  bond  The result is that ethylene and HBr combine to form bromoethane  All polar reactions occur by combination of an electron-rich site of a nucleophile and an electron-deficient site of an electrophile 52 First Step in Addition

 The  bond between carbons begins to break  The C–H bond begins to form  The H–Br bond begins to break

53 Formation of a Carbocation Intermediate  If a reaction occurs in more than one step, it must involve species that are neither the reactant nor the final product  These are called reaction intermediates or simply “intermediates”  HBr, a Lewis acid, adds to the  bond  This produces an intermediate with a positive charge on carbon - a carbocation  This is ready to react with bromide

54 Carbocation Intermediate Reactions with Anion

 Bromide ion adds an electron pair to the carbocation  An alkyl halide produced  The carbocation is a reactive intermediate

55 Describing a Reaction: Equilibria, Rates, and Energy Changes aA + bB cC + dD

c d a b Keq = [Products]/[Reactants] = [C] [D] / [A] [B]

 If Keq > 1, at equilibrium most of the material is present as products

 If Keq is 10, then the concentration of the product is ten times that of the reactant

 If Keq < 1, at equilibrium most of the material is present as the reactant

 If Keq is 0.10, then the concentration of the reactant is ten times that of the product 56 Magnitudes of Equilibrium Constants

7  Since Keq = 7.5 x10 then the reaction proceeds as written.  For a reaction to proceed as written, the energy of the products must be lower than the energy of the reactants, energy must be released.

57 Free Energy and Equilibrium

 The ratio of products to reactants is controlled by their relative Gibbs free energy  This energy is released on the favored side of an equilibrium reaction  The change in Gibbs free energy between products and reacts is written as “DG”

 If Keq > 1, energy is released to the surrounding (exergonic reaction)

 If Keq < 1, energy is absorbed from the surroundings (endergonic reaction) 58 Numeric Relationship of Keq and Free Energy Change

 The standard free energy change at 1 atm pressure and 298 K is DGº  The relationship between free energy change and an equilibrium constant is:

DGº = - RT ln Keq where R = 1.987 cal/(K x mol) T = temperature in Kelvin

ln = natural logarithm of Keq

59 Changes in Energy at Equilibrium

 Free energy changes (DGº) can be divided into  a temperature-independent part called entropy (DSº) that measures the change in the amount of disorder in the system

 - disorder decreases

 + disorder increases  a temperature-dependent part called enthalpy (DHº) that is associated with heat given off (exothermic, - ) or absorbed (endothermic, + )  Overall relationship: DGº = DHº - TDSº

60 Describing a Reaction: Bond Dissociation Energies

 Bond dissociation energy (D): Heat change that occurs when a bond is broken by homolysis  The energy is mostly determined by the type of bond, independent of the molecule  The C-H bond in methane requires a net heat input of 105 kcal/mol to be broken at 25 ºC.  Table 5.3 lists energies for many bond types  Changes in bonds can be used to calculate net changes in heat

61 Calculation of an Energy Change from Bond Dissociation Energies

 Addition of Cl-Cl to CH4 (Table 5.3)  Breaking: C-H D = 438 kJ/mol Cl-Cl D = 243 kJ/mol  Making: C-Cl D = 351 kJ/mol H-Cl D = 432 kJ/mol Energy of bonds broken = 438 + 243 = 681 kJ/mol Energy of bonds formed = 351 + 432 = 783 kJ/mol

DHº = 681 – 783 kJ/mol = -102 kJ/mol exothermic 62 Describing a Reaction: Energy Diagrams and Transition States

 The highest energy point in a reaction step is called the transition state  The energy needed to go from reactant to transition state is the activation energy (DG‡)

63 Reaction Diagram for Addition of HBr to Ethylene

 Two separate steps, each with a own transition state  Energy minimum between the steps belongs to the carbocation reaction intermediate . reactant product

64 Biological Reactions

 Reactions in living organisms follow reaction diagrams too  They take place in very controlled conditions  They are promoted by catalysts that lower the activation barrier  The catalysts are usually proteins, called enzymes  Enzymes provide an alternative mechanism that is compatible with the conditions of life

65 CHEM 212 Exam 2 Part 3

Professor Kelly Boebinger 6. Alkenes: Structure and Reactivity Alkene - Hydrocarbon With Carbon- Carbon Double Bond

 Also called an olefin but alkene is better  Includes many naturally occurring materials  Flavors, fragrances, vitamins  Important industrial products

 Cracking of natural gas alkanes (C1-C4) and gasoline (C4- C8)

n = 0 - 6

68 Degree of Unsaturation (DOU)

 The number of rings or multiple bonds in a compound:  Each ring or multiple bond replaces 2 H's

1. Add up carbons = n, then calculate 2n+2 for the number of hydrogens if saturated. 2. Add up hydrogens in formula. 3. Find the difference, divide by 2. This is the DOU. 4. How to handle halogens, oxygen and nitrogen; a. Add halogens to number of hydrogens b. Ignore oxygens c. Subtract nitrogens from hydrogens

69 Problems: Calculate DOU in the following:

7 x 2 + 2 = 16 DOU 1. C7H12 16 – 12 = 4 1. 4/2 = 2 6 x 2 + 2 = 14 2. C H 6 8 14 – 8 = 6 2. 6/2 = 3 3 x 2 + 2 = 8 3. C3H4 8 – 4 = 4 3. 4/2 = 2

4. 4. DOU = 2 Bonus Problem:

7. CH3CH2NH2

5. 5. DOU = 2

7. DOU = 0

6. 6. DOU = 1

70 Naming of Alkenes

1. Name the longest chain, WHICH CONTAINS THE DOUBLE BOND. 2. Number the carbon atoms in the main chain, give the double bond the lowest possible number from the closest end. 3. Identify substituents in the main chain. 4. Alkenes have an –ene ending. 5. Write the full name as one word.

1-pentene H3C CH2

71 Naming Alkenes

1. 2. 1. ethene 2. propene 3. 3. 2-methylpropene 4. 4. 2-methyl-1,3-butadiene

72 Many alkenes are known by their common names

-CH2- and

73 Problems: Give the IUPAC name for each of the following:

2,4,6-octatriene 2-methyl-3-hexene

H3C CH CH2 CH3 CH CH CH3 H3C

CH3

4,7-dimethyl-2,5-octadiene 3,4,4-trimethyl-1-pentene

H3C CH3 CH3 H3C C CH3 CH3 H2C CH CH 3 CH CH3 3-bromo-3-hexene 3-methyl-1-pentene 2-ethyl-1-pentene CH Br 3 H3C CH2 H CH CH C C CH2 C 3 HC CH CH CH H H3C CH CH2CH3 2 2 2 H3C CH2

74 Alkene Nomenclature

a) CH 3 b) c) CH3

H3C CH3

e) CH3 d) CH3

CH3 a) 1,5-dimethylcyclohexene b) 1,4-cyclohexadiene

f) CH(CH3)2 c) 1,5-dimethylcyclopentene d) 1-methylcyclohexene e) 4,4-dimethylcycloheptene

f) 3-isopropylcyclopentene 75 Draw the condensed structural formula (Line OK for ring) for each of the following: 2-methyl-1,4-pentadiene 1,3-cyclopentadiene

CH3

H2C C CH2 CH CH2

2-ethyl-1-hexene 3,6-diethyl-1,4-cyclohexadiene

H3C CH3 H2C C CH2CH2 CH2 CH3 CH2 CH2 CH 2CH 3

4-propyl-2-heptene

H3C CH CH CH CH2 CH2 CH3

CH 2CH 2CH 3

76 What is substitution in Info I found online an alkene?

Kahn Academy Link to How to Make Alkene Compounds

Naming Alkenes - Nomenclature Tutorial for Double Bound Organic Compounds

Leah Fisch

STEREOISOMERISM - GEOMETRIC ISOMERISM CIs & Trans Isomers Guide ( You are ready for this now)

E & Z Isomer Guide ( You will need this at the end of these slides ) 77 6.4 Electronic Structure of Alkenes

 Carbon atoms in a double bond are sp2-hybridized  Three equivalent orbitals at 120º separation in plane  Trigonal planar geometry  With alkanes there is free rotation around a single bond. Not with alkenes, the geometry around the carbon-carbon double bond is fixed.

Rotation of  Bond Is Prohibitive

78 6.5 Cis-Trans Isomerism in Alkenes

 The presence of a carbon- carbon double can create two CH CH=CHCH possible structures 3 3  cis isomer - two similar 2-butene groups on same side of the double bond  trans isomer similar groups on opposite sides  Each carbon must have a hydrogen for these isomers to occur, the alkyl groups can be the same or different.

79 Bonus Slide! Draw cis and trans. Cis and trans is used relative to hydrogen, R groups can be different

 cis-4-methyl-2-pentene

H H C C

H3C CH CH3 Cis and trans H3C CAN be shown with line structures  trans-4-methyl-2-pentene

H3C H C C

H CH CH3

H3C 80 Cis-Trans Isomerism in Alkenes

 You cannot have cis or trans isomers in terminal alkenes

H H C C

H CH CH3

H3C

81 6.6 Sequence Rules: The E,Z Designation  Neither compound is clearly “cis” or “trans”  Cis, trans nomenclature only works when each E carbon has one hydrogen  higher priority group has a higher atomic number.  E -entgegen, opposite Z sides  Z - zusammen, together on the same 82 side System for Comparison of Priority of Substituents

 Ranking Priorities:  Must rank atoms that are connected at comparison point  Higher atomic number gets higher priority  If there is a tie, then look at the second, third, etc. until a difference is found.  Multiple bonds are equivalent to the same number of single In this case,The higher priority bonds. C=O is equivalent to 2 groups are opposite: (E )-1-bromo-1-chloro-propene bonds.

83 Extended Comparison

 If atomic numbers are the same, compare at next connection point at same distance  Compare until something has higher atomic number  Do not combine – always compare

84 Examples

L O H H H H CH3 H3C C CH3 C Cl H3C C C E C C OH H Z H I C L L C H H H H L Cl2HC CH2

L L H3C CH2 OH H Cl CN H C C Z C C Z H2C Cl O CH NH H H 2 2 CH3 L CH3 L

85 Name the following

L H

L H H L

(E)-3-methyl-1,3-pentadiene (E)-1-bromo-2-isopropyl-1,3-butadiene

H2C=C- To count attachments to the last carbon, “switch back” and count the carbon on the other side of the double bond as 2 C’s 86 87