Chemistry 51, Organic 1 Prerequisite: Chemistry 2 Grading Scheme Recitation Grade: 100 pts average 75 Laboratory Grade 100 pts average 75 Lecture Exams 100 pts Final 100 pts Total 400 pts Old exams and quizzes: academic.brooklyn.cuny.edu/chem/howell/jhowell.htm Safety: goggles, pregnancy Cheating: F email@example.com
Goals of the course: • Structure of organic molecules, relation of structure to reactivity • Organic reaction patterns • Mechanism of reactions • Synthesis of organic compounds • Techniques of the trade
Scaling of Raw Grades for Recitation and Lab 1. Omit the student, B, that dropped the course. 2. Get the average to 75 Multiply these sums by 4.9342 Note that student D was absent from two labs and will be low for that reason. We will deal with D later. or Multiply these sums by 0.49342 In both cases get the same result…….
Now we have to spread the grades out so that there is a reasonable distribution about the desired average of 75. This is done by expanding or contracting the set of grades around the average so that the maximum turns out to be 95. Now for the student D who is outside of the desirable limits (50 – 95). We move his raw grade up to 70 so that D does not skew the scaling process. D receives a scaled grade of about 51.
Atomic Structure • Electrons occupy orbitals which are grouped into subshells and then shells. • Carbon atom: 1s22s22p2 • Nitrogen: 1s22s22p3 Valence Shell Electrons: Used in chemical bond formation C N
Electron Configuration of Atoms • Pauli Exclusion Principle: • No more than two electrons may be present in an orbital. If two electrons are present, their spins must be paired. • Hund’s Rule: • When orbitals of equal energy are available but there are not enough electrons to fill all of them, one electron is added to each orbital before a second electron is added to any one of them; the spins of the electrons in degenerate orbitals should be aligned. • Aufbau (“Build-Up”) Principle: • Orbitals fill in order of increasing energy from lowest energy to highest energy.
Atomic Structure • 1 shell consists of • 1s subshell holding the • 1s orbital • 2 shell consists of • 2s subshell holding the • 2s orbital and the • 2p subshell holding the • 2px, • 2py, and • 2pz orbitals. Each orbital can hold two electrons. • An orbital can be • an atomic orbital (on an atom) or • a molecular orbital (on a molecule)
Octet Rule, Lewis Structures Electrons are stabilized by bond formation. H atom can stabilize up to two electrons in the valence shell. CF can stabilize up to 8 electrons in the valence shell. Complete octet: two electrons around H; eight electrons for C F in the valence shell. Now we are providing maximum stabilization.
Completing the Octet Ionic Bonding: Electrons can be transferred to an atom to produce an anion and “complete the octet”. Covalent Bonding: Electrons can be shared between atoms providing additional stabilization. The shared electrons typically “complete the octet” for each atom.
Number of Bonds Define: Unused stabilization of atom = (number of electrons that can be held in valence shell) – (number of electrons already present) Bonds make use of the unused stabilizing capability of the atoms. # Bonds in molecule= (Sum of unused stabilizing capability)/2
Formal Charge: the charge an atom in a Lewis bonding diagram. Assign each valence shell electron to an atom: • Non-bonding electrons are assigned to the atom on which they reside. • Bonding electrons are divided equally between the atoms of the bond. Recipe to calculate the Formal Charges. Formal charge = (# valence shell electrons in neutral atom) - (# nonbonding electrons) - ½ (# bonded electrons)
Formal Charge and Bonding Patterns. If the octet rule obeyed… What is “wrong” with the yellow box? Give examples of common molecules.
Lewis Diagrams H C (3 * 4 + 6 * 1) / 2 = 9 bonds How many bonds left to draw? 9 – 8 = 1 bond left Put remaining bond(s) any place where the octet rule is not violated.
Another Example Draw Lewis structure for CH2N2 with formal charges. # bonds = ½ (4 + 2 *1 + 2*3) = ½ x 12 Let’s characterize the valence electrons # electrons present = 4 + 2*1 + 2*5 = 16 # bonding electrons = 12 # non-bonding electrons = 16 – 12 = 4 = 2 lone pairs Unused stabilizing capacities.
CH2N2 (continued) 6 bonds, 2 lone pairs Skeleton uses 4 bonds: OR Which of these two structures is more important? Which provides the better description? More important due to negative charge being on electronegative N
Curved Arrows to Reposition Electrons Curved arrows are the standard way of showing electron reorganizations, repositioning of lone pairs and bonds. Transforms the lone pair into a bonding pair. Still counts towards terminal N but also towards middle N. Too many electrons around middle N. Bonding pair must retreat from the middle N and become a lone pair on the C.
Some principals of using curved arrows. • Reactions: Restructure the molecular system, creating bonds between unbonded atoms. • Resonance: Moving pi bonds and lone pairs but leaving single bond framework intact. An overly simplified presentation involves simply the motion shown below and its reverse. • A lone pair on B is converted to bonding electrons. • What happens at B: The octet count at B is not changed but it does become more positive by 1. The actual charge depends on the initial charge. • What happens at A: Two more electrons are added to the octet count of A. We cannot exceed the octet count of 8. • If A originally had an octet count of only 6 then we have simply completed the octet. • If the octet of A was originally complete at 8 then we must withdraw some other pi electrons from A to make room for the newly formed bond. A becomes more negative by 1. Now for its reverse
The reverse motion, conversion of bond into a lone pair, is shown below. Here we convert a pi bond to a lone pair. What happens at B: The pi electrons move away from B and no longer count towards either the octet of B or its formal charge. B becomes more positive by 1 and no longer satisfies the octet rule (electron deficient). What happens at A: The octet count is not changed at A but A does become more negative by 1.
Example: resonance in amides Formamide, NH2CHO, could be written with two electron diagrams as shown below. Each structure obeys the octet rule (provides maximum stabilization by bonding). The structure on the right is less stable because of the charges that are present. The problem is how to smoothly interconvert one to the other.
We have to make room for the incoming electrons by retreating the pi electrons of the CO double bond onto the O. Proceeding with our elementary steps But this structure violates the octet rule on the high side. 10 ELECTRONS AT THE CARBON!! FORBIDDEN!!
These two steps would be combined into one (always) and avoid that repugnant structure which violates the octet rule on the high side. As the lone pair on the N moves in towards the central C the pi electrons in the C=O double bond retreat onto the O.
Maximum number of bonds permitted using valence shell orbitals: N is positive to allow for overall positive charge. Could have made any atom positive. 14/2 = 7 bonds
Set-up single bond framework. 6 bonds used in single bond framework. 1 left to assign. It will be a pi bond, part of a double bond. The additional bond is used to provide maximum stabilization; completing the octet. Where does it go? Only two possibilities…. How are they interconverted?
So what is the third structure? It won’t have the double bond. Will not obey the octet rule. Can collapse the pi bond into a lone pair. Could we have collapsed in other direction? No lone pair on this N. High energy structure; not very important.
Our three structures….. Does not obey octet rule. Least stable, least important. Intermediate stable with positive charge on more electronegative oxygen. Most stable with positive charge on the electropositive atom, N.
Examine one of the interconversions shown earlier. Looking at it in reverse. Higher energy structure Lower energy, one more bond. This is a common and significant resonance motion. A supply of electrons moving towards a place where needed and providing a new bond in the process.
Exceptions to Octet Rule But the S violates the octet rule by having more than 8 electrons around the S. Valence shell is 3s and 3p. S now would have to use additional atomic orbitals This means S would have to use the 3d shell orbitals. Luckily, they are low in energy and can be utilized. Octet is expanded due to use of 3d orbitals.
Donation from O to S. Observe that this is another example of donation of available electrons (lone pair) into a neighboring acceptor orbital. Empty 3d Filled 2p
Survey of Classes of Molecules • Hydrocarbons, C & H A Alkanes: only single bonds • Acyclic (no rings) Alkanes: CnH2n+2 suffix: -ane 2. Cyclic alkanes: example cyclohexane CnH2n+2-2(# rings) Presence of a ring eliminates two hydrogens
Alkenes: double bonds, pi bonds • Suffix: -ene i acyclic ii cycloalkenes A pi bond eliminates two hydrogens C4H10 (butane) C4H8 (butene) + 2 H • general formula: CnH2n+2-2(# rings) -2(#pi bonds)
Oxygen Containing Molecules Incorporation of an oxygen into a molecule does not change the relationship between the number of carbons and number of hydrogens. CnH2n+2-2(#pi bonds) – 2(# rings)Oa Watch how this makes sense.
Hydrogen Deficiency • - 2(#pi bonds) – 2(# rings) CnH2n+2 hydrogen deficiency. 2n + 2: Number of hydrogens for an acyclic compound with n carbons and no pi bonds. If a compound has a different number of hydrogens it must be due to the presence of pi bonds or rings. # pi bonds + # rings = 1/2 ( (2n + 2) - (actual number of hydrogens present) )
Singly bonded Oxygen Alcohols, ROH, -ol Two OH groups: a diol
Useful Classiification of Alcohols (and other things): Primary, Secondary, Tertiary Alcohols and many other groups may be classified as Primary: one carbon directly bonded to the C bearing the –OH. CH3CH2CH2OH Secondary: two carbons directly bonded to the C bearing the –OH. (CH3CH2)2CHOH Tertiary: three carbons directly bonded to the C bearing the –OH. (CH3CH2)3COH 1o, 2o, 3o
Ethers: ROR, isomeric with alcohols Methoxy group (alkoxy)
Doubly Bonded Oxygen Carbonyl Group This is a pi bond, resulting in the elimination of two hydrogens.
Aldehydes, RCHO, -al 4 3 2 1 Still more complex, now we put a double bond in the parent chain. A more complex example Start numbering the chain at CHO Double Bonds part of Parent Chain 2-methylbutanal
Ketones, R(CO)R, -one Both carbonyls must be included in parent chain –dione. Butyl Side chain. Simple example More complex example: Double bond included as part of parent chain 1
Carboxylic Acids, RCO2H, -oic acid Another more complex problem Methyl group Both carboxylic groups must be in parent chain C O H C O H Ethyl Group 2 2 Vinyl side chain Start numbering here
Attached to or part of the Parent Chain are: • Functional Groups: -OH, carbonyl (aldehydes, ketones), carboxylic acids • Substituents: alkyl groups, halogens, alkoxy groups
Some Simple Substituents • Alkyl, -R: An alkane minus one hydrogen (providing the point of attachment to the parent chain). -CH2CH3: ethyl (-Et); -CH2CH2CH3: propyl; -CH(CH3)2: isopropyl • Alkoxy, -OR. –OCH2CH3: ethoxy (-OEt) • Halo, -X. –Cl: chloro
More complex substituents Systematically named. Note that the atom directly attached to the parent chain is atom 1 in the substituent. Substituents on the butyl substituent!! 1 3 2 Substituent 4 6 Parent chain 6-(2-(chloromethyl)-4-methoxy-1-methylbutyl) Indicates position of attachment on parent chain
Functional Groups Non-Functional Groups (alkyl, alkoxy, halide) will always appear as substituent and as a prefix. Functional Groups (pi bonds, alcohol, aldehyde, ketone, carboxylic acid) will determine the suffix of parent chain. If more than one functional group then we must prioritize. Highest priority determines suffix. The rest ( alcohol, aldehyde, ketone, carboxylic acid) appear a prefix. But note unsaturation (pi bonding) in the parent chain is always specified as a suffix.
Functional Group Priorities Highest priority at bottom 1
2-(but-2-enyl)-3-(1-hydroxyethyl)-4-(2-oxoethyl)pent-2-enedioic acid One more example. 1 2 4 3 5 Highest Priority Functional Group: two carboxylic acids. Pentanedioic acid Double bond in parent chain: Pent-2-enedioic acid Ethyl side chain bearing a hydroxy at position 1: 3-(1-hydroxyethyl) Ethyl side chain bearing an oxo at position 2: 4-(2-oxoethyl) Butyl side chain with double bond at position 2: 2-(but-2-enyl)
Polar Bonds and Electronegativity H Electronegativity increases towards the upper right. Electronegativity differences results in polar bonds. Example: