chem232 UIUC unit 1

valence electrons

electrons in outermost shell, used for bonding

duet rule

H and He need 2 electrons for their valence shell

octet rule

8 electrons needed to fill the valence shell C, N, O, halogens

rules for condensing structures into formula

1. focus on the carbon atoms

2. list groups attached to each carbon after their carbon, with H atoms going first

3. add parentheses around clusters

   1. O2 implies presence of double bond for one of the oxygens (yields C-O and C=O)

formal charge patterns

more bonds= positive

more lone pairs= negative

neutral= c:4 bonds, N: 3 bonds and 1 lone pair, O: 2 bonds and 2 lone pairs, halogens: 1 bond and 3 lone pairs

formal charge equation

[# valence electrons]- [# non-bonded electrons] - [# bonds]

atomic size trend

increases left and down (gain valence shell)

polarizability

ability of electron cloud to distort, caused by increase in valence shells and distance from nucleus. increases down

electronegativity

tendency for atom to attract electrons towards itself. increases up and right. fluorine is most electronegative

van der Waals forces

temporary dipole created by increasing polarizability in growing carbon-hydrogen molecule. electrons can surge in one direction all at once causing temporary dipole

dipole-dipole interactions

unequal electron sharing/permanent dipole caused by differences in electronegativity. causes partial charges (pos/neg)

hydrogen bonding

permanent dipole. found in O-H, N-H, or F-H bonds due to large differences in electronegativity

resonance structures/contributors

occurring when pi bonds and lone pairs can be assigned in different ways for same molecule. arrow showing movement starts at electron pair, and drawing should reflect change in formal charge as result

how to draw resonance hybrid

1. make sigma bond skeleton

2. add dashed lines wherever pi bonds appear

3. add partial (𝛿) charge wherever formal charges appear

criteria to determine dominant resonance contributor

1. all atoms satisfy octet/duet rule (never exceed but can be deficient if carbocation) biggest contributor

   1. number of charges is minimized

   2. charges agree with electronegativity trends (more electronegative is negative and vice versa)

   3. opposite charges close together and same charges far apart

(bottom 3 in no particular order)

conformational changes

molecules “dancing” but no change to structure of molecule

atoms and bonds stay the same

rotations, dial turns (180 or 120), or pancake flips

isomers

molecules with same molecular formula but different arrangement of atoms. considered different molecules

constitutional isomer

positions or connections between atoms change “detach and reattach”

stereoisomer

same connections between atoms but spatial position/3d structure changes. (ask if molecules overlap, would the structure change?)

nodes

where no electron density occurs as a result of destructive overlap

number of atomic orbitals used

number of hybridized orbitals that form

build a molecular orbital diagram

1. what kind of bonds (pi or sigma)?

2. what is the hybridization of each bonded atom? (list out all four)

3. which atomic orbitals overlap to form the bond?

4. plot relative energies of orbitals on energy diagram

bond order

[bonding electrons-anti bonding electrons]/2

resonance

shows electron delocalization by shifting pi bonds and lone pairs. true structures are resonance hybrids

conjugation

shows electron delocalization by overlapping p orbitals (acts as passageway for electrons to move freely)

criteria for conjugated systems

1. composed of 3 or more p orbitals

2. p orbitals must be on adjacent atoms

3. p orbitals must be aligned parallel

   1. if conjugation and resonance present, bond geometry/hybridization may change

criteria for aromatic system

1. fully conjugated (cyclic system/3 or more adjacent p orbitals)

2. each atom in the ring has p orbital and each p orbital are parallel and aligned and flat

3. contains a multiple of 4n+2 pi electrons (huckels rule)

   1. stable and unreactive!

huckel’s rule

aromatic systems contain 4n+2 = π electrons

steps to drawing a frost circle

1. draw circle

2. inscribe shape with vertex pointing downward

3. put MO at each vertex

4. draw a straight line across the circle marking ½ (any MO above line are anti bonding and any MO below are bonding)

5. fill the orbitals with pi electrons
   

arrow pushing mechanisms

show electron movement. must start from bond or lone pairs

equilibrium favors/shifts to

the side with most stabilized base

aka of HCl

-7

aka of carboxylic acid (RCO2H)

5

pka of H20 and alcohols

16

pka of amines

38

pka of alkanes

50

4 ways to weaken base

1. atom effect [periodic trends: electronegativity (across) and atom size (down)]

2. resonance

3. inductive effect

4. orbital effect

inductive effect

pull of electron density through sigma bonds due to electronegativity differences between atoms. (another method of electron delocalization)

orbital effect

orbitals with more s characteristic have lower energy/more stable

naming non branched alkanes

1. find longest chain with most substituents and name it

2. number the main chain starting from the end closest to a substituent

3. name the substituents (halogens:bromo, iodo, chloro) and alkyls will end in yl

4. list substituents in alphabetical order based on their name

   1. with duplicates, use prefixes. but don’t alphabetize prefixes

naming branched substiuents

1. find longest chain starting from point of attachment

2. c1 is automatically the point of attachment

3. name substituents

4. compile name of branch and surround with parentheses

   1. for branched groups, use tris prefixes (bis, tris, tetrakis

linear alkanes

boiling point increases with chain length

melting point increases with chain length, and even number (symmetric molecules) have bigger jumps in melting point bc symmetry means more surface area, thus harder to break

more surface area = more difficult to break bonds

staggered

• spaced out and low energy

• can have anti (lower energy) and gauche/steric interactions (higher energy)

  • gauche and anti interactions involve non hydrogen atoms being opposite or adjacent

eclipsed

packed together and highest energy

how to draw Newman projection

• if planar atoms are pointing same direction= eclipsed

• if planar atoms are opposite= staggered

• pair wedged and dashed together on same side of projection

dihedral angle

angle between substituents on adjacent carbons

degrees of unsaturation

number of rings or pi bonds in molecule

formula given in exam

naming cycloalkanes

1. count number of carbons in the ring and add cyclo to front and one to end

2. number the substituents (first look for substiutent cluster

   1. for monosubstituted rings, carbon of attachment is c1

   2. for polysubstituted rings, using lowest numbering scheme

3. list substituents alphabetically followed by cycloalkane

4. if chain has more carbons than cyclic molecule, the cyclo becomes the substituent

   1. point of attachment is C1

trans

2 substituents on opposite faces (one wedged and one dashed)

cis

2 substituents on same face (both wedged or dashed)

always implies gauche interactions (which raise energy)

ring flip/chair flip

axial becomes equatorial and vice versa

ranking stability/energy of cycloalkanes

fewer carbons means higher in energy/less stable

ring strain

increase in energy of molecule due to deviance from sp3 and 109.5 bond angle

angle, torsional, transannular

angle strain

occurring in cyclobutane and cyclopropane (small rings) due to bond angle significantly below 109.5

main reason for raising energy

torsional strain

occurring in cyclobutane and cyclopropane (small rings) lack of flexibility and locked in high energy conformation also can occur in cyclopentane

less impact on high energy

transannular strain

occurring in medium sized rings (7-12) due to clashing and cramming of molecules across ring

equatorial groups

more space and in low energy positiondi

diaxial interactions

result in higher energy

to push groups from equatorial to axial

need a positive delta g because going from low to high energy

delta g of chair conformations

delta g= (sum of axial/equatorial changes) + (sum of gauche changes)

enantiomers

type of stereoisomer that are non-superimposable mirror images

can’t overlay or rotate to make look alike but if reflect mirror, images would be identical

share physical properties

diastereomers

stereoisomers that are not mirror images

chiral

has multiple stereo centers and has a non superimposable mirror image

asymmetric

optically active

achiral

internal mirror plane (if chopped in half, the two halves reflect)

symmetric

optically inactive

meso

form of achiral but has stereocenters