CHEM 0310 Lecture Notes - Lecture 1: Isopropyl Alcohol, Amine, Trigonal Pyramidal Molecular Geometry

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O Chem 1 Lecture 1
Fortney email subject line: 0310 SU18 Last Name Topic
What is OChem: Carbon based compounds: important b/c makes 4 bonds and is catanate: can
make bonds to itself more than any other element. What bonds C makes, how they’re made, are
they reactive.
Functional groups: learn in book pg: 70-71
Carbons bonds are strong and not reactive. It's not the carbon carbon bonds were interested in but
the functional groups, whats attached to carbon. Where the functional group is, is where the rxn
group is going to be.
Alkanes: single-bonded carbon compounds: know first 20 of them. Formula: CnH2n +2
Methane CH4, Ethane CH3CH3 (C2H6), Propane CH3CH2CH3 (C3H8), Butane
CH3CH2CH2CH3, Pentane
Halogens/Haloalkanes: CH3CH2X: X=F,Cl, Br, I. Recognize that its a halogen, but
mostly what it does to the carbon. Rule: the strongest acid has the weaker more stable
conjugate base.
The more stable the halogen is on its own, the better a leaving group it is from
carbon and the better we can replace it with something else. I is the strongest
leaving group because it has the weakest more stable conjugate base.
Alcohols: R-OH hydroxyl group. (a derivative of water)
Diff Types of alcohols: look at the carbon that has the OH attached to it, and how many
carbons are attached to it. (Note: methanol is NOT a primary alcohol b/c has 0 C attached
to it)
Primary alcohol: Propyl alcohol: has an alcohol group, but the attached carbon
has 1 other carbon and 2 H attached.
Secondary alcohol: Could have isopropyl alcohol which has same formula as
propan-1-ol but alcohol is on middle C so that C has 2 C attached to it.
Tertiary alcohol: three carbons on the C-OH.
Carboxylic Acids: RCOOH double bonded O on top, OH on the right.
Amine: derivative of ammonia- NH3
Primary: RNH2
Secondary: R2NH . Imagine the trigonal pyramidal structure, the 2 C groups are not
connected, they are each directly connected to the N.
Tertiary: R3N. Same for tertiary amine, all connected to N but the Cs not connected to
each other.
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A base has lone pair of electrons and often (-) charge. Picture ammonia has trigonal pyramidal
with 1 lone pair on top. ORBITALS ARE IMPORTANT. As add alkyl groups (R groups) change
the acidity and solubility.
Ether: derivative of an alcohol.
Symmetrical ether: R-O-R (CH3OCH3)
Asymmetrical: R-O-R’ (Ex: CH3CH2OCH3)
Ester: mix an alcohol and a carboxylic acid (with a strong acid catalyst)= Esterification
RCOOR double bonded O on top, O on the right connecting to the next R.
Amide: RCONH2(R) double bonded O on top, amine group on the right. The R on the
parenthesis represent potential subsitutants on the N instead of 2 Hs.
Amide ion (with long i pronounced) is the conjugate base to ammonia. Bent structure with NH2
and 2 lone pairs.
NH3 -- H+ → NH2
Heteroatoms: make more polar bonds.
IR: infrared spectroscopy: for functional group determination. To confirm that your esterification
worked, see a specific identifiable spectrum.
NMR: Nuclear Magnetic Resonance Spectroscopy: for structural determination. Added a carbon
so didn’t change the function but changed the structure.
Bonding: take into account: ionic vs covalent, shielding down group (inner electrons shield outer
electrons from pull of protons), effective nuclear charge down group (b/c of distance from
nucleus (periods=shells)), also across period consider: electrons being added but size not
changing b/c more protons have stronger pull on the electrons within same e- shell.
Effective nuclear charge: influence of nucleus on its electrons (this is strictly within a single
atom)
Electronegativity: that atoms tendency to draw atoms towards itself when its engaged in a bond
(this incorporate effect on other atoms)
Hybridization is important. Carbon has to change itself before it can make a bond with anything
else.
Technically with 4 valence electrons C should only be able to make 2 bonds b/c only 2 electrons
in 2p level. But it hybridizes to 4 half filled orbitals between s and p orbitals.
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Stereoisomers: same connectivity and structure but different relative positions of the functional
groups in space
The two main causes of the energy release associated with bonding are based on Coulomb’s law
of electric charge:
1. Opposite charges attract each other (electrons are attracted to protons).
2. Like charges repel each other (electrons spread out in space).
Steric number= # sigma bonds + # lone pairs
Alkanes=4 sigma bonds=sp3 orbitals
Alkenes= 3 sigma bonds= sp2 orbitals
Alkynes= 2 sigma bond= sp (1 s and 1p orbital being used)
In ethene, carbon only bonded to three atoms so only need 3 of our orbitals. So we promote the s
orbital up and bring 2 p orbitals down to an E level somewhere between the s and p as drawn
before. So now you have 3 orbitals a moderate E, 3 sp2 hybrid orbitals, and 1 higher E p orbital
left. So when drawn, trigonal pyramidal, 3 sp2 orbitals on bottom with one electron in each, and
one unhybridized p orbital sticking out the top and bottom also with one electron in it.
Remember: unhybridized p orbitals are dumbbell shaped, they stick out both sides. When
hybridized still a dumbbell but we ignore the smaller lobe so appears as sticking out in just one
direction, but the unhybridized one still goes out the top AND the bottom.
The reason double bonds are shorter is because when you have sp2 orbitals, you now have 33% s
character and 67% p (as opposed to 25% s and 75%p in sp3 orbitals). SO you have greater
concentration of s in sp2 orbitals which means everything is pulled a little closer to the nucleus
so when overlap, the bonds are shorter and stronger.
When we draw diagrams, if sp3 draw it 25% down from p, 75% higher than s
If sp2, lower the hybridized orbitals on the diagram to about ⅔ above s and ⅓ below p. The E of
the hybrid orbitals changes depending on alkanes, alkenes, alkynes.
--------------------------------------------------
Lecture 2 6/26/18
Reading for 6/17/18: 11.8 IR Spectra. Ph 456-460
10.2-10.8 pg 383-404
Homework prob for Ch 1: 25-31, 33, 35, 36, 38, 39, 40-51 + pre professional problem.
Hydrogen: can give up or take one an e-
H- = hydride anion =lewis base
H+ =proton=lewis acid
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Document Summary

Fortney email subject line: 0310 su18 last name topic. What is ochem: carbon based compounds: important b/c makes 4 bonds and is catanate: can make bonds to itself more than any other element. What bonds c makes, how they"re made, are they reactive. It"s not the carbon carbon bonds were interested in but the functional groups, whats attached to carbon. Where the functional group is, is where the rxn group is going to be. Alkanes: single-bonded carbon compounds: know first 20 of them. Methane ch4, ethane ch3ch3 (c2h6), propane ch3ch2ch3 (c3h8), butane. Recognize that its a halogen, but mostly what it does to the carbon. Rule: the strongest acid has the weaker more stable conjugate base. The more stable the halogen is on its own, the better a leaving group it is from carbon and the better we can replace it with something else. I is the strongest leaving group because it has the weakest more stable conjugate base.

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