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Uploaded on
May 21, 2025
Number of pages
39
Written in
2024/2025
Type
Class notes
Professor(s)
Pete punthasee
Contains
All classes

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Second
Notes

, Chapter 1- Structure and Bonding
·

Octect rule is ESSENTIAL to understanding bonding 3 molecules

-Group number informs you about the bonding capacity = now much covalent bonds can be formed
↳ 4 bonds (Tetravelent) ,
3 bonds (trivalent) ,
2 bonds (divalent) , I bond (monovalent)

·

Hydrocarbons-made from CBH-carbon bond to another carbon forms carbon chains



FUNCTIONAL GROUPS Constitutional isomer (Structural isomer) -

Alkane all single bonds c -
c -
C compounds w/ the same formula but dif. Structure

Alkene one or more double bond C= c -
C


Alkyne one or more triple bond C= C -
C When figuring out the dif . Isomer start w the

bond at the end of the chain
your way down
Alcohol OH C-C-OH longest chain work

Ether 0 the middle of the chain c- c C Make sure you don't repeat any (symmetrical
-

in
-
-




.
Aldenyde O double bond at the end of the chain C -
c -

C -
H chains)
G


Ketone O double bond i n the middle of the c- Try dif. functional (alcohol Bester ,
c
groups
-
-




chain aldenyde Ketones)

Alkyl halide Any halogen bond to the carbon chain C -
c -
C
-

Y


Arene Carbon circle , with 3 double 33

single bonds C formal charges
11


Esters carbon double bond to an oxygen c
-
c -
o -
C formal charge = valence e-dots-lines

single bond to an o in the middle -
Number assign to an atom to reflect the

Carboxylic carbon double bond to an0 3 C -
C -

OH number of valence electron

acid single bond to of lat the end) If
you take away a H the formal charge becomes

Amine carbon bond to any N C -
C -

Ny negative
Amide Carbon double bond to an bond C -
- NHe
C If you add the
a formal charge become positive
to NHZ Carbocation -
Carbon with a positive formal charge

Carbanion-Carbon with a negative formal charge


Molecular Representation

Full-write out all the , H
C , est with bond lines Bond-line Structure

partial condensed -
wo need to show bond lines w/H make sure that is goes with the molecular

Bond line- Each angle is one carbon ,
don't write H
geometry
but write out all dif .
atoms eX . Hz) =
CHCHz((CH3) 3

eX .




Full or


- only way to draw because its
In
-




H etrahedral (can't
+ have both

condensed branches pointing down
CH3COzCH3 ex .
CH3C = CCHzCH3

Bond line

i Try to make the its linear so theres no


- angles 1200 angle from carbon 1- 4




Electronegativity EN Difference

the relative ability to attract shared e 4 EN <O . S nonpolar covalent bund

EN : F > 0xC1 > N > Br > S > . . . ) H >>> metals -EN o S-1
.
. 7
polar covalent bund

Atom w/ the stronger EN gets 5- and the DEN) 1 7
.
IOniC bond

weaker gets St in a
polar covalent bond
-
label the atom w/ a + or-based on the


Non polar covalent -
equal En con sign

lonic bond-an atom lose an e ibond between metal nonmetal

,eX . Molecular Polarity Polar v .
Nonpolar solvents
A molecule will be nonpolar Dielectric constant (5) the



-
-




1) When a molecule has no polar bonds polarity per unt volume (size)
2) When a molecule has polar bond , but it is The smaller the molecule , the




symmetrical dipoles cancel out more powerful the polar group
ex-
so the
-




-
Polar solvents-water , acetone
-11)
H eX 0 = c = 0 Medium Polar solvents -
H H
.




nonpolar polar ethyl acetate , acetic acid

Hydrocarbons are usually nonpolar Nonpolar solvent-diethyl ether ,

nexane




Dissolving compounds in a solvent

Dissolving NaCl in water

Na has a positive charge so attracted to the 8 8

3
it
-

is on
6 molecules usually surround the o n
-
CI has a
negative charge so it is attracted to the Ston H

-
the bond between them is on-dipole interactions

H -




o. HH ci----
*
molecules surround
Dissolving NaOCH3 In acetone
08
CHzCOCH3
#
H20 the lons
+
- I

"H
* "
5
Na ---- St
>


Na + 0
- -
-
H
they disappear a
o
CH3
-




until -




* 'd

o !--- -
O




o
Na---
=0


%g

, Chapter 2- Molecular
How

Representation
Valence

1
do atoms form



covalent
Bond

bond
Theory
a covalent bond




is formed when
?


two half
Hybridization
ex .
CH4-shape
Sp3
is tetrahedral DC USEPR

filled orbitals overlap unhybridized C Hybridized C

2 . The greater the overlap , the stronger F
1 1




I
Energy levels is between
more stable the bond
Sp
PX Py P2 HV



+
-
O bond : head to head relatively stronger
No place 1
-
it bond : side to side
for 4H + O
Sp3 Sp3 Sp3 Sp3
Doesn't explain molecular geometry , bonding pattern 2S gs
same energy Shape
-
It CH4-carbon doesn't have 4 half filled for 4
space
.


& has H

orbitals there would be an H bonded to 25


B3 bonded to 2p Sp3


ex .
H20- >
sp3 hybridized
Hybridization sp2
molecular geometry : Trigonal planar unhybridized o Hybridized O
>
ex .
C2H
-




Two lone HL
pairs
E -1177
unhybridized C Hybridized C Hybridized C -
These make a it bond PX Py P2 1uju
1v 15
1
Sp3 Sp3
11 H H
1 HL HV Sp3 Sp3
+
1
I
PX Py P2 1v1 -
7 P2
11 HV
Sp2 Sp2 Spz Sp2 Spa
Sp2
11 1V
2S T 1V
11
O
V
~ These two 11


ja
Orbitals bond H
3) make o bond
1 -
i bond

H11 Hybridization sp
H
C
C
H1)
H ex-C2Hz-molecular geometry : linear


o bond
unhybridized C


E1
11
PX Py PL
characteristic of sp , Sp
>
&Sp3
11
2S

S orbital = short 3 fat Hybridized C Hybridized C

porbital =
long & thin
1 1
11 11
PY P2 H Py
H P2
S +
p
=
Sp (50 % S) closes to nucleus

14 1
spa (33 % 5)
P
+ =
S p
+
p
Sp Sp
s
+
p +
p +
p
=
sp3(25y S) .
farthest to nucleus




single bond-longest bond length weakest , -bond/ V

triple bond-shortest bond length , strongest
H C 11 C 11 H


L
n

. abond
1 ↓
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