CH203 Fall 2014 Lecture 11 September 26 Slides

8/11/2019 CH203 Fall 2014 Lecture 11 September 26 Slides

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!"#$ &'('#)*+

Exam 1 covers Chapters 1 3Dr. Zhou"s extra office hour

Monday/Tuesday discussions

,- ./, 0-1 2*+ 34--#$ 0-4--#$

,0 567 ,89 :;(


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/@


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The three moves of resonance

1. pi bond electrons become a lone pair

O

O

O

O


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2. lone pair electrons become a pi bond

O O


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3. pi bond electrons become a different pi bond

CH2 CH2


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Evaluating resonance contributors

0S Q(C('$H+( HP #OO #C*$F H+ CT( FC';BC;'( T#G( #+ *BC(C *P (O(BC'*+FS 6C';BC;'(F H+ UTHBT

#OO #C*$F V("B(&C TJI'*L(+W T#G( # B*$&O(C( *BC(C #'( (F&(BH#OOJ FC#DO( #+I $#X( O#'L('

B*+C'HD;)*+F C* CT( TJD'HIS


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Hyperconjugation

CH3+

HH

H

=


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Hyperconjugation

Hyperconjugation stabilizes a carbocationby delocalization of electron density from

a sigma bonding molecular orbital into anadjacent empty p atomic orbital


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Hyperconjugation

The combination of two atomic orbitals gives two molecular orbitals.


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Hyperconjugation

Anti-periplanar: Two bondsor groups with a 180o

dihedral angle (i.e., lying inthe same plane but pointing

in opposite directions.)

E(I4 # /[@ FHL$# D*+IH+L *'DHC#O B*+C#H+H+L

CU* (O(BC'*+F

\'((+4 CT( FHL$# #+)D*+IH+L V($&CJW

*'DHC#O *P CT( #IK#B(+C /[@ D*+I:THF I(O*B#OH]#)*+ *P (O(BC'*+ I(+FHCJ

FC#DHOH](F CT( FC#LL('(I '*C#$(' *G(' CT(

(BOH&F(I '*C#$('Y #F CT( #+)[&('H&O#+#'

#OHL+$(+C *P CT( !#+I CT( !^ HF $#"H$H](I

H+ CT( FC#LL('(I '*C#$('S


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Hyperconjugation


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IUPAC rules: alkanes

1. Identify the longest chain. This gives the root name.2. Identify and name the substituent(s). (Prefix + -yl)

3. If there is one substituent, assign it a number (or locant - where it is attachedto the chain). Choose the lower possible number.

4. If there are two or more identical substituents,a. Number the chain so the lower number goes to the substituent encountered

first.b. Use di-, tri-, tetra-, etc. to show the number of identical substituents.

5. If there are two or more different substituents,

a. List them in alphabetical order.

b. Number the chain so the lower number goes to the substituent encounteredfirst on the chain.

c. In case of a tie, give the lower number to the substituent of loweralphabetical order (comes before the other alphabeticallly).

6. The prefixes di-, tri, tetra, etc. are not included in the alphabetizing of

substituents. Determine the order first, then insert these prefixes.


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IUPAC rules: alkanes


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IUPAC rules: cycloalkanes

1. Prefix the name of the open-chain alkane with cyclo-2. Name each substituent on the ring

3. If there are two different substituents, list them in alphabetical order.Assign the first to carbon 1. Number the ring to give the second

substituent the lowest possible number.

4. If there are three or more substituents, number the ring so that one ofthe substituents is at carbon 1 and the other locations have the

lowest possible numbers.5. Assemble the name by listing groups in alphabetical order and giving

each group (if there are two or more) a location number. The

prefixes di, tri, tetra etc., are not considered when alphabetizing.


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IUPAC rules: cycloalkanes


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Index of Hydrogen Deficiency (IHD)

C6H14 (CnH2n+2) is the molecular formula of hexane.Take away one pair of hydrogens and you have C6H12 (CnH2n)

An-Hexane

C6H14

BCyclohexane

C6H12

C3-Hexene

C6H12

Each pair of hydrogens removed from the ideal alkane molecular formulaCnH2n+2 represents the presence of either one ring or one double bond.

Each pair of hydrogens removed = an increase in the index of hydrogendeficiency of 1

A had an IHD = 0. B and C each have an IHD = 1


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IHD = 4 (1 ring, 3 double bonds) IHD = 3

Nicotine IHD = 5 THC IHD = 7

IHD = 2


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Calculation of IHD from the molecular formula by referencing the ideal alkane(the straight-chain unsaturated alkane with the same number of carbons):

IHD = #(number of hydrogens in the ideal alkane number of hydrogens in

the actual molecule)

For benzene, C6H6, the ideal alkane is C6H14

IHD = #(14 6) = #(8) = 4. True by inspection.

How to calculate the IHD when the actual molecular formula contains otherelements?


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For every Group 17 element in the actual MF, add one HFor every Group 16 element in the actual MF, ignore it

For every Group 15 element in the actual MF, subtract one H

Example: C6H6ClNO (ideal alkane would be C6H14)

Actual molecular formula: C6H6ClNO

Add one H for the one Cl: C6H6HNOIgnore the O: C6H6HN

Subtract one H for the one N: C6H6HH-1 or C6H6

IHD = #(number of hydrogens in the ideal alkane number of

hydrogens in the actual molecule)= #(14 6) = #(8) = 4

Matches the structure.


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Put together as a formula, this becomes and is often seen expressed as:

IHD = #((2C + 2) (H + X N))

Where C = number of carbons (in the actual MF)H = number of hydrogens

X = number of halogensN = number of nitrogens (or phosphori)

This makes sense when you recognize that 2C + 2 is the number of

hydrogens in the ideal alkane MF and that (H + X N) are the adjustments tothe number of hydrogens in the actual MF we arrived at by inspection:

(Start with H)

For every Group 17 element in the actual MF, add one H

For every Group 16 element in the actual MF, ignore itFor every Group 15 element in the actual MF, subtract one H


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CH203 Fall 2014 Lecture 11 September 26 Slides