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Orgo 1 > Ch 2 - Molecular Representations > Flashcards

Ch 2 - Molecular Representations Flashcards

1
Q

4 types for small molecules

A
  • Lewis structure
  • partially condensed structure
  • condensed structure
  • molecular formula
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2
Q

Lewis Structure

A
  • all atoms and bonds are explicitly drawn

- only practical for very small molecules(large is hard to draw)

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3
Q

partially condensed structure

A
  • drawing style in which CH bonds are not drawn explicitly but all other bonds are
    - only practical for small molecules
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4
Q

condensed structures

A
  • drawing style in which none of the bonds are drawn
    - groups of atoms are clustered together when possible
    - (CH3)2CHOH
    - only practical for small molecules
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5
Q

molecular formulas do not

A

provide enough information

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6
Q

Bond-Line Structures

A
  • most common drawing style employed by organic chemist
    • all carbon atoms and most hydrogen atoms are implied but not explicitly drawn
  • each corner or end point is a carbon atom
    • triple bonds are drawn in a linear format(sp hybridized carbon) and will have linear geometry
    • each carbon in a bond line is assumed to have enough Hydrogen to complete 4 bonds so the hydrogens are not drawn
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7
Q

How to drawn Bond-Line structures

A
  • carbon atoms in a straight chain should be drawn in a zigzag format
    - when drawing double bonds, draw all bonds as far apart as possible
    - when drawing single bonds, the direction in which the bonds are drawn is irrelevant
    - all heteroatoms(atoms other than C and H) must be drawn, and any hydrogen atoms attached to a heteroatom must be drawn
    - never draw a carbon atom with more than four bonds. Carbon only has four orbitals in its valence shell and therefore carbon can only form four bonds
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8
Q

converting Lewis structure to Bond-line

A
  • delete hydrogen atoms(except those connected to heteroatoms)
    - draw in zigzag formal keeping triple bonds linear
    - delete carbon atoms
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9
Q

functional group

A

a characteristic group of atoms/bonds that possess a predictable chemical behavior

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10
Q

C=C is a functional group

A

Alkene

  • typically react with molecular hydrogen(H2) in the presence of a catalyst
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11
Q

the chemistry of every organic compound is determined by

A

the functional groups present in the compound

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12
Q

R-X

A
  • X = Cl,Br,or I

- alkyl halide

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13
Q

R-(R-)C=C-R(-R)

A

alkene

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14
Q

R-C-=C-R

A

Alkyne

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15
Q

R-OH

A

Alcohol

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16
Q

R-O-R

A

Ether

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17
Q

R-SH

A

Thiol

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18
Q

R-S-R

A

Sulfide

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19
Q

Benzene ring of 6 carbon(double bond every other)

A

Aromatic(or arene)

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20
Q

R-C(=O)-R

A

Ketone

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21
Q

R-C(=O)-H

A

Aldehyde

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22
Q

R-C(=O)-O-H

A

Carboxylic Acid

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23
Q

R-C(=O)-X

A

Acyl halide

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24
Q

R-C(=O)-O-C(=O)-R

A

Anhydride

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25
Q

R-C(=O)-N(-R)-R

A

Amide

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26
Q

R-C(=O)-O-R

A

Ester

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27
Q

R-N(-R)-R

A

Amine

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28
Q

a carbon atom will generally have four bonds only when it does not have a formal charge

A
  • either a + or – charge on carbon will result in 3 bonds
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29
Q

a + on carbon means 3 bonds total so H will

A

fill in missing spots(0,1,2)

30
Q

a – on carbon will still form 3 bonds total and have

A

1 extra lone pair(5 valence electrons total)

31
Q

C+ the 4th orbital is _____ and C- the fourth orbital holds ______.

A

empty

a lone pair of electrons

32
Q

three bond categories

A
  • covalent(sharing)
    - polar covalent(uneven sharing)
    - ionic(electron transfer)
33
Q

the bond type is based on

A

the electronegativity(measure of an atoms ability to attract electrons) difference of two atoms

34
Q

General guideline for electronegativity(not definitive)

A
  • 0.0-0.5 is covalent(C-C)
    - 0.5-1.7 is polar covalent bond(C-O)
    - 1.7+ is ionic(NaOH)
35
Q

induction

A

the withdrawal of electron density that occurs when a bond is shared by to atoms of differing electronegativity

36
Q

to determine the formal charge on an atom we must know how many lone pairs it has

A
  • or to determine lone pairs we must know the formal charge
37
Q

a bond line structure will only be clear if it contains either all of the lone pairs or all of the formal charges

A
  • conventionally formal charges will be drawn(less of them to draw usually)
38
Q

formal charges must always be drawn, unlike

A

lone pairs which can be omitted

39
Q

2 steps for determining the number of lone pairs(per section 1.4)

A
  • determine the appropriate number of valence electrons for the atom
    - O should have 6 valence electrons
    - determine if the atom actually exhibits the appropriate number of electrons
    - C-O means O exhibits 7 valence electrons which means negative and is must ave 3 lone pairs
40
Q

Oxygen atom formal charge patterns

A
  • negative charge corresponds with one and three lone pairs
    - absence of charge corresponds with 2 bonds and 2 lone pairs
    - positive charge corresponds with three bonds and one lone pair
41
Q

Nitrogen atom formal charge patterns

A
  • negative charge corresponds with two bonds and two lone pairs
    - absence of charge corresponds with three bonds and one lone pair
    - positive charge corresponds with four bonds and no lone pairs
42
Q

wedges and dashes

A
  • wedge is coming out of the page(toward you)

- dash is going behind the page(away from you)

43
Q

resonance

A

draw reciprocal bond line structures and think of them as being melded together

44
Q

resonance structure

A

drawing showing how positive charge is spread over two locations at the same time

45
Q

resonance hybrid

A

the resonance structure is placed in brackets with between the two parts indicating resonance
- the entity is not flipping back between the two drawn figures

46
Q

no single drawing will be adequate to describe resonance

A
  • there must be a melding of images in your mind
47
Q

resonance does not describe something that is happening but

A

is a way to deal with the inadequacy of bond-line structures

48
Q

delocalization

A

the spreading of positive or negative charge of an over two locations

49
Q

resonance stabilization

A

the delocalization of either positive or negative charge to stabilize a molecule
- plays a major role in the outcome of many reactions

50
Q

curved arrows

A

tool required to draw resonance structures properly

	- always a tail and head to the arrow
	- do not represent the motion of electrons
		- they treat electrons as if they were moving, even though not usually moving
51
Q

the arrow must be drawn precisely in the right location

A
  • the tail is where the electrons are coming from

- head is where the electrons are going(not actually but as a representation)

52
Q

never draw curved arrows:

A
  • avoid breaking a single bond
    - never place the tail on a single bond
    - never exceed an octet for second-row elements.
    - CNOF can only have 4. Never 5 or 6
53
Q

when multiple arrows are used it is acceptable for one to violate the rules i

A

f the end result does not

54
Q

2 steps to formal charge in resonance structures

A
  • carefully read what the curved arrows indicate

- assign formal charges

55
Q

the total charge of both resonance structures MUST

A

be the same

56
Q

five patterns in resonance patterns

A
  • allylic lone pair
    - allylic positive charge
    - lone pair adjacent to a positive charge
    - pie bond between two atoms of differing electronegativity
    - conjugated pie bonds in a ring
57
Q

allylic lone pair

A
  • vinylic position – C=C the two carbons are in these positions
    - allylic position – atoms connected directly to the vinylic positions
    - requires two curved arrows for resonance
    - the first goes from the lone pair to forma pie bond
    - the second goes from the pie bond to form a lone pair
    - when the atom with the lone pair has a negative charge then the negative charge is transferred to the atom that ultimately receives the lone pair
    - if the atom with the lone pair does not have a negative charge then it will incur a positive charge
58
Q

vinylic position

A

C=C the two carbons are in these positions

59
Q

allylic position

A

atoms connected directly to the vinylic positions

60
Q

allylic positive charge

A
  • only one curved arrow will be used
    • conjugated – when there are two pie bonds separated by exactly one sigma bond
    • when there is a positive charge we move the positive from one end of the other end of the system
    • never place the tails of an arrow on a + sign
      • put them on the pie bond
61
Q

lone pair adjacent to a positive charge

A
  • if the atom with lone pairs has a negative charge then the + and – cancel
    - not true for Nitro groups(NO2 where N is + and O is -)because it will violate the octet rule for N
    - requires two curved arrows resulting in – switching from one O to the other and N staying +
    • if the atom with lone pairs is neutral then it will become + by the end
    • one curved arrow
62
Q

pie bond between two atoms of differing electronegativity

A
  • when the pie bond in a double bond is split then one atom gets the – while the other gets +
    • one curved arrow
63
Q

conjugated pie bonds in a ring

A
  • in a closed ring if one pie bond in a double ring is pushed then all are
    - rotates an Arene one bond over for each bond
    • three curved arrows
64
Q

3 factors to resonance importance

A
  • minimizes charge
    - electronegative atoms like N,O,Cl can bear a positive charge but only if they possess an octet of electrons
    - avoid drawing a resonance structure in which two carbon atoms bear opposite charge
65
Q

3 factors to resonance importance

minimizes charge

A
  • the best resonance structure is one without charge
    - structures with more than 2 charges should be avoided
    - the Nitro(NO2) group always has a minimum of two charges so they are considered 0(a 4 charge would actually be considered 2 and thus significant)
66
Q

3 factors to resonance importance

electronegative atoms like N,O,Cl can bear a positive charge but only if they possess an octet of electrons

A
  • O can have 3 bonds + a lone pair which = 5 electrons(1 less than normal) and thus positive
    - the most significant factor in resonance structures is whichever structure gives all atoms a complete octet
67
Q

3 factors to resonance importance

avoid drawing a resonance structure in which two carbon atoms bear opposite charge

A

resonance structures in which there are C= and C- are insignificant

68
Q

delocalized lone pair

A

a lone pair that is allylic to a pie bond and will participate in resonance

69
Q

when an atom possesses a delocalized lone pair the geometry of the atom is

A

affected by the lone pair

An amide is expected to be an sp^3 hybrid but is actually sp^2 with the lone pair creating the pie bond in C=N

70
Q

localized lone pair

A

a lone pair that does not participate in resonance(not allylic to a pie bond)

71
Q

whenever an atom possesses both a pie bond and a lone pair, they will not both participate in resonance

A
  • in general the pie bond will and the lone pair will not

- the pie bond takes the p orbital disallowing the lone pair to fill the same space

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