Ch 2 - Molecular Representations Flashcards
(71 cards)
1
Q
4 types for small molecules
A
- Lewis structure
- partially condensed structure
- condensed structure
- molecular formula
2
Q
Lewis Structure
A
- all atoms and bonds are explicitly drawn
- only practical for very small molecules(large is hard to draw)
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
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
5
Q
molecular formulas do not
A
provide enough information
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
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
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
9
Q
functional group
A
a characteristic group of atoms/bonds that possess a predictable chemical behavior
10
Q
C=C is a functional group
A
Alkene
- typically react with molecular hydrogen(H2) in the presence of a catalyst
11
Q
the chemistry of every organic compound is determined by
A
the functional groups present in the compound
12
Q
R-X
A
- X = Cl,Br,or I
- alkyl halide
13
Q
R-(R-)C=C-R(-R)
A
alkene
14
Q
R-C-=C-R
A
Alkyne
15
Q
R-OH
A
Alcohol
16
Q
R-O-R
A
Ether
17
Q
R-SH
A
Thiol
18
Q
R-S-R
A
Sulfide
19
Q
Benzene ring of 6 carbon(double bond every other)
A
Aromatic(or arene)
20
Q
R-C(=O)-R
A
Ketone
21
Q
R-C(=O)-H
A
Aldehyde
22
Q
R-C(=O)-O-H
A
Carboxylic Acid
23
Q
R-C(=O)-X
A
Acyl halide
24
Q
R-C(=O)-O-C(=O)-R
A
Anhydride
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R-C(=O)-N(-R)-R
Amide
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R-C(=O)-O-R
Ester
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R-N(-R)-R
Amine
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a carbon atom will generally have four bonds only when it does not have a formal charge
- either a + or – charge on carbon will result in 3 bonds
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a + on carbon means 3 bonds total so H will
fill in missing spots(0,1,2)
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a – on carbon will still form 3 bonds total and have
1 extra lone pair(5 valence electrons total)
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C+ the 4th orbital is _____ and C- the fourth orbital holds ______.
empty
a lone pair of electrons
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three bond categories
- covalent(sharing)
- polar covalent(uneven sharing)
- ionic(electron transfer)
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the bond type is based on
the electronegativity(measure of an atoms ability to attract electrons) difference of two atoms
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General guideline for electronegativity(not definitive)
- 0.0-0.5 is covalent(C-C)
- 0.5-1.7 is polar covalent bond(C-O)
- 1.7+ is ionic(NaOH)
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induction
the withdrawal of electron density that occurs when a bond is shared by to atoms of differing electronegativity
36
to determine the formal charge on an atom we must know how many lone pairs it has
- or to determine lone pairs we must know the formal charge
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a bond line structure will only be clear if it contains either all of the lone pairs or all of the formal charges
- conventionally formal charges will be drawn(less of them to draw usually)
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formal charges must always be drawn, unlike
lone pairs which can be omitted
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2 steps for determining the number of lone pairs(per section 1.4)
- 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
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Oxygen atom formal charge patterns
- 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
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Nitrogen atom formal charge patterns
- 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
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wedges and dashes
- wedge is coming out of the page(toward you)
| - dash is going behind the page(away from you)
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resonance
draw reciprocal bond line structures and think of them as being melded together
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resonance structure
drawing showing how positive charge is spread over two locations at the same time
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resonance hybrid
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
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no single drawing will be adequate to describe resonance
- there must be a melding of images in your mind
47
resonance does not describe something that is happening but
is a way to deal with the inadequacy of bond-line structures
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delocalization
the spreading of positive or negative charge of an over two locations
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resonance stabilization
the delocalization of either positive or negative charge to stabilize a molecule
- plays a major role in the outcome of many reactions
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curved arrows
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
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the arrow must be drawn precisely in the right location
- the tail is where the electrons are coming from
| - head is where the electrons are going(not actually but as a representation)
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never draw curved arrows:
- 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
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when multiple arrows are used it is acceptable for one to violate the rules i
f the end result does not
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2 steps to formal charge in resonance structures
- carefully read what the curved arrows indicate
| - assign formal charges
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the total charge of both resonance structures MUST
be the same
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five patterns in resonance patterns
- 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
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allylic lone pair
- 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
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vinylic position
C=C the two carbons are in these positions
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allylic position
atoms connected directly to the vinylic positions
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allylic positive charge
- 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
lone pair adjacent to a positive charge
- 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
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pie bond between two atoms of differing electronegativity
- when the pie bond in a double bond is split then one atom gets the – while the other gets +
- one curved arrow
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conjugated pie bonds in a ring
- 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
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3 factors to resonance importance
- 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
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3 factors to resonance importance
minimizes charge
- 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
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
- 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
3 factors to resonance importance
avoid drawing a resonance structure in which two carbon atoms bear opposite charge
resonance structures in which there are C= and C- are insignificant
68
delocalized lone pair
a lone pair that is allylic to a pie bond and will participate in resonance
69
when an atom possesses a delocalized lone pair the geometry of the atom is
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
localized lone pair
a lone pair that does not participate in resonance(not allylic to a pie bond)
71
whenever an atom possesses both a pie bond and a lone pair, they will not both participate in resonance
- 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
