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Chem 351 > Midterm > Flashcards

Midterm Flashcards

1
Q

the lower the pKa ….

interms of eq and acidity

A

the larger the equilibrium constant and the stronger the acid.

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2
Q

At a pH above the pKa the acid exists largely as

A

its conjugate base (A−)

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3
Q

At a pH below the pKa the acid largely exists as

A

HA

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4
Q

the pKa of HCl is around –7, this tells us that in solution Ka for hydrogen chloride is

A

10^7 mol dm−3.

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5
Q

a weak acid is reluctant to ionize because it has

A

an unstable conjugate base

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6
Q

unstable anions A− make — bases and their conjugate acids AH are — acids

A

unstable anions A− make strong bases and their conju- gate acids AH are weak acids

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7
Q

the lower down the periodic table we go, the — the acid.

A

stronger

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8
Q

oxygen acids are — than nitrogen acids

A

stronger

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9
Q

The closer the electron density is to the nucleus, the — it is.

A

more stable

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10
Q
A

s-orbital is closer to nucleus than p-orbital comparing with same shell.
Moreover, more the s-character, more electronegative of that (carbon) atom.

Therefore, sp hybridized C withdraws the bonding e- closer to itself and it is easier to deprotonate.

It all depends on the s character of the bond. Think about the molecular orbitals. The electrons in the s orbital are more tightly bound to the nucleus, while electrons in the p orbitals are more inclined to be shared between atoms (e.g. in pi bonds). Therefore, electrons in molecular orbitals with a high s character are less inclined to be shared between atoms. If electrons are less inclined to be shared, the bond formed by those electrons is weaker and more easily broken. So, in a terminal alkene, which contains a bond between an sp-hybridized carbon (50% s character) and a hydrogen, the bond is easily broken because of the carbon’s high s character, resulting in a relatively high acidity (pKa ~20). However, in a bond between an sp3-hybridized carbon (25% s character) and a hydrogen, the bond is much stronger, resulting in very low acidity (pKa ~50-60).

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11
Q

Electronegativity in terms of acidity:

A

When comparing atoms within the same row of the periodic table, the more electronegative the anionic atom in the conjugate base, the better it is at accepting the negative charge.

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12
Q

Size in acidity

A

When comparing atoms within the same group of the periodic table, the easier it is for the conjugate base to accommodate negative charge (lower charge density). The size of the group also weakens the bond H-X (note this trend should be applied with care since it only works within a group).

HI > HBr > HCl > HF

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13
Q

Resonance in acidity

A

In the carboxylate ion, RCO2- the negative charge is delocalised across 2 electronegative oxygen atoms which makes it more stable than being localised on a specific atom as in the alkoxide, RO-.

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14
Q

What 6 properties affect acidity

A
  1. Resonance
  2. Hybridization
  3. Electronegativity
  4. Polarizability
  5. Inductive effects
  6. Charge balance
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15
Q

Why is F low in the orbital energy diagram

A

Higher Zeff and lower atomic size, nucleus pulls more strongly on electron

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16
Q

Methane PKA

A

~50

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17
Q

Amonia pka+formula

A

pka= 35
Nh3

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18
Q

HF pka

A

3.2

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19
Q

sp3->sp2 how much increase in acidity?

A

10^6
or 6 units in PKA

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20
Q

sp2->sp how much increase in acidity?

A

10^19
or 20 units

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21
Q

More e- = —- acidic

A

more

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22
Q

a big molecule can—–pka

A

have many pka

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23
Q

Down a periodic table, —–polarizable

A

the more

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24
Q

S and O acidicty in benzene rings

A

S one more acidity bc more polarizable
OH- pka 10
S- pka 6.6

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25
adding a ch3 group and its inductive effects:
In a molecule that has a charge, it wants something to pull away the neg charge. But carbon can actually add more e- density if its electronegativity is lower.
26
What side of the equilibrium is favoured?
higher PKA one that produces weaker acid and base
27
Position of equilibrium formula
Eq= 10^(change in PKA)
28
Bond strength: Bond order
Mostly wins. Nothiing is going to overcome bond order. As bond length decrease, bond strength increase
28
Moving down column while keeping one side constant or symmetric:
as you move down the column more electron = less bond strength
29
In atoms where there are lone pairs in second row, the bond strength
decrease
30
Hybridization in bond strength
Big change in sp->sp2 but small change from sp2->sp3 in bond strength
31
More stable radical come from ---- CH bond
weaker
32
Resonance always makes bonds
weaker
33
IHD Formula
IHD = 0.5 * [2c+2-h-x+n]
34
Allotrope
pure elements existing in different forms (Isomers of pure elements)
35
IHD Shortcut
Count number of rings + number of pi-bonds
36
What to add to a 5 carbon ring to make it saturated:
1 methyl group
37
What to add to a 4 carbon ring to make it saturated:
2 methyl groups (can stack to make ethyl etc)
38
What to add to a 3 carbon ring to make it saturated:
3 methyl groups (play around, can stack to make propane etc)
39
What happens to Hf in endothermic? | +very high Hf means=?
It is positive less stable
40
What happens to Hf in exothermic | +very low Hf means=?
Negative more stable
41
Hf
amount of energy it takes to form a molecule from starting elements in their standard states
42
Hc
the heat released when one mole of a substance is completely burned
43
More branches= --- thermodynamically stable
more
44
change in Hc is more neg means
more unstable | bc there is less energy released when it makes into more stable
45
In Alkenes and Alkynes more substitued mean
more stable
46
Trans bonds are ---- stable except in a ----
more ring
47
Phenol is a
weak acid
48
Radical halogenation | What it does+purpose
introduces a functional group into otherwise inert alkanes
49
In initaitaion, the total number of unpaired electrons
increases
50
In propaagation, the total number of unpaired electrons
stays the same
51
How do we know the pupose of chain reaction without words?
The net reaction can be solved by the propagation formula
52
selectivity calculation formula for Br, Cl
To determine % of desired product from radical rxn: (Relative factor X # of H of that type)/(Sum of same as top but for all type H) 1. find all the possible pleaces the cl can bind 2. count at that place how many H there is and determine the amount of H type that that H in the molecule 3. if there is 3 types of that H (type= where cl binded) times that with the relative factor of the degree of H 4. Divide by sum of all the diff structures (repeated 1-3 but withh all structure and sum it)
53
1 degree H
CH3
54
2nd degree hydrogen
CH2
55
3rd Degree Hydrogen
CH
56
1st degree cl in relative rate for CL
1
57
2nd degree cl in relative rate for CL
3.9
58
3rd degree cl in relative rate for CL
5.2
59
Br relative rate 3 degree
1640
60
1st degree Br in relative rate for Br
1
61
2nd degree Br in relative rate for Br
82
62
Resonance make bonds
weaker
63
How to find what shape a structure is with diff types of Carbon?
(Number of type of carbon)/(Total of number of carbon) If this formula is 1/2= mirror if the number is 1/2 or a good 1/X (ex: x= 4, 3) then rotational symmetry If number is fewer or more than 1/2 then branch symmetry or mirror with bisection
64
Resonance makes radicals ---and makes a bond -----
more stable weaker
65
Mirror+ bisect means it
goes thru the bonds
66
If 2-4 branches are identical, then each carbon atom walking away from branch is
identical type of carbons
67
reduced mass
((m1 X m2))/(m1+m2)
68
Bond strength and fequency relationship
Stronger bonds require more force to compress or stretch, which means that they will also vibrate faster than weaker bonds. Thus, frequency increases as bond strength increases.
69
Intensity of IR band is affected by
Polarity (Bond dipole) While electronegativity does not affect frequency, it does effect the intensity of the peak. This is because bond polarity effected absorption intensity. For example, C-C single and double bonds are either weak or moderate in intensity, while C-O single and double bonds are quite strong. This is because as infrared light causes stretching vibrations, bond dipoles will change. If a C=O double bond lengthens, the bond dipole will be stronger. This increases the intensity of the IR band.
70
IR spectrum scale for single bonds to H
>2500
71
IR spectrum scale for triple bond
2000-2500
72
IR spectrum scale for double bond
1550-1800
73
IR spectrum scale for single bonds
<1550
74
OH bond describe them on the IR graph
- 3300 to 3400 cm - broad and not pointy
75
CH bond in IR spectatory descirbe them
- >2500 - usually below 3000 - sharp and not broad
76
Having responance in IR diagram would
ship all the peaks onto the right side a bit
77
Why is 3000 a important number
- >3000 is sp2/sp (Left is double bond carbon with a bonded H) - <300 is sp3
78
Terminal triple bond | C=-C-H
Long and pointy in the 3000
79
Internal triple bond | -c-C=-C-c
small bumps
80
In a IR spectrumn Ester has
three bumps(CH, C=O, C-O)
81
What is the PKA
15
82
What is the PKA
20
83
What is the PKA
25
84
What is the PKA
30
85
H-H PKA
~35
86
WHat is the PKA?
1. 44 2. 25
87
Nh2 on benzene ring pka
35
88
PKA of ester
25
89
Terminal alkyne pka
25
90
Why is branching ideal?
The more branched isomer is more stable because branching increases the number of primary C atoms and hence the number of primary C-H bonds. Since, in terms of bond strength, primary C-H bonds > secondary C-H > tertiary C-H, a molecule with more primary C-H (more branching) is more stable (1 mark)
91
A tertiary carbon atom
A tertiary carbon atom is a carbon atom bound to three other carbon atom
92
secondary carbon atom
A carbon atom that is bonded to two other carbon atoms is a secondary carbon atom
93
What is the pka?
94
Whats the PKA
43
95
Whats the PKA
10
96
What is the PKA
-10
97
what is the PKA
-10
98
NH4 PKA
9.2
99
What does primary, secondary and teritary carbon mean
100
Being adjacent to the alkene ---- the C-H bonds
weakens
101
the primary allylic is --- than the secondary allylic
stronger
102
Category 1: Sp3 substitution only H labelling
1. primary 2. secondary 3. tertiary
103
Category 2: Hybridization
1. Aromatic 2. Acetylenic 3. vinylic
104
Category 3: Resonance
1. benzylic 2. Propargylic 3. allylic
105
Keytone function group +PKA
r-c(=o)-c Pka= 20
106
When asked something like "How many numebr of different monochlorinated constitutional isomers can be formed" on the stricture how do you approach this?
Each different monochlorination product comes from a different type of H. Count the H
107
Oxonium ion
108
soluble in NaOH solution but not NaHCO3
a phenol

Chem 351 (3 decks)

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