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Chem 351 > Midterm > Flashcards

Midterm Flashcards

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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
Q

adding a ch3 group and its inductive effects:

A

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.

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26
Q

What side of the equilibrium is favoured?

A

higher PKA
one that produces weaker acid and base

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27
Q

Position of equilibrium formula

A

Eq= 10^(change in PKA)

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28
Q

Bond strength: Bond order

A

Mostly wins. Nothiing is going to overcome bond order.
As bond length decrease, bond strength increase

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28
Q

Moving down column while keeping one side constant or symmetric:

A

as you move down the column more electron = less bond strength

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29
Q

In atoms where there are lone pairs in second row, the bond strength

A

decrease

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30
Q

Hybridization in bond strength

A

Big change in sp->sp2 but small change from sp2->sp3 in bond strength

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31
Q

More stable radical come from —- CH bond

A

weaker

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32
Q

Resonance always makes bonds

A

weaker

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33
Q

IHD Formula

A

IHD = 0.5 * [2c+2-h-x+n]

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34
Q

Allotrope

A

pure elements existing in different forms (Isomers of pure elements)

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35
Q

IHD Shortcut

A

Count number of rings + number of pi-bonds

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36
Q

What to add to a 5 carbon ring to make it saturated:

A

1 methyl group

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37
Q

What to add to a 4 carbon ring to make it saturated:

A

2 methyl groups (can stack to make ethyl etc)

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38
Q

What to add to a 3 carbon ring to make it saturated:

A

3 methyl groups (play around, can stack to make propane etc)

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39
Q

What happens to Hf in endothermic?

+very high Hf means=?

A

It is positive
less stable

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40
Q

What happens to Hf in exothermic

+very low Hf means=?

A

Negative
more stable

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41
Q

Hf

A

amount of energy it takes to form a molecule from starting elements in their standard states

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42
Q

Hc

A

the heat released when one mole of a substance is completely burned

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43
Q

More branches= — thermodynamically stable

A

more

44
Q

change in Hc is more neg means

A

more unstable

bc there is less energy released when it makes into more stable

45
Q

In Alkenes and Alkynes more substitued mean

A

more stable

46
Q

Trans bonds are —- stable except in a —-

A

more
ring

47
Q

Phenol is a

A

weak acid

48
Q

Radical halogenation

What it does+purpose

A

introduces a functional group into otherwise inert alkanes

49
Q

In initaitaion, the total number of unpaired electrons

A

increases

50
Q

In propaagation, the total number of unpaired electrons

A

stays the same

51
Q

How do we know the pupose of chain reaction without words?

A

The net reaction can be solved by the propagation formula

52
Q

selectivity calculation formula for Br, Cl

A

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
Q

1 degree H

A

CH3

54
Q

2nd degree hydrogen

A

CH2

55
Q

3rd Degree Hydrogen

A

CH

56
Q

1st degree cl in relative rate for CL

A

1

57
Q

2nd degree cl in relative rate for CL

A

3.9

58
Q

3rd degree cl in relative rate for CL

A

5.2

59
Q

Br relative rate 3 degree

A

1640

60
Q

1st degree Br in relative rate for Br

A

1

61
Q

2nd degree Br in relative rate for Br

A

82

62
Q

Resonance make bonds

A

weaker

63
Q

How to find what shape a structure is with diff types of Carbon?

A

(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
Q

Resonance makes radicals —and makes a bond —–

A

more stable
weaker

65
Q

Mirror+ bisect means it

A

goes thru the bonds

66
Q

If 2-4 branches are identical, then each carbon atom walking away from branch is

A

identical type of carbons

67
Q

reduced mass

A

((m1 X m2))/(m1+m2)

68
Q

Bond strength and fequency relationship

A

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
Q

Intensity of IR band is affected by

A

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
Q

IR spectrum scale for single bonds to H

A

> 2500

71
Q

IR spectrum scale for triple bond

A

2000-2500

72
Q

IR spectrum scale for double bond

A

1550-1800

73
Q

IR spectrum scale for single bonds

A

<1550

74
Q

OH bond describe them on the IR graph

A
  • 3300 to 3400 cm
  • broad and not pointy
75
Q

CH bond in IR spectatory descirbe them

A
  • > 2500
  • usually below 3000
  • sharp and not broad
76
Q

Having responance in IR diagram would

A

ship all the peaks onto the right side a bit

77
Q

Why is 3000 a important number

A
  • > 3000 is sp2/sp (Left is double bond carbon with a bonded H)
  • <300 is sp3
78
Q

Terminal triple bond

C=-C-H

A

Long and pointy in the 3000

79
Q

Internal triple bond

-c-C=-C-c

A

small bumps

80
Q

In a IR spectrumn Ester has

A

three bumps(CH, C=O, C-O)

81
Q

What is the PKA

A

15

82
Q

What is the PKA

A

20

83
Q

What is the PKA

A

25

84
Q

What is the PKA

A

30

85
Q

H-H PKA

A

~35

86
Q

WHat is the PKA?

A
  1. 44
  2. 25
87
Q

Nh2 on benzene ring pka

A

35

88
Q

PKA of ester

A

25

89
Q

Terminal alkyne pka

A

25

90
Q

Why is branching ideal?

A

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
Q

A tertiary carbon atom

A

A tertiary carbon atom is a carbon atom bound to three other carbon atom

92
Q

secondary carbon atom

A

A carbon atom that is bonded to two other carbon atoms is a secondary carbon atom

93
Q

What is the pka?

A
94
Q

Whats the PKA

A

43

95
Q

Whats the PKA

A

10

96
Q

What is the PKA

A

-10

97
Q

what is the PKA

A

-10

98
Q

NH4 PKA

A

9.2

99
Q

What does primary, secondary and teritary carbon mean

A
100
Q

Being adjacent to the alkene —- the C-H bonds

A

weakens

101
Q

the primary allylic is — than the secondary allylic

A

stronger

102
Q

Category 1: Sp3 substitution only H labelling

A
  1. primary
  2. secondary
  3. tertiary
103
Q

Category 2: Hybridization

A
  1. Aromatic
  2. Acetylenic
  3. vinylic
104
Q

Category 3: Resonance

A
  1. benzylic
  2. Propargylic
  3. allylic
105
Q

Keytone function group +PKA

A

r-c(=o)-c
Pka= 20

106
Q

When asked something like “How many numebr of different monochlorinated constitutional isomers can be formed” on the stricture how do you approach this?

A

Each different monochlorination product comes from a different type of H. Count the H

107
Q

Oxonium ion

A
108
Q

soluble in NaOH solution but not NaHCO3

A

a phenol

Chem 351 (3 decks)

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