FINAL Exam

Question 1

pKa of Water

Answer 1

16

Question 2

pKa of Ammonia

Answer 2

35–38

Question 3

pKa of Methane

Answer 3

50

Question 4

What does a high pKa value indicate?

Answer 4

• A weak acid
• A strong base

Question 5

What does a low pKa value indicate?

Answer 5

• A strong acid
• A weak base

Question 6

What does a high Ka value indicate?

Answer 6

• A strong acid
• A weak base

Question 7

What does a low Ka value indicate?

Answer 7

• A weak acid
• A strong base

Question 8

Components of Resonance

Answer 8

• Octet Rule
• Charge Compatibility
• Charge Separation

Question 9

Features that Stabilize Negative Charge

Answer 9

• Electronegativity
• Atom Size (Polarizability)
• Hybridization
• Resonance

Question 10

pKa of Hydrochloric Acid

Answer 10

–7

Question 11

pKa of Hydrobromic Acid

Answer 11

–9

Question 12

pKa of Hydroiodic Acid

Answer 12

–10

Question 13

How does bulkyness/branchedness impact the boiling point of an alkane?

Answer 13

• A more bulky/branched alkane has a lower boiling point.
• A less bulky/branched alkane has a higher boiling point.

Question 14

How does molecule size impact the boiling point of an alkane?

Answer 14

• A heavier alkane will have a higher boiling point.
• A lighter alkane will have a lower boiling point.

Question 15

What determines the boiling point of a compound?

Answer 15

The strength/quantity of london-dispersion forces within the compound.

A molecule will more london-dispersion forces will have a higher boiling point.

Question 16

What determines the melting point of a compound?

Answer 16

The ability of the compound to pack together into crystalline structures.

The more closely a compound can pack together, the higher its melting point will be.

Question 17

Why are staggered Newton conformations more stable than eclipsed Newton conformations?

Answer 17

• In the staggered Newton conformation, electron-electron repulsion between adjacent substituents are minimized.
• In the eclipsed Newton conformation, electron-electron repulsion between adjacent substituent causes significant torsional strain.

Question 18

Torsional Strain

Answer 18

The barrier to free rotation caused by electron-electron repulsion between nearby substituents.

Question 19

Why does the anti-form Newton projection result in the most stable conformation?

Answer 19

The anti-form conformation places the largest substituents (with the greatest number of electrons) at the maximum distance from one another.

Question 20

Steric Strain

Answer 20

The result of two substituents/molecules attempting the occupy the same region in space.

Question 21

Why is a tertiary carbon radical more stable than a primary carbon radical?

Answer 21

• The tertiary carbon radical experiences more hyperconjugational inductive donation from neigboring σ bonds.
• The primary carbon radical possessese fewer adjacent C—R σ bonds that can inductively donate electrons.

Question 22

Why does the Bromine radical only cleave the weakest bond of the substrate?

Answer 22

• The bromine radical is more stable due to the greater size of the atom, so it is less likely to cleave bonds.
• Creation of the bromine radical from Br—Br homolytic cleavage releases less heat, so less energy is available to cleave bonds of the substrate.

Question 23

Why will the Chlorine radical potentially cleave all bonds of the substrate?

Answer 23

• The chlorine radical is less stable due to the smaller size of the atom, so it is more likely to cleave bonds.
• Creation of the chlorine radical from Cl—Cl homolytic cleavage releases more heat, so more energy is available to cleave bonds of the substrate.

Question 24

Hammond’s Postulate

Answer 24

The more stable an intermediate structure is, the faster the product will form.

Question 25

What is the rate-determining step of radical halogenation?

Answer 25

The removal of hydrogen from the alkane (by the halogen radical) to form the carbon radical and the H—X compound.

Question 26

Why does radical Chlorination occur faster than radical Bromination?

Answer 26

• The cleavage of the C—H bond (of the alkane) with the Chlorine radical results in the formation of the more stable H—Cl bond.
• The cleavage of the C—H bond with the Bromine radical results in the formation of the less stable H—Br bond.

Hammond’s postulate states that the more stable compound will form faster, so the more stable H—Cl (moderate orbital overlap) compound will form faster than the less stable H—Br compound (poor orbital overlap).

Question 27

Why is radical Fluorination extremely reactive?

Answer 27

• The cleavage of the C—H (of the alkane) by the Fluorine radical releases a large amount of energy due to the formation of the highly stable H—F bond.
• The formation of the C—F bond (after the creation of the carbon radical) releases a large amount of energy due to great orbital overlap.

Question 28

Why does radial Iodination tend not to occur?

Answer 28

• The cleavage of the C—H (of the alkane) by the Iodine radical releases is endothermic due to the formation of the unstable H—I bond.
• The formation of the C—I bond (after the creation of the carbon radical) releases a minimal amount of energy due to poor orbital overlap.

Overall, the radical Iodination reaction is endothermic.

Question 29

Angle Strain

Answer 29

The strain caused by non-ideal bonding angles.

Question 30

Ring Strain

Answer 30

The sum of angle strain and torsional strain.

Question 31

Transanular Strain

Answer 31

The strain between the 1-substituent and the 4-substituent of a boat-conformation cyclohexane.

Question 32

1,3–Diaxial Interaction

Answer 32

The steric hindrance between the axial 1-substituent and the axial 3-substituent of a chair-confirmation cyclohexane.

Question 33

pKa of Hydrofluoric Acid

Answer 33

3.2

Question 34

pKa of Alkyne

Answer 34

24

Question 35

pKa of Alkene

Answer 35

42

Question 36

20-Carbon Alkane

Answer 36

Eicosane

Question 37

pKa of Alcohol

Answer 37

18

Question 38

pKa of Carboxylic Acid

Answer 38

5