Polarity of Bonds

Which of the following bonds is most polar? Solution

1. Mg-Br
2. Na-F
3. H-Cl
4. B-F
5. O-H

Polarity is directly proportional to the difference in electronegativity

O-H
Oxygen electronegativity = 3.5, Hydrogen electronegativity = 2.1
Electronegativity Difference: 3.5 - 2.1 = 1.4

Na-F
Sodium electronegativity = 0.9, Fluorine electronegativity = 4.0
Electronegativity Difference: 4.0 - 0.9 = 3.1
The highest polarity is between the metal and non-metal with the greatest electronegativity difference.

H-Cl
Hydrogen electronegativity = 2.1, Chlorine electronegativity = 3.0
Electronegativity Difference: 3.0 - 2.1 = 0.9

Mg-Br
Magnesium electronegativity = 1.2, Bromine electronegativity = 2.8
Electronegativity Difference: 2.8 - 1.2 = 1.6

B-F
Boron electronegativity = 2.0, Fluorine electronegativity = 4.0
Electronegativity Difference: 4.0 - 2.0 = 2.0

Units: Mass and Volume

1.0 kg equals: Solution

Hint Clear Info

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g

Hint Unavailable

1.0 kg = 1,000 g

1 mL equals: Solution

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cm³

Hint Unavailable

1 mL = 1 cm³.

(1,000 L = 1 m³)

Naming Multivalent Metals

Iron has two different charge states, iron(II) and iron(III). When referring to the charges on an element, the name can be ferrous, in which the suffix -ous refers to: Solution

1. A +2 charge
2. A +3 charge
3. A lower charged state
4. A higher charged state
5. None of the above

-ous is a lower charged state, whereas -ic is a higher charged state. Be careful, "-ous" does not always refer to a +2 charge. It is a relative term, and the magnitude depends on the element.

Naming Compounds

Name the hydrate: CuSO₄·5 H₂O Solution

1. copper sulfide pentawater
2. copper sulfide pentahydrate
3. copper(II) sulfate pentahydrate
4. copper(III) sulfate(II) hydrate
5. cuprous sulfate pentahydride

Copper is multivalent, therefore name using the stock naming system to indicate the charge with a roman numeral...
The amount of waters is named with the prefix system... 5 is penta...
copper(II) sulfate pentahydrate

Periodic Trends Review

On the periodic table, atomic radius decreases Solution

1. Down a group, and left-to-right across a period.
2. Up a group, and left-to-right across a period.
3. Down a group, and right-to-left across a period.
4. Up a group, and right-to-left across a period.
5. None of the above.

Up a group (because less energy levels), and left-to-right across a period (because higher nuclear/valence charge).

Molecular Weight

Calculate the molecular weight of ammonium sulfate in the following reaction, rounded to the nearest whole number. Solution $Cr_3(PO_4)_2 \ + \ 3(NH_4)_2SO_4 \ \longleftrightarrow \ 3CrSO_4 \ + \ 2(NH_4)_3(PO_4)$

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g/mol

Hint Unavailable

Ammonium sulfate, (NH₄)₂SO₄:

N × 2
H × 8
S × 1
O × 4
= 132 g/mol

Review Chemical Formulas: Criss-Cross Method

The chemical formula of aluminum chloride is: Solution

1. AlCl
2. AlCl₂
3. Al₂Cl
4. AlCl₃
5. ClAl₃

Balance the charges, Al³⁺ & Cl^1-...

AlCl₃

Empirical and Molecular Formula

A molecule contains 85.7% carbon and 14.3% hydrogen.

Determine the empirical formula. Solution

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

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CH₂ (Solution not shown)

If the molar mass of the compound is 56 g/mol, determine the molecular formula. Solution

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

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Hint Unavailable

C₄H₈ (Solution not shown)

Determining the Limiting Reagent

Given 100g of HCl and 100g of CaCO₃, which of the following will be consumed first as the limiting reagent (reactant)? (Careful: don't forget to balance your equation first.) Solution $HCl + CaCO_3 \longleftrightarrow CaCl_2 + H_2O + CO_2$

1. HCl
2. CaCO₃
3. CaCl₂
4. CO₂

Don't forget to balance first. $2HCl + CaCO_3 \longleftrightarrow CaCl_2 + H_2O + CO_2$ Second: convert mass to moles.
$\begin{align} \text{moles} & = \dfrac{mass}{MM} \\ \\ & = \dfrac{100g}{36.46\ g/mol } \\ \\ & = 2.74mol \ HCl \\ \\ \end{align}$ $\begin{align} \text{moles} & = \dfrac{mass}{MM} \\ \\ & = \dfrac{100g}{100.09\ g/mol } \\ \\ & = 1.00mol \ CaCO_3 \\ \\ \end{align}$ Third: Divide moles by the stoichiometric coefficient - the lower number is from our limiting reagent.
$\begin{align} & = \dfrac{2.74mol \ HCl}{2 mol \ HCl} \\ \\ & = 1.37 \\ \\ \end{align}$ $\begin{align} & = \dfrac{1.00mol \ CaCO_3}{1mol \ CaCO_3} \\ \\ & = 1 \\ \\ \end{align}$ Since CaCO₃ has a lower ratio, then it is the limiting reagent.

Stoichiometry: Theoretical Yield

If O_2(g) is in excess, and 42.0g of C₂H_4(g) is consumed in the following reaction, determine the theoretical yield of CO_2(g). Solution $C_2H_{4(g)} + 3O_{2(g)} \longleftrightarrow 2CO_{2(g)} + 2H_2O_{2(g)}$

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g

Hint Unavailable

Since there are only two reactants and one of them is in excess, then the theoretical yield has to be based on the only other reactant → C₂H_4(g).

1st, convert mass to moles: $\begin{align} & = \dfrac{42.0g}{28.06 g/mol} \\ \\ & = 1.5\ mol \\ \\ \end{align}$ 2nd, use the stoichiometric ratio of C₂H₄ to CO₂. $\begin{align} \dfrac{mol\ CO_2}{coefficient\ CO_2} & = \dfrac{mol\ C_2H_4}{coefficient\ C_2H_4} \\ \\ \dfrac{mol\ CO_2}{2} & = \dfrac{1.5\ mol}{1} \\ \\ mol\ CO_2 & = \dfrac{1.5\ mol}{1}(2) \\ \\ & = 3.0\ mol \\ \\ \end{align}$ 3rd, convert moles to mass: $\begin{align} mass & = (3.0\ \cancel{mol} )(44\ g/\cancel{mol}) \\ \\ & = 132\ g \\ \\ \end{align}$

Percent Yield

If 50.0 g of K₂S reacts with excess zinc chloride to produce 40.1 g of zinc sulfide, determine the percent yield of the precipitate. Solution $K_2S_{(aq)} + ZnCl_{2(aq)} \longleftrightarrow 2KCl_{(aq)} + ZnS_{(s)}$

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%

Hint Unavailable

Since there are only two reactants and one of them is in excess, then the theoretical yield has to be based on the only other reactant → K₂S.

1st, convert mass to moles: $\begin{align} & = \dfrac{50.0g}{110.27 g/mol } \\ \\ & = 0.45\ mol \\ \\ \end{align}$ 2nd, use the stoichiometric ratio of K₂S to ZnS. $\begin{align} \dfrac{mol\ ZnS}{coefficient\ ZnS} & = \dfrac{mol\ K_2S}{coefficient\ K_2S} \\ \\ \dfrac{mol\ ZnS}{1} & = \dfrac{0.45\ mol}{1} \\ \\ mol\ ZnS & = \dfrac{0.45\ mol}{1}(1) \\ \\ & = 0.45\ mol \\ \\ \end{align}$ 3rd, convert moles to mass: $\begin{align} mass & = (0.45\ \cancel{mol} )(97.46\ g/\cancel{mol}) \\ \\ & = 43.86\ g\ ZnS \\ \\ \end{align}$ 4th, calculate percent yield $\begin{align} & = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ \\ & = \dfrac{40.1\ g}{43.86\ g} \times 100\% \\ \\ & = 91\% \\ \\ \end{align}$

Units: Solutions

In solutions chemistry, the unit 'M' refers to Solution

1. mass, g
2. mol
3. volume, L
4. concentration, mol/L
5. molar mass, g/mol

Capital 'M' stands for molarity, which is the unit of concentration. This is the same as mol/L.

Calculating Molarity

Determine the molarity (molar concentration) of 1.0 mol NaCl dissolved in 250 mL of water. Solution

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mol/L

Hint Unavailable

Although 1 mol NaCl dissociates into 2 mols of ions: Na⁺ + Cl^-, the molar concentration uses the mols of the solute, or compound instead of the total ion equivalent. $\begin{align} C & = \dfrac{n}{V} \\ \\ & = \dfrac{1.0 \text{mol}}{0.250 \text{L}} \\ \\ & = 4.0 \text{mol/L} \\ \\ \end{align}$

The concentration of chloride anions (Cl^-) in 1.0 mol of MgCl₂, dissolved in 1.0 L of water is 2 mol/L Solution

1. True
2. False

MgCl₂ dissociates into 2 chloride (Cl^-) ions: $MgCl_2 \xrightarrow{} Mg^{2+} + 2Cl^{-}$ 1.0 mol of MgCl₂ dissolved in 1.0L of water = 1 mol/L MgCl₂.

Since there are every 1 MgCl₂ forms 2 Cl^- ions, then the molarity of chloride anions (Cl^-) = 2(1 mol/L)
= 2 mol/L

Net Ionic Equation

Determine the net ionic equation for the reaction below. Show your work in your own notes. Solution $Pb(NO_3)_{2(aq)} + (NH_4)_2CO_{3(aq)} \xrightarrow{} PbCO_{3(s)} + 2NH_{3(g)} + 2HNO_{3(aq)}$

Separate the ions of all soluble compounds (leaving the solid, liquid, and gases).
$Pb^{2+}_{(aq)} + 2NO^-_{3(aq)} + 2NH^+_{4(aq)} + CO^{2-}_{3(aq)} \xrightarrow{} PbCO_{3(s)} + 2NH_{3(g)} + 2H^+_{(aq)} + 2NO^-_{3(aq)}$
Cancel ions on the left and right side, leaving:
$Pb^{2+}_{(aq)} + 2NH^+_{4(aq)} + CO^{2-}_{3(aq)} \xrightarrow{} PbCO_{3(s)} + 2NH_{3(g)} + 2H^+_{(aq)}$

Conjugate Pair of Acids and Bases

Which of the following is a conjugate acid-base pair? Solution $H_2CO_3 + NaOH \xrightarrow{} NaHCO_3 + H_2O$

1. H₂CO₃ and H₂O
2. NaOH and NaHCO₃
3. H₂CO₃ and NaOH
4. H₂CO₃ and NaHCO₃
5. NaHCO₃ and H₂O

The carbonic acid H₂CO₃ dissociates into the bicarbonate anion, which forms a salt with the available sodium ions, as sodium bicarbonate: NaHCO₃

Solutions of Acids and Bases: Monoprotic Neutralization

A 3.33 L sample of 8.88 mol/L nitric acid (HNO₃) is used to neutralize what volume of 4.44 mol/L potassium hydroxide (KOH)? Solution

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L

Hint Unavailable

Neutralization equation: $\begin{align} mol_{acid} & = mol_{base} \\ \\ C_a V_a & = C_b V_b \\ \\ (8.88\text{mol/L})(3.33\text{L}) & = (4.44\text{mol/L})(V_b) \\ \\ V_b & = \dfrac{(8.88\text{mol/L})(3.33\text{L})}{4.44\text{mol/L}} \\ \\ & = 6.66 \text{L} \\ \\ \end{align}$

Thermochemistry: Endothermic and Exothermic

A reaction that causes the temperature of the system to decrease from 12.˚C to 5.4˚C is endothermic. Solution

1. True
2. False

Endothermic reactions absorb heat energy, causing the temperature of the system to decrease. Exothermic reactions release heat energy, causing the temperature of the system to increase.

Converting Kelvin and Celsius

Convert 273K to Celsius: Solution

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˚C

Hint Unavailable

Celsius = Kelvin - 273

Kelvin = 273K - 273

= 0˚C

Definition of Specific Heat

Specific heat is defined as Solution

1. The energy required to raise the temperature of one gram by 1˚C
2. The energy required to raise the temperature of one kilogram by 1˚C
3. The energy required to raise the temperature of one kilogram by 1K
4. The temperature required to raise the energy of one gram by 1˚C
5. The energy required to raise the temperature of pure water by one gram by 1˚C

Specific heat is

The energy required to raise the temperature of one gram by 1˚C

Specific Heat

12 kJ of energy is added to water causing the temperature to increase from 296K to 343K. If the specific heat capacity of water is 4.18 J/(g·˚C) determine the mass of water. Solution

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g

Hint Unavailable

q = mc(T₂ - T₁)
Make sure to use Joules and not Kilojoules in the equation...
Change in temperature can use either Kelvin or ˚C because the magnitude of the change is the same for both. $\begin{align} m & = \dfrac{q}{c(T_2 - T_1)} \\ \\ & = \dfrac{12,000\ J}{4.18 J/(g·˚C)(343K - 296K)} \\ \\ & = 61.08\ g \\ \\ \end{align}$

Gas Laws

Which of the following is Charles' Law? Solution

1. $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$
2. PV = nRT
3. $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$
4. P₁V₁ = P₂V₂
5. $\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}$

Charles' Law: $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$

Gay-Lussac's Law: $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

Avogadro's Law: $\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}$

Boyle's Law: P₁V₁ = P₂V₂

Ideal Gas Law: PV = nRT

Dalton's Law of Partial Pressure

If 12 kPa of O₂ gas is mixed with the following gases in a 20 L container, determine the total pressure if they do not react: 20 kPa CO₂, and 30 kPa N₂. Solution

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kPa

Hint Unavailable

$P_{total} = P_1 + P_2 + P_3 \\ \\ = 12 + 20 + 30$

Ideal Gas Law

Determine the temperature of 10.0 g of carbon dioxide gas (CO₂) in a 500mL container at 1.5 atm. (R = 8.31 kPa·L/mol·K) Solution

T =

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K

Hint Unavailable

First convert all units to those used in 'R': kPa·L/mol·K
500mL = 0.5L
1.5atm = $1.5 atm × \dfrac{101.3 kPa}{1 atm} = 151.95 kPa$
Convert 10.0g of CO₂ to moles: $\dfrac{10.0 g}{44.0 g/mol} = 0.23 mol$

Plug into equation to solve for T. $\begin{align} PV & = nRT \\ \\ T & = \dfrac{PV}{nR} \\ \\ & = \dfrac{(121.95 kPa)(0.5 L)}{(0.23 mol)(8.31 kPa \cdot L/mol \cdot K)} \\ \\ & = 31.9\ K \\ \\ \end{align}$

Molar Volume

At STP (0˚C and 101.3 kPa), the molar volume of a gas is 22.4 mol/L. Determine the volume that 50.0 g of fluorine gas occupies. Solution

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L

Hint Unavailable

Convert to mol, $\dfrac{50.0\ g}{2(18.99\ g/mol)} = 1.32\ mol$ Use molar volume at STP: $\begin{align} Volume & = \dfrac{Moles}{Molar\ Volume} \\ \\ & = \dfrac{1.32\ mol}{22.4\ mol/L} \\ \\ & = 0.06\ L \\ \\ \end{align}$

Stoichiometry with Gases

100 g of nitrogen gas combines with excess hydrogen in the following synthesis reaction at (0˚C, 101.3 kPa, 22.4 mol/L), determine the mass of ammonia produced. Solution $N_{2(g)} + 3H_{2(g)} \xrightarrow{} 2NH_{3(g)}$

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g

Hint Unavailable

First: Convert mass to moles $\begin{align} mol & = \dfrac{mass}{molar \ mass} \\ \\ & = \dfrac{100g}{2(14 g/mol)} \\ \\ & = 3.57 mol N_{2(g)} \\ \\ \end{align}$ Second: Use the stoichiometric ratio $\begin{align} \dfrac{\text{moles} \ NH_{3(g)}}{\text{coefficient} \ NH_{3(g)}} & = \dfrac{\text{moles} \ N_{2(g)}}{\text{coefficient} \ N_{2(g)}} \\ \\ \dfrac{x \ mol}{2} & = \dfrac{3.57 mol}{1} \\ \\ x & = 2(3.57 mol) \\ \\ & = 7.14 mol \\ \\ \end{align}$ Third: Convert moles to mass $\begin{align} mass & = moles × molar \ mass \\ \\ & = (7.14 mol) × (17.0 g/mol) \\ \\ & = 121.38 g NH_{3(g)}\\ \\ \end{align}$

Naming Alkanes, Alkenes, Alkynes, and Alkyl Halides

Name the following hydrocarbons.

With lots of branching. Solution

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Hint Unavailable

3-ethyl-2-methyl-4-propylnonane

With multiple double bonds. Solution

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Hint Unavailable

2,4-dimethylpent-1,5-diene

With branching, alkyne, and a halide. Solution

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Hint Unavailable

3-chloro-3-iodobut-1-yne

(Start numbering from the alkyne side)

With equal numbers of carbon in the chain and cyclic. Solution

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Hint Unavailable

The cyclic group takes priority as the backbone when the cyclic carbon number is 'greater than or equal to' the number of carbons in the straight chain.

The straight chain has 4C and the cyclic has 4C, so the backbone is cyclobutane.

sec-butyl cyclobutane

With a longer straight chain than the cyclic group. Solution

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The 8C straight chain takes priority over the 3C cyclic group...

4-cyclopropyl-octane

With more carbons in the cyclic group. Solution

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The longest continuous carbon structure will take priority as the backbone... this is the 6C cyclohexane...

When listing alphabetically, 'cyclo-' counts!

1-cyclobutyl-4-ethyl-cyclohexane

With branching on cyclic. Solution

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Hint Unavailable

3-chloro-1-ethyl-6-methylcyclohexene

(Priority: Alkene > Alkyl > Halogen)

(Lowest numbering combination, must fully encompass double bond at start: 3-1-6 < 2-3-6)

Unsaturated Hydrocarbons

What does unsaturated mean? [1] Solution

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Unsaturated means it has double or triple bonds

Addition Reactions of Acid Halides with Alkenes (Markovnikov)

2-methyl-1-butene reacts with HBr to form what major product? Solution Video

1. 1,1-dibromo-2-methylbutane
2. 1,2-dibromo-2-methylbutane
3. 1-bromo-2-methylbutane
4. 2-bromo-2-methylbutane

Markovnikov Addition
In the major product, the hydrogen (H) atom attaches to the carbon with fewer alkyl (C) substituents, or the larger number of hydrogen atoms. The halide (X) group becomes attached to the carbon with more alkyl substituents.

The bromine attaches to the 3˚ carbon and the hydrogen attaches to the 1˚ carbon across the double bond. The major product is 2-bromo-2-methylbutane (and the minor product is 1-bromo-2-methylbutane).

What reaction type results in the major product of the reaction shown below? Solution

1. Addition
2. Substitution
3. Hydrohalogenation
4. Halogenation

Markovnikov Addition in Hydrohalogenation
In the major product, the hydrogen (H) atom attaches to the carbon with fewer alkyl (C) substituents, or the larger number of hydrogen atoms. The halide (X) group becomes attached to the carbon with more alkyl substituents.

Adding water to which of the following will make an equal mixture of two different products? Solution Video

1. 2-methyl-2-butene
2. Ethene (Ethylene)
3. 2-pentene
4. None of the above
5. All of the above

Markovnikov Addition
In the major product, the hydrogen (H) atom attaches to the carbon with fewer alkyl (C) substituents, or the larger number of hydrogen atoms. The hydroxide (OH) group becomes attached to the carbon with more alkyl substituents.

2-methyl-2-butene ✘

• Major product will form with -OH group attached to carbon 2, forming 2-methyl-2-butanol
• OH will add to the 3˚ carbon more than the 2˚ carbon
• (Minor product formed is 3-methyl-2-butanol)

Ethene (Ethylene) ✘

• Ethene is symmetrical about double bond, therefore will form 100% same molecule (highest yield)
• OH group adds to either of the 1˚ carbons across double bond
• Forms ethanol

2-pentene ✓

• Both carbons are 2˚, therefore there will be equal amounts of the two different products in a 50-50% mix of: 3-pentanol and 2-pentanol
• (The minor product on one is the major product of the other so it is only 50-50)

Similar to alkenes, addition of acid halides to alkynes follows Markovnikov's rule. Solution

1. True
2. False

The halogen bonds to the more highly substituted carbon... And in the presence of excess acid halide, a further addition reaction will occur to the alkene to produce a di-halide carbon.

Isomers

Isomers must always have the same chemical formula. Solution

1. True
2. False

All isomers have the same chemical formula, but either the bonding or the spatial arrangement is/are different.

Two isomers have the same bonding, but different arrangements in across carbon-carbon double bonds. Determine the classification of isomer. Solution

1. Geometric
2. Constitutional
3. Enantiomer
4. Conformational

Constitutional/Structural Isomers
Different bonding

Stereoisomers Isomers
Same bonding, dfferent arrangement... Eg)

• Geometric Diastereomers (different arrangement across C=C bonds)
• Enantiomers (mirror image, non-superimposable)

CC: Yassine Mrabet, 2008

Types of Isomers (Chart in Answer)

Classify the types of isomers below.

Alkenes... Solution Video

1. Geometric Diastereomer (Stereoisomers)
2. Constitutional (Structural) isomers
3. Conformational (Diastereomer) isomers
4. Enantiomers (Stereoisomers)

cis-2-Butene and trans-2-Butene are geometric (cis-trans) isomers, which are under the category of diastereomer and stereoisomer.

CC: Yassine Mrabet, 2008

Branched alkanes... Solution Video

1. Geometric
2. Constitutional (Structural)
3. Conformational
4. Enantiomers

Conformational isomers are a type of stereoisomer in which the bonding groups differ by rotation around a single C-C bond.

CC: Yassine Mrabet, 2008

Cyclic molecules... Solution Video

1. Conformational (Diastereomer) isomers
2. Geometric Diastereomer (Stereoisomers)
3. Enantiomers (Stereoisomers)
4. Constitutional (Structural) isomers

These cyclic ether molecules are structural isomers (aka constitutional isomers) because they would have the same chemical formula (C₃H₆O₃), but they have a different bond structure.

CC: Yassine Mrabet, 2008

Cyclic, with side groups... Solution

1. Geometric Diastereomer (Stereoisomers)
2. Constitutional (Structural) isomers
3. Conformational (Diastereomer) isomers
4. Enantiomers (Stereoisomers)

These diols (molecules containing two hydroxyl groups) are conformational isomers because the hydoxy group is rotated or flipped-up around a chiral carbon.

These are not enantiomers because they are not non-superimposable mirror images.

CC: Yassine Mrabet, 2008

(Not a diastereomer because does not have double bonds)

(Not constitutional/structural because carbons still have same things bonded to them)

Chiral Centers (Stereocenters)

How many stereoisomers are possible with 3 chiral centers in a molecule? Solution

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$\begin{align} & = 2^{chiral\ centers} \\ \\ & = 2^{3} \\ \\ & = 8 \\ \\ \end{align}$

The number of chiral centers in this Cholesterol molecule is Solution

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Chiral centers have different groups on all 4 bonds around a carbon. (Remember that chiral carbons are asymmetric such that the mirror image is non-superimposable.)

Naming Alcohols

Name the following alcohols using IUPAC naming.

Which of the following synonyms is the correct IUPAC name? Solution

1. butan-2-ol
2. butane-2-ol
3. 2-butanol
4. 2-hydroxy butane
5. None of the above

IUPAC naming places the number next to the parent/functional group suffix... butan-2-ol Remember to always drop the last "e" from the hydrocarbon name.

With one alcohol... Solution

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2-methyl-pentan-2-ol

With two alcohols... Solution

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2-methyl-pentan-2,3-diol

With three alcohols... Solution

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2-methyl-hexan-2,4,5-triol

Cyclic with two hydroxide groups Solution

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cyclopropan-1,2-diol

Cyclic with three hydroxide groups, and more... Solution

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Hint Unavailable

Priority: alcohol > alkene > alkyl Steps:

1. Find the longest continuous carbon chain
2. Start numbering from the side with the highest priority functional group
3. Number the side groups with the lowest combination of numbers
4. Add carbon numbers, prefixes, and drop the 'e' from root name.
5. Order alphabetically

4-methyl-cyclohex-1-en-1,3,5-triol

Naming Alcohols

The following name is incorrect. Determine the correct name. Solution 1-methyl-cyclohexane-3-ol

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Hint Unavailable

The alcohol takes priority so you should start numbering from the side with the alcohol... Also you should drop the 'e' at the end of the root name. 3-methyl-cyclohexan-1-ol

Name the following saturated, branched alcohol molecule. Solution

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Hint Unavailable

Steps...

1. Find the longest continuous carbon chain
2. Start numbering from the side with the highest priority functional group
3. Number the side groups with the lowest combination of numbers
4. Add carbon numbers, prefixes, and drop the 'e' from root name.
5. Order alphabetically

4,7-diethyl-3,7-dimethyl-decan-2-ol

Name the following unsaturated alcohol molecule. Solution

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The alcohol has the highest priority, so start numbering from that end... 3,7-dimethyl-oct-6-en-1-ol

Physical Properties: Melting & Boiling Points of Alcohols

Alcohols are composed of a non-polar, hydrophobic alkyl group (C_XH_Y), and a polar, hydrophilic hydroxide group (OH). The electrostatic dipole in the hydroxide group allows the non-polar chain to dissolve in water.

Melting and boiling point of alcohols depends on what two main factors? [2] Solution

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Molecular weight and side group branching.

Which of the following alcohols has the highest boiling point? Solution

1. ←
2. ←
3. ←
4. ←

Because it can form the most hydrogen bonds, which makes the intermolecular force strong and increases the boiling point (the point at which intermolecular forces are overcome in the liquid phase to spread apart into the gas phase).

Reaction: Hydration of Alkene Produces Alcohol

Hydration of propene mainly produces Solution Video

1. propan-2-ol
2. propan-1-ol
3. propyne
4. propane

Markovnikov's Rule
The hydrogen (H) attaches to the carbon atom with the most hydrogens, while the hydroxide (OH) attaches to the carbon with the highest alkyl substitution

Reactions: Addition (Hydration) Reactions

Complete the reaction (draw in your notes) and name the product formed. Solution Video

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Markovnikov's Rule
The hydrogen (H) attaches to the carbon atom with the most hydrogens, while the hydroxide (OH) attaches to the carbon with the highest alkyl substitution.



3-methyl-hexan-3-ol

Reaction: Elimination (Dehydration) of Alcohol

Treating an alcohol with concentrated sulfuric acid (H₂SO₄) produces the following molecule. What would the starting molecule have been? Solution Video

The concentrated sulfuric acid (H₂SO₄) catalyzes the dehydration of the alcohol in the starting molecule.

A hydrogen (H) and a neighboring hydroxyl (OH) group is then removed, resulting in the formation of a double bond (C=C).

Answer choice 'C' is incorrect because the dehydration of this alcohol would yield a major product with an internal double bond on more highly substituted carbons.

1. 
2. 
3. 
4. 

Reactions: Combustion of Alcohols

Complete combustion of one mole of 3-ethyl-2-methylhexan-1-ol would yield how many moles of carbon dioxide?
Solution Video

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mol

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The moles of carbon dioxide equals the number of carbons in one mole of the alcohol molecule.

C_xH_YO_Z + mO₂ --> xCO₂ + nH₂O

Naming Aldehydes

The name of the following molecule is Solution Video

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Recognize the aldehyde functional group; aldehydes take the suffix -al
The backbone is 5 carbons long, so it takes the root name pentan-... = pentanal.

Naming Aldehydes

Name the following molecules

Solution

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4-methylpentanal
Start numbering from the higher priority functional group - the aldehyde.

Solution

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5-bromo-7-hydroxy-4-methylheptanal
Start numbering from the higher priority functional group - the aldehyde.

Structural Isomers: Aldehyde

Which of the following is a structural isomer of 1-propanal? Solution

1. CH₃CH₂COOH
2. CH₃CH₂CH₂OH
3. CH₃C(O)CH₃
4. CH₃CH(OH)CH₃

Structural isomers (aka constitutional isomers) have the same chemical formula, but have different bond connectivity in their structures.

1-propanal is an aldehyde, and aldehydes have structural isomers with ketones (with same chemical formula).

Reactions: Oxidation of Alcohols

In the presence of a strong oxidizing agent, propan-1-ol is fully oxidized into propanal as the end product. Solution

1. True
2. False

A strong oxidizing agent will not stop at propanal, and will continue oxidation to form propanoic acid.

(To make aldehydes, a primary alcohol must be oxidized with a weak oxidizer like PCC).

Reactions of Aldehydes

Determine the name of the product of the following reaction. Solution

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Hydrogenation of an aldehyde will reduce it to an alcohol

Hydrogenation of an aldehyde will reduce it to an alcohol... propanol

Naming Ketones

Name the following ketone molecules.

Solution

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Accept butanone (or 2-butanone), when the molecule is superimposable when rotated in 3D space, you don't have to indicate the carbon number. There is no other possible position for the internal carbonyl carbon in a butane chain.

Solution

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4-methyl-2-pentanone

Start numbering from the end with the higher priority functional group - the ketone.

Solution

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3-methyl-2,6-heptdione

Start numbering from the end with the higher priority functional group - the ketone.

Properties of Ketones

Ketones have boiling points slightly higher than aldehydes of the same carbon chain length. Solution

1. True
2. False

The molecular weight of ketones is slightly higher than that of aldehydes.

Terminology of Carboxylic Acids

The standard name of the -COOH group in the diagram below is: Solution

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The -COOH group is a carboxyl. (The -C=O group is a carbonyl).

Naming Carboxylic Acids

Name the following carboxylic acid molecules.

With a halide... Solution

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4-bromopentanoic acid

With lots of branching... Solution

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2-ethyl-3,5-dimethylheptanoic acid

With two carboxyl groups... Solution

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butandioic acid

(Notes: drop the 'e' and don't have to indicate 1,4 because carboxylic acids have to be terminal.)

Reactions of Carboxylic Acids

Oxidation of which molecule will produce the following carboxylic acid? Solution

1. 5-hydroxy-1-pentanal
2. 2-methyl-pentane-1,5-diol
3. 5-hydroxy-3-methyl-2-pentanone
4. 2-methyl-pentane-2,4-diol

Oxidation of alcohols with a strong oxidizing agent like potassium permanganate (KMnO₄) will make carboxylic acids.

Naming Esters

Name the following ester molecules.

With an alkane chain on the oxygen... Solution

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butyl methanoate

With an alkane chain on the carbonyl carbon... Solution

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methyl pentanoate

With chains all over... Solution

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3-pentyl 2-ethylbutanoate

Naming Esters

Name the following ester molecule. Solution

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3-chlorobutan-1-ol 4-bromo-3-methylbutanoate

Properties of Esters

Different ester molecules have a wide range of characteristic odours. Smells are detected by olfactory receptors in the nose when an ester molecule lands on one of the receptors. Which molecule would have the strongest smell at room temperature? Solution

1. CH₃(CH₂)₆COOCH₃
2. CH₃CH₂CH₂COO(CH₂)₁₂CH₃
3. CH₃(CH₂)₂₂COOCH₃
4. CH₃CH₂COOCH₂CH₂CH₃

To detect the smell, the ester molecule must be evaporated into the air as a gas. The molecule that evaporates most will have a relatively low boiling point; with a lower molecular weight. The long chain molecules will not vaporize at room temperature and not be easily detected.

Ester Reactions: Esterification

In the presence of heat and an acid, ethanol + acetic acid will produce an ester with four carbons. Solution

1. True
2. False

Ethanol + Acetic Acid --> Ethylacetate

Ester Reactions: Synthesis of Esters

Esters are formed in a condensation reaction when hydrogen (-H) from a carboxylic acid and a hydroxyl (-OH) group from alcohol combine to form a water molecule, in the presence of heat. Solution

1. True
2. False

Esters are formed when the hydroxyl (-OH) from a carboxylic acid and a hydrogen (-H) from alcohol combine to form a water molecule, in the presence of heat.

State the reactants required to synthesize the following molecule and provide the name for this Ester. Solution

1. Butanol + Ethanoic Acid   →   butyl ethanoate
2. Ethanol + Butanoic Acid   →   ethyl butanoate
3. Ethanol + Butanol   →   Name: ethyl ether
4. Ethanal + Butanal   →   Name: butyl ether

Alcohol + Carboxylic acid → Ester

(This is a dehydration reaction)

Ester Reactions: Hydrolysis

Sodium hydroxide is added to a propylmethanoate ester, shown below. Determine (in your notes) the molecular structure of the conjugate base and conjugated acid products. Solution

The following process is called... Solution $\text{triglyceride} \xrightarrow[\Delta]{NaOH} \text{soap} + \text{glycerol}$

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The ester bonds in triglycerides (fats, lipids) are cleaved with the base (NaOH), which creates an alcohol (glycerol) and a carboxylate anion salt with the sodium. This is basically how soap is made... saponification

Naming Ethers

Name the following ether. Solution CH₃CH₂CH₂CH₂CH₂-O-CH₂CH₂CH₂CH₃

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Take the longest carbon chain (-pentane) and that is the backbone. Then add the other chain on the oxygen (but-) as the prefix...

butoxypentane

Properties of Ethers

The boiling point of butane is higher than the boiling point of ethoxyethane (diethyl ether). Solution

1. True
2. False

The boiling point of butane is lower than the boiling point of ethoxyethane (diethyl ether). This is due to the polarity of the C-O bonds in ethers being higher than the polarity of C-C bonds in the alkane, resulting in greater intermolecular forces and an increase in boiling point.

Reactions: Ether Dehydration Reaction

Fill in the missing reactant, X, for the following condensation reaction. Solution $X \ + \ CH_3CH_2CH_2CH_2OH \ \xrightarrow{H_2SO_4} \ CH_3CH_2\,O\,CH_2CH_2CH_2CH_3 \ + \ H_2O$

1. 
2. 
3. 
4. 

This is a condensation reaction between a hydroxyl, OH group from one alcohol and a hydrogen, H from another alcohol.

Naming Amines

Name the following molecule with IUPAC nomenclature. Solution

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N-ethyl-N-methylaminopropane The highest carbon chain gets the root name, while the remaining substituents are numbered with N and listed in alphabetical order. (Sometimes can be named with non-Iupac: ethylmethylpropylamine, or when CA naming convention: N-ethyl-N-methylpropanamine)

Naming Amines

Name the following amine molecules.

Solution

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N-ethyl-2-aminobutane
(or with CA naming convention: N-ethyl-2-butanamine)

Solution

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1,3-diaminopropane
(or with CA naming convention: 1,3-propanediamine)

Solution

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N-ethyl-N-methylaminoethane
(or with CA naming convention: N-ethyl-N-methyl-1-ethanamine)

Solution

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2-amino-3-methylhexane
(or with CA naming convention: 3-methyl-2-hexanamine)

Solution Video

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The ethyl group on the amine (N-ethyl) gets named first. The longest carbon-chain backbone is 10 carbons, decane... N-ethyl-5-amino-2,9-dimethyl-5-propyldecane (or CA naming convention: N-ethyl-2,9-dimethyl-5-propyl-5-decylamine)

Properties of Amines

All pure primary, secondary, and tertiary amines can hydrogen-bond. Solution

1. True
2. False

Only primary and secondary amines have N-H bonds that can hydrogen bond. The N-C bonds in tertiary amines cannot hydrogen bond.

Properties of Amines

Order the following molecules in order of increasing boiling point (from bottom to top) by dragging them. Solution Video

Hexanol

Hexanamine

Hexane

Hexanoic acid

Hexanoic acid has the strongest hydrogen bonding and highest molecular weight, so it has the highest boiling point. Hexanol can hydrogen bond. Hexanamine can hydrogen bond but has a lower boiling point than hexanol because the N-H dipole is weaker than the O-H dipole in the alcohol.

Functional Groups

Name the following functional group. Solution

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You should recognize the functional group, amide.

Naming Amides

Name the following amide molecules.

Solution Video

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N-ethyl butanamide

Solution

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N,N-diethyl ethanamide

Solution Video

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N,N-diethyl-3-methylpentanamide

Physical Properties of Amides

N-propyl-3-methyl butanamide can hydrogen bond. Solution

1. True
2. False

The N-H bond in the amide is capable of hydrogen bonding.

Amide Condensation Reactions

Which of the following molecules cannot react with ethanoic acid in a condensation reaction? Solution

1. Ammonia
2. 3-aminohexane
3. N-ethyl-1-aminobutane
4. N,N-dimethylaminopropane

Tertiary (3˚) amines like N,N-dimethylaminopropane cannot react in a condensation reaction with ethanoic acid because this reaction requires at least one N-H bond. All the other amines listed can react.

Synthesizing Amides

N,N-dimethyl butanamide can be synthesized from Solution Video

1. Ammonia plus dimethyl amine
2. A primary amine plus butanoic acid
3. A secondary amine plus butanoic acid
4. A tertiary amine plus butanal

Draw the N,N-dimethyl butanamide amide molecule first to visualize the bonding.

To form the amide bond (C-N), there must have been a condensation (dehydration) reaction between a secondary (2˚) amine and carboxylic acid.

Amides

Protein molecules can be split by adding water, separating into amino acids. Solution

1. True
2. False

Protein molecules are polypeptide chains (polymers) joined together with amide bonds. These amide bonds can be hydrolyzed, broken with water.

Carbonyl Groups

Carbonyl (-C=O) groups are present in all of the following molecules: Solution Ester, Ketone, Amide, Carboxylic Acid, Aldehyde

1. True
2. False

Carbonyl groups are carbon atoms that are double-bonded to oxygen.

Naming Simple Aromatic Hydrocarbons

The common name for this aromatic molecule is: Solution

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...phenol (this is a benzene ring with one alcohol side group)

Benzene Structures

What is ortho, meta, and para and when are they used? [1] Solution

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The ortho, meta, and para naming system is an alternative system that is used when benzene rings are substituted with exactly 2 side groups. This naming system cannot be used when benzene has 1 or 3+ side groups.

Ortho side groups are on adjacent carbons. Meta side groups are on carbons that have one carbon between them. Para side groups are on carbons with two other carbons between them.

Solution

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1-ethyl-3-methylbenzene

(aka m-methyl ethyl benzene) -- the ortho/meta/para nomenclature is interchangeable with the other naming system, but can only be used with two side groups. The other system can be used with two or more side groups.

Solution

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4-methyl-1-propylbenzene

The higher molecular weight propyl takes priority over methyl.

The substituents should be in alphabetical order with 'm' before 'p'. So 1-propyl-4-methylbenzene is not entirely correct.

(aka p-methyl propyl benzene) -- the ortho/meta/para nomenclature is interchangeable with the other naming system, but can only be used with two side groups. The other system can be used with two or more side groups.

Aromatic Hydrocarbons: Alyklbenzene

Name the following, which have benzene as a substituent.

Solution

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4-methyl-2-phenyl-1-hexene.

(Benzene becomes phenyl as a side group).

The substituents should be in alphabetical order with 'm' before 'p'.

Solution

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4-ethyl-5-methyl-3-phenyl-1,8-octadiene

Reactions of Aromatics

Determine the product when phenylmethanol is oxidized in a controlled oxidation. Solution

The controlled oxidation of phenylmethanol will produce benzaldehyde...

Polymers

Polymers are made of identical monomer subunits. Solution

1. True
2. False

It is true that polymers are made of many monomer subunits, but the monomer subunits do not have to be identical when joined by addition or condensation polymerization.

Which one of the following common molecules is not a polymer? Solution

1. DNA
2. Glucose
3. Protein
4. Polysaccharides

Glucose is not a polymer because it is a monomer! Glucose is a monosaccharide like the monomers that make polysaccharide polymers like starch or cellulose...

Examples of polymers...

• Polysaccharides (starch, cellulose...)
• Deoxyribonucleic Acid (DNA)
• Synthetic polymers (like Nylon, Teflon, Polyeser...)
• Proteins
• Rubber (vulcanized)

Carbohydrates and Amino Acids

Carbohydrates and amino acids are polymers. Solution

1. True
2. False

Amino acids are monomers, the polymer of amino acids is a protein (polypeptide).

Amino Acid Polymerization

Fill in the blank: Polymerization of amino acids through a _____________ reaction creates a bond between ______________ atoms. Solution

1. Decarboxylation, Carbon-Oxygen
2. Polymerization, Carbon-Carbon
3. Condensation, Carbon-Nitrogen
4. Addition, Carbon-Hydrogen

Polysaccharides

Sucrose is a polysaccharide made from the monosaccharide subunits Solution

1. Lactose and Cellulose
2. Glucose and Maltose
3. Fructose and Galactose
4. Glucose and Fructose

Polysaccharides are polymers of simple sugar molecules. In this case, Sucrose is a polysaccharide made from Glucose and Fructose.

Polymerization

The condensation reaction to polymerize the following molecules will form what monomer subunit? Solution Video

1. [—CH₂CH₂OCCH₂CH₂C—]_n
2. [—OCH₂CH₂OCH₂CH₂CO—]_n
3. [—CH₂CH₂OOCCH₂CH₂C—]_n
4. [—OCH₂CH₂OOCCH₂CH₂CO—]_n

A carboxylic acid on one molecule and an alcohol on another molecule condense (dehydrate) to form the ester linkage (...CH₂OOCCH₂...) plus a water molecule H-OH. = [—OCH₂CH₂OOCCH₂CH₂CO—]_n

Polymerization of Alkenes

Draw the monomer subunit when 3-bromo-5-methyl-1,6-heptadiene undergoes polymerization Solution

Draw (in your own notes) and name the monomer subunit in the following polymer. Solution

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The monomer subunit is... BUT, it's considered best form to name the monomer before it polymerizes... 1-bromo-2-chloro-ethene (not 1-bromo-2-chloro-ethane)

Other: Free Radical Mechanism

The steps shown in the following chain mechanism of free radicals is called Solution

1. Initiation
2. Radicalization
3. Propagation
4. Termination

In these steps you can see the free radical propagating through to the next molecules in a chain reaction.

Naming Functional Groups

Name all of the functional groups present in this molecule. [5] Solution

1. Nitrogen, Oxygen, Hydrogen, Carbon
2. Amine, Ketone, Ether, Alkene
3. Benzene, Amine, Amide, Alkene, Ether
4. Alcohol, Ketone, Amine

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Benzene, Amine, Amide, Alkene, Ether

Priority of Naming with Multiple Functional Groups

Name the following molecule given the priority table below. Solution

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Priority Ranking	Functional Group
1 (Highest Priority)	Carboxylic acid
2	Ester
3	Amide
4	Aldehyde
5	Ketone
6	Alcohol
7	Amine
8	Ether
9	Alkene
10	Alkyne
11	Halide
12 (Lowest Priority)	Alkane branch

5-amino-3-hydroxy-oct-7-enoic acid


Carboxylic acids rank highest on the list. Start numbering from the carbonyl carbon on the carboxylic acid functional group.

Synthesis

Determine the structure of each unknown molecule in the pathways below.

The two possible molecules that would make the given alcohol in the reactions below. Solution

The molecules along the pathway to form the ester. Solution

Starting with an alkene. Solution

Starting with an alkene and ending with an amide. Solution

Atomic Structure

The region around an atom where an electron is most likely to be found is called an energy level. Solution

1. True
2. False

The energy levels are discrete energetic regions where electrons are located, with a high probability.

Electron Configurations

The shorthand, ground state electron configuration for Silicon is [Ne] 3s²4p² Solution

1. True
2. False

The shorthand, ground state electron configuration for Silicon is [Ne] 3s²3p²

Electron Configurations: Half-filled d Orbital

The ground-state electron configuration 1s² 2s² 2p ⁶ 3s² 3p ⁶ 4s¹ 3d⁵ is for the element... Solution

1. Manganese, Mn
2. Iron, Fe
3. Chromium, Cr
4. Cobalt, Co

Chromium (Cr) takes this configuration because there is extra stability from the half-filled d orbital.
This occurs with the other elements in the same group: Molybdenum (Mo) and Tungsten (W).

(The 6 elements with unexpected configurations are Cu, Ag, Au, and Cr, Mo, W).

Electron Configurations: s Vs. d Orbital

The ground state electron configuration for Copper is Solution

1. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹
2. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
3. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
4. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

This same thing occurs for the other elements in the same group as Copper (Cu): Silver (Ag) and Gold (Au).

The expected configuration: 4s² 3d⁹
The actual, more stable configuration: 4s¹ 3d¹⁰

The valence electrons are more stable when the s sublevel is half full and the d sublevel is full.

Electron Configurations: Transition Metal Ions

The Iron atom Fe³⁺ is formed by when iron loses Solution Video

1. 6 d electrons
2. 3 d electrons
3. 2 s electrons and 1 d electron
4. 1 s electron and 2 d electrons

Iron is multivalent, most common is Fe³⁺. Ground state iron has 6 electrons in the 3d shell, and 2 electrons in the 4s.

The higher energy 3d loses 1 electron, to become half full. The lower energy 4s loses both electrons to become empty.
1s²2s²2p⁶3s²3p⁶4s²3d⁶ ... 1s²2s²2p⁶3s²3p⁶4s⁰3d⁵ = 1s²2s²2p⁶3s²3p⁶3d⁵ Writing 4s² before 3d⁶ follows the Aufbau principle order, however some other sources show the other way with 3d before 4s.

Sublevels and Orbitals

The p sublevel contains how many orbitals? Solution

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There are 3 orbitals in the 'p' sublevel. (1 orbital in the 's' sublevel, and 5 orbitals in the 'd' sublevel).

Orbital

One orbital can not have which of the following? Solution

1. 0 electrons
2. 1 electron
3. 2 electrons
4. 8 electrons
5. none of the above

One orbital can hold up to 2 electrons with opposite spins. They can be empty, with zero electrons. Note that orbitals should not be confused with 'octet'. '

Electron Configuration: Orbitals

The number of orbitals in the 3^rd energy level is Solution

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Number of orbitals = (n energy level)² = (3)² = 9

Rutherford's Gold Foil Experiment

Ernest Rutherford's observations of alpha particle deflection from gold foil allowed him to conclude that the nucleus was a large proportion of the atom. Solution

1. True
2. False

The reflection of any alpha particles at all was surprising proof that the atom did contain something inside it.

However, only a small number of alpha particles were deflected, so the atom was mostly empty space with a very small nucleus. The large majority of an atom is the electron cloud. Rutherford coins the term 'proton' in 1914.

Geiger–Marsden (Rutherford's) Gold Foil Experiment

If Thomsons's atomic model (1897) was correct, Rutherford would have observed which of the following during the gold foil experiment? Solution

1. Alpha particles (α) would not pass through the layer of gold foil
2. Alpha particles (α) would all pass through the layer of gold foil
3. Alpha particles (α) would melt the layer of gold foil creating a small hole
4. Alpha particles (α) would deflect and pass through the layer of gold foil
5. None of the above

Thompson's model spaced the negatively charged particles evenly throughout a positively charged substance in the atom, like a "plum pudding", or chocolate chips in a cookie. The alpha particles need a mass like a nucleus to reflect, and would not be reflected from these tiny, lone subatomic particles in Thomson's model.

The Photoelectric Effect

The photoelectric effect is one of the experiments that demonstrates light behaves not just as waves, but as particles too. Solution

1. True
2. False

Light clearly behaves as electromagnetic waves, and the photoelectric effect shows that light behaves is discreet quanta of energy as photons or particles. The experiment involved shining light at a certain frequency on a metal to cause electrons to be released. Although Hertz did this experiment in 1887, it was Einstein who used it in 1905 to explain the quantum (packet) energy of light depends on its frequency. Einstein later won the Nobel Prize for this explanation in 1921.

Isotopes and Mass Spectrometer

A mass spectrometer measures the Solution

1. mass of neutral, elemental particles
2. exact molar mass of charged particles placed into a vacuum chamber
3. degree of ionization of an unknown charged particle from the radius of curvature
4. mass of ions from deflection in a magnetic field
5. none of the above

A mass spectrometer accelerates the particles (charges) through a uniform magnetic field, and the amount of deflection or curvature indicates the mass.

Different masses of ions were explained by different numbers of neutrons: isotopes!

Isotopes and Mass Spectrometer

Heavier isotopes are deflected more than lighter isotopes in a mass spectrometer. Solution

1. True
2. False

A mass spectrometer accelerates the particles (charges) through a uniform magnetic field, and the amount of deflection or curvature indicates the mass.

Heavier isotopes are deflected less in a mass spectrometer. Their radius of curvature is larger.

Bohr Theory

Bohr's theory of atomic structure was based on the idea that electrons orbit the nucleus on energy levels and when discreet quanta of energy is absorbed, electrons can transition or jump to higher energy levels. Conversely when electrons return or fall back down to lower energy levels, discreet quanta of energy, called photons are released. Solution

1. True
2. False

When discreet quanta of energy is absorbed, electrons can transition or jump to higher energy levels.

When electrons return or fall back down to lower energy levels, discreet quanta of energy, called photons are released.

The discrete quanta is based on the distance between energy levels and partly shows the existence of energy levels in atoms.

Quantum Numbers

What does the secondary quantum number represent? Solution

Principal Quantum Number	n	Energy Level, or Shell, n = 1, 2, 3, 4...
Secondary, Azimuthal Quantum Number	l	Sublevel/Subshell, (0)s, (1)p, (2)d, (3)f
Third, Magnetic Quantum Number	ml	Orientation, _, _ _ _, _ _ _ _ _, _ _ _ _ _ _ _.
Fourth, Spin Quantum Number	ms	Electron Spin, +½, -½

1. The energy of an electron
2. The shape of the orbit
3. The direction of the electron
4. The spin of the electron
5. None of the above

Get the range of l... ±l = n-1 The second, azimuthal quantum number allows the third quantum number to be determined. For example, for
n = 1 →→ l = 0 →→ 0 →→ s sublevel
n = 2 →→ l = 0, 1 →→ -1, 0, +1 →→ p sublevel
n = 3 →→ l = 0, 1, 2 →→ -2, -1, 0, +1, +2 →→ d sublevel

S (shart), P (principal), D (diffuse), F (fundamental)... these sublevels or subshells have different shapes.

Quantum Numbers

For the principle quantum number of n = 3, which of the following are the values of secondary azimuthal, l and magnetic, m_l quantum numbers, respectively? Solution

1. 1 only
2. 2 only
3. 3 only
4. 0, or 1
5. 0, or 1, or 2
6. 1, or 2
7. 1, or 2, or 3
8. -1, 0, 1
9. -2, -1, 0, 1, 2
10. -3, -2, -1, 0, 1, 2, 3

1. III, and VII
2. VII, and IX
3. V, and IX
4. III, and V
5. VI, and VIII

Secondary quantum number, l = (n-1)
l = 3 - 1
l = 2 (l can be a maximum of 2: 0, or 1, or 2)
for the highest l = 2, then m_l is from -2 --> +2

Magnetic quantum number is from -l to +l. The secondary quantum number indicates the shape of the electron orbit on subshells from the number of values for l. The magnetic quantum number is the number of orientations of orbitals that can exist. E.g.) A secondary quantum number = (1)p would have the following number of orbitals (_ _ _) according to (-1, 0, 1).

Quantum Numbers

An atom with the quantum numbers, n = 2 and l = 2, has 8 electrons in the shell. Solution

1. True
2. False

The following combination of quantum numbers is not possible: n = 2 and l = 2.

The secondary, azimuthal quantum number (l) is (n-1) = 1. So n = 2 and l = 1.

(It is true that there are 8 electron in the 2s, 2p shell).

The quantum number responsible for explaining the interaction of atoms with magnetic fields is called the magnetic quantum number, m_l. Solution

1. True
2. False

This is misleading. The magnetic quantum number actually describes the electron's orientation/location in space, with one, three, or five possible orientations according to the following positions:

m_l = 0
m_l = -1, 0, or 1.
m_l = -2, -1, 0, 1, or 2.

It is actually the different spin of electrons (+½, -½) from the spin quantum number, m₂ that explains if the atom will be attracted or repelled by a magnetic field. Paired electrons are diamagnetic and are not attracted to magnetic fields, rather slightly repelled. Unpaired electrons are paramagnetic and are attracted to magnetic fields.

Which of the following symbols corresponds to the quantum numbers: n = 2, l = 1? Solution

1. 1s
2. 2s
3. 1p
4. 2p
5. 3d

The principal quantum number (n)
The second, azimuthal quantum number (l)
The third, magnetic quantum number (m_l)
The fourth, spin quantum number (m_s).
n = 2 → this tells us it's in the 2nd energy level.
l = 1 → this tells us the number of orbitals in the sublevel from -l to +l, which would be -1, 0, +1 → which is 3 orbitals → the p sublevel has 3 orbitals.

State the maximum number of electrons in an atom that can have the following quantum numbers in total in all principle quantum rows (n) and any sublevel: n = 2, m_l = 0. Solution

1. 1
2. 2
3. 4
4. 6
5. 8

Using the first quantum number (n = 2), there are 2 azimuthal values (l) from 0 to (n-1): 0, 1

The two azimuthal values are written from -l to +l, giving us the number of orbitals in a sublevel:
0 = 1 orbital in s
-1, 0, +1 = 3 orbitals in s.

2s has 2e^- with M_l = 0
2p has 2e^- with M_l = 0.

2e^- + 2e^- = 4e^-

Think of this like a notation for each orbital containing electrons:

1. n = 2, l = 0, m_l = 0
2. n = 2, l = 1, m_l = -1
3. n = 2, l = 1, m_l = 0
4. n = 2, l = 1, m_l = +1

See that there are 4 possible combinations, with 2 m_l values.

(Note that we don't count the n=1 sublevel 1s, which has a m_l=0

Quantum Numbers: f-block

The f sublevel has 7 orbitals and the d sublevel can hold 10 electrons. Solution

1. True
2. False

True, the f sublevel has 7 orbitals and the d sublevel can hold 10 electrons.

What energy levels are the f-blocks in? [2] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

4-f & 5-f

(Not 6 & 7 as many people think)

Pauli Exclusion Principle

Two electrons in the same orbital have the same spin directions. Solution

1. True
2. False

Wolfgang Pauli, Austrian physicist (1900-1958). The Pauli exclusion principle states that no two electrons in an atom have the same four quantum numbers: principle n, secondary l, magnetic m_l, and spin m_s.

Hund's Rule

The 2p subshell in an oxygen atom contains the following electron configuration Solution $2p: \quad \underline{\ ↑ ↓ \ } \ \underline{\ ↑ ↓ \ } \ \underline{\ \ \ \ \ _ \ \ \ \ }$

1. True
2. False

Each of the three orbitals must fill with at least one electron before the electrons begin to pair up. $2p: \quad \underline{\ ↑ ↓ \ } \ \underline{\ ↑ \ } \ \underline{\ ↑ \ }$

An orbital can hold up to 8 electrons Solution

1. True
2. False

An orbital can only hold up to 2 electrons. Orbitals are not to be confused with subshells, which contain orbitals, and have the same values of n and l, e.g. 2s, 3p...

Heisenberg's uncertainty principle

"The more precisely the position [of a subatomic particle] is determined, the less precisely the ____________ is known in this instant, and vice versa." Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

"The more precisely the position [of a subatomic particle] is determined, the less precisely the momentum is known in this instant, and vice versa." -Heisenberg

Max Planck's Explanation of Blackbody Radiation

Max Planck's experiments with blackbodies (perfectly black objects that emit, rather than reflect light) showed that energy of atoms in solids are not quantized, rather the energy is continuous. Solution

1. True
2. False

The experiment(s) showed that energy of atoms in solids are quantized, and not continuous.

The spectrum curves appear continuous when presented on a simple line graph, but this line actually misrepresents the very small differences in discreet, quantized energies. These 'quanta' of light energy are photon particles.

Electronegativity

Which of the following covalent bonds has the highest polarity? Solution

1. B-H
2. N-H
3. C-H
4. O-H
5. Br-H

Oxygen has the highest electronegativity of all the elements included here. Therefore the electronegativity difference with hydrogen will be highest, giving the O-H bond the highest polarity.

Hydrogen Bonding

Hydrogen bonding is possible whenever a molecule contains hydrogen bonds Solution

1. True
2. False

Hydrogen bonding is possible whenever the molecule contains hydrogen atoms bonded to N, O, and F atoms

Central Atoms

In addition to carbon, sulfur and phosphorus can act as central atoms in a compound. Solution

1. True
2. False

And they can form more than an octet on the central atom: SF₆ and PCl₅

Bonding and Electron Configurations

Determine the compound that will form by combining an element that has the electron configuration [Ne]3s², plus Oxygen. Solution

1. Mg₂O
2. MgO
3. Mg₂O₂
4. MgO₂

[Ne]3s² is Magnesium, Mg and forms a Mg²⁺ cation because it is in group two and can easily donate the two electrons to become isoelectronic with the noble gas Neon.
Oxygen forms an O^2- anion because it is in group 6A (16) and can easily gain two more electrons to get a full octet.

The net charges are balanced in the electron transfer: Mg²⁺O^2-

Since the charges are balanced, the compound takes one of each cation and anion: MgO.

Lewis Diagram

The Lewis structure of an atom with the electron configuration 1s²2s²2p⁶3s²3p⁴ would have Solution

1. 2 pairs of electrons and 2 unpaired electrons in the third sublevel orbitals
2. 1 pair of electrons and 2 unpaired electrons in the third sublevel orbitals
3. 1 pair of electrons and 4 unpaired electrons in the third sublevel orbitals
4. 2 pairs of electrons and 2 unpaired electrons in the second sublevel orbitals

One pair in the 3s sublevel, one pair and two unpaired in the 3p sublevel.

Lewis Diagram

The lewis structure of a molecule of carbon dioxide, CO₂ would display how many total electrons? Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

CO₂ has 4 valence electrons in carbon, and 6 valence electrons in each oxygen.

4 + 6 + 6 = 16.

Lewis Diagram for Ions

A lewis structure for the Scandium ion, Sc³⁺ would contain how many electrons? Solution

1. 21
2. 18
3. 3
4. 0

The scandium has given away three electrons to become the Sc³⁺ ion, leaving the valence shell with zero electrons remaining.