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MS 4525 - C420 Unit 2 Homework

MS 4525 - C420 Unit 2 Homework

Submitted by brucezenner on January 29, 2015.

Keywords: elementary acid-base, formal charge, introduction, resonance, review

Description: Second homework set for Chem 420

Total Problems: 13

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Problem # 1287

Rank each set of compounds in order of decreasing boiling point (1 = highest boiling point):

a) ethane, n-octane, n-pentane

b) n-butane, 1-butanol, 1-chlorobutane.

c) n-octane, 2-methylheptane, 2,5-dimethylhexane

(Note that the n- prefix before an alkane just means that it's one chain, without any branching.)
Solution
Remember the rules of what gives compounds a higher boiling point (and melting point):

Van der Waals Forces

More electrons = more Van der Waals interactions = higher boiling point

So in general, the more "stuff" in a molecule (the higher the molecular weight), the higher its b.p. Example: C₁₀H₂₂ will have a higher b.p. than C₅H₁₂.

How to remember this trend? What do you grill with? propane gas. (C₃H₈)

What's in a lighter? butane liquid (C₄H₁₀)

Hydrogen Bonding

Compounds that can hydrogen bond have higher b.p./m.p/ than those that don't.

Compounds can hydrogen bond is they have a N, O, or F bonded directly to an H.

In organic chemistry, the best examples are alcohols (ROH) and amines (RNH₂).

Branching

Branching makes it harder for molecules to pack together, which makes it harder to form Van der Waals interactions, and so tends to lower b.p./m.p.

a) (highest b.p.) n-octane > n-pentane > ethane (lowest b.p.)

Reason: octane has the most "stuff" (higher molecular weight).

b) 1-butanol > 1-chloroethane > n-butane

Reason: 1-butanol can hydrogen bond. 1-chloroethane has a higher molecular weight than n-butane.

c) n-octane > 2-methylheptane > 2,5-dimethylhexane

Reason: Branching. n-octane has no branching. 2,5-dimethylhexane has the most branching. Notice that each compound has the same molecular formua- C₈H₁₈. (and therefore the same weight).

Problem Details
MendelSet practice problem # 1287 submitted by Matt on October 2, 2011.
Subject: Organic Chemistry
Subject Topic(s): Intermolecular Forces and Functional Groups
Question Type: Concept (explain why this is so..)
Keyword(s): boiling point, intermolecular forces
Problem # 315

Draw all possible resonance forms for each structure below. Use curved arrows.

Note that some structures only show charge, and not implied protons or lone pairs!
Solution
Notice that when drawing resonance forms with positive charges, the arrows never come from the positive charge. Arrows only come from π (pi) electrons- lone pairs or double/triple bonds.

Problem Details
MendelSet practice problem # 315 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Lewis Structures, Formal Charges, and Resonance
Question Type: Concept (explain why this is so..)
Keyword(s): resonance, carbocation
Problem # 310

For each molecule, determine the formal charge of the indicated atom.
Solution
Remember that when calculating formal charge, you count both electrons in a lone pair but only half of the electrons in a bonding pair. This is why a helpful formula is:

formal charge = (# of valence e^-) - ("sticks + dots")

For example, in the first compound (the protonated oxygen), the oxygen has one lone pair ("2 dots") and three bonding pairs ("3 sticks"). Oxygen has a valence of 6, so its formal charge in this species is 6 - 5 = 1 or +1.

Problem Details
MendelSet practice problem # 310 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Lewis Structures, Formal Charges, and Resonance
Question Type: Concept (explain why this is so..)
Keyword(s): formal charge
Problem # 311

For each molecule below, draw in all implied lone pairs and/or protons (hydrogens) based on the formal charge shown.
Solution
Being able to determine implied lone pairs and/or hydrogens is a good skill to have, as its likely that much of the time your professor or TA will not write in all lone pairs, and will almost never draw in the implicit hydrogens. Until you are very comfortable with formal charges, you should always draw in all lone pairs and hydrogens on each atom.

Problem Details
MendelSet practice problem # 311 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Lewis Structures, Formal Charges, and Resonance
Question Type: Concept (explain why this is so..)
Keyword(s): formal charge
Problem # 313

Draw all possible resonance forms for each structure below. Use curved arrows.

Note that some structures only show charge, and not implied protons or lone pairs!
Solution

Problem Details
MendelSet practice problem # 313 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Lewis Structures, Formal Charges, and Resonance
Question Type: Concept (explain why this is so..)
Keyword(s): resonance, carbanion
Problem # 569

Draw all resonance forms for each species.

For the anion and cation species, used curved arrows. For the radical species, use hooks.
Solution
Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.

(Radical resonance uses hooks instead of arrows and so is a little different.)

Problem Details
MendelSet practice problem # 569 submitted by Matt on July 8, 2011.
Subject: Organic Chemistry
Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control)
Question Type: Concept (explain why this is so..)
Keyword(s): resonance, allylic position
Problem # 580

Draw all resonance forms for each species.

For the anion and cation species, used curved arrows. For the radical species, use hooks.
Solution
(This is similar to problem 569, which involved the allylic position.)

Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.

(Radical resonance uses hooks instead of arrows and so is a little different.)

Problem Details
MendelSet practice problem # 580 submitted by Matt on July 9, 2011.
Subject: Organic Chemistry
Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Benzene Side Chain Reactions and the Benzylic Position
Question Type: Concept (explain why this is so..)
Keyword(s): resonance, benzylic position
Problem # 697

Rank the following compounds in order of decreasing boiling point.

Also, make a guess about their relative solubilities in water. Explain your reasoning.
Solution
Water is very polar, so the more polar a molecule is, the more soluble it will be in water.

Alcohols can hydrogen bond, and so will be the most polar, have the highest boiling point, and be the most soluble in water.

Aldehydes (carbonyls) can't hydrogen bond but the C=O double bond is very polar, so aldehydes are somewhat soluble in water.

The ether is the least polar of the three functional groups, and so will have the lowest boiling point, and be the least soluble in water.

So the overal trend for polarity, water solubility, and boiling point is:

alcohol > aldehyde > ether

Problem Details
MendelSet practice problem # 697 submitted by Matt on July 20, 2011.
Subject: Organic Chemistry
Subject Topic(s): Ethers and Epoxides, Intermolecular Forces and Functional Groups
Question Type: Concept (explain why this is so..)
Keyword(s): boiling point, polarity, solubility
Problem # 535

Rank the following anions in order of decreasing stability (1 = most stable)
Solution
Charged molecules are generally less table than neutral ones, and each of the molecules below has a negatively charged oxygen.

But not all negative charges are equal; some oxygens are "closer to neutral" than others. How? Because resonance stabilizes charges by sharing electron density over multiple atoms (this is called electron delocalization).

Hydroxide (HO^-) doesn't have resonance, so the oxygen has 100% of the negative charge to itself. On the other hand, the sulfonate ester (SO₃R^-) has three resonance forms, so each oxygen only has ~33% of a negative charge. So the sulfonate ester is the most stable anion.

In general, the more resonance forms a molecule has, the more stable it is.

Problem Details
MendelSet practice problem # 535 submitted by Matt on July 2, 2011.
Subject: Organic Chemistry
Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): resonance
Problem # 307

Rank the group of molecules below in in order of decreasing basicity. (1 = most basic)

Explain your reasoning.
Solution
The alkyne (triple bond) is the most stable and therefore the least basic. This is because it is sp hybridized.

An s-orbital is closer to the nucleus and more electronegative than a p-orbital, and an sp hybridized atom has 50% s-character.

Alkenes are sp² hybridized and have 33% s-character, and alkanes are sp³ hybridized and have 25% s-character.

Because sp hybridized carbons have the most s-character, they are more electronegative and are better at stabilizing negative formal charges than sp² or sp³ carbons are.

Therefore, the alkane (sp³) is the least stable/strongest base, while the alkyne (sp) is the most stable/weakest base.

Problem Details
MendelSet practice problem # 307 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Terminal Alkyne Acidity and Acetylide
Question Type: Concept (explain why this is so..)
Keyword(s): base strength, hybridization
Problem # 540

Rank the following anions in order of decreasing stability (1 = most stable)
Solution
Each of these ions has a charge of -1, and none of them has any resonance forms. So the two things to consider are electronegativity and size. Electronegativity is relevant as we go from left to right across the periodic table. But here we are going up to down on the periodic table, so size is relavant. Larger ions are more stable than smaller ions (due to a smaller charge:size ratio), so I^- is the most stable of the group.

Problem Details
MendelSet practice problem # 540 submitted by Matt on July 3, 2011.
Subject: Organic Chemistry
Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): periodic trends, leaving groups
Problem # 286

Draw the conjugate base forms of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).

Explain your reasoning.
Solution
Electronegativity increases as you travel from left to right along the periodic table, so the fluorine anion is more stable than a negative oxygen, which is more stable than a negative nitrogen, etc.

Because F^- is the most stable, it is the weakest base, and its conjugate acid (HF) is the most acidic.

Problem Details
MendelSet practice problem # 286 submitted by Matt on June 6, 2011.
Subject: Organic Chemistry
Subject Topic(s): Acid Strength Trends and Conjugate Base Stability
Question Type: Concept (explain why this is so..)
Keyword(s): periodic trends, acid strength
Problem # 288

Draw the conjugate base form of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).

Explain your reasoning.
Solution
Size increases as you go down the periodic table, so iodine is the larger than bromine, which is larger than chlorine, etc.

Because I^- is the largest anion, it is best able to "handle" its negative charge (due to its small charge:size ratio), and so is the most stable conjugate base, and therefore the weakest conjugate base. So HI is the strongest acid.

Problem Details
MendelSet practice problem # 288 submitted by Matt on June 6, 2011.
Subject: Organic Chemistry
Subject Topic(s): Acid Strength Trends and Conjugate Base Stability
Question Type: Concept (explain why this is so..)
Keyword(s): resonance, acid strength

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