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MS 4553 - SCC Chem 420 Unit 4 Homework Set

MS 4553 - SCC Chem 420 Unit 4 Homework Set

Submitted by brucezenner on February 12, 2015.

Description: Please complete ALL the problems in Ch 4 of "Second Language", and the ODD problems in Chapters 5 and 6 of "Second Language", as well as the following problem set.

Total Problems: 12

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Problem # 320

Let's perform conformational analysis on 2-methylbutane along the C2-C3 bond. We'll use the energy chart given below.

First, draw out the Newman projections along the C2-C3 bond, rotating the front carbon (C-2) by 60 degrees clockwise each time while keeping the back carbon (C-3) stationary.

According to the table above, how much energy does each conformation "cost?"

Second, make a plot of the total energy value for each Newman projection versus its dihedral angle.
Solution

Problem Details
MendelSet practice problem # 320 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Conformational Analysis and Steric Strain
Question Type: Concept (explain why this is so..)
Keyword(s): conformational analysis, Newman projections
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 # 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 # 309

Convert each formula to a carbon skeleton diagram, or vice-versa.
Solution

Problem Details
MendelSet practice problem # 309 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): Lewis structures
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 # 312

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 # 312 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
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 # 314

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 # 314 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
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 # 317

α-D-Glucose is shown below. Draw its two chair forms. Which conformation is more stable? Explain.

I recommend using the common convention wedge = "up" and dash = "down."
Solution
The chair form on the left is more stable because it has no axial substituents, which lead to 1,3-diaxial interactions and higher energies. Glucose is the most stable of the hexose carbohydrates because it is the only one that has al substituents at C2-C5 in the equatorial position.

It's probably no coincidence that the most abundant sugar in the world (glucose) is also the sugar that is most stable!

Problem Details
MendelSet practice problem # 317 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms, Carbohydrates
Question Type: Concept (explain why this is so..)
Keyword(s): cyclohexane chair forms
Problem # 318

Two stereoisomers of trimethylcyclohexane are shown below (compounds A and B). Compare cyclohexane chair forms to determine which isomer has a lower heat of combustion. Explain your reasoning.
Solution
A has a chair form that avoids having any substituents in the axial position.

B's most stable chair form has one axial substituent.

So A is more stable than B. Therefore, B is higher eneregy and has a higher heat of combustion than A.

Problem Details
MendelSet practice problem # 318 submitted by Matt on June 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): Cycloalkanes and Cyclohexane Chair Forms
Question Type: Concept (explain why this is so..)
Keyword(s): cyclohexane chair forms, heats of combustion

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