My Problems

Submitted by Matt

Published

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List of user's published problems.

Problem # 306

Rank each group of acids in order of decreasing acidity. (1 = most acidic)

Explain your reasoning. You will have to use more than one rule in your explanation (resonance, electronegativity, atomic radius, etc.).

Solution

Compounds 1 and 2 are more acidic than compound 3 because their conjugate bases have more resonance forms than that of compound 3.

Compound 1 is more acidic than compound 2 because the resonance form of its conjugate base has two oxygen atoms and is more stable than that of compound 2, which has one oxygen and one nitrogen; a negative oxygen atom is more stable than a negative nitrogen atom.

Problem Details

MendelSet practice problem # 306 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Concept (explain why this is so..)

Keyword(s): resonance, acid strength

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 # 308

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

Explain your reasoning.

Solution

The periodic trend for acidity is increasing acid stength as you move from left to right or from up to down on the periodic table, so the trend for basicity will be opposite; amines (which contain nitrogen) are the most basic neutral compounds, and oxygen is most basic than sulfur.

Problem Details

MendelSet practice problem # 308 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability

Question Type: Concept (explain why this is so..)

Keyword(s): base strength, periodic trends

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 # 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 # 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 # 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 # 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 # 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 # 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 # 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.