Textbook: Wade 7th Ed. (2009)

Chapter 1: Introduction and Review

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Individual Problems

Individual Problems

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 # 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.)

(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 # 755

The nitrosyl cation is shown below. Also shown are several proposed resonance arrows, only one of which is correct.

Draw the resonance forms that would follow from each set of arrows, and include formal charges. Which one is the correct resonance form? Explain your reasoning.

Solution

When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).

Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.

Problem Details

MendelSet practice problem # 755 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Diazonium Chemistry (NaNO2, Sandmeyer, Nitrosyl)

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

Keyword(s): resonance, Lewis structures

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

Problem # 303

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

Perchlorate (ClO₄^-) has the most resonance forms and therefore has the most electron delocalization, so it is the most stable and weakest base, which makes its conjugate acid (HClO₄) the strongest acid.

Problem Details

MendelSet practice problem # 303 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): resonance, acid strength

Problem # 304

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

The ammonium species is the most acidic because its conjugate base is the most stable (because its neutral).

Neutral amines aren't acidic because their conjugate bases are very unstable- a nitrogen with a negative formal charge is a very strong base.

Problem Details

MendelSet practice problem # 304 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): 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 # 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 # 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 # 305

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

Phenol is more acidic than cyclohexanol because the conjugate base of phenol (phenolate) has resonance while the conjugate base of cyclohexanol does not.

Thiophenol is more acidic than phenol because sulfur is larger than oxygen, and so RS^- is more stable than RO^-.

Problem Details

MendelSet practice problem # 305 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Phenols (Claisen Rearrangement, Kolbe-Schmitt, Phenol EAS)

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

Keyword(s): resonance, acid strength

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 # 745

The molecule below has five different types of hydrogens (A through E). Rank each in order of decreasing acidity.

(1 = most acidic). Explain your reasoning.

Solution

Hydrogens D and E are alpha to two carbonyls and so will be the more acidic than C, which is only alpha to one carbonyl. This is because the enolates that arise from deprotonate at D and E have more resonance forms than the enolate that arrises at C.

Because ketones are more electron withdrawing than esters, D will be more acidic than E. So far we have:

D > E > C > (other)

The carbanions that would arrise from deprotonation of carbons A and B both do not have any resonance forms, so we don't expect either to be acidic. But because A is next to a fluorine atom, which is an electron withdrawing group, A will be more acidic than B. So the overall order is:

(most acidic) D > E > C > A > B (least acidic)