Textbook: Vollhardt 6th Ed. (2010)

Chapter 1: Structure and Bonding in Organic Molecules

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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 # 1286

For the molecule shown below, indicate the hybridization (sp³, sp², sp, etc.) of atoms A through H, and the bond angles of X, Y, and Z.

Solution

The trick to this problem is to look for implicit hydrogens- hydrogens (protons) that are not drawn in, but assumed to be there because there are no charges.

Also remember these rules:

If an atom has no multiple bonds, the total number of "things" (bonds and lone pairs) surrounding the atom is probably four, so it's sp³ hybridized (s¹ + p³ = 4).
If an atom has one double bond, the total number of "things"surrounding the atom is probably three, so it's sp² hybridized (s¹ + p² = 3).
If an atom has one triple bond or two double bonds, the total number of "things" surrounding the atom is probably two, so it's sp hybridized (s¹ + p¹= 2).

A sp² This oxygen has one double bond and two lone pairs. Three things total = sp².

B sp² This carbon has one double bond and two single bonds, to a carbon and an oxygen (so two bonds). Three things total = sp².

C sp³ This nitrogen has one long pair, two implicit hydrogens (two bonds), and one single bond to a carbo (another bond). Four things total = sp³.

D sp³ This carbon has one implicit hydrogen and three single bonds. Four things total = sp³.

E sp³ This oxygen has two lone pairs and two bonding pairs. Four things total = sp³.

F sp² This carbon has one double bond (one bond), one implicit hydrogen (another bond), and is bonded to one other carbon (a third bond). So three bonds in total. Three things total = sp².

G sp This carbon has a triple bond (one bond) and a single bond (another bond). So two things in total = sp.

H sp This nitrogen has a lone pair and a triple bond. Two things total = sp.

For angles, remember that electrons repel each other, and so will try to be as far apart from each other as possible:

Best way to separate two things is a straight line (180°).
Best way to separate three things is a "trifecta" (120°).
Best way to separate four things in three dimensional space is a tetrahedron (109.5°).

X 120° The carbon between the two oxygens is sp²hybridized, so the shape is trigonal planar and the bond angle is 120°.

Y 180° The carbon bonded to the nitrogen is sp hybridized, so the shape is linear and the bond angle is 180°.

Z 109.5° The carbon is sp³ hybridized (the most common type of carbon). So the shape is tetrahedral and the bond angle is 109.5°.

Problem Details

MendelSet practice problem # 1286 submitted by Matt on October 2, 2011.