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!
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.
MendelSet practice problem # 315 submitted by Matt on June 7, 2011.
For each molecule, determine the formal charge of the indicated atom.
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:
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.
MendelSet practice problem # 310 submitted by Matt on June 7, 2011.
For each molecule below, draw in all implied lone pairs and/or protons (hydrogens) based on the formal charge shown.
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.
MendelSet practice problem # 311 submitted by Matt on June 7, 2011.
For the anion and cation species, used curved arrows. For the radical species, use hooks.
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.)
MendelSet practice problem # 569 submitted by Matt on July 8, 2011.
For the anion and cation species, used curved arrows. For the radical species, use hooks.
(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.)
MendelSet practice problem # 580 submitted by Matt on July 9, 2011.
Rank the following anions in order of decreasing stability (1 = most stable)
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.
MendelSet practice problem # 535 submitted by Matt on July 2, 2011.
Rank the group of molecules below in in order of decreasing basicity. (1 = most basic)
Explain your reasoning.
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.
MendelSet practice problem # 307 submitted by Matt on June 7, 2011.
Rank the following anions in order of decreasing stability (1 = most stable)
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.
MendelSet practice problem # 540 submitted by Matt on July 3, 2011.
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.
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.
MendelSet practice problem # 286 submitted by Matt on June 6, 2011.
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.
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.
MendelSet practice problem # 288 submitted by Matt on June 6, 2011.