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Short link: mendelset.com/chapters/265
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.
Convert each formula to a carbon skeleton diagram, or vice-versa.
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.
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.
Draw all resonance forms for each species.
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.)
(This is similar to problem 569, which involved the allylic position.)
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.
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.
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.
Draw the conjugate base form of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).
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.
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.
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.
Rank the group of molecules below in in order of decreasing basicity. (1 = most basic)
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.
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.
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.
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.
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.).
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-.
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.
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.
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)
Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 1
Keyword: acid strength
Description: This mendel set goes through rules that affect acid and base strength:
"What makes a compound stable?" problems are key to understand for the MCAT.
Total Problems: 6
Keywords: formal charge, Lewis structures, resonance
Description: This mendel set goes through the basics of structure:
Total Problems: 7
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