Organic Chemistry Practice Problems and Problem Sets
EDG and EWG
Rank the amines A through D below in order of decreasing basicity (1 = most basic). Explain your reasoning.
sp2 carbons are more electronegative (electron withdrawing) than sp3 carbons due to increased s-character. Electron withdrawing groups (EWG) make acids more acidic and bases less basic.
For this reason, piperidine (compound A) is the more basic than pyridine (compound B):
(more basic) A > B (less basic)
Nitro (-NO2) is an EWG, so 4-nitropyridine (compound C) will be less basic than compound A.
Dimethylamine is an electron donating group (EDG), so it adds electron density to pyridine and increases basicity. So dimethylamino pyridine (DMAP, compound D) will be more basic than compound A:
(more basic) D > B > C (less basic)
So how does compound A compare with compound D? There is no way of predicting this based on EWG/EDG rules alone. After all, what should be more basic: piperidine, which has sp3 carbons (increases base strength), or DMAP, which has sp2 carbons (decreases base strength) but also an EDG (increases base strength) ?
There's no way to tell. But it turns out that Piperidine is a little more basic than DMAP. So the overal order is:
(more basic) A > D > B > C (least basic)
MendelSet practice problem # 752 submitted by Matt on July 27, 2011.
Alpha bromination is usually carried out under acidic conditions via the enol intermediate.
Alpha bromination is uncontrollable under basic conditions, which goes through the enolate intermediate. Let's explore why.
a) Rank each carbonyl below in order of decreasing alpha-proton acidity (1= most acidic). Explain.
b) based on a), why does the reaction below lead to polyhalogenation?
a) Bromine is an electron withdrawing group (EWG), which makes nearby protons more acidic. So the carbonyl with two bromines is the most acidic.
b) To perform alpha bromination, the enolate (or enol) must first be formed, which then attacks Br2.
The problem with doing this under basic conditions is that each successive bromination leads to a carbonyl that is more acidic, and so forms an enolate even more easily. So once this reaction starts, it's difficult to control.
Enols are less nucleophilic than enolates, so this react can be controlled under acidic conditions. (Br2/H3O+ instead of Br2/NaOH).
MendelSet practice problem # 748 submitted by Matt on July 27, 2011.
Rank the carbonyls A-D below in order of decreasing electrophilicity (reactivity with nucleophiles).
(1 = Most reactive). Explain your reasoning.
The carbonyl carbon is electrophilic because it has a partial positive charge.
Are there any groups that make a carbon more positive? Yes, electron withdrawing groups (EWG). They pull away electron density, which increases electrophility. So C is the most reactive.
Conversely, electron donating groups (EDG) add electron density, and so make the carbonyl carbon less positive, and less electrophilic. Alkyl groups (carbon chains) are mildly EDG so ketone B will be less reactive than C.
Hydrogen is neither a EDG or EWG, so the aldehyde A will be in between the B and C. Also, the hydrogen is very small, so the carbonyl carbon is easily to get to (less steric bulk to block an attack).
Ester D is the least reactive because it has a resonance (the lone pair on the oxygen gets involved).
So overall, the order form most reactive to least reactive is C > A > B > D.
MendelSet practice problem # 710 submitted by Matt on July 22, 2011.
Let's draw resonance forms to see why some groups are EDG or EWG. (I've started you off)
Where are the positive or negative charges placed in EDG/EWG? (ortho/meta/para) Why would this affect EAS reactions?
Note: EDG = electron donating group, EWG = electron withdrawing group
The resonance forms for EDG add electron density to the ring (and add a negative charge), while the resonance forms for EWG remove electron density from the ring (and leave a positive charge). This is where the terms electron donating group and electron withdrawing group come from.
EAS reactions require benzene to attack something positive (an electrophile), so the more electron density, the better. This is why EDG tend to speed up EAS reactions, while EWG slow down EAS reactions.
Notice that for EDG, the ortho and para positions are partially negative. In EAS reactions benzene attacks a positively charged electrophile, so its not too surprising that the electrophile will want to add at o/p if a EDG is present.
But for EWG, the o/p positions pick up a partial positive charge. So for EAS reactions with a EWG, it's not surprising that the electrophile will avoid the o/p positions, and add meta instead.
This is one explanation for the general trend:
EDG are ortho/para directors and activating.
EWG are meta directors and deactivating.
MendelSet practice problem # 588 submitted by Matt on July 9, 2011.