base strength

Problem # 753

Rank the amines W through Z below in order of decreasing basicity (1 = most basic). Explain your reasoning.

Electron withdrawing groups (EWG) reduce electron density and reduce basicity. Resonance delocalizes electron density and so also decreases basicity (the electrons are less available to pick up a proton).

Electron donating groups (EDG) add electron density and increase basicity.

Alkyl groups are weakly EDG (induction), and hydrogen is neither EDG nor EWG. So cyclohexylamine (compound X) is a stronger base than ammonia (compound W).

Aniline (compound Y) has resonance, so it is less basic than compound X, which does not have resonance.

N-acyl aniline (compound Z) has more resonance forms than aniline, so it will be the least basic.

So the overall over is:

(most basic) X > W > Y > Z (least basic)

Problem Details

MendelSet practice problem # 753 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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

Keyword(s): base strength, amine bases, EDG and EWG

Problem # 752

Rank the amines A through D below in order of decreasing basicity (1 = most basic). Explain your reasoning.

Solution

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)

Problem Details

MendelSet practice problem # 752 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

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

Keyword(s): base strength, amine bases, EDG and EWG

Problem # 584

Imidazole (shown below) has two nitrogen atoms, N-1 and N-3. Which nitrogen is more basic?

To answer this problem, draw the product after each nitrogen protonates, and compare their stabilities. Explain your reasoning.

Solution

Imidazole is aromatic. When N-3 protonates, the product is still aromatic.

But when N-1 protonates, the product is no longer aromatic (and therefore significantly less stable).

Because of this, N-3 is much more basic than N-1. Another way of thinking of this is that the lone pair on N-1 is involved in the aromatic circuit, and so is not available to pick up a proton.

Problem Details

MendelSet practice problem # 584 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Benzene and Aromaticity, Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis), Pyridine and Heteroaromatic EAS

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

Keyword(s): base strength, aromaticity, amine bases

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 # 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.