Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

Problem # 754

Show how each amine can be prepared from a carbonyl and an amine via reductive amination.

Reductive amination is a two step process. First, an imine or enamine is formed from a carbonyl and an imine.

Then, a the imine (or enamine) is reduced to an amine (using reducing agents such as H₂, NaBH₄, NaBH₃(CN), NaBH(OAc)₃, etc.). For practice on forming imine and enamines, see problem 712).

Problem Details

MendelSet practice problem # 754 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): imines, enamines, reductive amination

Problem # 753

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

Solution

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 # 728

The acyl group is a protecting group for amines. Amines can be acylated using acetic anhydride, and deacylated with base.

Propose a mechanism for each reaction.

Solution

These reactions are both nucleophilic acyl substitution under basic conditions. See problem 705 for the mechanism in detail.

Problem Details

MendelSet practice problem # 728 submitted by Matt on July 24, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives), Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): nucleophilic acyl substitution

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.