Chem 345 PSet 02 Key

1.) When you mix ammonia (NH3) and water, does it stay mainly ammonia and + - water, does it form ammonium (NH4 ) and hydroxide (HO ), or does it form + - - hydronium (H3O ) and amide (NH2 ). (Unfortunately, NH2 is spelled amide and pronounced uh-mid). Use pKa's to figure out this problem.

NH3 H2O NH4 HO pka ~ 15 pKa ~ 10

NH3 H2O NH2 H3O pka ~ 35 pKa ~ 0

When you mix ammonia and water, a majority of it stays as ammonia and water. But some of it becomes ammonium and hydroxide. None of it becomes amide and hydronium. Ammonia, water, ammonium, and hydroxide can all coexist. However, amide and hydronium can NEVER exist together because they will obliterate each other.

2.) Butanol and diethyl ether are constitutional isomers and are common solvents. Only one can be used when butyl lithium (a very common chemical reactant) is used in a reaction. Which one and why?

Butyl lithium is a really strong base (conj. pKa ~ 45). Butanol has an acidic proton (pKa ~ 15). There is a fast and irreversible reaction between them. As a result, butyl lithium cannot coexist with butanol. Diethyl ether, on the other hand, does not have an acidic proton, so it will remain unreactive to butyl lithium.

~ Li conjugate pKa ~45 butyl lithium REALLY, REALLY, REALLY, REALLY Strong base

+ OH Li + OLi H butanol has an acidic proton butyl lithium pKa ~ 15 Reaction is fast and irreversible butane pKa ~ 45 3.) Alanine is one of the twenty amino acids. What is a more realistic depiction of alanine? This is another acid base problem. Amino acids have an acidic site and a basic sight. SO you can just set this up like an acid base problem except that the two functional groups are tethered together.

O O pKa ~ 5

H3C H3C OH O

NH2 NH3

Alanine pKa ~ 10 More realistic depiction of alanine

4.) The pKa of a molecule is represented by the pKa of its most acidic functional group. On the line underneath each structure, write the approximate pKa value. Circle the most acidic compound of the five.

OH O N

H N H2 45 15 10

O

O O

20 45

5.) Why don’t the following molecules exist? This is a lot like question 2 and question 3. These molecules do not exist because they have an acidic proton and a STRONGLY basic group attached to them. As a result, if they were made, then they would rapidly react to form a more stable version.

= N N N H H Li H pKa ~ 25 pKa ~ 35

=

HO MgBr HO O H pKa ~ 15 pKa ~ 45

6.) The pKa for protonated aniline is about 5. The pKa for ammonium is 10. Protonated aniline is a stronger acid (105x) than ammonium. Therefore, aniline is a weaker base than ammonia. Explain why. NH2 NH3 aniline ammonia When you pull off the proton off of protonated aniline, you get a lone pair that is involved with resonance. You actually gain resonance. If you pull off the proton from ammonium, you get ammonia. That lone pair is not involved in any resonance. Therefore, it is more profitable to pull the proton off of protonated aniline then it is to pull the proton off of ammonia.

7.) pKa Table R=H or sp3 C

-5 0 5 10 15 20 25 35 45

HCl H O R H O H H H H O N O H N R H R OH R R R R R R R R H OH R N

SH R F H

For the acids listed in the pKa table above, draw their conjugate bases.

-5 0 5 10 15 20 25 35 45

Cl O R O O N O N R H R O R R R R R R R R R O N

O S R R F

8.) Identify which compound is acting as an acid on each side of the equilibrium by writing acid underneath it. Then using the pKa table provided, determine the direction and extent of the reaction by using one of these arrows. ( )

A.

+ + N H N H2 NH3 NH2 acid acid pKa ~ 5 pKa ~ 10

B.

+ + N H N 2 H O OH

acid acid pKa ~ 10 pKa ~ 10

C.

+ + N H N O OH

acid acid pKa ~ 35 pKa ~ 10

9.) Predict the splitting patterns of the indicated H’s. a.) b.) O c.) OH O

triplet

septet singlet

d.) e.) f.) O O O

nonet doublet doublet

g.) h.) i.)

Br Cl

Cl pentet quartet doublet of quartets J~ 14,7,7,7 10.) The following 1H NMR spectra are from the aromatic regions from four isomers (A-D) of dinitrophenol (C6H4N2O5). For each spectrum, draw the structure of the isomer and assign the signals (Ha, Hb, or Hc) on the structure. (The signal for the OH is not shown).

11.) Sketch the following patterns:

a.) Two coupling constants, J=7, 7 Hz J=7,7 Hz 2 equal coupling constants so triplet

4

2 2 7Hz 1+1=2 1 1 7Hz 7Hz

2

1 1

7Hz 7Hz

b.) Two coupling constants, J=14, 7 Hz J=14,7 Hz 2 unequal coupling constants so doublet of doublets 4

2 2 14 Hz 1 1 1 1 7Hz 7Hz

1 1 1 1

7Hz 7Hz 7Hz

c.) Three coupling constants, J=7, 7, 7 Hz J=7,7,7 Hz

3 equal coupling constants so quartet

8=23

4 4 7Hz

2 4 2 7Hz 7Hz 1 3 3 1

7Hz 7Hz 7Hz

3 3

1 1

7Hz 7Hz 7Hz

d.) Three coupling constants, J=14, 7, 7 Hz J=14,7,7 Hz 3 coupling constants with the smaller two being the same doublet of triplets 8

4 4 14 Hz 2 2 2 2 7Hz 7Hz 1 2 2 2 1

7Hz 7Hz 7Hz 7Hz

2 2 2

1 1

7Hz 7Hz 7Hz 7Hz

e.) Three coupling constants, J=10, 4, 2 Hz J=10,4,2 Hz 3 coupling constants all of them different doublet of doublets of doublets

8

4 4 10 Hz 2 2 2 2 4Hz 4Hz

1 1 1 1 1 1 1 2Hz 2Hz 2Hz 2Hz

1 1 1 1 1 1 1

2Hz 2Hz 2Hz 4Hz 2Hz 2Hz 2Hz

f.) Three coupling constants, J=10, 10, 4 Hz J=10,10,4 Hz 3 coupling constants The two larger ones the same Triplet of doublets

8=23

4 4 10Hz 2 4 2 10Hz 10Hz

1 1 2 2 1 1

4Hz 4Hz 4Hz

2 2

1 1 1 1

4Hz 6Hz 4Hz 6Hz 4Hz

g.) Three coupling constants, J=10, 4, 4 Hz J=10,4,4 Hz 3 coupling constants The two smaller ones the same doublet of triplets 8

4 4 10 Hz

2 2 2 2 4Hz 4Hz 2 2 1 1 1 1 4Hz 4Hz 2Hz 4Hz 4Hz

2 2

1 1 1 1

4Hz 4Hz 2Hz 4Hz 4Hz