solventfront

I. (9 points) Silica gel thin layer chromatography (TLC) is often used to monitor the progress of an organic reaction. For the following ester hydrolysis reaction, a solvent system is selected to give the starting material an Rf value of about 0.5. (1) Provide in the box below the structure of the expected product 2. (2) Fill in the spots that would be expected when the reaction is 50% compete and 100% complete, each after acidic work-up. Make sure to assign each spot you draw to the corresponding compound number (1, 2, or 3). Consider only the compounds that can be visualized as a spot on TLC upon exposure of each solvent-dried plate to a 254 nm-UV lamp.

Name ___key_____________________

On each TLC plate, a student has placed a sample of the starting material (1) as areference on the left of the plate, a spotof the reaction mixture on the right, and a co-spot in the center of each. Co-spotting is where some of 1 and some of the reaction mixture are spotted together in order to make better comparisons.

50% completion

co-spot co-spot

reaction mixture

100% completion

1

216 S09 Exam #1 - Page 2.

1

H2SO4 (cat) H2O

Δ (heat)

1

HO

O

23

+

3

O

H

H H

H

O

O

II. (8 points) What would be the expected order of Rf values for each of the following sets of organic compounds when subjected to TLC analysis on silica gel? For each set, circle the compound that is expected to give the higher Rf value. Assume dichloromethane is used as the developing solvent for the TLC analysis.

(1) OH

H3C

OCH3

OCH3

NH2

O

OH

O

(2)

CH3

O(3) (4)

OH

A B A B

A B A B

CH3 OH

reaction mixture

O

H

H H

H

HO

1

2

3

216 S09 Exam #1 - Page 3 Name_________key_______________ III. (15 points) N-Benzylaniline (6) can be prepared by a reaction of benzyl chloride (4) and aniline (5) in the presence of aqueous sodium bicarbonate (NaHCO3).

ClNH2

NaHCO3 NH

+NaCl + H2O+ CO2

4 5 6

++H2O90 °C

The procedure calls for 372 g of aniline (5), 127 g of benzyl chloride (4), 105 g of NaHCO3, and 100 mL of water. A typical reaction produces 160 g of pure benzylaniline (6). Use the following atomic weights: H, 1; C, 12; N, 14; Na, 23; Cl, 35.5. (1) (2 points) What is the molecular weight of benzylaniline (6)? Mol formula of 6 is C13H13N . So, the molecular weight is: 183 (g/mol). (2) (3 points) What is the molar ratio of aniline (5): benzyl chloride (4) used in this reaction? Show your work. 4: 127 / 126.5 = 1.004 mol 5: 372 / 93 = 4.0 Therefore, the molar ration of 5 : 4 is 4 / 1.004 = 4 (3) (2 points) What is the liming reagent in this reaction? Benzyl chloride (4): 1.004 mol; aniline (5): 4.0 mol; NaHCO3: 1.25 mol Therefore, the limiting reagent is benzyl chloride (4). (4) (3 points) What is the theoretical yield of benzylaniline (6)? Show your work. 1 mol x 183 g / mol = 183 g Theoretical yield is 183 g. (5) (3 points) What is the percentage yield of benzylaniline (6) if 160 g is produced? Show your work. (160 g / 183 g) x 100 = 87.4 Yield: 87.4% (6) (2 points) What volume of benzyl chloride (density 1.10 g/mL) is used in this reaction? Show your work. 127 g / (1.10 g/mL) = 115.5 mL Volume of benzyl chloride used: 115.5 mL

216 S09 Exam #1 - Page 4 Name ______key__________________

IV. (8 points) The pKa values of the conjugate acids of aniline and 4-cyanoaniline are determined to be, respectively, 4.60 and 1.57. Explain in the box below why the latter is considerably more acidic than the former using pertinent resonance structures of both the conjugate acid of 4-cyanoaniline and 4-cyanoaniline itself.

C NH2N

4-cyanoaniline

4-cyanoaniline

C NH3N C NH3N C NH2N C NH2N

charge-chargerepulsion, making the conjugate acid less stable thanC6H5NH3

+. Therefore, thisconjugate acid is more acidic than C6H5NH3

+

The NH2 lone-pair electronsare delocalized into the CN group,i.e., more stable than C6H5NH2.This makes the NH2 nitrogen less basic. Therefore, the conjugate acid of 4-cyanoaniline is more acidic than that of aniline.

For the conjugate acid of 4-cyanoaniline: For 4-cyanoaniline:

V. (10 points) For the following reaction, provide in the boxes below the structures of the expected products and a step-by-step mechanism through the use of the curved arrow convention.

NH3O

O O

benzoic anhydride

+CH2Cl2

room temperature

2

+

2

HO

OHN

O

Reaction mechanism:

6

NH2 O

O O

HN

O

NH2

OO

O

N

O

H HO

O

OH

O

+

intermediate str: 1 pt eacheach set of mechanistic arrows: 1 pt each

216 S09 Exam #1 – Page 5 Name ______key_________________ VI. (10 points) Treatment of aldehyde ester 7 with 1-mole equivalent of NaBH4 in tetrahydrofuran-dimethyl sulfoxide at room temperature results in, upon aqueous acidic work up, the formation of compound 8 (C9H10O3). Compound 8 shows strong IR peaks at 3400 (broad) and 1735 cm-1. Draw in the box below the structure of compound 8 and provide the step-by-step mechanism for its formation using the curved arrow convention.

7

H

O

O

O

H

O

O

O

1. NaBH4

2. H3O+ (protonation)

738

7

Mechanism for the formation of 8:

BH

H

HH

Na

H

O

O

O

H

H

O

H

O

OO

O

O

O

OH O

O

OH O

OH

H H

8

intermediate str: 1 pt eacheach set of mechanistic arrows: 1 pt each

VII. (8 points) For each of the following pairs of compounds, match the expected IR wavenumbers for the C=O bond stretching vibration absorption to the wavenumbers given.

(1) 1700 or 1660 cm-1

SPh

O

N(CH3)2

O

(2) 1685 or 1655 cm-1

N(CH3)2OO

(3) 1760 or 1710 cm-1 (4) 1715 or 1695 cm-1

OCH3

OO

O

OO

1660 cm-11700 cm-1

1710 cm-1 1760 cm-1

1685 cm-1 1655 cm-1

1695 cm-1 1715 cm-1

216 S09 Exam #1 – Page 8 Name ___key_____________________ VIII. (continued)

E

O

OCH3

B

C

O

OHF

G

H

NH2

O

D

N

I

J

K

L

M

N

A

OH

O

NCH3

H

N CH3

H

O

H3C CH3HO

CH3

H3CCH3

O

O

OH

OH

OH

O

HOOCH3

H

NH2