216 s11-e2 page 2 name key - university of michiganchem216/216 s11 e2-key.pdf · 216 s11-e2 page 2...

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216 S11-E2 Page 2 Name ________Key______________ I. (9 points) Answer in the boxes below the following questions for the Grignard reagent CH 3 -MgBr. (1) (2 points) Is the carbon atom associated with magnesium electrophilic or nucleophilic? electrophilic nucleophilic [circle one] (2) (3 points) Explain why the Grignard reaction has to be carried out in an anhydrous solvent. The Grignard reagent produced would be destroyed by the reaction with water present in a non-anhydrous solvent. (3) (4 points) Write the conjugate carbon acid (i.e., C-H) of the CH 3 – MgBr and explain why Grignard reagents are strong bases. structure: explanation: The pKa of CH 4 is ca. 50. Therefore, its conjugate base is highly basic. HC H H H II. (6 points) Ethyl alcohol (C 2 H 5 OH) quite often is present in technical-grade diethyl ether. If this grade of diethyl ether were used, what effect, if any, would the ethyl alcohol have on a Grignard reagent used? Explain using a chemical equation using methylmagnesium bromide as an example. explanation based upon the reaction mechansim: A Grignard reagent formed would deprotonate H of the ethyl alcohol OH. R MgBr H OCH 2 CH 3 R H BrMg OCH 2 CH 3 + III. (8 points) For each of the reactions below, draw in the boxes provided the structure of the expected major organic product. 4 Br CH 3 O Mg (1.1 mol equiv) ether aq NH 4 Cl (acidic workup) 4 Br H 3 C CH 3 O aq NH 4 Cl (acidic workup) + enantiomer + enantiomer 2. 1. Mg (1.1 mol equiv) ether (1.1 mol equiv) (1) (2) OH OH MgBr O

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216 S11-E2 Page 2 Name ________Key______________ I. (9 points) Answer in the boxes below the following questions for the Grignard reagent CH3-MgBr. (1) (2 points) Is the carbon atom associated with magnesium electrophilic or nucleophilic?

electrophilic nucleophilic

[circle one] (2) (3 points) Explain why the Grignard reaction has to be carried out in an anhydrous solvent.

The Grignard reagent produced would be destroyed by the reaction with water present in a non-anhydrous solvent.

(3) (4 points) Write the conjugate carbon acid (i.e., C-H) of the CH3– MgBr and explain why Grignard reagents are strong bases.

structure: explanation:The pKa of CH4 is ca. 50. Therefore, its conjugate base is highly basic.

H C HH

H II. (6 points) Ethyl alcohol (C2H5OH) quite often is present in technical-grade diethyl ether. If this grade of diethyl ether were used, what effect, if any, would the ethyl alcohol have on a Grignard reagent used? Explain using a chemical equation using methylmagnesium bromide as an example.

explanation based upon the reaction mechansim:

A Grignard reagent formed would deprotonate H of the ethyl alcohol OH.

R MgBr

H OCH2CH3R H BrMg OCH2CH3+

III. (8 points) For each of the reactions below, draw in the boxes provided the structure of the expected major organic product.

4

Br

CH3

O

Mg (1.1 mol equiv) ether

aq NH4Cl(acidic workup)

4

Br

H3CCH3

O

aq NH4Cl(acidic workup)

+ enantiomer

+ enantiomer

2.

1. Mg (1.1 mol equiv) ether

(1.1 mol equiv)

(1)

(2)

OH

OH

MgBr

O

216 S11-E2 Page 3 Name _______Key_______________ IV. (16 points) For the following reaction sequence, draw the correct structures in the boxes provided.

Ph3P: Br CH3

O+

note: Ph = C6H5

benzene, reflux

"salt"

4

NaOC2H5 /C2H5OH

+ NaBr + C2H5OH

O

H

H

H

dichloromethaneroom temperature

4major isomer: C12H12O

4minor isomer: C12H12O

+

+

(C18H15P)

C21H19PO (ylide)

Ph3P=O

4

Ph3P CH3

OBr

Ph3PCH3

O

H

H

H

CH3

O

H

H

H

HH

O CH3

trans

cis

V (12 points) Show how many peaks you would expect to observe in the proton-decoupled 13C NMR spectrum of each of the following compounds. Indicate your answers in the boxes provided.

HO

CH3

COOH

COOH

(2) (3)

(4) (5)

-OOC S

S

NH3+

(1)

(6)O

OO

CH3

CH3

H3C

H3CCOOH

COOH

CH3H3C

O

O

7

55

6 7

6

180° rotation

chiral center!

CO2HCO2Hcis

216 S11-E2 Page 4 Name ________Key_____________ VI. (5 points) The proton NMR spectrum of a mixture of acetonitrile (CH3C≡N), methylene chloride, (CH2Cl2) and benzene (C6H6) is integrated and results in step heights of 15, 6, and 18 mm for the signals at δ 2.10, 5.30, and 7.26 ppm, respectively. In what mole ratio are the three substances present? No explanation required. No partial credit will be given to this problem.

acetonitrile : methylene chloride : benzene = : :153

62

186

: :5 3 3

VII. (8 points) For each of the reactions given below, draw in the box the structure of the expected product. Make sure to indicate the stereochemistry of the product whenever applicable.

4

OCH3

endo product

+ enantiomerO

O

O

+CH2Cl2

(solvent)room temperature

(1)

4endo product

+ enantiomer+CH2Cl2

(solvent)room temperature

(2)

H

H3C

O

O

O

O

CH3OHH

H

H

CH3O

H

H H

H OHCH3

or

VIII. (10 points) Provide in the box below a step-by-step, curved-arrow mechanism for the formation of hydroxy acid 2 from diketone 1. Make sure to include the mechanism for the acidic workup as well.

O

O

OH

OH

O

mixture of stereoisomers here

1 2

KOH ( 4 mol equiv)H2O

reflux

H3O+

to pH = 1.0

O

O1 OH

O

OHO

OO H

O

OH

OO

O H OH

H

OO

OH

HOH

H

2

O

O1

OHThis is fine, too.

mech arrows: 1 ptmech arrows: 2 pts

mech arrows: 1 pt

mech arrows: 1 pt

mech arrows: 1 pt

interm str: 1 pt each

216 S11-E2 Page 5 Name ________Key_____________

216 S11-E2 Page 6. Name __________________________ IX. (continued)

O

O

O

O

O

O

Br O

OO

O

A B C D

E F G H

X. (15 points) On the basis of its spectroscopic information provided below, propose the structure for the compound, C5H12O. Draw the proposed structure in the box on page 7 and give a brief explanation as to how the structure is deduced from the spectroscopic information provided.

10

Explanation (just assign the IR peak at 3368 cm-1 and all H-1 peaks right next to thecorresponding functional group/H-1 nuclei):

5

Structure:

H3C

C

H3C

C

CH3

OHH

H

chiral center!diastereotopic

methyls

0 unit of unsaturation

H3C

C

H3C

C

CH3

O

H

H

1.92 ppm

1.3 ppm (doublet)

1.6 ppm (octet)

3.55 ppm(multiplet)

0.88 ppm(doublet)

0.92 ppm(doublet)

H

3368 cm-1

216 S11-E2 Page 8 Name ________Key_____________ XI. (15 points) On the basis of its spectroscopic information provided below, propose the structure of the compound, C10H16O4. Draw the proposed structure in the box on page 9 and give a brief explanation as to how the structure is deduced from the spectroscopic information provided. Make sure to assign all proton and C-13 NMR peaks.

216 S11-E2 Page 9 Name ________Key_____________ XI. (continued)

10

Explanation (just assign the strongest IR peak and all H-1 and C-13 peaks right next to thecorresponding functional group/H-1, C-13 nuclei):

5

Structure:

H3C

H3C

OCH2CH3

OCH2CH3

O

O

CH3H3C

O OOCH2CH3CH3CH2O

CH3

H3CO

OCH2CH3

OCH3CH2O

These should show only one sp2

alkene C-13 peak. => -1 point

H3C

H3C

OCH2CH3

OCH2CH3

O

O

1725 cm-1

Units of unsaturation: 3; only 6 C-13 peaks!

125 ppm156 ppm

22 ppm

167 ppm

14 ppm

60 ppm

1725 cm-1

1.2 ppm

14 ppm2.0 ppm

2.0 ppm

4.2ppm

4.2 ppm