DISCUSSION 8 RESPONSE ORGANIC 1

1

Your Name:

Synthesis and Spectroscopy Part I 1. Provided below is the IR and 1H NMR for a compound having the molecular formula of

C7H14O2. Using this information, determine the degrees of unsaturation for this structure and solve and draw the structure responsible for these spectral data. To validate your answer, you need to assign the type of bond (ie. stretch) responsible for each peak highlighted by a box in the IR. You must also assign each 1H NMR peak (a)-(e) to the correct set of protons in your proposed structure. The integration for each peak has been provided for you. You must label the multiplicity observed for each peak. Is your structure consistent with all of this data? If not, chances are you do not have the correct structure, or the peaks have been assigned incorrectly. In your peer review, offer feedback concerning the accuracy of the submitted structures and spectral assignments. If you believe the answers are incorrect, then you need to provided explanation. In the end, you should all be getting the same correct structure. Please draw the final structure in the box provided underneath the NMR.

IR

2

1H NMR – for clarity, expansions have been provided for peaks that are not a singlets.

Proposed Structure: Remember to assign it to the NMR per instructions.

2. Below is a synthetic scheme that is missing the structure of the desired product. This product

is the same structure that you solved for in prompt 1 using the molecular formula, IR and NMR data. Draw the structure of the desired product (hint, it has one degree of unsaturation but is not an alkene or a ring). Starting from 1-bromo-2-methylbutane, design a multi-step synthesis for this desired product. Your synthetic plan may not use any SN1 or E1 reactions. You may only use SN2 reactions, reactions that involve the use of PBr3, SOCl2, or TsCl, E2 reactions, and any necessary alkene addition reaction from chapter 8 (including the reaction discussed in chapter 10.5).

H3C O

O CH3

CH3 Formula for degree of unsaturations or HDI: [2+2x(# of carbon atoms)-(# of hydrogen atoms) + (# of nitrogen atoms) - (# of halogen atoms)]/2 [2+2(7)-14]/2 = 1 degree of unsaturation

Molecular formula: C7H14O2 7-14/2+1=1 Structure has 1 double bond

Sharp peak at 1745/1750 cm -1 is due to C=O stretching of the carbonyl group. Strong and sharp peak in region of 2800 to 3000 cm -1 is due to aliphatic C-H stretching of alkene. Peak at around 1459 to 1379 is due to C-H bending vibrations of alkane.

IR Spectrum Analysis:

Frequency (cm-1) Functionality Cause for Vibration of bond

Peak shape/ Intensity

2980 & 2860

1745

1460 & 1385

1245 & 1020

Aliphatic C-H

Carbonyl ester

Aliphatic C-H

Acyl & alkoxy

Sp3 C-H

Sp2 C=O

Sp3 C-H

Sp3 C-O

Sharp/strong

Sharp/strong

Sharp/medium

Sharp/strong

Triplet at 4.5 ppm is due to methylene proton attached to an oxygen. Singlet at 2.1 ppm is due to a methly proton attached to the carbonyl group. The quartet at around 1.55 ppm is due to a methylene proton attached to another methylene group. The multiplet at 1.5 ppm is due to a CH proton attached to 2 methyl groups and 1 methylene group. The doublet at around 0.9 ppm is due to 2 methyl protons attached to CH.

Chemical shift Integration Types of protons

Splitting pattern n=neighboring proton

6 protons Methyl (type-1)

(n+1)= (1+1); doublet0.90 ppm

1 proton CH (type-2) (n+1)=(8+1); 9 peaks 2 protons 3 protons

2 protons

CH2 (type-2) (n+1)=(3+1); quartet

Methyl (type-1) (n+1)=(0+1); singlet

CH2 (type-2) (n+1)= (2+1); triplet

1.55ppm

1.5

2.1

4.5

3

Separate each individual reaction in your proposal with a reaction arrow. For each step, write in the reagents necessary to carry out each step either above or below the reaction arrow. Also draw the expected product for each step after each reaction arrow. For reactions types where more than one regioisomer product is possible, you must use reagents that are regiospecific. For example, in E2 reactions, the choice of bulky base or non-bulky base can dictate which alkene is formed preferentially. Also, several (but not all) of the alkene addition reactions learned in chapter 8 and 10.5 are regiospecific and are either Markovnikov or Anti- Markovnikov addition reactions. Please indicate which steps this was important to consider. Your reagent choices should address these concerns appropriately. Discuss with your peers alternative approaches that would be better should you feel that a step is incorrect. Furthermore, discuss with your peers any other synthetic routes that would have led to the same product.

Synthetic Transformation Desired

Proposed Synthesis:

O

O

Isopentyl acetate

Br

1-bromo-2-methylbutane

NaOEt, EtOH

2-methylbut-1-ene

HBr Br

2-bromo-2-methylbutane

NaOEt,EtOH

2-methylbut-2-ene

BH3, H2O2, OH-

OH

3-methylbutan-2-ol

Pyridine, TsCl

OTs

KtOBu, THF

3-methylbut-1-ene

BH3, H2O2, OH-

OH

3-methylbutan-1-ol H+, THF

O

OH

O

O

Isopentyl acetate

1-bromo-2-methylbutane reacts with sodium ethoxide to form alkene 2-methylbut-1-ene which reacts with HBR to form 2-bromo-2-methylbutane through a carbocation intermediate. Then, borane reacts with alkene to form a secondary alcohol which is tosylated with pyridine and tosyl chloride. Then, potassium tert butoxide is used for elimination to form alkene. At the end, borane is converted into 3-methylbut-1-ene to 3-methylbutan-1-ol and reacts with acetic acid to form isopentyl acetate.

4

IR FREQUENCY TABLE (in wavenumber)

5

NMR CHEMICAL SHIFT TABLE (in ppm)