Peer-Reviewed Journal with EAS reaction

Resol and novolac are two types of phenolic resins; they are widely used in industry because of their chemical resistance, electrical insulation, and dimensional  stability. Phenolic resin-bonded textile felts can be considered to be fiber-reinforced plastic with high levels of fibers; these fibers are derived from textile scraps recycled from the textile industry. Due to the increase of phenol cost, researchers have been working to partially substitute this monomer by natural polymers that present similiar structure close to that of phenolic resin.

The best possible substitutes of phenol based on cost and availabilty are lignosulfonates, which are obtained from the sulphite process. The ammonium lignosulfonate is the best canidate to substitute phenol, because final properties of phenolic resins are better. Lignosulfonates are not very reactive with phenol and formaldehyde, so it is usual to modify their structure by methyloletion and phenolation. Although phenolation is more expensive.

The reaction mechnism associated with the formulation of lignin-novolac is explained in the following text. The alpha-carbon of the lateral chain of the phenyl-propane units is occupied by the sulfonate group. Lignosulfonates may be reacting by itself or with the phenol adding to the Beta-carbon of the lateral chain. The first step consists of the condensation between lignin fragments and the phenol present with formaldehyde. In the second step, vacuum distillation is employed to adjust the content of free phenol in accordance with the required specifications for textile felt applications. The addition of a curing agent (HMTA) with methylene groups is also necessary for the crosslinking. (1)

On page 3/14, of the following link is Figure 2: Mechanism of lignin-novolac resins synthesis substituted with lignosulfonates. Since this peer-reviewed journal is protected, I could not copy the reaction scheme onto my blog. My apologies.

http://www.ncsu.edu/bioresources/BioRes_02/BioRes_02_2_270_283_Perez_RAOE_Novolac_Resin_Lignosulfonate.pdf

References:
1. Characterization of a Novolac; Jaun Manuel Perez; Bioresources; http://www.ncsu.edu/bioresources/BioRes_02/BioRes_02_2_270_283_Perez_RAOE_Novolac_Resin_Lignosulfonate.pdf; Web; (2007)

Aromaticity

For a compound to be considered aromatic, it must satisfy the following four rules:

1.       The molecule must be cyclic.

2.       The molecule must be planar.

3.       The molecule must be completely conjugated.

4.       The molecule must satisfy Huckel’s rule.

Determining if a compound is cyclic is rather simple. If the compound is a ring and all bonds are connected to form this ring, then the compound satisfies the first rule.

 A planar molecule has all atoms in one plane; basically the molecule is “flat”.

Conjugation means the compound must have alternating double bonds. Benzene is considered completely conjugated because each pi orbital (double bond) overlaps and is separated by a sigma bond (single bond). However, 1,3-cyclohexadiene is not conjugated because there are three sigma bonds (single bonds) with no pi orbital (double bond).

Benzene                                                                                   

1,3-cyclohexadiene

Huckel’s rule applies to the number of pi electrons in the compound.  The equation used to find the acceptable number of pi electrons is 4n+2, (n=1, 2, 3, etc.). The acceptable numbers of pi electrons are 2, 6, 10, 14, 18, etc. The trend is adding 4 to each number. A pi electron is usually shown in the compound by a dot around the molecule with the electrons, or by a double or triple bond. Each double bond has two pi electrons. For example, the benzene molecule above has 3 double bonds, which means it has 6 pi electrons.

Let’s look at a few examples of compounds and find out if they are considered aromatic.

1.

Let’s begin with rule number one; the molecule must be cyclic. This molecule is considered cyclic, it is a ring and all bonds are connected to form that ring.
Rule 2: The molecule must be planar. This compound is not flat due to the carbons that are sp³ hybridized. The two hydrogen atoms connected to the carbon are in two different planes.
Rule 3: The molecule is not completely conjugated, if there were one more double bond, the molecule would satisfy this rule.
Rule 4: The molecule has 3 double bonds which means it has 6 pi electrons.
This molecule is NOT considered aromatic because it does not satisfy rule 2 or 3.

2.  

This molecule is commonly known as Pyrrole. Let’s begin with rule 1: the molecule must be cyclic. This molecule is considered cyclic; all bonds are connected to form a ring.
Rule 2: The molecule must be planar. All carbon atoms are sp2 hybridized and the hydrogen on the nitrogen is in the same plane as all the other atoms. This molecule is considered planar.
Rule 3: The molecule must be conjugated. This molecule has alternating double bonds, each pi orbital overlaps, so this molecule is considered to be conjugated.
Rule 4: Huckel’s Rule; there are 2 double bonds which equals 4 pi electrons, but there are also two pi electrons on nitrogen. Therefore there are six pi electrons, satisfying Huckel’s rule.
This molecule IS considered to be aromatic.

Exam 1

I expected to see a question like the following example on the exam:
A 1H NMR signal is observed at 157.0-Hz (downfeild of TMS) on a 135-MHz instrument.
a.) What is the observed chemical shift in ppm? 1.16 ppm
The chemical shift is found by dividing the downfield signal (Hz) by the spectrometer frequency (MHz). (157.0-Hz/135-MHz=1.16 ppm)
b.)What is the chemical shift in ppm, if the sample is analyzed with a 300.0 MHz instrument? 1.16 ppm
Chemical shift is a ratio, the value in hertz increases proportionally to the value in megahertz. The shift observed at 135-MHz will be 1.16 ppm at any other operating frequency.
c.)Where will the proton signal appear, in hertz, if the sample is analyzed with a 300.0-MHz instrument? 349 Hz
The downfield signal (Hz) is the chemical shift (ppm) multiplied by the spectrometer frequency (MHz). (1.16 ppm*300-MHz=300-Hz)

This type of problem was assigned for the Sapling Homework, and we discussed this topic in class. Dr. Mullins even spent time with an example, although there was no question even similar to this on the exam.

I felt mislead after taking the exam. In the Monday-Wednesday lab section, we did an example where we had to draw an IR spectrum for a compound provided by Dr. Mullins. I specifically remember asking Dr. Mullins aloud, if we were going to be required to draw an IR or MS spectrum on the test and his answer to my question was NO. There were multiple problems on the exam where we had to draw the spectrum for a compound. I am not implying whatsoever that we should know specifics about our exam prior to the test date. I am however saying that when ANY teacher says that something will not be on the exam, they should hold to their word. Because I know I am not alone when I say, I don't even bother to study material that is not going to be on the exam.

Muddiest Point to Date

I have read Chapter 13, and I am confident that I understand the material from this chapter. I have yet to read chapter 14 over NMR spectra thus, NMR spectra is my muddiest point to date. I plan to read the chapter and lecture on Friday will be over this subject. Hopefully this will help me to better understand Nuclear Magnetic Resonance. The homework assigned in lab found on WebSpectra should give me some practice on reading an NMR spectrum. The necessary resources to understand Nuclear Magnetic Resonance are chapter 14, WebSpectra, and lecture on Friday. I plan to take complete advantage of the previous resources to understand Nuclear Magnetic Resonance.