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Here's my explanation.
Melting points depend, not only on the strength of intermolecular forces, but also on the ability of the molecules to pack into a crystalline array.
This, in turn, depends on their shape.
Cyclohexane has a high degree of symmetry, and this enables the molecules to pack nicely into a crystalline structure with strong London dispersion forces between them..
Cyclohexene exists in a half-chair conformation.
It has less symmetry than hexane, so it does not pack as well into a crystal structure.
Thus, cyclohexene has a much lower melting point than cyclohexane (-103.5 °C vs. 6.5 °C).
Permanganate Oxidation (Von Baeyer test) and the addition of bromine are useful test to identify unsaturation.
Permanganate Oxidation (Von Baeyer test)
Under neutral conditions
Note that in the course of this reaction under alkaline conditions, the purple color of the aqueous potassium permanganate is changed first to green as a result of the formation of manganate(VI) ions . . . Then the color changes to brown due to the formation of a brown solid which is manganese (IV) oxide (
The color change is a positive test for the presence of a double bond (or triple bond). Although this test has limitations, it is a good test to differentiate alkanes and aromatics from alkenes.
Addition of Bromine
The bromine test is another qualitative test for the presence of unsaturation, like the Baeyer's test, it has some limitations. The disappearance of the deep brown coloration of bromine indicates a positive test.
Keep in mind that modern spectroscopic methods ( NMR and IR) are better methods for determining the structural features and identity of unknown compounds.
I would recommend using note cards.
First familiarize yourself with the rules of nomenclature for organic compounds.
Then draw the structural formula of a compound (I'm assuming you have been given a list) on one side of a note card, and the name on the other side, which is a tactile learning style.
Study them, look for the application of the rules to the names of the compounds. Then quiz yourself until you have them memorized. While studying, look for patterns, such as similarities and differences, which involves both a visual learning style and critical thinking.
You can also say the names orally, which will touch on the auditory learning style, and it's much easier to remember a word if you can pronounce it (even if your pronunciation is atypical).
My technique is to go through the cards and set aside the ones that I already know. Then I study the rest and add back the ones I already know. Then I go through all the cards again, and set aside the ones I got right (hopefully there will be more). Then I study the remaining cards, and add back the cards I already know. I keep doing this until I am able to get all of the names correctly from their structural formulas, and understand why they are named that way.
This will take some investment of your time, but it is a very useful studying tool and works for many different subjects.
A spectrophotometer works by shining light of a specific wavelength through a liquid sample.
A spectrophotometer is essentially a very simple piece of equipment in that it consists of a light source, a monochromator (this selects the correct wavelength of light), a sample holder, and a detector.
The wavelength is set to a specific value for the compound being measured. This wavelength is determined by the maximum absorbance of that compound.
The graph below shows the absorption spectra for Nicotinamide adenine dinucleotide (NAD).
At 260 nm both oxidised (NAD+) and reduced (NADH) absorb the light. However, at 340 nm only the reduced NADH wil absorb. This means that in a reaction that is producing NADH you could record the rate of the reaction by measuring the appearance of NADH at 340 nm. However, if you carried out the measurement at 260 nm you may see no or very little change because as the absorbance due NAD+ would decreases as it is converted to NADH, and the absorbance due to NADH would increase.
Here's my proposal for the mechanism for the Fischer indole synthesis of 2-phenylindole.
Note: For convenience, I am using
Step 1. Protonation of the phenylhydrazone
Step 2. Tautomerization to an enamine
Step 3. Cyclic [3,3]-sigmatropic rearrangement
This forms a diimine and results in the loss of aromaticity.
Step 4. Re-aromatization
Step 5. Protonation of the imine
Step 6. Cyclization
Step 7. Deprotonation of the ammonium ion
Step 8. Protonation of the primary amine
Step 9. Elimination of ammonia
This forms a stable, aromatic indole.
And that is my proposed mechanism for the formation of 2-phenylindole.
Aromatic rings are stable because they are cyclic, conjugated molecules.
Conjugation of π orbitals lowers the energy of a molecule.
Thus, the order of stability is ethene < buta-1,3-diene < hexa-1,3,5-triene ≪ benzene.
(From Kshitij IIT JEE Online Coaching)
Benzene is different because it is a cyclic conjugated molecule.
In a cyclic conjugated molecule, each energy level above the first occurs in pairs.
Thus, the total energy of the six π electrons in benzene is much less than the energy of three separate ethene molecules or even of a hexa-1,3,5-triene molecule.
A Frost circle reflects the pattern of orbital energies.
You place the ring inside a circle with one of its vertices pointing downwards.
Then you draw a horizontal line though the vertices. This represents the orbital energy levels.
If there isn't one already, you draw a horizontal dashed line through the circle's (and molecule's) centre. This represents the nonbonding energy level.
Qualitatively, a Frost circle explains the
You need 2 electrons to fill the first level and 4 electrons to fill each level above the first.
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