The high melting point is caused by π-π stacking of the aromatic rings.
In organic chemistry, π–π stacking refers to attractive interactions between the π clouds of aromatic rings.
There are various types of stacking.
Neither benzene nor hexafluorobenzene has a dipole moment.
However, they have strong quadrupole moments, caused by the π clouds above and below the rings.
For example, in benzene, the π clouds are negatively charged and the plane of the ring is positively charged.
The situation is reversed in hexafluorobenzene, because the electronegative fluorine atoms withdraw electron density from the ring.
You can see the charge distribution better in this image:
Both benzene and hexafluorobenzene are destabilized by sandwich stacking, because areas with the same charge are placed next to each other.
However, benzene and hexafluorobenzene are strongly stabilized by sandwich stacking, because areas with opposite charge are placed next to each other.
Theoretical calculations put the stabilization energy at about 20 kJ/mol.
That makes the attractions as strong as many hydrogen bonds and dipole-dipole interactions.
Thus, the strong intermolecular quadrupole attractions cause a 1:1 mixture of benzene and hexafluorobenzene to have a high melting point.
Do you mean with respect to NMR spectrometers.......?
This is typically the residual proton residue of the deuterated solvent that is used to acquire the spectrum (i.e. because we typically want to observe protons, we use a solvent such as
The residual protons in the sample (rarely can we afford to use 100% deuterated solvents) act as an internal reference with which we can calibrate the observed
And thus in the
The use of deuterons rather than protons, is important in another respect. Modern NMR spectrometers usually have THREE channels; (i) for observing
I am told these days, the modern operators sometimes do not even bother to use the locking channel. The magnetic fields generated are so stable, that they do not vary so much, and
There will be two neighbouring hydrogens.
A doublet of doublets (dd) occurs when a hydrogen atom is coupled to two non-equivalent hydrogens.
An example is the NMR spectrum of methyl acrylate.
(From Chemistry LibreTexts)
Each of the vinyl protons
Let's examine the
There are four separate peaks because
The result is a doublet of doublets.
The splitting diagram shows that
These peaks are each split into doublets by coupling with
Because lactose contains a free anomeric carbon whereas sucrose does not, enabling it to equilibriate with the anomeric forms of the cyclic sugar.
Let me first quickly go over a little bit about
Let's look at sucrose first. Sucrose is made by the condensation reaction of a glucose molecule and a fructose molecule. What essentially happens is the hemiacetal of the glucose molecule reacts with the hydroxyl group of the fructose which forms the
During the formation of sucrose, both the glucose and fructose molecules' anomeric carbons get involved to form the O-glycosidic bond. An
Looking at the straight chain form, the carbonyl carbon reacts internally with the hydroxyl group to form a 5 membered ring. The carbon labeled 1 (C1) is the anomeric carbon in the ring form
The cool thing about anomeric carbons is that depending on their stereochemistry two
In order for mutarotation to occur, there needs to a
Here's what I think.
The Hofmann elimination is a process in which an amine is converted to a quaternary ammonium salt by treatment with excess methyl iodide and then reacted with silver oxide and water to form an alkene.
The silver oxide and water form hydroxide ions which eliminate a β-hydrogen.
The major product is the least substituted alkene.
Hofmann elimination of 1-azabicyclo[4.3.0]nonane
I would expect the quaternary ammonium salt to give a mixture of two alkenes.
One is N-methyl-2-(prop-2-en-1-yl)piperidine, formed by elimination from
the 5-membered ring.
The other is 2-(but-3-en-1-yl)-N-methylpyrollidine, formed by elimination from
the 6-membered ring.
I would not expect indole to undergo Hofmann elimination because it has no aliphatic β-hydrogens to be eliminated.
All of these involve bulky substrates, so these will all contain some mixture of
The major product forms from
#S_N1#; the alkyl bromide is tertiary, so it is bulky/sterically-hindered, and is more likely to lose #"Br"^(-)#as a leaving group before managing to get attacked by a nucleophile (ethanol). The proton on the attached ethanol then dissociates on its own.
That is characteristic of a first-order process, where the slow step is the departure of the leaving group and a carbocation intermediate forms (here, a tertiary carbocation).
There is a minor product from
#E1#, but this is very minor unless the temperature is sufficiently high. This would be a #beta#elimination, i.e. the alcohol extracts a proton from the carbon adjacent to #"Br"#and patches up the carbocation intermediate.
Basically the same idea as in
The only difference is a mixture of elimination minor products. Either alkene is equally-substituted (both have one doubly-substituted and one singly-substituted
#sp^2#carbon), so by Zaitsev's rule, either minor product is likely to occur.
Basically the same idea as in
#(1)#and #(2)#, except now the so-called nucleophile, acetic acid, does not do its job well at all; it's quite bulky, and is better at being a Lewis base (ugh, a base? That'll take gusto, as it's an acid! Gotta wait for it to dissociate as much as possible!).
#E1#is likely, while #S_N1#pretty much won't happen (i.e. some ridiculously bulky ester won't form).
By Zaitsev's rule, we expect the more substituted alkene to form, which is the left one (because it has two disubstituted
#sp^2#carbons, compared to the right-hand one).
Reaction with deuterated compounds is not that different from the undeuterated forms. It is just easier to tell which
#a)#is a mixture of #S_N1#and #E_1#since the alkyl halide is bulky, as in #(1)#. #E1#at a sufficiently high temperature.
#b)#The difference with respect to #(a)#is that now the reactant added is more bulky. In this case, the major and minor products switch, because substitution would require the approach of a big molecule towards a bulky molecule.
Thus, the kinetically-slow monodeuterated cyclohexanol prefers to be a base and perform
#E1#to give the thermodynamically-favored product.
WARNING! Long answer! Inductive effects are the effects on rates or positions of equilibrium caused by the polarity of the bond to a substituent group.
A -I effect or negative inductive effect occurs when the substituent withdraws electrons.
A +I effect or positive inductive effect occurs when the substituent donates electrons.
The highly electronegative
The bond will be polarized, with the
The α-carbon will in turn withdraw some electron density from the β-carbon, giving it a smaller partial positive (
The inductive removal of electron density is passed with diminishing effect through the chain of
Again, the effect is passed along the chain of carbon atoms but it dies out rapidly with distance.
The strength of a carboxylic acid depends on the extent of its ionization: the more ionized it is, the stronger it is.
As an acid becomes stronger, the numerical value of its
Thus, for example. the order of acidity as shown by the
Since this effect is caused by the inductive removal of electrons, it is called
a -I effect
In the same way, an electron-donating alkyl group decreases the acidity of a carboxylic acid.
Thus, for example. the order of acidity as shown by the
Thus, acetic acid is weaker than formic acid, and propionic acid is weaker than acetic acid.
Since this effect is caused by the inductive donation of electrons, it is called
a +I effect
Let's break this down a bit.
First, your job is to identify the parent chain, which is the longest carbon chain.
That's easy as the last part of the structure's name gives it away:
Next, you have to identify the substituents. Well, knowing that cyclopentane is the parent chain, whatever else that we have left are treated as the substituents. Let's write them down.
If we take our cyclopentane and first number all the carbons,
we can then, starting in numerical order, attach the substituents one-by-one.
Next, we take a look at
Lastly, looking at
I would think that it's because pyrene has less resonance stabilization than benzene does (increasing its HOMO-LUMO gap by less), due to its sheer size causing its energy levels to be so close together.
A smaller HOMO-LUMO gap means a more reactive system, despite it having resonance throughout.
DISCLAIMER: THOROUGH/LONG ANSWER!
EXAMINING THE EXTENSIVITY OF RESONANCE STABILIZATION
Consider napthalene, anthracene, and phenanthrene (if you add one benzene ring to the upper-right of phenanthrene, you have pyrene):
The resonance stabilization that one benzene ring gets is
#"Total Resonance Energy"#
#~~ "Number of Benzene Rings" xx "Resonance Energy"#
But you can see in the above diagram that it isn't:
From this, we could postulate that in general, the more extended the
ENERGY GAPS AS A FUNCTION OF VOLUME (AND ENTROPY)
The energy gaps (and thus the HOMO-LUMO gap) in any molecule are a function of the system volume and entropy.
At constant entropy though (which means at a constant distribution of states amongst the energy levels), the trend of volume vs. energy gap can be examined.
To illustrate this, the following graph was generated and derived from Huckel MO Theory, for which we have:
#E_k = alpha + 2betacos((2kpi)/n)#,
#k#is the energy level index and #n#is the number of fused rings. #alpha#is the nonbonding energy and #beta#is the negative difference in energy from the nonbonding level.
We can see then that the HOMO-LUMO gap converges as the number of rings increases, i.e. as the system volume increases.
PARTICIPATION OF HOMO & LUMO IN ELECTROPHILIC ADDITION
How is this relevant?
Well, the HOMO and LUMO are both required in electrophilic addition reactions. For example, with adding
And this forms the so-called bromonium complex:
(Here, the HOMO contained the
Since the HOMO-LUMO gap gets smaller when the system gets larger, it's very likely that the gap is so small for pyrene that the resonance stabilization (which increases this gap) isn't enough to make it unreactive towards electrophilic addition.
Well, these distribution graphs should correlate with the titration curve.
If we know the first
#"pH"_("1st half equiv. pt.") = "pKa"_1 + cancel(log\frac(["HA"^(-)])(["H"_2"A"]))^("Equal conc.'s, "log(1) = 0)#
#"pH"_("2nd half equiv. pt.") = "pKa"_2 + cancel(log\frac(["A"^(2-)])(["HA"^(-)]))^("Equal conc.'s, "log(1) = 0)#
We represent each stage of a diprotic acid as:
#"H"_2"A"(aq) rightleftharpoons overbrace("HA"^(-)(aq))^"singly deprotonated" + "H"^(+)(aq)#
#rightleftharpoons overbrace("A"^(2-)(aq))^"doubly deprotonated" + "H"^(+)(aq)#
The two midpoints shown are the first and second half-equivalence points, respectively.
At midpoint 1, we have that
At midpoint 2, we have that
A distribution graph shows the change in concentration of each species in solution as the
See below for an overlay of both:
Each species in solution is tracked in the bottom graph.
The cross-over points on the distribution graph are the half-equivalence points on the titration curve.
The maximum concentration for each species after the starting