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Here's how I would do it.
A doublet of doublets (dd) is a pattern of four lines of approximately equal intensity that results from coupling to two different protons.
I can think of two situations in which I would expect to see dd splitting patterns:
A. Vinyl groups
Typical coupling constants in alkenes are
(From organic spectroscopy international)
Each proton on the vinyl group is split into a doublet of doublets by its neighbours.
Typical coupling constants in substituted benzenes are
The proton that has ortho and meta neighbours (on carbon 6) will be a doublet of doublets.
It appears at δ 7.95 (J = 8.5, 2.3 Hz).
The three factors that determine carbocation stability are adjacent (1) multiple bonds; (2) lone pairs; and (3) carbon atoms.
(1) Adjacent multiple bonds
An adjacent π bond allows the positive charge to be delocalized by resonance.
Resonance delocalization of the charge through a larger π cloud makes the cation more stable.
(2) Adjacent lone pairs
A lone pair on an adjacent atom stabilizes a carbocation.
The cation is more stable because the charge is spread over two atoms.
(3) Adjacent carbon atoms
The stability of carbocations increases as we add more carbon atoms to the cationic carbon.
The stabilization is explained by a type of resonance called hyperconjugation.
The delocalization of charge stabilizes the carbocation.
Thus, the more alkyl groups there are surrounding the cationic carbon, the more stable it becomes.
That gives us the order of stability:
Sawhorse projections generally show when something is antiperiplanar or synperiplanar, more easily than something like a Newman projection or a basic line structure can.
Take ethane as an example.
An antiperiplanar conformation has a
A synperiplanar conformation has a
A sawhorse projection approximates this 3D structure extremely well, and allows one to judge whether an
A Newman projection would depict the dihedral angle correctly, but because one would be viewing the important atoms from the front instead of an aerial view, you might actually be looking at an octahedral molecule (with a main-chain bond length of
I suspect the reason involves steric hindrance.
The structure of piperidine is
Piperidine is a weak base with
It has a cyclohexane ring structure in which the lone pair on the nitrogen atom is quite accessible to an attacking acid.
We can see this in both the ball-and-stick model
and the space-filling model.
The structure of morphine is
Ring D in its structure is a piperidine ring.
The methyl group on the nitrogen group should make morphine more basic than piperidine.
However, morphine is a weaker base, with
Morphine has a rigid pentacyclic ring structure, and the nitrogen atom is "buried" in the interior of the molecule, where it is less accessible to an attacking acid.
This becomes even clearer in a space-filling model of morphine.
Attack by an acid is strongly hindered on one side by Ring A.
We find basicities by measuring the position of an equilibrium:
If access to the base is hindered, the position of equilibrium lies further to the left, and we say that the base is weaker.
Thus, morphine is a weaker base than piperidine.
Tetrachloromethane or Carbon tetra chloride (
So testing only the inflammability they can easily be distinguished
Lucas' reagent is a solution of anhydrous zinc chloride in concentrated hydrochloric acid. This is used to distinguish between primary,secondary and tertiary alcohol lower molar.
They all produce alkyl halides when treated with this reagent through
In case of 2-methylpropan-2-ol the turbidity appears almost immediately but in case of 1-propanonl turbidity appears after long time.
It is soluble in dilute
It is insoluble in dilute
In carbileamine test the comound under test is treated with chloform and alcoholic
Both the compounds are unsaturated but the second one being aromatic one it is more stable and does not respond to test of un-saturation like dicolorization of red bromine water as done by the first one.
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
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