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## How can you know if there is a doublet of doublets by looking at a structure?

Ernest Z.
Featured 6 months ago

Here's how I would do it.

#### Explanation:

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:

• vinyl groups
• 1,3,4-trisubstituted benzenes

A. Vinyl groups

Typical coupling constants in alkenes are ${J}_{\text{trans" ≈ "16 Hz}}$, ${J}_{\text{cis" ≈ "10 Hz}}$, and ${J}_{\text{gem" ≈ "2 Hz}}$.

Consider the $\text{^1"H}$-NMR spectrum of methyl acrylate:

(From organic spectroscopy international)

Each proton on the vinyl group is split into a doublet of doublets by its neighbours.

1,3,4-Trisubstituted benzenes

Typical coupling constants in substituted benzenes are ${J}_{\text{ortho" ≈ "8 Hz}}$, ${J}_{\text{meta" ≈ "2 Hz}}$, and ${J}_{\text{para" ≈ "0 Hz}}$.

Consider the $\text{^1"H}$-NMR spectrum of 3,4-dichlorobenzoyl chloride:

(From www.chem.wisc.edu)

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).

## What determines carbocation stability?

Ernest Z.
Featured 6 months ago

The three factors that determine carbocation stability are adjacent (1) multiple bonds; (2) lone pairs; and (3) carbon atoms.

An adjacent π bond allows the positive charge to be delocalized by resonance.

Thus, $\text{H"_2"C=CHCH"_2^"+}$ is more stable than $\text{CH"_3"CH"_2"CH"_2^"+}$.

Resonance delocalization of the charge through a larger π cloud makes the cation more stable.

A lone pair on an adjacent atom stabilizes a carbocation.

Thus, $\text{CH"_3"OCH"_2^"+}$ is more stable than $\text{CH"_3"CH"_2"CH"_2^"+}$ because of resonance.

The cation is more stable because the charge is spread over two atoms.

${\text{CH"_3"O"stackrel(+)("C")"H"_2 ⟷ "CH"_3stackrel(+)("O")"=CH}}_{2}$

The stability of carbocations increases as we add more carbon atoms to the cationic carbon.

For example, $\text{CH"_3"CH"_2^"+}$ is more stable than $\text{CH"_3^"+}$

The stabilization is explained by a type of resonance called hyperconjugation.

${\text{H-CH"_2"-"stackrel("+")("C")"H"_2 ⟷ stackrel("+")("H") " CH"_2"=CH}}_{2}$

The ${\text{sp}}^{3}$ orbitals of the adjacent $\text{C-H}$ bonds overlap with the vacant $\text{p}$ orbital on the carbocation.

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: #3° > 2° >1° > 0°#

## When is it better to draw Sawhorse projections in comparison with other projections?

Truong-Son N.
Featured 3 months ago

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 ${180}^{\circ}$ dihedral angle, i.e. the atoms of interest across one bond are on opposite sides along the vertical molecular plane.

A synperiplanar conformation has a ${0}^{\circ}$ dihedral angle, i.e. the atoms of interest across one bond are on the same side along the vertical molecular plane.

A sawhorse projection approximates this 3D structure extremely well, and allows one to judge whether an $E 2$ reaction is likely to occur or not (it requires an antiperiplanar conformation).

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 $0$), instead of, say, a two-carbon organic molecule.

Furthermore, a ${180}^{\circ}$ dihedral angle does not really tell you what the bond angles are of the $\text{X"-"C"-"C"-"Y}$ bond, where $\text{X}$ is the atom at the top middle of a Newman projection and $\text{Y}$ is the atom at the bottom middle of the same Newman projection.

## Why is piperidine a stronger base than morphine?

Ernest Z.
Featured 1 month ago

I suspect the reason involves steric hindrance.

#### Explanation:

The structure of piperidine is

Piperidine is a weak base with #"pK_text(b) = 2.80#.

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.

Morphine

The structure of morphine is

(From ResearchGate)

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 #"pK_text(b) = 6.13#.

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:

$\text{B + HX ⇌ BH"^"+" + "X"^"-}$

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.

## What simple test would be used to distinguish between the following compounds? A). Tetrachloromethane and butane B). 1-propanol and 2-methylpropan-2-ol C). Pentanamine and pentane D). Pent-1-ene and benzene

dk_ch
Featured 1 month ago

A) $\text{ } C C {l}_{4}$ and ${C}_{4} {H}_{10}$

Tetrachloromethane or Carbon tetra chloride (#C Cl_4)#has molar mass $154 g \text{/} m o l$ whereas butane ${C}_{4} {H}_{10}$ has molar mass 58 $g \text{/} m o l$. So due to higher inter molecular force the first one is liquid at normal temperature and pressure whereas the second one is a gas under the same condition. The second is highly inflammable gas and commonly used as a gaseous fuel. The carbon in (#C Cl_4)# is in its highest oxidation state (+4) as in $C {O}_{2}$. So this compound is not at all inflammable , it is used as good fire extinguisher (trade name- pyrene )

So testing only the inflammability they can easily be distinguished

B) 1-propanonl($C {H}_{3} C {H}_{3} C {H}_{2} O H \to$ pimary alcohol) and

2-methylpropan-2-ol (#CH_3)_3COH->#tertiary alcohol)

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 $S {N}_{1}$ mechanism and the alkyl halides formed being insoluble in the reaction mixture it makes the mixture turbid

In case of 2-methylpropan-2-ol the turbidity appears almost immediately but in case of 1-propanonl turbidity appears after long time.

C) Pentanamine $\left({C}_{5} {H}_{11} N {H}_{2}\right) \to$ primary amine

It is soluble in dilute $H C l$ and it responds to carbylamine reaction.

and pentane $\left({C}_{5} {H}_{12}\right) \to$ alkane

It is insoluble in dilute $H C l$ and it does respond to carbylamine reaction.

In carbileamine test the comound under test is treated with chloform and alcoholic $K O H$. the formation of carbile amine is detected by its foul smell.

Reaction

D). Pent-1-ene $\left(C {H}_{3} C {H}_{2} C {H}_{2} C H = C {H}_{2}\right) \to$alkene

and benzene $\left({C}_{6} {H}_{6}\right) \to$aromatic.

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.

Reaction

## How to draw distribution graph if pKa of acid is 4,4 and pKa of base is 6,7?

Truong-Son N.
Featured 2 weeks ago

Well, these distribution graphs should correlate with the titration curve.

If we know the first $\text{pKa}$ is $4.4$ and the second $\text{pKa}$ is $6.7$, then we have an idea of where the half-equivalence points are (i.e. where the concentrations of acid and conjugate base are equal), because the $\text{pH}$ $=$ $\text{pKa}$ at those points:

#"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:

${\text{H"_2"A"(aq) rightleftharpoons overbrace("HA"^(-)(aq))^"singly deprotonated" + "H}}^{+} \left(a q\right)$

#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 $\left[{\text{H"_2"A"] = ["HA}}^{-}\right]$, and that $\text{pH} \approx 4.4$.

• At midpoint 2, we have that $\left[{\text{HA"^(-)] = ["A}}^{2 -}\right]$, and that $\text{pH} \approx 6.7$.

A distribution graph shows the change in concentration of each species in solution as the $\text{pH}$ increases. It correlates well with a base-into-diprotic-acid titration curve.

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 $\text{pH}$ correlate with the first few equivalence points, and the last species to show up dominates at high $\text{pH}$.

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