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## 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.

## What would the product be if I add Br2 as step one and CH3OH in step 2?

Martin M.
Featured 1 month ago

For determining the product in these type of questions, we have to look at the functional groups of the reactants. In this case, we have a $B {r}_{2}$ molecule and a cyclohexane ring with a double bond side group. As we know, double bonds are pretty reactive, so we expect something to happen there.

Now the $B {r}_{2}$ molecule can line up with this double bond, creating partial charges. This is indicated in the figure below. Because of the partial positive charge (created on the side of the double bond), the electrons will attack there and create a cyclic structure with one of the $B r$ atoms.

The other $B r$ will be released.

Now we add $C {H}_{3} O H$. The free electrons on the $O$ can react with the structure. This is showed in the mechanism below.

The carbon atom of the cyclic bromine group is partial positively charged, as indicated in the image above. This is created by hyperconjugation. This carbon atom is attached to 3 other carbon atoms (tertiary carbon atom), which stabilises a charge on the carbon atom more easily. Therefore the electrons from methanol will attack at that carbon atom, 'pushing away' the electrons from the bromine bond.

In the last step, the $H$ atom will be released. This causes the oxygen to be neutral instead of positively charged.

And there you have your product!

## 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 is the difference between acetoacetate , acetyl-CoA , and acetoacetyl-CoA?

Ernest Z.
Featured 1 month ago

Here's what I get.

#### Explanation:

Acetoacetate

Acetate is the carboxylate ion of acetic acid.

In acetoacetate, an α-hydrogen has been replaced by an aceto or acetyl group, $\text{CH"_3"CO}$.

Acetyl-CoA

Acetyl-CoA is Coenzyme A in which the $\text{H}$ atom in the thiol group has been replaced by an acetyl group.

This is Coenzyme A:

And this is acetyl-CoA:

(From Wikipedia)

Acetoacetyl-CoA

Acetoacetyl-CoA is Coenzyme A in which the $\text{H}$ atom in the thiol group has been replaced by an acetoacetyl group, $\text{CH"_3"COCH"_2"CO}$.

(From Wikipedia)

## 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.

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