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11

## What are the intermolecular forces of CHF3, OF2, HF, and CF4?

Dr. Cawas K.
Featured 1 month ago

The relative magnitude of the inter molecular forces are:
$C {F}_{4} < O {F}_{2} < C H {F}_{3} < H F .$

#### Explanation:

All molecules will have London dispersion forces which get stronger as the molecule gets heavier (more electrons causes a shift in electron cloud distribution resulting in a temporary dipole).

$C H {F}_{3}$ is a polar molecule. So it will have dipole - dipole interaction along with the weaker dispersion forces.

$O {F}_{2}$ is a polar molecule with a bent shape just like ${H}_{2} O$. There are two pairs of bonded electrons and two pairs of unbonded lone pair electrons. These lone pairs repel the bonded electrons resulting in a bent shape with an angle less than ${105}^{0}$. The dominant inter molecular forces would be dipole-dipole.

HF is a polar molecule. Hence the primary inter molecular forces would be dipole - dipole and hydrogen bond which is a special type of dipole - dipole interaction between the hydrogen atom and electronegative F atom.

$C {F}_{4}$ has a tetrahedral structure. It is non-polar molecule. The dominant inter molecular force would be London dispersion force.

2

## How do you name alkynes?

Ernest Z.
Featured 1 month ago

Here are some of the rules for naming alkynes.

#### Explanation:

Alkynes are unsaturated hydrocarbons containing a triple bond, $\text{R-C≡C-R}$.

You name them the same way as you name alkanes, but you replace the ending -ane with -yne.

When there is more than one triple bond, you use the multiplying prefixes to
get -diyne, -triyne, etc.

Step 1. Find the longest continuous chain of carbons containing the triple bond.

For example, ${\text{CH"_3"CH"_2"CH"_2"C≡CCH}}_{3}$ has six carbon atoms, so its base name is hexyne.

Step 2. Number the carbon atoms in the main chain from the end closer to the triple bond.

stackrelcolor(blue)(6)("C")"H"_3stackrelcolor(blue)(5)("C")"H"_2stackrelcolor(blue)(4)("C")"H"_2stackrelcolor(blue)(3)("C")≡stackrelcolor(blue)(2)("C")stackrelcolor(blue)(1)("C")"H"_3

Insert the first number of the alkyne carbons as close as possible before the
ending -yne.

The name becomes hex-1-yne.

Step 3. If there are other substituents, list them alphabetically with their locating numbers.

${\text{H"stackrelcolor(blue)(1)("C")"≡"stackrelcolor(blue)(2)("C")stackrelcolor(blue)(3)("C")"H"("CH"_3)stackrelcolor(blue)(4)("C")"≡"stackrelcolor(blue)(5)("C")stackrelcolor(blue)(6)("C")"H}}_{3}$

For example, the compound above is 3-methylhexa-1,4-diyne.

2

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

Ernest Z.
Featured 1 month 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).

2

## Explain the Michael reaction?

Truong-Son N.
Featured 1 month ago

The Michael reaction is an addition reaction, usually of an enolate to an $\alpha , \beta$-unsaturated ketone (a conjugated enone), to form a diketone.

The case you're familiar with is this one:

where I've highlighted the molecular fragment that gets attached. The mechanism is a bit visually challenging, but it's not too bad:

You can think of it as a variation on the aldol addition, where we've replaced the aldehyde with a conjugated enone.

1. A base deprotonates the $\boldsymbol{\alpha}$-carbon on the ketone, which forms the enolate. Sometimes, LDA, lithiumdiisopropylamide, is used instead to give a better yield (though it is too reactive for aldehydes).
2. The enolate then acts as a good nucleophile to attack the $\beta$ position on the enone. Draw the resonance structure, and you should find that the carbonyl oxygen, which is electron-withdrawing and ${\delta}^{-}$, causes the $\beta$-carbon to be ${\delta}^{+}$.
3. The reaction finishes by tautomerizing the enolate back into the ketone by deprotonating the water molecule that was made in step 1.

This is usually the first step in a Robinson Annulation. Here is an example of an annotated Robinson Annulation (notice the heat added in step 7):

2

## How would you explain the halogenation of benzene?

Ernest Z.
Featured 1 month ago

The halogenation of benzene is an electrophilic aromatic substitution reaction.

#### Explanation:

Electrophilic aromatic substitution

Electrophilic aromatic substitution is a reaction in which an atom on a aromatic ring is replaced by an electrophile.

A typical halogenation reaction is

The electrophile is an ion that is generated by the catalyst.

Mechanism

Step 1. Generation of the electrophile

A Lewis acid catalyst, usually ${\text{AlBr}}_{3}$ or ${\text{FeBr}}_{3}$, reacts with the halogen to form a complex that makes the halogen more electrophilic.

Step 2. Electrophilic attack on the aromatic ring

The nucleophilic π electrons of the aromatic ring attack the electrophilic $\text{Br}$ atom .

This forms $\text{FeBr"_4^"-}$ and generates a cyclohexadienyl cation intermediate, destroying the aromaticity of the ring.

Step 3: Loss of $\text{H"^"+}$ and restoration of aromaticity

The $\text{FeBr"_4^"-}$ removes the $\text{H"^"+}$ from the ring.

This re-forms the aromatic ring, produces $\text{HBr}$, and regenerates the catalyst.

2

## Why are aromatic rings stable?

Ernest Z.
Featured 1 month ago

Aromatic rings are stable because they are cyclic, conjugated molecules.

#### Explanation:

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 $\left(4 n + 2\right)$ rule.

You need 2 electrons to fill the first level and 4 electrons to fill each level above the first.

2

## What determines carbocation stability?

Ernest Z.
Featured 1 month 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°

5

## Ortho,para-bromoanisole + NaNH_2 + Liquid NH_3 =? How do you predict the product?

Ernest Z.
Featured 3 weeks ago

Here's how I would do it.

#### Explanation:

These are the reaction conditions for generating benzyne intermediates.

The reaction of ortho-bromoanisole with potassium amide in liquid ammonia (b. p. -33 °C) is extremely rapid.

Step 1. The amide ion attacks the $\text{H}$ atom that is ortho to $\text{C3}$, generating a carbanion.

Step 2. Loss of $\text{Br"^"-}$ to form a benzyne intermediate.

The elimination is by an E2cb pathway.

Step 3. Addition of $\text{NH"_2^"-}$

The strain caused by a triple bond in a benzene ring can be relieved by a nucleophilic addition ($\text{Ad"_"N}$) of $\text{NH"_2^"-}$.

The methoxy group is electron-withdrawing by induction, so the nucleophile will attack $\text{C3}$ to place the carbanion as close as possible to the methoxy group.

Step 4. Protonation of the carbanion.

The product is meta-methoxyaniline.

para-Bromoanisole

The reaction with para-bromoanisole also follows a benzyne mechanism.

The nucleophile can attack either end of the triple bond, and the methoxy group is far enough away that its inductive effects are minimal.

The product is a mixture of para- and meta-methoxyaniline.

3

## What is the difference between an oxyacid and an organic acid? What are examples of each?

Dwight
Featured 3 weeks ago

The difference lies in the structure of the molecules and the manner in which the oxygen and hydrogen atoms are arranged. See below...

#### Explanation:

An oxyacid is one that will contain, in addition to hydrogen and another element (such as nitrogen, sulfur or phosphorus), a number of oxygen atoms.

Examples would include sulfuric acid ${H}_{2} S {O}_{4}$, and also sulfurous acid ${H}_{2} S {O}_{3}$, phosphoris acid ${H}_{3} P {O}_{4}$, nitric acid $H N {O}_{3}$ to name just a few.

An organic acid is one which contains, in its structure, the particular arrangement of atoms called a carboxyl group $- C O O H$

Examples include formic acid $H C O O H$, acetic acid (in vinegar) $C {H}_{3} C O O H$ and many more.

The difference comes in the actual structure of the molecule. In the oxyacid, the oxygen atoms are all bonded to the nitrogen or sulfur, or whatever it happens to be, with hydrogen atoms bonded to one or more of these oxygens.

In a carboxylic acid (the organic variety), a carbon is doubly bonded to one oxygen atom and singly bonded to a second oxygen. This second oxygen has the H atom bonded to it. So, a very particular structure.

"R" just represents the rest of the molecule.

3

## How do you tell "S"_"N"1 and "S"_"N"2 reactions apart?

Morgan
Featured 3 weeks ago

See below.

#### Explanation:

There are several key differences between ${S}_{N} 1$ and ${S}_{N} 2$ reactions. I've outlined a comparison below, with the assumption that the reader has some basic knowledge of both reaction types.

${S}_{N} 2$

${S}_{N} 2$ stands for "substitution, nucleophilic, bimolecular," or bimolecular nucleophilic substitution. This implies that there are two molecules (bimolecular) involved in the transition state or rate-determining (slow) step. This also tells us that the rate of the reaction depends upon both reactants (nucleophile and electrophile); if you double the concentration of one of the reactants, you double the rate of the reaction.

${S}_{N} 2$ reactions require a strong nucleophile. Strong nucleophiles are strong bases, so it may be easier to identify them this way at first. For example, strong nucleophiles bear a negative charge. $N a O C {H}_{3}$ is a strong nucleophile, as it breaks apart into $N {a}^{+}$ and $O C {H}_{3}^{-}$ in solution.

This strong nucleophile forces what is called a backside attack. This is fairly literal. The nucleophile attacks the carbon opposite the leaving group as the two repel each other.

${S}_{N} 2$ products show inversion of stereochemistry, a result of the backside attack. For example, if the leaving group was once represented as a wedge in the perspective drawing of the molecule, the nucleophile which replaces it will now be shown as a dash. The stereochemistry of any other substituents are left alone.

${S}_{N} 2$ reactions are concerted, which means that the nucleophile attacks the electrophilic carbon at the same time that the leaving group leaves. There is no intermediate.

${S}_{N} 2$ reactions prefer polar aprotic solvents, where polar protic solvents hinder ${S}_{N} 2$ reactions. Examples include $D M S O$ and acetone.

${S}_{N} 2$ reactions favor electrophilic carbon atoms which are least highly substituted, so ${1}^{o} > {2}^{o} > {3}^{o}$. You won't see a tertiary carbon undergo an ${S}_{N} 2$ reaction. This is the big barrier for ${S}_{N} 2$ reactions. It is due to the fact that the reaction is concerted, and a backside attack must take place. The steric hinderance on a tertiary carbon is too great to allow this.

This is a reaction diagram for a general ${S}_{N} 2$ reaction, with the reaction coordinate on the $x$-axis and energy on the $y$-axis. The reactants are represented by $\textcolor{b l u e}{N {u}^{-}} + R \textcolor{g r e e n}{L G}$. This symbolizes the nucleophile $\left(\textcolor{b l u e}{N {u}^{-}}\right)$ plus the leaving group $\left(\textcolor{g r e e n}{L G}\right)$ attached to some $R$. The transition state or rate-determining step is represented by ${\left[\textcolor{b l u e}{N u c} - - - R - - - \textcolor{g r e e n}{L G}\right]}^{-}$, which symbolizes the concerted reaction, the backside attack of the nucleophile while the leaving group leaves. Note that the transition state cannot be isolated! This is just a visualization of what is happening at this point. We see two molecules involved here: the nucleophile and the compound it attacks (electrophile). Finally, we have the products all the way to the right of the diagram.

${S}_{N} 1$

${S}_{N} 1$ stands for "substitution, nucleophilic, unimolecular," or unimolecular nucleophilic substitution. This implies that there is only one molecule (unimolecular) involved in the transition state or rate-determining (slow) step. This also tells us that the rate of the reaction depends upon only one reactant at a time.

${S}_{N} 1$ requires a weak nucleophile. This is because the ${S}_{N} 1$ reaction is step-wise, or occurs in two steps. First, the slow step: the leaving group leaves and the formation of the carbocation. If a strong nucleophile is present, this slow step does not occur because the nucleophile quickly attacks the electrophile. Additionally, because the carbocation formed is such a reactive electrophile, a weak nucleophile is all that is required. We also see solvolysis in ${S}_{N} 1$ reactions, meaning that the nucleophile and the solvent are the same. A common example is $C {H}_{3} O H$. You might also see heat is used, given by $\Delta$.

${S}_{N} 1$ products show both inversion and retention of stereochemistry. You will usually get a mixture of stereoisomers in your products.

Because a carbocation is formed in ${S}_{N} 1$ reactions, rearrangement is possible. Rearrangement will only occur if a more substituted product is possible through a hydride or alkyl shift (must be adjacent). Because no carbocation is formed in ${S}_{N} 2$ reactions, no rearrangement is possible.

${S}_{N} 1$ reactions prefer polar protic solvents. Examples include $C {H}_{3} O H$ and acetic acid.

${S}_{N} 1$ reactions favor electrophilic carbon atoms which are most highly substituted, so ${3}^{o} > {2}^{o} > {1}^{o}$. You will not see a primary carbon undergo an ${S}_{N} 1$ reaction.

In this diagram for an ${S}_{N} 1$ reaction, we see that there are two separate transition states. One represents the formation of the carbocation, which is the slow, rate-determining step (notice it requires more energy), while the second represents the nucleophilic attack on the newly-formed carbocation. This is the fast step. Note that $S M$ stands for starting material and $P$ for products.

Here is a comparison chart: