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

Dr. Cawas K.
Featured 12 months ago

#### Answer:

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.

## Are alkyl halides sterically hindered?

Morgan
Featured 10 months ago

Sometimes.

#### Explanation:

This depends on the specific alkyl halide. For example, take tert-butylbromide and bromopentane.

tert-butylbromide:

We see that this alkyl halide is tertiary $\left({3}^{o}\right)$, making it very sterically hindered. This makes a backside attack, as seen in the ${S}_{N} 2$ mechanism, virtually impossible. Inversely, the ${S}_{N} 1$ mechanism, for example, would favor this alkyl halide well under the appropriate reaction conditions.

Bromopentane:

This is a primary $\left({1}^{o}\right)$ alkyl halide, which has minimal sterical hinderance. This makes a backside attack very possible, and an ${S}_{N} 2$ mechanism would be favored.

Backside attack:

For an $S {N}_{2}$ mechanism to occur, a backside attack must be able to take place. We can see below that this would much more difficult on the tert-butylbromide than the bromopentane. I've used a strong base, $N a O C {H}_{3}$ (sodium methoxide) for the ${S}_{N} 2$ mechanism. I've shown the mechanism for bromopentane first.

Note this stereochemistry could be flipped. I simply chose one possible orientation for the example.

Note the inversion of stereochemistry, a product of the backside attack. Of course, $N {a}^{+}$ is present in solution. It is a counter ion and has not been shown.

For tert-butylbromide, the backside attack isn't plausible, as there are too many other substrates bonded to the same carbon as the halide. It is literally too crowded for this mechanism to take place.

Note that I only considered ${S}_{N} 1$ and ${S}_{N} 2$ mechanisms in my examples. The same rules for elimination mechanisms still apply, and they are in competition with the corresponding substitution mechanisms.

Note:

It is possible to have a primary alkyl halide which is too sterically hindered for an ${S}_{N} 2$ mechanism to occur. Neopentyl bromide is an example:

The sterical hinderance of the adjacent carbon is enough to render an ${S}_{N} 2$ mechanism highly improbable.

## Why is the dipole moment of 1,4-dichloro-anthracene not 0?

Andy Wolff
Featured 10 months ago

#### Answer:

Because it is not symmetric so the bond dipoles do not cancel out.

#### Explanation:

You only get a zero dipole moment for an organic molecule when the bond dipoles cancel each other out. For example, in 1,4-dicholorbenzene the two chlorines will pull electrons from the aromatic ring, but those two pulls exactly cancel each other:

However, take a look at 1,4-dicholoranthracene:

Yes--across the molecule the two chlorines will cancel each other out, but the electrons in the rest of the aromatic rings will still be pulled toward the ring with the two chlorines on it.

You could fix this by substituting the three ring:

Now there is a zero dipole moment because all the forces on the electrons of the aromatic bond are the same!

## What are examples of dipoles?

Andrew B.
Featured 9 months ago

#### Answer:

Any molecule that has a nonzero vector sum of dipole moments is said to be polar and have a dipole moment

#### Explanation:

A dipole moment refers to slight opposite charges on opposite sides of a bond. The resulting bond is said to be polar; it has a positive pole and a negative pole, much like a bar magnet.

In order to determine if a particular bond is polar or not, one must look for the electronegativity of each atom. Pauling's electronegativity is a measure of how strong a particular atom pulls electrons towards it in a bond. The value of the difference between their electronegativities ($\Delta E N$) determines how polar a bond is.

If:
$0 \le \Delta E N \le 0.4$, the bond is nonpolar.

$0.4 <$$\Delta E N$$\le 1.8$, the bond is polar

$\Delta E N > 1.8$, the bond is ionic

Consider the bonds in ${H}_{2} O$,

There is one oxygen bonded to two hydrogens in one water molecule. Based on the difference in electronegativites for the bonds, it is clearly a polar molecule
$E {N}_{O} = 3.44$
$E {N}_{H} = 2.20$
$\Delta E N = 3.44 - 2.20 = 1.24$

In the figure above, the $\delta$ symbol indicates an area of partial charge on the atom. Note that they are not full charges as in ions, but partial charges due to a difference in electron density at each "pole". The arrow in the figure indicates the direction of electron density and the slight negative charge ${\delta}^{-}$and the cross indicates an area of electron deficiency and the slight positive charge ${\delta}^{+}$. This difference in charges is called a dipole moment and it is a vector quantity; it has magnitude and direction.

Notice that the water molecule has an overall dipole moment that points straight up towards the oxygen. This is because a dipole moment of a molecule depends on the vector sum of the bond dipoles.

Consider $C {O}_{2}$,

As you can see, the $\Delta E N$ for the $C - O$ bond is within the polar range. However, since $C {O}_{2}$ is a linear molecule, the dipoles point in opposite directions and the vector sum of the two is equal to zero. $C {O}_{2}$ is nonpolar.

## What is the major product of the Stork enamine synthesis, that is, the reaction of cyclopentanone with pyrrolidine, heat, #CH_3CH_2Br#, #H^+#, and #H_2O#?

Ernest Z.
Featured 8 months ago

#### Answer:

The major product is 2-ethylcyclopentanone.

#### Explanation:

The reaction is often called the Stork enamine synthesis.

Step 1. Formation of an enamine.

Step 2. ${\text{S}}_{\textrm{N}} 2$ alkylation

The double-bonded carbon in the enamine has a partial negative charge, so it can act as a nucleophile and displace the $\text{Br}$ from ethyl bromide and form an iminium salt.

Step 3. Hydrolysis of the iminium salt

The iminium salt is hydrolyzed to regenerate the ketone and the amine.

Note: The reaction is done this way because direct alkylation of the cyclopentanone enolate gives multiple alkylation, but the enamine stops at one alkylation.

## 1-Chloro-3-methylcyclopentane is heated under reflux with aqueous sodium hydroxide. what are the products formed?

Ernest Z.
Featured 4 months ago

#### Answer:

The cis isomer will form 3- and 4-methylcyclopentene in roughly equal amounts. The trans isomer will form 4-methylcyclopentene as the major product.

#### Explanation:

What type of reaction?

I predict an $\text{E2}$ elimination because:

• The substrate is a secondary alkyl halide.
• The substrate has at least one β-hydrogen: possible elimination.
• The nucleophile ($\text{OH"^"-}$) is a strong base: possible elimination.
• There is (presumably) a high base concentration: favouring $\text{E2}$.
• Water is a polar protic solvent: favouring elimination.
• The high temperature favours elimination.

cis-1-Chloro-3-methylcyclopentane

The structure of the substrate is

Elimination requires a trans arrangement of the β-hydrogen and the leaving group.

We see appropriate β-hydrogens at $\text{C2}$ and $\text{C5}$.

Elimination of the hydrogen from $\text{C2}$ forms 3-methylcyclopentene.

Elimination of the hydrogen from $\text{C5}$ forms 4-methylcyclopentene.

These two isomers would probably be formed in roughly equal amounts.

trans-1-Chloro-3-methylcyclopentane

The structure of the substrate is

Again, we see trans β-hydrogens at $\text{C2}$ and $\text{C5}$.

However, elimination will be slower in each case because of steric hindrance by the methyl group.

Elimination of the hydrogen from $\text{C5}$ will be faster than from $\text{C2}$ because of its greater distance from the methyl group.

Thus, the major product will be 4-methylcyclopentene.

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