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## What are examples of dipoles?

Andrew B.
Featured 1 year 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.

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

Truong-Son N.
Featured 1 year 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 are the #-"I"# and #+"I"# inductive effects?

Ernest Z.
Featured 1 year ago

#### Answer:

WARNING! Long answer! Inductive effects are the effects on rates or positions of equilibrium caused by the polarity of the bond to a substituent group.

#### Explanation:

A -I effect or negative inductive effect occurs when the substituent withdraws electrons.

A +I effect or positive inductive effect occurs when the substituent donates electrons.

Inductive effects

Consider a $\text{C-F}$ bond.

The highly electronegative $\text{F}$ atom will draw the electrons in the $\text{C-F}$ bond more closely toward itself.

The bond will be polarized, with the $\text{F}$ atom getting a partial negative (#δ^"-"#) charge and the α-carbon atom getting a partial positive (#δ^"+"#) charge.

#underbrace(stackrelcolor(blue)(δδδδ^"+")("C"))_color(red)(δ)-underbrace(stackrelcolor(blue)(δδδ^"+")("C"))_color(red)(γ)-underbrace(stackrelcolor(blue)(δδ^"+")("C"))_color(red)(β)-underbrace(stackrelcolor(blue)(δ^"+")("C"))_color(red)(α)-stackrelcolor(blue)(δ^"-")("F")#

The α-carbon will in turn withdraw some electron density from the β-carbon, giving it a smaller partial positive (#δδ^"+"#) charge.

The inductive removal of electron density is passed with diminishing effect through the chain of $\text{C-C}$ σ-bonds until it is almost negligible at the δ-carbon.

$\text{C}$ is less electronegative than $\text{H}$, so alkyl groups are electron releasing.

#underbrace(stackrelcolor(blue)(δδδδ^"-")("C"))_color(red)(δ)-underbrace(stackrelcolor(blue)(δδδ^"-")("C"))_color(red)(γ)-underbrace(stackrelcolor(blue)(δδ^"-")("C"))_color(red)(β)-underbrace(stackrelcolor(blue)(δ^"-")("C"))_color(red)(α)-stackrelcolor(blue)(δ^"+")("R")#

Again, the effect is passed along the chain of carbon atoms but it dies out rapidly with distance.

-I effect

The strength of a carboxylic acid depends on the extent of its ionization: the more ionized it is, the stronger it is.

As an acid becomes stronger, the numerical value of its $\text{p} {K}_{\textrm{a}}$ drops.

Thus, for example. the order of acidity as shown by the $\text{p} {K}_{\textrm{a}}$ values is

${\underbrace{\text{Br-CH"_2"CH"_2"CH"_2"COO-H")_color(red)(4.59) < underbrace("Br-CH"_2"CH"_2"COO-H")_color(red)(4.01) < underbrace("Br-CH"_2"COO-H}}}_{\textcolor{red}{2.86}}$

The electronegative $\text{Br}$ atom removes electron density from the atoms next to it, eventually weakening the $\text{O-H}$ bond at the other end of the chain and making the compound more acidic.

Since this effect is caused by the inductive removal of electrons, it is called
a -I effect

+I effect

In the same way, an electron-donating alkyl group decreases the acidity of a carboxylic acid.

Thus, for example. the order of acidity as shown by the $\text{p} {K}_{\textrm{a}}$ values is

${\underbrace{\text{CH"_3"-CH"_2"COO-H")_color(red)(4.87) < underbrace("CH"_3"-COO-H")_color(red)(4.76) < underbrace("H-COO-H}}}_{\textcolor{red}{3.74}}$

Thus, acetic acid is weaker than formic acid, and propionic acid is weaker than acetic acid.

Since this effect is caused by the inductive donation of electrons, it is called
a +I effect

## How do you name ethers, esters, and nitriles? Which IUPAC naming method is right?

Ernest Z.
Featured 10 months ago

#### Answer:

Here's what I get.

#### Explanation:

When there is more than one permissible IUPAC name, the first name is preferred.

Ethers

There are two systems of IUPAC names for ethers.

1. Substitutive names.

The ethers are named as alkoxyalkanes, with the senior component selected as the parent compound.

Thus, ${\text{CH"_3"OCH"_2"CH"_2"CH}}_{3}$ is 1-methoxypropane.

2. Functional group names.

The ethers are named as alkyl alkyl ethers, with the alkyl groups in alphabetical order followed by the class name ether, each as a separate word.

Thus, ${\text{CH"_3"OCH"_2"CH"_2"CH}}_{3}$ is methyl propyl ether.

Esters

Esters are named as alkyl alkanoates.

The name of the alkyl group is written first, followed by the name of the acid with the ending -ic acid replaced by the ending -ate.

Thus, ${\text{CH"_3"CH"_2"COO-CH}}_{3}$ is methyl propanoate (two words).

Cyanides (nitriles)

There are three systems of IUPAC names for nitriles.

1. As alkanenitriles.

The ending -nitrile is added to the name of the alkane with the same number of carbon atoms.

Thus, $\text{CH"_3"CH"_2"C≡N}$ is propanenitrile.

2. If the compound is considered to be formed from a carboxylic acid with a "trivial name" ($\text{RCOOH → RC≡N}$), the ending -ic acid is changed to -onitrile.

Thus, $\text{CH"_3"CH"_2"C≡N}$ is also called propionitrile (from propionic acid).

3. As alkyl cyanides.

The name of the alkyl group precedes the class name cyanide.

Thus, $\text{CH"_3"CH"_2"-C≡N}$ is ethyl cyanide (two words).

## Why is phenoxide more stable than phenol?

Ernest Z.
Featured 9 months ago

It isn't! Phenol is more stable than phenoxide ion.

Consider the equilibrium between the two species:

$\text{C"_6"H"_5"OH(aq)" +"H"_2"O(l)" ⇌ "C"_6"H"_5"O"^"-""(aq)" + "H"_3"O"^"+""(aq)}$

You would expect two oppositely charged ions to attract and neutralize each other, if possible.

Thus, phenol is more stable than phenoxide ion.

The equilibrium constant, #K_text(a) = 1.6 × 10^"-10"#.

This shows that the position of equilibrium lies far to the left.

The ${K}_{\textrm{a}}$ value corresponds to a free energy difference of $55.9 \textcolor{w h i t e}{l} \text{kJ·mol"^"-1}$.

## Could someone please explain (ii)* to me?

Truong-Son N.
Featured 3 months ago

ISOMER 1: n-bromobutane

The primary (${1}^{\circ}$) bromobutane ($\text{C"_4"H"_9"Br}$) has the electrophilic carbon marked below.

$\text{H"_3"C"-"CH"_2-"CH"_2-stackrel(delta^+)(stackrel("*")"C")"H"_2-stackrel(delta^-)"Br}$

You said that you are using ${\text{OH}}^{-} \left(a q\right)$, which would do a nucleophilic backside-attack on the primary carbon marked $\text{*}$.

Due to the low steric hindrance around that carbon (no non-H substituents are on $\text{C"^"*}$), it is susceptible to a second-order substitution (${\text{S}}_{N} 2$).

Therefore, the nucleophile and substrate will both participate in a reaction that has no intermediate, giving a rate law of:

$r \left(t\right) = k \left[\text{OH"^(-)]["C"_4"H"_9"Br}\right]$

The reason why an intermediate doesn't form is that primary carbocations are very unstable.

The electron density from a $\text{C"-"H}$ bond would spread out via hyperconjugation to stabilize the positive charge on the central carbon as shown below by sending electron density into an empty $2 {p}_{z}$ orbital:

The more alkyl groups you have surrounding the $\text{C"^"*}$, the more stabilized it is because alkyl groups are electron-releasing (electron-donating, giving a so-called "inductive effect").

Instead of an intermediate, a transition state forms, which is simply going to look halfway between the reactant and the product. It will be a trigonal bipyramidal geometry around the central carbon!

$\text{ "" "" "" "" "" "" "" "color(white)(.)"Br}$
$\text{ "" "" "" "" "" "" "" } \textcolor{w h i t e}{.} \vdots$
${\text{H"_3"C"-"CH"_2-"CH"_2-stackrel("*")"C"^(delta^+)"H}}_{2}$
$\text{ "" "" "" "" "" "" "" } \textcolor{w h i t e}{.} \vdots$
$\text{ "" "" "" "" "" "" "color(white)(....):ddot"O} {:}^{{\delta}^{-}}$
$\text{ "" "" "" "" "" "" "" "" ""|}$
$\text{ "" "" "" "" "" "" "color(white)(.....)"H}$

Of course, you should draw it with the proper bond angles. Be sure to include the partial charges.

ISOMER 2: tert-butyl bromide

You can tell then that tert-butyl bromide (#("H"_3"C")_3"CBr"#) is more substituted.

${\text{ "" "" ""CH}}_{3}$
$\text{ "" "" ""|}$
$\text{H"_3"C"-"C"-"Br}$
$\text{ "" "" ""|}$
${\text{ "" "" ""CH}}_{3}$

It has a tertiary (${3}^{\circ}$) central carbon, so it has more steric hindrance, i.e. it has more things blocking a nucleophile from coming in for a backside-attack, so only one molecule can participate in the reaction (the substrate).

Hence, a first-order substitution (${\text{S}}_{N} 1$) will occur, and a planar carbocation intermediate will form, giving a rate law of:

#r(t) = k_1[("CH"_3)_3"CBr"]#

(where the first step is slow)

while the intermediate looks like this:

${\text{ "" "" ""CH}}_{3}$
$\text{ "" "" ""|}$
${\text{H"_3"C"-"C}}^{+}$
$\text{ "" "" ""|}$
${\text{ "" "" ""CH}}_{3}$

(Of course, you should be drawing this with the proper bond angles!)

This is stabilized by the three ${\text{CH}}_{3}$ electron-releasing alkyl groups around the central carbon, spreading electron density out to stabilize the positive charge via hyperconjugation as before.

This stabilizing effect is much stronger than in a primary carbocation.

[Being planar, the intermediate will lead to a racemic mixture of $R / S$ stereoisomers if the alkyl group(s) around the central carbon are all different.]

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