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

Ernest Z.
Featured 10 months 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.

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

Morgan
Featured 10 months 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:

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

Andy Wolff
Featured 9 months ago

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 products of these Hoffman eliminations?

Ernest Z.
Featured 5 months ago

Here's what I think.

#### Explanation:

The Hofmann elimination is a process in which an amine is converted to a quaternary ammonium salt by treatment with excess methyl iodide and then reacted with silver oxide and water to form an alkene.

The silver oxide and water form hydroxide ions which eliminate a β-hydrogen.

For example,

$\text{CH"_3"CH"_2"CH"("CH"_3)"NH"_2 stackrelcolor(blue)("MeI"color(white)(mm))(→) "CH"_3"CH"_2"CH"("CH"_3) stackrelcolor(blue)("+")("N")"Me"_2 stackrelcolor(blue)("Ag"_2"O", "H"_2"O", Δ color(white)(mm))(→ )"CH"_3"CH"_2"CH=CH"_2 + "Me"_2"NH}$

The major product is the least substituted alkene.

Hofmann elimination of 1-azabicyclo[4.3.0]nonane

I would expect the quaternary ammonium salt to give a mixture of two alkenes.

One is N-methyl-2-(prop-2-en-1-yl)piperidine, formed by elimination from
the 5-membered ring.

The other is 2-(but-3-en-1-yl)-N-methylpyrollidine, formed by elimination from
the 6-membered ring.

I would not expect indole to undergo Hofmann elimination because it has no aliphatic β-hydrogens to be eliminated.

## What is steric hindrance?

Truong-Son N.
Featured 2 months ago

It can be considered as basically spatial crowding.

Steric hindrance is a kinetic factor that limits the ease to which a nucleophile (electron pair donor) can approach an electrophile (electron pair acceptor).

As an example, consider the reaction seen below, which one might hope is ${\text{S}}_{N} 2$:

Here, the nucleophile is cyanide ($\text{^(-):"CN}$) and the electrophile is the central carbon on the alkyl halide (tert-butyl bromide).

We should predict that reaction does not work via an ${\text{S}}_{N} 2$ mechanism, based on the idea of steric hindrance---the three ${\text{CH}}_{3}$ groups are blocking the cyanide from performing its backside-attack.

(LEFT: steric hindrance; RIGHT: reduced steric hindrance)

That reduces the ease to which a successful collision can occur between two reactants, and slows down the first step in a given substitution mechanism (making it the rate-limiting step).

Hence, this reaction proceeds more easily as a first-order mechanism (e.g. ${\text{S}}_{N} 1$ or $E 1$), since the ${\text{Br}}^{-}$ has the time (during the slow step) to come off on its own.

The rate law for this could then be approximated as first-order based on the slow step, since it dominates the extent of the reaction time:

$r \left(t\right) \approx {k}_{1} \left[{\text{Br}}^{-}\right]$

(Depending on the choice of solvent, either ${\text{S}}_{N} 1$ or $E 1$ could occur.)

## Draw the structure of 5,5-diethyl-2,2,3-trimethylheptane? Please help me understand how you came up with the structure!

Truong-Son N.
Featured 1 month ago

I would start from the parent name, heptane. The prefix "hept" means $7$, so the main chain contains $7$ carbons:

$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$
${\text{H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH}}_{3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$

Then I would work from right to left in the name. The $2 , 2 , 3 - \text{trimethyl}$ suggests that:

• $\text{tri} \to$ three duplicate branching groups
• $\text{methyl" -> "H"_3"C} -$ groups
• The first two are on carbon-2 and the third one is on carbon-3.

So far, we fill it in as:

$\textcolor{red}{\textcolor{w h i t e}{\ldots \ldots \ldots .} {\text{CH"_3color(white)(.)"CH}}_{3}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$
${\text{H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH}}_{3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$
$\textcolor{red}{\textcolor{w h i t e}{\ldots \ldots \ldots .} {\text{CH}}_{3}}$

Lastly, we proceed to the leftmost part of the name. The $5 , 5 - \text{diethyl}$ suggests that:

• $\text{di} \to$ two duplicate branching groups
• ${\text{ethyl" -> "H"_3"CCH}}_{2} -$ groups
• These are both on carbon-5.

So we get:

$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{red}{{\text{CH}}_{3}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots . .} |$
$\textcolor{red}{\textcolor{w h i t e}{\ldots \ldots \ldots .} {\text{CH"_3color(white)(.)"CH"_3color(white)(.....)"CH}}_{2}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$
${\text{H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH}}_{3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$
#color(red)(color(white)(..........)"CH"_3color(white)(..............)color(red)("CH"_2))#
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots . .} |$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{red}{{\text{CH}}_{3}}$

Lastly, since this is an alkane, the most saturated kind of hydrocarbon (relative to alkenes and alkynes), just fill in the rest of the branches with $\text{H}$.

$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{red}{{\text{CH}}_{3}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots . .} |$
$\textcolor{red}{\textcolor{w h i t e}{\ldots \ldots \ldots} \text{CH"_3color(white)(.)"CH"_3color(white)(.)"H"color(white)(...)"CH"_2color(white)(.)"H}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$
${\text{H"_3"C"-"C"-"C"-"C"-"C"-"C"-"CH}}_{3}$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} | \textcolor{w h i t e}{\ldots \ldots} |$
$\textcolor{red}{\textcolor{w h i t e}{\ldots \ldots \ldots .} \text{CH"_3color(white)(.)"H"color(white)(..)"H"color(white)(...)color(red)("CH"_2)color(white)(.)"H}}$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots . .} |$
$\textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{red}{{\text{CH}}_{3}}$

And this in reality looks like:

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