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## Please explain an easy way to convert from the straight-chained structure of a carbohydrate into its ring form?

Ernest Z.
Featured 8 months ago

Here's one way to do it.

#### Explanation:

Draw the Fischer projection of your sugar, e.g. glucose.

Now, rotate the Fischer projection 90° clockwise.

Next, draw a template for a Haworth structure.

Look at the $\text{OH}$ groups on carbons, 2, 3, and 4 of the Fischer projection.

Those on the right will be "down" in the Haworth structure. Those on the left will be "up".

If the sugar is a D-sugar, the $\text{CH"_2"OH}$ group will be "up".

The OH on carbon 1 can be "down" for the α-anomer or "up" for the β-anomer.

And you have the Haworth structure of glucose.

## Why are aromatic rings stable?

Ernest Z.
Featured 8 months 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.

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

Ernest Z.
Featured 7 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.

## Are alkyl halides sterically hindered?

Morgan
Featured 6 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.

## What is meant by +I & -I effects??

Ernest Z.
Featured 2 months ago

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

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

Ernest Z.
Featured 4 days ago

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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