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1

## What is the frequency range in "cm"^(-1) of -OH in water using infrared spectrometry?

Truong-Son N.
Featured 1 month ago

I assume you mean for the $\text{OH}$ stretch of water in the IR region. Organic $\text{OH}$ stretches are typically around the 3200 ~ 3500 ${\text{cm}}^{- 1}$ range.

For water, it's a combination band of the symmetric stretch (${\nu}_{1}$), asymmetric stretch (${\nu}_{3}$), and symmetric bend (overtone of ${\nu}_{2}$) motions:

This is a Raman spectrum, but this particular vibrational motion is active in both Raman and IR, so this is a suitable proof of the frequency range.

You can see that its range is somewhere around bb(3200 ~ 3600) $\boldsymbol{{\text{cm}}^{- 1}}$. This website reports it as $3404.0$ ${\text{cm}}^{- 1}$.

2

## 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 3 weeks ago

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.

3

## Explain why only one of the following is correct? a) 3-methyl-4-isopropylpentane b) 2-ethyl-4-tertiary-butylpentane c) 2,2,3,5-tetramethylheptane

Martin M.
Featured 3 weeks ago

The naming of C is correct.

#### Explanation:

To answer this kind of question, you should draw the molecular structure. To do this we start with the end of the names. So for A, we start with pentane, which indicates a chain of 5 carbon atoms.
Then we add all the other stuff to this chain. We got a methyl group at carbon atom 3 and an isopropyl group at position 4 at the carbon chain. The same method applies to the other structures

The drawings are shown below.

We can now name this structures using the IUPAC rules. These rules are showed below this answer.

Now have a look at the structures above. We see that with A and B the longest carbon chain wasn't correctly chosen. I have provided the correct names of the structures in the figure. Also, the correct carbon chain is shown in the blue circle.

For the C the naming of the structure was correct.

Simplified IUPAC Rules

1. Identify and name the longest carbon chain. If there are two chains with an equal amount of carbon atoms, than the one with the most substitutions is the base chain.
2. Name all the groups attached to this chain as alkyl substituents.
3. Number the carbons of the longest chain beginning with the end that is closest to a substituent. (Use alphabetical order if there are 2 substituents at equal distance of the chain). Number the chain in the direction that gives the lower number at the first difference. (so for example instead of 1,3,5 we use 1,7,5)
4. Write the name of the alkane by first arranging all the good substituents in alphabetical order (preceded by carbon number with -): di, tri, tetra, penta, etc
4

## What would the product be if I add Br2 as step one and CH3OH in step 2?

Martin M.
Featured 6 hours ago

For determining the product in these type of questions, we have to look at the functional groups of the reactants. In this case, we have a $B {r}_{2}$ molecule and a cyclohexane ring with a double bond side group. As we know, double bonds are pretty reactive, so we expect something to happen there.

Now the $B {r}_{2}$ molecule can line up with this double bond, creating partial charges. This is indicated in the figure below. Because of the partial positive charge (created on the side of the double bond), the electrons will attack there and create a cyclic structure with one of the $B r$ atoms.

The other $B r$ will be released.

Now we add $C {H}_{3} O H$. The free electrons on the $O$ can react with the structure. This is showed in the mechanism below.

The carbon atom of the cyclic bromine group is partial positively charged, as indicated in the image above. This is created by hyperconjugation. This carbon atom is attached to 3 other carbon atoms (tertiary carbon atom), which stabilises a charge on the carbon atom more easily. Therefore the electrons from methanol will attack at that carbon atom, 'pushing away' the electrons from the bromine bond.

In the last step, the $H$ atom will be released. This causes the oxygen to be neutral instead of positively charged.

And there you have your product!

2

## Are D and L carbohydrates mirror images of each other?

Maxwell
Featured 6 hours ago

Yes

#### Explanation:

D-carbohydrates and L-carbohydrates are visually best seen in their straight chain forms.

Let's take $\textcolor{\mathmr{and} a n \ge}{\text{D-glucose}}$ and $\textcolor{m a \ge n t a}{\text{L-glucose}}$, for example.

When numbering, the $\text{C1}$ carbon is the Carbon that is part of the $\textcolor{b l u e}{\text{aldehyde}}$ functional group because of highest priority. C2 and on are counted going down the straight chain molecule.

Both of these molecules are glucose. The only difference is that they both differ in their absolute configurations at their $\text{chiral}$ Carbon centers. Since at each chiral center- $\textcolor{b l u e}{\text{C2, C3, C4, and C5}}$ - their absolute configurations are opposite to the mirror image, they are considered $\textcolor{b l u e}{\text{enantiomers}}$ of each other.

$\textcolor{b l u e}{\text{Enantiomers}}$ differ in their absolute configurations at every chiral center. This means if you were to assign $\text{(R)}$ or $\text{(S)}$ at $\text{C2, C3, C4, and C5}$, the $\text{D form}$ would have either the $\text{(R)}$ or $\text{(S)}$ absolute configuration and its mirror image, the $\text{L form}$, would have the exact opposite absolute configuration.

Another property of enantiomers is that they are $\textcolor{b l u e}{\text{optically active}}$. This means, they can rotate plane polarized light to a certain angle. If one rotates light to an angle of $+ {34}^{\circ}$ (dextrorotary), its enantiomer would rotate it at $- {34}^{\circ}$ (levorotarty). Same magnitude but different directions.

Note: The D and L in the carbohydrates just refers to the direction the $\text{C5}$ hydroxyl group faces (D = OH faces right side, L= OH faces left side). d, l, D, L, R, and S have no relation to each other.

2

## Why is piperidine a stronger base than morphine?

Ernest Z.
Featured 6 hours ago

I suspect the reason involves steric hindrance.

#### Explanation:

The structure of piperidine is

Piperidine is a weak base with "pK_text(b) = 2.80.

It has a cyclohexane ring structure in which the lone pair on the nitrogen atom is quite accessible to an attacking acid.

We can see this in both the ball-and-stick model

and the space-filling model.

Morphine

The structure of morphine is

(From ResearchGate)

Ring D in its structure is a piperidine ring.

The methyl group on the nitrogen group should make morphine more basic than piperidine.

However, morphine is a weaker base, with "pK_text(b) = 6.13.

Morphine has a rigid pentacyclic ring structure, and the nitrogen atom is "buried" in the interior of the molecule, where it is less accessible to an attacking acid.

This becomes even clearer in a space-filling model of morphine.

Attack by an acid is strongly hindered on one side by Ring A.

We find basicities by measuring the position of an equilibrium:

$\text{B + HX ⇌ BH"^"+" + "X"^"-}$

If access to the base is hindered, the position of equilibrium lies further to the left, and we say that the base is weaker.

Thus, morphine is a weaker base than piperidine.

1

## What makes so many epoxides so carcinogenic?

Ernest Z.
Featured 6 hours ago

Epoxides are carcinogenic because they disrupt our DNA.

#### Explanation:

Epoxides are three-membered cyclic ethers.

They are highly strained because the nominal bond angles are 60° instead of 109.5°.

They tend to react with other molecules to open the ring and reduce the strain.

A common pro-carcinogen is benzo[a]pyrene.

It is a constituent of automobile exhaust, smog, cigarette smoke and charcoal-cooked foods.

When the body absorbs benzo[a]pyrene, it tries to make it water-soluble so it can be excreted in the urine or feces.

(From researcgate.net)

It converts the benzo[a]pyrene into benzo[a]pyrene-7,8-epoxide.

This is hydrolyzed to the 7,8-dihydrodiol, which is further converted into benzo[a]pyrene-7,8-dihydrodiol-9,10-epoxide.

Once formed, the epoxide can π stack with the bases in DNA where the epoxide group reacts with guanine residues.

This distorts the structure of DNA and causes errors in DNA replication.

If these mutations occur in a gene that encodes a molecule that regulates the production of cells, the result may be cancer.

1

## What is the difference between acetoacetate , acetyl-coa , and acetoacetyl-coa?

Ernest Z.
Featured 6 hours ago

Here's what I get.

#### Explanation:

Acetoacetate

Acetate is the carboxylate ion of acetic acid.

In acetoacetate, an α-hydrogen has been replaced by an aceto or acetyl group, $\text{CH"_3"CO}$.

Acetyl-CoA

Acetyl-CoA is Coenzyme A in which the $\text{H}$ atom in the thiol group has been replaced by an acetyl group.

This is Coenzyme A:

And this is acetyl-CoA:

(From Wikipedia)

Acetoacetyl-CoA

Acetoacetyl-CoA is Coenzyme A in which the $\text{H}$ atom in the thiol group has been replaced by an acetoacetyl group, $\text{CH"_3"COCH"_2"CO}$.

(From Wikipedia)

3

## What are two characteristics of a molecule that make it suitable for analysis by infrared spectroscopy?

Truong-Son N.
Featured 5 hours ago

You are right that it has to have a dipole (not necessarily a net dipole). I'm not sure what other specific requirements you may be wondering about, but I can list several, and maybe one of them is what you are looking for.

REQUIREMENTS FOR VISIBILITY IN THE IR REGION

The main properties of a molecule that allow it to be analyzable via IR spectroscopy are:

1) It can experience a change in dipole moment, whether it is induced or permanent. We then must have that it is heteronuclear, if it is diatomic (therefore, ${\text{N}}_{2}$, ${\text{F}}_{2}$, etc. are invisible in the IR).

2) It should have resonant frequencies that are in the infrared frequency range of $100 - 4000$ ${\text{cm}}^{- 1}$.

3) Ideally, it should be not overly soluble in a nonpolar solvent, which is ideal since many analyses are carried out using a solvent such as ${\text{CS}}_{2}$ or ${\text{CCl}}_{4}$. If something is too soluble, one may saturate or overload the spectrometer.

CHANGE IN DIPOLE MOMENT

Requirement $\left(1\right)$ is usually easy to check; if you can imagine the molecule stretching or bending in a way that enforces asymmetry, you've more than likely changed its dipole moment, and thus that kind of vibrational motion is IR-active.

A somewhat tricky example is ${\text{CO}}_{2}$.

Even though it is symmetrical, it can stretch both oxygens in the same direction (as in $B$) to generate a nonzero dipole moment along the molecular axis. It can also bend its $\text{O"-"C"-"O}$ angle (as in $C$ and $D$) so that it generates a nonzero dipole moment through the oxygen.

RESONANT FREQUENCIES IN THE IR RANGE

This is usually satisfied automatically, just due to the general strengths of chemical bonds, but it couldn't hurt to check. The fundamental frequency can be estimated using the force constant $k$ of the bond.

$\boldsymbol{t i l \mathrm{de} \omega = \frac{1}{2 \pi c} \sqrt{\frac{k}{\mu}}}$,

where:

• $t i l \mathrm{de} \omega$ is the fundamental frequency in ${\text{cm}}^{- 1}$.
• $c$ is the speed of light, $2.998 \times {10}^{10} \text{cm/s}$.
• $k$ is the force constant in ${\text{kg/s}}^{2}$.
• $\mu = \frac{{m}_{1} {m}_{2}}{{m}_{1} + {m}_{2}}$ is the reduced mass of the two atoms in the bond, where each mass is in $\text{kg}$.

As an example, I randomly found on a quick Google search that $k \approx \text{2385 N/m}$ for ${\text{CO}}_{2}$. So:

tildeomega = 1/(2pi*2.998 xx 10^(10) "cm/s") sqrt(("2385 kg/s"^2)/((12.011*15.999)/(12.011 + 15.999) xx 10^(-3) "kg"/"mol" xx "mol"/(6.0221413 xx 10^(23) "molecules"))

$\approx {\text{2429 cm}}^{- 1}$

and the literature value for the fundamental frequency is ${\text{2349 cm}}^{- 1}$ from NIST.

This is found to be the strongest peak in its IR spectrum:

which you can estimate to be near ${\text{2350 cm}}^{- 1}$, in agreement with the literature value.

SOLUBILITY IN A NONPOLAR SOLVENT

Nice IR solvents that are commonly used are ${\text{CCl}}_{4}$ and ${\text{CS}}_{2}$. Since both are nonpolar, you would have the most "control" over the IR intensity if you had a reasonably soluble analyte. This is why IR tends to be suitable for organic molecules.

This kind of consideration is not necessary, but being too soluble may make it frustrating to get a spectrum without overloading the spectrometer. Being "somewhat" soluble in nonpolar solvents is probably fine.

1

## What simple test would be used to distinguish between the following compounds? A). Tetrachloromethane and butane B). 1-propanol and 2-methylpropan-2-ol C). Pentanamine and pentane D). Pent-1-ene and benzene

dk_ch
Featured 7 hours ago

A) $\text{ } C C {l}_{4}$ and ${C}_{4} {H}_{10}$

Tetrachloromethane or Carbon tetra chloride (C Cl_4)has molar mass $154 g \text{/} m o l$ whereas butane ${C}_{4} {H}_{10}$ has molar mass 58 $g \text{/} m o l$. So due to higher inter molecular force the first one is liquid at normal temperature and pressure whereas the second one is a gas under the same condition. The second is highly inflammable gas and commonly used as a gaseous fuel. The carbon in (C Cl_4) is in its highest oxidation state (+4) as in $C {O}_{2}$. So this compound is not at all inflammable , it is used as good fire extinguisher (trade name- pyrene )

So testing only the inflammability they can easily be distinguished

B) 1-propanonl($C {H}_{3} C {H}_{3} C {H}_{2} O H \to$ pimary alcohol) and

2-methylpropan-2-ol (CH_3)_3COH->tertiary alcohol)

Lucas' reagent is a solution of anhydrous zinc chloride in concentrated hydrochloric acid. This is used to distinguish between primary,secondary and tertiary alcohol lower molar.

They all produce alkyl halides when treated with this reagent through $S {N}_{1}$ mechanism and the alkyl halides formed being insoluble in the reaction mixture it makes the mixture turbid

In case of 2-methylpropan-2-ol the turbidity appears almost immediately but in case of 1-propanonl turbidity appears after long time.

C) Pentanamine $\left({C}_{5} {H}_{11} N {H}_{2}\right) \to$ primary amine

It is soluble in dilute $H C l$ and it responds to carbylamine reaction.

and pentane $\left({C}_{5} {H}_{12}\right) \to$ alkane

It is insoluble in dilute $H C l$ and it does respond to carbylamine reaction.

In carbileamine test the comound under test is treated with chloform and alcoholic $K O H$. the formation of carbile amine is detected by its foul smell.

Reaction

D). Pent-1-ene $\left(C {H}_{3} C {H}_{2} C {H}_{2} C H = C {H}_{2}\right) \to$alkene

and benzene $\left({C}_{6} {H}_{6}\right) \to$aromatic.

Both the compounds are unsaturated but the second one being aromatic one it is more stable and does not respond to test of un-saturation like dicolorization of red bromine water as done by the first one.

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