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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.

How can we distinguish the following compounds from each other?

dk_ch
Featured 11 months 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.

Reaction

When you have a doublet of doublets how many neighboring Hydrogens will there be?

Ernest Z.
Featured 10 months ago

There will be two neighbouring hydrogens.

Explanation:

A doublet of doublets (dd) occurs when a hydrogen atom is coupled to two non-equivalent hydrogens.

An example is the NMR spectrum of methyl acrylate.

(From Chemistry LibreTexts)

Each of the vinyl protons ${\text{H"_text(a), "H}}_{\textrm{b}}$ and ${\text{H}}_{\textrm{c}}$ is a dd.

Let's examine the ${\text{H}}_{\textrm{c}}$ signal at 6.21 ppm.

There are four separate peaks because ${\text{H}}_{\textrm{c}}$ is coupled to both ${\text{H}}_{\textrm{a}}$ and ${\text{H}}_{\textrm{b}}$, but with different coupling constants for each.

The result is a doublet of doublets.

The splitting diagram shows that ${\text{H}}_{\textrm{c}}$ is split into a doublet by ${\text{H}}_{\textrm{a}}$ with a large coupling constant ${J}_{\textrm{t r a n s}}$.

These peaks are each split into doublets by coupling with ${\text{H}}_{\textrm{a}}$ with a much smaller coupling constant ${J}_{\textrm{\ge m}}$ = 1.5 Hz.

Why is phenoxide more stable than phenol?

Ernest Z.
Featured 6 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 $55.9 \textcolor{w h i t e}{l} \text{kJ·mol"^"-1}$.

What is mass spectroscopy?

yumean1119
Featured 6 months ago

Mass spectroscopy is a method to find a molecule weight.

Explanation:

I'm not so familiar, sorry...

[Step 1] Molecules are ionized and crashes into fragments.
[Step2] Then, the ions are accelerated and induced into magnetic field.
[Step3] In the magnetic field, the ions are affected by Lorentz force $F \left(N\right)$ and their obrit is bended.

$F = q \left(v \times B\right)$ ($q \left(C\right)$ the charge of ion: $v \left(m \text{/} s\right)$: velocity of ion, $\times$: cross product, $B \left(T\right)$: magnetic flux density).

And of course...
$F = m a$ (m($k g$): mass of ion, #a(m"/"s^2#): acceleration) and you can conclude:
$a = \frac{q \left(v \times B\right)}{m}$

The point is that $\textcolor{red}{\text{orbit of lighter ions or fractions are bended}}$
$\textcolor{red}{\text{more easily than that of hevier ions.}}$
Acceleration rate $a$ is proportional to $\frac{q}{m}$.

[Step4] Sort ions or fractions by $\frac{m}{q}$(i.e. the reciprocal of $\frac{q}{m}$) and record the relative abundance of ions. This is how you can find a mass of the molecule or mass of fragments.

This is a mass spectrum for 1-phenyl-1-butanone(${C}_{6} {H}_{5} C O {C}_{3} {H}_{7}$). Cited from http://www2.odn.ne.jp/had26900/topics_&_items2/about_MS1.htm

What you can find from the result…
148: The mass of the whole molecule, and this is exactly the same as ${C}_{6} {H}_{5} C O {C}_{3} {H}_{7}$.
77: The mass is same as the phenyl group, ${C}_{6} {H}_{5} -$. This implies that the molecule might have benzene ring.
105: $105 - 77 = 28$ is the mass of carbonyl group($- C O -$). Thus we can expect the molecule to have ${C}_{6} {H}_{5} C O -$ group.

Could someone please explain (ii)* to me?

Truong-Son N.
Featured 2 weeks 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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