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9

Answer:

The relative magnitude of the inter molecular forces are:
#CF_4 < OF_2 < CHF_3 < HF.#

Explanation:

All molecules will have London dispersion forces which get stronger as the molecule gets heavier (more electrons causes a shift in electron cloud distribution resulting in a temporary dipole).

#CHF_3# is a polar molecule. So it will have dipole - dipole interaction along with the weaker dispersion forces.

#OF_2# is a polar molecule with a bent shape just like #H_2O#. There are two pairs of bonded electrons and two pairs of unbonded lone pair electrons. These lone pairs repel the bonded electrons resulting in a bent shape with an angle less than # 105^0#. The dominant inter molecular forces would be dipole-dipole.

HF is a polar molecule. Hence the primary inter molecular forces would be dipole - dipole and hydrogen bond which is a special type of dipole - dipole interaction between the hydrogen atom and electronegative F atom.

#CF_4# has a tetrahedral structure. It is non-polar molecule. The dominant inter molecular force would be London dispersion force.

2

Answer:

The realtion between #K_A# and #pK_a# is #K_a = 10^(-pK_a)#. This value provides an essential constant for keeping track of concentration of #[H^+]# in solutions.

Explanation:

Much like the way that pH is written using the logarithm function to convert the actual #[H^+]# of a solution into a simpler value, so the #K_a# of an acid can be converted into a #pK_a# value.

Once you have the #K_a# value, the applications that can be made from #K_a# or #pK_a# (whichever you prefer to use), include the ability to determine the #[H^+]# in a solution of that acid, regardless of the concentration of the solution, or whether it is buffered for example.

#K_a# is an equilibrium constant, and as such is independent of the concentration of a solution. This makes it the best value for establishing the strength of that acid relative to others.

2

Answer:

Here's one way to do it.

Explanation:

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

4.bp.blogspot.com

Now, rotate the Fischer projection 90° clockwise.

Rotated
(Adapted from http://4.bp.blogspot.com)

Next, draw a template for a Haworth structure.

Template

Look at the #"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 #"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.

upload.wikimedia.org

1

Answer:

You can carry out four symmetry operations on trans-1,3-dichlorocyclobutane.

Explanation:

The structure of trans-1,3-dichlorocyclobutane is

Structure

The ring is puckered, but it undergoes a rapid ring flip between the two bent forms.

We can treat the ring as if it were planar. However, at low temperatures the two conformations would be "frozen out" and the ring would have a different symmetry.

The planar molecule has #"C"_(2h)# symmetry.

This point group contains four symmetry operations:

#bbhat"E" color(white)(ml)"the identity ('do nothing') operation"#
#bbhat"C"_2 color(white)(m)"a rotation about a two-fold axis"#
#bbhat"i" color(white)(mm)"inversion about a single point"#
#bbhatsigma_h color(white)(m)"reflection about the mirror"#
#" "" ""plane perpendicular to the C"_n#
#" "" "color(white)(l)"axis of the highest n"#

Let's apply these operations to the molecule.

#bbhat"E"#, the identity operation (symmetry element: the entire molecule)

This one is easy. It amounts to doing nothing to the molecule.

#bbhat"C"_2#, rotation by #(360^@)/2 = bb(180^@)# about an axis (a #C_n# rotation with #n = 2#; symmetry element: the #C_2# rotation axis)

Rotation

The #"C"_2# axis passes between carbons 2 and 4, and lines up with the plane of the ring (it is not through the ring).

Rotation by 180° about this axis interchanges the two #"Cl"# atoms, the two #"H"# atoms, and carbons 1 and 3. A front-facing atom is now facing the rear and vice versa.

#bbhat"i"#, the inversion operation (symmetry element: a dot at the center of the molecule)

Inversion

There is a centre of inversion at the mid-point of the ring.

Inversion through this point interchanges the two #"Cl"# atoms, the two #"H"# atoms, and carbons 1 and 3. In other words, it takes #(x,y,z)# and transforms each coordinate into #(-x,-y,-z)#.

#bbhatσ_"h"#, reflection through a mirror plane perpendicular to the #"C"_2# axis (symmetry element: the mirror plane itself)

Reflection

A mirror plane passes through the #"H, C"# and #"Cl"# atoms at carbons 1 and 3.

It divides the molecule into two front and back mirror-image halves, and bisects the ring, while being coplanar with the explicit #"H"# and #"Cl"# atoms.

The operation reflects carbons 2 and 4, but the remainder of the atoms stay in place.

2

Answer:

Here are some of the rules for naming alkynes.

Explanation:

Alkynes are unsaturated hydrocarbons containing a triple bond, #"R-C≡C-R"#.

You name them the same way as you name alkanes, but you replace the ending -ane with -yne.

When there is more than one triple bond, you use the multiplying prefixes to
get -diyne, -triyne, etc.

Step 1. Find the longest continuous chain of carbons containing the triple bond.

For example, #"CH"_3"CH"_2"CH"_2"C≡CCH"_3# has six carbon atoms, so its base name is hexyne.

Step 2. Number the carbon atoms in the main chain from the end closer to the triple bond.

#stackrelcolor(blue)(6)("C")"H"_3stackrelcolor(blue)(5)("C")"H"_2stackrelcolor(blue)(4)("C")"H"_2stackrelcolor(blue)(3)("C")≡stackrelcolor(blue)(2)("C")stackrelcolor(blue)(1)("C")"H"_3#

Insert the first number of the alkyne carbons as close as possible before the
ending -yne.

The name becomes hex-1-yne.

Step 3. If there are other substituents, list them alphabetically with their locating numbers.

#"H"stackrelcolor(blue)(1)("C")"≡"stackrelcolor(blue)(2)("C")stackrelcolor(blue)(3)("C")"H"("CH"_3)stackrelcolor(blue)(4)("C")"≡"stackrelcolor(blue)(5)("C")stackrelcolor(blue)(6)("C")"H"_3#

For example, the compound above is 3-methylhexa-1,4-diyne.

2

Answer:

Here's how I would do it.

Explanation:

A doublet of doublets (dd) is a pattern of four lines of approximately equal intensity that results from coupling to two different protons.

chemistry.oregonstate.edu

I can think of two situations in which I would expect to see dd splitting patterns:

  • vinyl groups
  • 1,3,4-trisubstituted benzenes

A. Vinyl groups

Typical coupling constants in alkenes are #J_"trans" ≈ "16 Hz"#, #J_"cis" ≈ "10 Hz"#, and #J_"gem" ≈ "2 Hz"#.

Consider the #""^1"H"#-NMR spectrum of methyl acrylate:

Methyl acrylate
(From organic spectroscopy international)

Each proton on the vinyl group is split into a doublet of doublets by its neighbours.

1,3,4-Trisubstituted benzenes

Typical coupling constants in substituted benzenes are #J_"ortho" ≈ "8 Hz"#, #J_"meta" ≈ "2 Hz"#, and #J_"para" ≈ "0 Hz"#.

Consider the #""^1"H"#-NMR spectrum of 3,4-dichlorobenzoyl chloride:

3,4-dichlorobenzoyl chloride
(From www.chem.wisc.edu)

The proton that has ortho and meta neighbours (on carbon 6) will be a doublet of doublets.

It appears at δ 7.95 (J = 8.5, 2.3 Hz).

2

The Michael reaction is an addition reaction, usually of an enolate to an #alpha,beta#-unsaturated ketone (a conjugated enone), to form a diketone.

The case you're familiar with is this one:

where I've highlighted the molecular fragment that gets attached. The mechanism is a bit visually challenging, but it's not too bad:

You can think of it as a variation on the aldol addition, where we've replaced the aldehyde with a conjugated enone.

  1. A base deprotonates the #bb(alpha)#-carbon on the ketone, which forms the enolate. Sometimes, LDA, lithiumdiisopropylamide, is used instead to give a better yield (though it is too reactive for aldehydes).
  2. The enolate then acts as a good nucleophile to attack the #beta# position on the enone. Draw the resonance structure, and you should find that the carbonyl oxygen, which is electron-withdrawing and #delta^(-)#, causes the #beta#-carbon to be #delta^(+)#.
  3. The reaction finishes by tautomerizing the enolate back into the ketone by deprotonating the water molecule that was made in step 1.

This is usually the first step in a Robinson Annulation. Here is an example of an annotated Robinson Annulation (notice the heat added in step 7):

2

Answer:

The halogenation of benzene is an electrophilic aromatic substitution reaction.

Explanation:

Electrophilic aromatic substitution

Electrophilic aromatic substitution is a reaction in which an atom on a aromatic ring is replaced by an electrophile.

A typical halogenation reaction is

upload.wikimedia.org

The electrophile is an ion that is generated by the catalyst.

Mechanism

research.cm.utexas.edu

Step 1. Generation of the electrophile

A Lewis acid catalyst, usually #"AlBr"_3# or #"FeBr"_3#, reacts with the halogen to form a complex that makes the halogen more electrophilic.

Step 2. Electrophilic attack on the aromatic ring

The nucleophilic π electrons of the aromatic ring attack the electrophilic #"Br"# atom .

This forms #"FeBr"_4^"-"# and generates a cyclohexadienyl cation intermediate, destroying the aromaticity of the ring.

Step 3: Loss of #"H"^"+"# and restoration of aromaticity

The #"FeBr"_4^"-"# removes the #"H"^"+"# from the ring.

This re-forms the aromatic ring, produces #"HBr"#, and regenerates the catalyst.

2

Answer:

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.

Energy levels
(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.

ursula.chem.yale.edu

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.

archive.uea.ac.uk

Qualitatively, a Frost circle explains the #(4n+2)# rule.

You need 2 electrons to fill the first level and 4 electrons to fill each level above the first.

2

The three factors that determine carbocation stability are adjacent (1) multiple bonds; (2) lone pairs; and (3) carbon atoms.

(1) Adjacent multiple bonds

An adjacent π bond allows the positive charge to be delocalized by resonance.

Thus, #"H"_2"C=CHCH"_2^"+"# is more stable than #"CH"_3"CH"_2"CH"_2^"+"#.

research.cm.utexas.edu

Resonance delocalization of the charge through a larger π cloud makes the cation more stable.

chem.libretexts.org

(2) Adjacent lone pairs

A lone pair on an adjacent atom stabilizes a carbocation.

Thus, #"CH"_3"OCH"_2^"+"# is more stable than #"CH"_3"CH"_2"CH"_2^"+"# because of resonance.

The cation is more stable because the charge is spread over two atoms.

#"CH"_3"O"stackrel(+)("C")"H"_2 ⟷ "CH"_3stackrel(+)("O")"=CH"_2#

(3) Adjacent carbon atoms

The stability of carbocations increases as we add more carbon atoms to the cationic carbon.

For example, #"CH"_3"CH"_2^"+"# is more stable than #"CH"_3^"+"#

The stabilization is explained by a type of resonance called hyperconjugation.

#"H-CH"_2"-"stackrel("+")("C")"H"_2 ⟷ stackrel("+")("H") " CH"_2"=CH"_2#

The #"sp"^3# orbitals of the adjacent #"C-H"# bonds overlap with the vacant #"p"# orbital on the carbocation.

images.tutorcircle.com

The delocalization of charge stabilizes the carbocation.

Thus, the more alkyl groups there are surrounding the cationic carbon, the more stable it becomes.

That gives us the order of stability: #3° > 2° >1° > 0°#

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