The relative magnitude of the inter molecular forces are:
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).
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.
The realtion between
Much like the way that pH is written using the logarithm function to convert the actual
Once you have the
Here's one way to do it.
Draw the Fischer projection of your sugar, e.g. glucose.
Now, rotate the Fischer projection 90° clockwise.
(Adapted from http://4.bp.blogspot.com)
Next, draw a template for a Haworth structure.
Look at the
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
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.
You can carry out four symmetry operations on trans-1,3-dichlorocyclobutane.
The structure of trans-1,3-dichlorocyclobutane is
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
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.
This one is easy. It amounts to doing nothing to the molecule.
Rotation by 180° about this axis interchanges the two
There is a centre of inversion at the mid-point of the ring.
Inversion through this point interchanges the two
A mirror plane passes through the
It divides the molecule into two front and back mirror-image halves, and bisects the ring, while being coplanar with the explicit
The operation reflects carbons 2 and 4, but the remainder of the atoms stay in place.
Here are some of the rules for naming alkynes.
Alkynes are unsaturated hydrocarbons containing a triple bond,
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.
Step 2. Number the carbon atoms in the main chain from the end closer to the triple bond.
Insert the first number of the alkyne carbons as close as possible before the
The name becomes hex-1-yne.
Step 3. If there are other substituents, list them alphabetically with their locating numbers.
For example, the compound above is 3-methylhexa-1,4-diyne.
Here's how I would do it.
A doublet of doublets (dd) is a pattern of four lines of approximately equal intensity that results from coupling to two different protons.
I can think of two situations in which I would expect to see dd splitting patterns:
A. Vinyl groups
Typical coupling constants in alkenes are
(From organic spectroscopy international)
Each proton on the vinyl group is split into a doublet of doublets by its neighbours.
Typical coupling constants in substituted benzenes are
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).
The Michael reaction is an addition reaction, usually of an enolate to an
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.
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):
The halogenation of benzene is an electrophilic aromatic substitution reaction.
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
The electrophile is an ion that is generated by the catalyst.
Step 1. Generation of the electrophile
A Lewis acid catalyst, usually
Step 2. Electrophilic attack on the aromatic ring
The nucleophilic π electrons of the aromatic ring attack the electrophilic
Step 3: Loss of
This re-forms the aromatic ring, produces
Aromatic rings are stable because they are cyclic, conjugated molecules.
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
You need 2 electrons to fill the first level and 4 electrons to fill each level above the first.
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.
Resonance delocalization of the charge through a larger π cloud makes the cation more stable.
(2) Adjacent lone pairs
A lone pair on an adjacent atom stabilizes a carbocation.
The cation is more stable because the charge is spread over two atoms.
(3) Adjacent carbon atoms
The stability of carbocations increases as we add more carbon atoms to the cationic carbon.
The stabilization is explained by a type of resonance called hyperconjugation.
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: