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The high melting point is caused by π-π stacking of the aromatic rings.


In organic chemistry, π–π stacking refers to attractive interactions between the π clouds of aromatic rings.

There are various types of stacking.

Neither benzene nor hexafluorobenzene has a dipole moment.

However, they have strong quadrupole moments, caused by the π clouds above and below the rings.

For example, in benzene, the π clouds are negatively charged and the plane of the ring is positively charged.

The situation is reversed in hexafluorobenzene, because the electronegative fluorine atoms withdraw electron density from the ring.


You can see the charge distribution better in this image:

Both benzene and hexafluorobenzene are destabilized by sandwich stacking, because areas with the same charge are placed next to each other.

However, benzene and hexafluorobenzene are strongly stabilized by sandwich stacking, because areas with opposite charge are placed next to each other.

Theoretical calculations put the stabilization energy at about 20 kJ/mol.

That makes the attractions as strong as many hydrogen bonds and dipole-dipole interactions.

Thus, the strong intermolecular quadrupole attractions cause a 1:1 mixture of benzene and hexafluorobenzene to have a high melting point.


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.


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 #"C-F"# bond.

The highly electronegative #"F"# atom will draw the electrons in the #"C-F"# bond more closely toward itself.

The bond will be polarized, with the #"F"# atom getting a partial negative (#δ^"-"#) charge and the α-carbon atom getting a partial positive (#δ^"+"#) charge.


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 #"C-C"# σ-bonds until it is almost negligible at the δ-carbon.

#"C"# is less electronegative than #"H"#, so alkyl groups are electron releasing.


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 #"p"K_text(a)# drops.

Thus, for example. the order of acidity as shown by the #"p"K_text(a)# values is

#underbrace("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")_color(red)(2.86) #

The electronegative #"Br"# atom removes electron density from the atoms next to it, eventually weakening the #"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 #"p"K_text(a)# values is

#underbrace("CH"_3"-CH"_2"COO-H")_color(red)(4.87) < underbrace("CH"_3"-COO-H")_color(red)(4.76) < underbrace("H-COO-H")_color(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

It isn't! Phenol is more stable than phenoxide ion.

Consider the equilibrium between the two species:

#"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_text(a)# value corresponds to a free energy difference of #55.9 color(white)(l) "kJ·mol"^"-1"#.

You'll have to be able to draw these out to name them. It is impossible to name by inspection of their chemical formulas.

  1. I've highlighted the longest hydrocarbon chain, which you would have had to identify. #(a)# is therefore a kind of hexane, while #(b)# is a kind of heptane (hex = six, hept = seven carbons in the main chain).

  2. Now you simply count from each end to determine the lowest possible set of carbon indices, and account for duplicate functional groups using prefixes. Alphabetize them afterwards.

NOTE: the smallest index should be the first point of difference between two sets of numberings.

There are four methyl groups in #(a)#, while there is one methyl and one ethyl group in #(b)#.

Counting from the left:

#(a)# is then denoted #ul"2,2,3,3-tetramethylhexane"#.

#(b)# is then denoted #"3-ethyl-6-methylheptane"#.

Counting from the right:

#(a)# is then denoted #"4,4,5,5-tetramethylhexane"#.

#(b)# is then denoted #ul"5-ethyl-2-methylheptane"#.

The first name for #(a)# is correct since its indices are minimized.

The second name for #(b)# is correct since the first encountered functional group with the lowest index is methyl, but since ethyl comes before it alphabetically, we rearrange these while keeping the indices as they are.

#(b)# is more difficult, so remember that example.


Warning! Long Answer. The compound is propionic anhydride.


Preliminary analysis

You know the formula is #"C"_6"H"_10"O"_3"#.

An alkane with six carbon atoms has the formula #"C"_6"H"_14"#.

The degree of unsaturation #U# is

#U = (14-10)/2 = 4/2 = 2#

Therefore, the compound contains two rings and/or double bonds.

#""^1"H NMR"#

The spectrum has 10 protons and only two peaks. The molecule must have a symmetrical structure.

A peak with 2 neighbours and aone with 3 neighbors corresponds to an ethyl group (#"C"_2"H"_5#, triplet-quartet pattern).

The #"6H:4H"# pattern tells us there are two ethyl groups.

However, I think you have the assignments reversed. The methyl group should have the smaller chemical shift.

A methyl group is normally at 0.9 ppm. Something is pulling it downfield
to 1.2 ppm.

A methylene group is normally at 1.3 ppm. Something is pulling it downfield
to 2.5 ppm.

We see from the table that a #"CH"_3# next to a #"C=O"# is shifted downfield to 2.2 ppm (a shift of 1.3 ppm).

We could expect a similar 1.3 ppm shift for a #"CH"_2# group; from 1.2 ppm to 2.5 ppm.

This is just where the #"CH"_2# group appears.

We now know that the partial structure is

#"CH"_3"CH"_2"(C=O)"# + #"(O=C)CH"_2"CH"_3#

These fragments add up to #"C"_6"H"_10"O"_2"#.

There is only one #"O"# atom left to insert. It must go in the middle:


The compound is propionic anhydride.


1. The compound has two double bonds.

2. Three #""^13"C"# NMR signals tell us there are three different carbon environments,

We should expect to see

  • #"CH"_3color(white)(mm)# at 10 ppm
  • #"CH"_2 color(white)(mm)# at 30 ppm
  • #"(C=O)O"# at 170 ppm

We see peaks at 8 ppm, 28 ppm, and 170 ppm. This is consistent with propionic anhydride.

3. The #""^13"C"# NMR spectrum of propionaldehyde is


In this compound there are 4 optical isomers; this is calculated by multiplying the number of chiral centres by two.


The number of optical isomers in a compound is determined by the number of chiral centres in it. A chiral centre is a carbon atom that is bonded to four different molecules or atoms. Each chiral centre will result in two different optical isomers.

So, to work out the number of chiral centres, draw out the displayed formula of the compound, and circle/highlight any carbons in the compound that have four different molecules attached to them (to check this, try drawing the molecule as a tetrahedral shape around your chosen chiral centre; each bond from the central C atom should go to a unique compound). Then, just multiply the number of chiral centres by two to give the number of optical isomers.


So below is your molecule (I have used MolView to sketch this):
To work out which carbons are chiral, you just have to check each one separately. A chiral carbon is usually not on a branched group, or on the end of a chain, so often you can eliminate those pretty quickly. Below is the molecule with the chiral centres highlighted in yellow:
Theory of optical isomerism:

Stereoisomers are compounds that have the same structural formulae, but due to the lack of rotation around a double carbon bond (aka #pi#-bond) will form slightly different compounds where a structure is above the molecular plane in one isomer, and below in the other.

Optical isomers (aka enantiomers) follow a similar principle, except instead of thinking about differences in location above and below a double bond, we are thinking about the orientation around a tetrahedral carbon centre.

The rule is that if each of the four substituents that are bonded to the central carbon are different, then it is possible for a stereoisomer of this compound to exist which is a mirror image of this compound. It is a form of stereoisomerism because both compounds have the same structural formulae, but cannot be superimposed on top of each other no matter how you rotate them.

The easiest way to get a feel for this is to draw a tetrahedral molecule, with (any) four different bonding groups; then imagine a mirror line next to the compound and draw its mirror image on the other side (this effect can be achieved by using a real mirror). Then, try and superimpose one molecule on top of the other - it should not be possible.

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