Question

9. A mixture (solution) of 3 different liquid compounds is formed. Each compound has the same condensed formula of C4H10O. Separate analyses of these compounds confirmed that none of the compounds has branching of C’s from the main chain of attached C atoms. Also one of the compounds experiences a weaker interparticle force as its dominant force than the other two.

(a) Draw 3 possible preferred Lewis structures consistent with the above formula and information.

It is desired to separate the compounds in this mixture by simple distillation. In simple distillation of an ideal solution (which we are assuming the above solution to be), a mixture of liquids is boiled over a temperature range to convert each compound to its gas phase. The generated gas phase of each compound is then re-condensed over a cooling column and collected as the pure liquid form for each compound present. Given that the boiling points of the liquids in the mixture are sufficiently different, the compound in the mixture with the lowest boiling point will be collected first, and the compound in the mixture with the highest boiling point will be collected last.

(b) Among the structures you drew in part (a), draw the structure of the compound you expect to be collected first in the distillation. Using applicable comparisons based on your structures, briefly explain your answer.

(c) Among the structures you drew in part (a), draw the structure of the compound you expect to be collected last in the distillation. Using applicable comparisons based on your structures, briefly explain your answer.

Answer 1

As per the information provided, there are no branching carbon chains. So this is a linear molecule having 4 carbon atoms. There are 4 carbon atoms and 10 hydrogen atoms. Next we need to check whether it is an alkane, alkene or an alkyne. The general formula for alkane is CnH2n+2, for alkenes CnH2n, for alkynes, CnH2n-2 ( n is the #of carbon atoms). In case of 4 carbon atoms, the hydrogens in alkane will be 2 x 4 +2 = 10. So this molecule is an alkane. There are no pi bonds.

1 Oxygen atom is present in the molecule. There are 3 ways the oxygen atom can be arranged around 4 carbon atom linear chain.

OH Hņi I-U- 1 Compound H-¢-¢-Ć-Ć-H Hн не HGH H H H-c-c-c-C-H I-U- I エー0- 10- 11 I-U- I I-U- Compound 2 H H H. I P- Hннн H-C-c-o-c-C-H Нинн m Compound 3 I- Olher possible structures: н: Он Н it I . equivalent to compo- ť I und 2 нин H H H. i I-U- ć I-U- - Ć-H und a re H I I-u- I I-U- ör - or Ć - ç - Equivalent to compou- nd 1 equivalen H-Ć - Ć- - ç ç je h H H. I I a I H H This is a possibility -c-o-c-4-4-H This compound &. ut the compound 3 are the possible structures for ether

The boiling point of a compound depends on various factors.

1. Increasing carbon chain length. As the length of carbon chain increases, the boiling point also increases if all ther factors remain same ( same number of hydrogens, same type of functional groups). In these three molecules, the carbon chain length is same. Only the functional groups differ.

2. Molecules with ionic bonds will have higer boiling point. This is because ionic bonds will display higher attractive forces between neighbouring molecules due to the tendency of opposite charges to attract each other. In all these 3 molecules, there are no ionic bonds, so this factor will not influence the boiling point.

3. Next factor is the hydrogen bonding. This is the weak bonding that occurs between electronegative atom of 1 molecule and hydrogen bonds of second molecule. More the hydrogen bonding, higher the boiling point. Alcohols will have higher degree of hydrogen bonding as compared to ethers. So compounds 1 and 2 will show higher hydrogen bonding and thus higher boiling point than compound 3.

4. Third force is a dipole dipole interaction. Dipole will be created whenever there is a electronegative atom in the molecule. Oxygen is highly electronegative. It will attract the electrons towards itself. So the electron density will be higher on Oxygen than on the carbon or hydrogen it forms a bond with. I. All 3 cases dipole dipole interactions will occur

5. Another factor that influences the boiling point is the Van der wall dispersion forces. These are attractive forces between two neighbouring molecules. As the size of the molecule increases, these forces become stronger. Also, these forces are stronger for linear molecules than for branched molecules. In the comparison between compound 1 and 2, 1 is more linear. Therefore 1 will have higher boiling point than 2.

So the order of collection of compounds will be

3 will be first, 2 will be second and 1 will be third

The order of boiling points is

1>2>3