IUPAC and optical activity questions

Question 3

1. Remember the rule: select the longest chain as the parent chain

- counting horizontally from left to right gives five carbons in the chain and the parent name Pent. Since only hydrocarbon groups are involved, the parent name will be Pentane.

2. Number from the end of the chain to give the smallest number in naming

- ie counting from  left to right places the two CH₃ groups at position 2 and not  4 if the chain was numbered from the right hand side.

3. Since one more similar group (methyl) is attached at position 4, this will be grouped with the others.

This gives trimethyl - two groups at position 2 and one group at position 4

Full name use option A

Question 4

1. If the molecule shows optical activity, it should form two non- superimposable isomers that are mirror images of each other. To do this the molecule must have a chiral centre. This is a carbon centre (atom) which has four different groups attached to it

2. Search the examples given for a carbon fitting this description.

3. To answer

i. Eliminate II this molecule is symmetrical.

ii. Eliminate III this will have similar (methyl) groups on the central carbon atom.

iii. I and IV fit the description and we have seen these in our text books before

Option  B is the answer

Calculation of empirical formula - combustion analysis

Combustion analysis  of 0.18g of an organic compound produces  0.396 g of carbon dioxide  and 0.216 g of water . The empirical formula of the compound, given that it contains carbon, hydrogen and oxygen only is

A.  C₃H₈O

B.  C₄H₈O

C.  C₂H₈O₄

D.  C₆H₁₆O

Collate data ie

0.396 g carbon dioxide

0.216 g water

Empirical formula can only have  H O C

Equation for combustion    M   + O2     =   CO2   +  H2O

  multiply values by 1000   180g                396g        216g

Divide values by Relative Formula Mass (RFM) to obtain moles of carbon dioxide and moles of water

     396/44    =   9 moles of carbon dioxide,   216/18   =  12 moles of water 

Note: Some oxygen is present in M and some added during burning

9 moles of carbon dioxide contains 9x12 = 108 g of Carbon

12 moles of water   contains          12x2  = 24 g of   Hydrogen

Therefore, the mass of (C+H) present in 180 g of M = 10-8+24 = 132 g

The remainder is oxygen ie, 180 -132  = 48 g of oxygen

Summary

moles of  Carbon atoms       =  9

moles of  Hydrogen atoms   =  24

moles of  Oxygen  48/16      = 3

Divide by smallest value to get ratios  to give      9/3 = C 3,  24/3 = H   8,  3/3 =  O  1

Empirical Formula  C3 H8 1

Hope this was clear. Look out for other questions and answers.