HYDROCARBONS AND THEIR HALOGEN DERIVATIVES. 471 



bon monoxide and phosphorus trichloride are un saturated, as they 

 combine directly with a number of substances ; for instance, with 



chlorine, thus : 



CO + 2C1 = COC1 2 , 



PC1 3 + 2C1 = PC1 5 . 



The hydrocarbons of the methane series are saturated ; they can- 

 not be made to enter directly into combination with other substances, 

 because there are no bonds left unprovided for. 



On the other hand, we have several homologous series of hydro- 

 carbons which are unsaturated. The olefins belong to this kind, 

 and the reason is found in the structure of the molecules. 



Looking at the graphic formulas of the normal hydrocarbons of the methane 

 series on page 465, we find all affinities completely saturated. The structure 

 of ethylene, C 2 H 4 , the first member of the olefines, may be represented by either 

 of the following formulas : 



H H H H 



H C C H 



Each of these representations shows that two bonds are left unsaturated, and 

 as certain considerations lead us to assume that two hydrogen atoms are in 

 combination with one carbon atom the second representation is the one agree- 

 ing with our views. Instead of leaving the affinities unsaturated in our for- 

 mulas as above, we use double linkage, and give to ethylene the formula 



H H 

 H C=C H or H 3 C=CH 2 . 



Whenever direct combination between ethylene and another substance 

 occurs the double linkage is broken and the bonds are utilized for holding 



the respective atoms, or radicals, thus : 



Br Br 



H 2 C=CH 2 + 2Br = H 2 C CH 2 . 



As the higher members of the ethylene series are obtained by replacement 

 of hydrogen atoms by hydrocarbon radicals in ethylene, which replacement 

 does not alter the double linkage of its carbon atoms, all members behave like 

 unsaturated compounds. 



In a similar manner we represent the unsaturated hydrocarbon acetylene 

 C 2 H 2 , by the formula HC=CH, showing triple linkage between the carbon 

 atoms. That this view is in keeping with the facts is shown by the action of 

 bromine or of hydrobromic acid on acetylene, thus : 



HC=CH + 4Br = Br 2 HC CHBr^ 

 HfeCH + 2HBr = BrH 2 C CH 3 Br. 



