46 PHYSIOLOGY 



Examples of the first two groups are ethylene CH 2 



II 

 CH 2 



and acetylene CH 



lil 

 CH 



Derivatives of all these groups occur in the body. 



THE ALCOHOLS. The first product of the oxidation of hydrocarbons is the 

 series of bodies known as the alcohols. Examples of these are : 



CH 3 OH methyl alcohol 



C 2 H 5 OH ethyl 



C 3 H 7 OH propyl ,-, 



C 4 H 9 OH butyl 



C 5 H n OH amyl 



C 6 H 13 OH capryl and so on, 



the general formula for the group being 



C n H 2n + xOH. 



In all these alcohols the OH group is, so to speak, more mobile than the other atoms 

 connected with the carbons, and can therefore be replaced by other substances or 

 groups with comparative ease. In this respect therefore an alcohol can be compared 

 to water HOH or to an alkaline hydroxide NaOH or KOH. The best-known example 

 of the group is ethyl alcohol, the ordinary product of fermentation of sugar. In these 

 alcohols the H of the OH group can be replaced by Na. Thus, water with metallic 

 sodium gives sodium hydroxide and hydrogen as follows : 



2HOH + 2Na = 2NaOH + H 2 . 



In the same way alcohol treated with metallic sodium gives off hydrogen, and the 

 remaining fluid contains sodium ethylate, thus : 



2C 2 H 5 OH + 2Na = 2C 2 H 5 ONa + H 2 

 (sodium ethylate) 



On the other hand, the OH group may be replaced by acid radicals. Thus, if ethyl 

 alcohol be treated with phosphorus pentachloride, ethyl chloride is formed together 

 with phosphorus oxychloride and hydrochloric acid. Thus : 



Et.OH + PC1 5 = POC1 3 + HC1 + Et.Cl 



(ethyl chloride) 



With concentrated sulphuric acid the reaction is similar to that which obtains between 

 sodium hydrate and this acid, and we have formed ethyl hydrogen sulphate and water. 

 Thus: 



Et.OH + H 2 S0 4 = Et.HS0 4 + HOH 



If alcohol be warmed with acetic acid and strong sulphuric acid, among the products 

 of the reaction is ethyl acetate, which is volatile, and therefore passes off. Thus : 



Et.OH + HC 2 H 3 2 = Et.C 2 H 3 2 + HOH. 



These compounds of the hydrocarbon group of the alcohol, such as methyl, ethyl, 

 propyl, &c., with an acid, in which the ethyl takes the part of a base, are known as 

 esters. 



An ester treated with an alkali is decomposed with the formation of an alkaline 

 salt of the acid, and the corresponding alcohol which, being volatile, is given off on 

 warming the mixture. Thus : 



Et.C 2 H 3 O 2 + NaHO = NaC 2 H 3 O 2 + Et.OH. 

 (ethyl acetate) (potassium acetate) (alcohol) 



This process of decomposition of an ester wMrthe formation of the alkaline salt of an 

 acid is often spoken of as saponification, i.e. soap formation, though the term ' soap ' 



