Predicting Organic Reactions

Members of a homologous series have the same functional group and undergo similar reactions. Alkanes contain only single carbon–carbon bonds. Alkenes have a C=C double bond, while alcohols have an –OH group. To predict a product, find the part of the given molecule that reacts, keep track of its carbon atoms, and then check that each atom and bond is accounted for. The same approach works for the first four members and for a larger member whose structure is supplied.

Combustion: reacting with oxygen

Alkanes, alkenes and alcohols can all burn. With enough oxygen, their carbon atoms form carbon dioxide and their hydrogen atoms form water. For example, complete combustion of the alkane propane is

$\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$

The same products are formed when an alkene such as ethene or an alcohol such as ethanol burns completely:

$\mathrm{C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O}$

$\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}$

An alcohol already contains oxygen, so include that atom when balancing its equation. If oxygen is limited, combustion may be incomplete: carbon monoxide, or carbon as soot, can form instead of some of the carbon dioxide. Carbon monoxide is poisonous.

Adding atoms across an alkene double bond

The C=C bond is the reacting part of an alkene. In an addition reaction, the double bond becomes a single C–C bond and each of its two carbon atoms gains a new bond. No carbon atoms are lost from the chain.

Bromine, $\mathrm{Br_2}$, supplies one bromine atom to each carbon. For ethene, the structural change is

$\mathrm{CH_2{=}CH_2 + Br_2 \rightarrow CH_2Br{-}CH_2Br}$

The product is 1,2-dibromoethane, with molecular formula $\mathrm{C_2H_4Br_2}$. Bromine water changes from orange to colourless when it reacts with an alkene. For a longer given alkene, place the two bromine atoms on the two carbons that were double-bonded: $\mathrm{CH_3{-}CH{=}CH_2}$ becomes $\mathrm{CH_3{-}CHBr{-}CH_2Br}$.

Hydrogen, $\mathrm{H_2}$, adds in the same pattern, one hydrogen atom to each carbon of the former double bond. This turns an alkene into an alkane; for example,

$\mathrm{CH_2{=}CH_2 + H_2 \rightarrow CH_3{-}CH_3}$

A nickel catalyst and heat can be used for this reaction. The positions of the added atoms follow the position of the double bond in the given alkene, so inspect its structure rather than relying on its molecular formula alone.

Oxidising an alcohol to a carboxylic acid

Potassium manganate(VII) is an oxidising agent. When acidified and heated with a primary alcohol, it can change the end group from –CH₂OH to –COOH, forming a carboxylic acid. The carbon chain remains in place: ethanol, $\mathrm{CH_3CH_2OH}$, becomes ethanoic acid, $\mathrm{CH_3COOH}$. An equation using $\mathrm{[O]}$ to represent oxygen supplied by the oxidising agent is

$\mathrm{CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O}$

The potassium manganate(VII) solution changes from purple to colourless as it reacts. Methanol similarly forms methanoic acid; a straight-chain primary propanol forms propanoic acid. The position of –OH matters for other given alcohols: a secondary alcohol does not oxidise to a carboxylic acid in this reaction.