Question

How many carbon atom(s) are present in the third homologue of methyl ether?
a.) 1
b.) 2
c.) 3
d.) 4

Answer 1

Hint: In order to solve this question, draw the structure of methyl ether first. After that, add two consecutive carbons to make the third homologue. Then, count the number of carbons for the answer.

Complete step by step solution:
Let us start this question by drawing the structure for methyl ether.
Methyl ether is also known as dimethyl ether. It is a compound composed of two alkyl groups (methyl) that is joined together by an oxygen atom.
The structure of methyl ether is –

\[C{{H}_{3}}-O-C{{H}_{3}}\].

A homologous series is defined as “the series in which each subsequent member differs from its previous member by \['-C{{H}_{2}}-'\]group”.
Therefore, according to the definition –
Second homologue will be: \[C{{H}_{3}}-O-C{{H}_{2}}-C{{H}_{3}}\]
The IUPAC name of this compound is methyl ethyl ether.
Third homologue will be: \[C{{H}_{3}}-O-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\]
The IUPAC name of this compound is methyl propyl ether.
The molecular formula of methyl propyl ether can be written as - \[{{C}_{4}}{{H}_{10}}O\].
As we can see there are four carbons in this compound.
Therefore, the answer is – option (d) – 4 carbon atoms are present in the third homologue of methyl ether.


Note: Methyl ether is also known by the names – methoxymethane or dimethyl ether. It is the simplest ether possible. It is an isomer of ethanol, because it has the same molecular formula as that of ethanol- \[{{C}_{4}}{{H}_{10}}O\].