Question

An organic compound with the formula ${{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{{\rm{O}}_{\rm{3}}}$ shows properties of ether and alcohol when treated with an excess HBr yields only one compound 1,2-dibromoethane. Write structural formulas of compounds.

Answer 1

Hint: We know that an organic compound is the compound where there is a covalent link of carbon with other atoms like hydrogen, nitrogen, oxygen etc. Although in carbonates, carbides, cyanides carbon atoms are present but they are not classified as organic compounds.

Complete step by step answer:
Let’s understand functional groups in detail. Functional groups are atoms or groups of atoms that decide the chemical properties of an organic compound. Some functional groups are alcohol, aldehyde, carboxylic acid, ether, amide etc. The structure of functional group ether is R-O-R and alcohol is R-OH (R is the alkyl group).

Now, come to question. We have to identify the molecule whose chemical formula is ${{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{{\rm{O}}_{\rm{3}}}$ and the compound shows properties of both alcohol and ether. So, the compound must contain both alcohol and ether functional groups. Also given that, the compound when reacts with excess HBr produces only one compound 1,2-dibromoethane. We know, symmetrical ether when reacts with excess hydrogen bromine, gives only one product. So, the compound must be symmetrical ether. So, the structure of the compound is ${\rm{OH}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{O}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{OH}}$.

The reaction of the compound with excess Hbr is,

${\rm{OH}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{O}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{OH}} + {\rm{Hbr}}\left( {{\rm{Excess}}} \right) \to {\rm{Br}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{Br}}$

Therefore, the structural formula of the compound is ${\rm{OH}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{O}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{OH}}$.

Additional Information:
Williamson synthesis is the method used in the laboratory for preparation of symmetrical and unsymmetrical ether. In this method, an alkyl halide undergoes reaction with sodium alkoxide.
${\rm{R}} - {\rm{X}} + {\rm{R}} - {\rm{O}} - {\rm{Na}} \to {\rm{R}} - {\rm{O}} - {\rm{R}} + {\rm{NaX}}$
Here, R is an alkyl group.

Note:
Always remember that ether is classified into two types, symmetrical and unsymmetrical If the two alkyl groups in ether are same, such as, ${\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{O}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}$ the ether is termed as symmetrical ether and if alkyl groups are different (${\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{O}} - {\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_{\rm{3}}}$) then the ether is termed as an unsymmetrical ether.