The increasing order of boiling points of the below
mentioned alcohols is
(I) 1,2 - dihydroxybenzene
(II) 1, 3 - dihydroxybenzene
(II) 1, 4 - dihydroxybenzene
(IV) Hydroxybenzene
A.
I < II < IV < III
B.
I < II < III < IV
C.
IV < II < I < III
D.
IV < I < II < III
Answer :
IV < I < II < III
Solution :
Among the given compounds, hydroxybenzene $$(IV)$$ has least molar mass and therefore possess least boiling point. Among the three isomeric dihydroxybenzenes, 1,2 - dihydroxybenzene (I) forms intramolecular $$H$$ - bonding with the result it will not form intermolecular $$H$$ - bonding leading to lowest boiling point. On the other hand 1,3 - dihydroxybenzene (II) and 1, 4 - dihydroxybenzene (III) do not undergo chelation,
hence they will involve extensive intermolecular $$H$$ - bonding leading to higher boiling point. Further intermolecular hydrogen bonding is stronger in the $$p$$ - isomer than the m - isomer hence former has highest b.p. Thus the decreasing order of boiling points is III > II > I >IV.
Releted MCQ Question on Organic Chemistry >> Alcohol, Phenol and Ether
Releted Question 1
Ethyl alcohol is heated with conc $${H_2}S{O_4}$$ the product formed is
A.
\[{{H}_{3}}C\underset{\begin{smallmatrix}
\parallel \\
O
\end{smallmatrix}}{\mathop{-C-}}\,O{{C}_{2}}{{H}_{5}}\]