Question

(a) What happens when chlorobutane is treated with Na metal in the presence of dry ether ?

(b) Arrange the following compounds in increasing order of their boiling points :

1-Bromobutane, 1-Iodobutane, Butane, 1-Chlorobutane

Hint:

Na in dry ether is the Wurtz reaction, giving n-octane. Boiling point rises with molecular mass and polarisability: butane is lowest and the iodide highest.

Answer 1

Step 1: Identify the reaction in part (a):
Alkyl halides react with sodium metal in dry ether. This is the Wurtz reaction.
Two alkyl groups join to form a longer alkane.

Step 2: Write the reaction for chlorobutane:
Two molecules of 1-chlorobutane lose their halogen to sodium.
The two butyl groups then join with a new C-C bond.
\[ 2\,\mathrm{CH_3CH_2CH_2CH_2Cl} + 2\,\mathrm{Na} \xrightarrow{\text{dry ether}} \mathrm{CH_3(CH_2)_6CH_3} + 2\,\mathrm{NaCl} \]
The product is n-octane, and sodium chloride is the by-product.

Step 3: Recall what fixes boiling point in part (b):
These molecules have no hydrogen bonding.
Boiling point depends on the strength of van der Waals forces, which grow with molecular mass and surface area.

Step 4: Compare the four compounds:
Butane is non-polar and has the weakest forces, so it boils lowest (about 273 K).
The three halobutanes are polar and heavier.
Molar masses are about 92.5 (Cl), 137 (Br) and 184 (I) g mol\(^{-1}\).
Larger halogen atoms are more polarisable, so the forces increase from Cl to I.

Final Answer:
(a) n-Octane is formed (Wurtz reaction).
(b) Boiling points increase in the order:
\[ \boxed{\text{Butane} < \text{1-Chlorobutane} < \text{1-Bromobutane} < \text{1-Iodobutane}} \]