Given below are two statements :
Statement I : Dimethyl ether is completely soluble in water. However, diethyl ether is soluble in water to a very small extent.
Statement II : Sodium metal can be used to dry diethyl ether and not ethyl alcohol.
In the light of  given statements, choose the correct answer from the options given below

The problem tests the basic properties of alcohols and ethers—particularly hydrogen-bonding, solubility in water, and reactivity toward active metals such as sodium.

Case 1: Verifying Statement I

Ethers are capable of forming hydrogen bonds with water through the lone pairs present on the oxygen atom. The extent of solubility depends on the size of the hydrophobic alkyl groups attached to oxygen.

• Dimethyl ether, $$CH_3-O-CH_3$$, has two very small $$CH_3$$ groups. The hydrophobic character is therefore minimal, so hydrogen bonding with water predominates. Hence it dissolves in water in all proportions (completely soluble).

• Diethyl ether, $$C_2H_5-O-C_2H_5$$, contains two larger $$C_2H_5$$ groups. The non-polar, hydrophobic part is now large enough to suppress extensive hydrogen bonding with water. Thus diethyl ether is only sparingly soluble (≈6-7 g per 100 mL at room temperature).

Therefore, Statement I is true.

Case 2: Verifying Statement II

Sodium metal reacts vigorously with any compound that can supply acidic protons (-OH in alcohols, -COOH in acids, etc.).

• Ethyl alcohol (ethanol) possesses an acidic hydrogen in the $$-OH$$ group and readily undergoes the reaction $$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \uparrow$$ producing sodium ethoxide and hydrogen gas. Because of this rapid chemical reaction, sodium cannot be used as a drying agent for ethanol.

• Diethyl ether lacks an $$-OH$$ group; its hydrogens are bonded to carbon, not oxygen. Those $$C-H$$ hydrogens are not acidic, so sodium metal does not react with diethyl ether under ordinary conditions. Hence sodium can safely remove traces of water from diethyl ether (it converts the water to $$NaOH$$ and $$H_2$$ while leaving the ether untouched).

Therefore, Statement II is also true.

Since both statements are correct, the appropriate choice is:

Option D: Both Statement I and Statement II are true.