Step-by-Step Solution

1. Analyze the Reactants and Reaction Conditions

   • The starting material is compound P, which is 4-(2-hydroxyethyl)phenol. Its molecular formula is $C_8H_{10}O_2$.
   • The reaction is conducted in an acidic medium ($H^+$).
   • The problem states that an ionic intermediate is formed. This strongly suggests a carbocation-mediated mechanism.

2. Formation of the Ionic Intermediate (Electrophile)

   • In the presence of acid, the primary alcohol group on the side chain gets protonated because its oxygen is more basic than the phenolic oxygen (whose lone pairs are delocalized into the aromatic ring). $HO-C_6H_4-CH_2-CH_2-OH + H^+ \rightleftharpoons HO-C_6H_4-CH_2-CH_2-OH_2^+$
   • The protonated hydroxyl group is a good leaving group (water). It departs to form a primary carbocation, which is the key ionic intermediate. $HO-C_6H_4-CH_2-CH_2-OH_2^+ \rightarrow HO-C_6H_4-CH_2-CH_2^+ + H_2O$ This carbocation acts as an electrophile ($E^+$).

3. Electrophilic Aromatic Substitution and Cyclooligomerization

   • The benzene ring of another molecule of P is electron-rich due to the activating effect of the -OH group. It acts as a nucleophile.
   • The carbocation ($E^+$) attacks the nucleophilic ring, primarily at the ortho position to the -OH group. This is an intermolecular Friedel-Crafts alkylation reaction.
   • This process of forming a carbocation and subsequent alkylation can repeat, leading to polymerization. Due to steric factors and stability, such reactions often lead to the formation of cyclic oligomers (macrocycles). A common and stable product in this type of condensation is a cyclic tetramer.

4. Formation of an Intermediate Cyclic Tetramer

   • Four molecules of compound P ($C_8H_{10}O_2$) react to form a cyclic tetramer. In this process, four new C-C bonds are formed, and four molecules of water are eliminated. $4 \, (C_8H_{10}O_2) \rightarrow (C_8H_8O)_4 + 4 \, H_2O$
   • The molecular formula of this intermediate cyclic tetramer is $C_{32}H_{32}O_4$.
   • Let's calculate the degree of unsaturation (DoU) for this intermediate. The formula for DoU is C + 1 - H/2. $DoU_{intermediate} = 32 + 1 - \frac{32}{2} = 33 - 16 = 17$ This DoU corresponds to 4 benzene rings (4 × 4 = 16) and 1 large macrocycle ring (1), for a total of 17.

5. Formation of the Final Product Q

   • The reaction often proceeds further to a more stable product. The intermediate $C_{32}H_{32}O_4$ has four phenolic -OH groups. Under acidic conditions, two of these hydroxyl groups can undergo intramolecular dehydration to form an ether bridge, creating an additional ring. $C_{32}H_{32}O_4 \xrightarrow{-H_2O} C_{32}H_{30}O_3$
   • The molecular formula of the final product, compound Q, is $C_{32}H_{30}O_3$.

6. Calculate the Degree of Unsaturation of Q

   • Now, we calculate the degree of unsaturation for the final product Q ($C_{32}H_{30}O_3$). $DoU_Q = C + 1 - \frac{H}{2} = 32 + 1 - \frac{30}{2} = 33 - 15 = 18$
   • Alternatively, the formation of the ether bridge creates one new ring. Each new ring increases the degree of unsaturation by one. Therefore, the DoU of Q is one more than the DoU of the cyclic tetramer intermediate. $DoU_Q = DoU_{intermediate} + 1 = 17 + 1 = 18$

Thus, the degree of unsaturation of compound Q is 18.