Question

Benzene is subjected to the following sequence of reactions:
\[ \text{Benzene} \xrightarrow[\text{}]{\text{1) } CH_3Cl/AlCl_3} \xrightarrow[\text{}]{\text{2) } Cl_2/AlCl_3} \xrightarrow[\text{}]{\text{3) } K_2Cr_2O_7/H^+} X \] If \(p\) gram of \(X\) reacts with \(NaHCO_3\), then \(11.2\ dm^3\) of gas is obtained at STP. Calculate \(p\).

Hint:

Whenever an alkyl benzene is oxidized with strong oxidizing agents like \(K_2Cr_2O_7/H^+\), the entire side chain is converted into \(COOH\), provided at least one benzylic hydrogen is present.

Answer 1

Correct Answer: 78.25

Step 1: Identify the product formed in each reaction.
In the first step, benzene reacts with \(CH_3Cl/AlCl_3\) by Friedel-Crafts alkylation and forms toluene.
In the second step, toluene reacts with \(Cl_2/AlCl_3\). Since the methyl group is an ortho-para directing group, chlorination takes place mainly at the ortho and para positions to form o-chlorotoluene and p-chlorotoluene.
In the third step, oxidation with \(K_2Cr_2O_7/H^+\) converts the side-chain \(CH_3\) group into \(COOH\). Therefore, \(X\) is a mixture of o-chlorobenzoic acid and p-chlorobenzoic acid.
Step 2: Write the reaction of \(X\) with sodium bicarbonate.
Both o-chlorobenzoic acid and p-chlorobenzoic acid are monocarboxylic acids, so each mole of acid reacts with one mole of \(NaHCO_3\) to produce one mole of \(CO_2\):
\[ RCOOH + NaHCO_3 \rightarrow RCOONa + H_2O + CO_2 \] Thus, 1 mole of \(X\) gives 1 mole of \(CO_2\).
Step 3: Calculate moles of gas evolved.
At STP, \(22.4\ dm^3\) of a gas corresponds to \(1\) mole.
Given volume of gas \(= 11.2\ dm^3\)
\[ \text{Moles of } CO_2 = \frac{11.2}{22.4} = 0.5 \] So, moles of \(X\) reacted \(= 0.5\) mole.
Step 4: Find the molar mass of \(X\).
The molecular formula of chlorobenzoic acid is:
\[ C_7H_5ClO_2 \] Its molar mass is:
\[ (7 \times 12) + (5 \times 1) + 35.5 + (2 \times 16) \] \[ = 84 + 5 + 35.5 + 32 = 156.5\ g/mol \] Step 5: Calculate the mass \(p\).
\[ p = \text{moles} \times \text{molar mass} \] \[ p = 0.5 \times 156.5 = 78.25\ g \] Final Answer:
\[ \boxed{p = 78.25\ g} \]