Acetylene is passed through red hot iron tube at 873 K to produce P. P reacts with chlorine gas in the presence of anhydrous $\text{AlCl}_3$ to produce Q. Q reacts with benzyl chloride and sodium to produce R as the major product. The structure of R is:
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The Wurtz-Fittig reaction is highly effective for coupling aryl halides with alkyl or aralkyl halides using sodium in dry ether to form alkyl-substituted aromatic compounds.
Step 1: Trace the first product P. Acetylene passed through a red hot iron tube at 873 K undergoes trimerization, meaning three acetylene molecules join into a ring. Three two carbon units make a six carbon ring, which is benzene. So P is benzene.
Step 2: Trace the second product Q. Benzene reacts with chlorine in the presence of anhydrous $\text{AlCl}_3$. This is an electrophilic substitution that places one chlorine on the ring, giving chlorobenzene. So Q is chlorobenzene.
Step 3: Set up the third step. Q (chlorobenzene) now reacts with benzyl chloride and sodium metal. Sodium with two organic chlorides is the Wurtz type (Wurtz-Fittig) coupling, which joins the two carbon fragments together by knocking out the two chlorines as NaCl.
Step 4: Decide what joins to what. In Wurtz-Fittig coupling, the aryl part from chlorobenzene (a phenyl group) couples with the benzylic part from benzyl chloride ($\text{C}_6\text{H}_5\text{CH}_2$). Joining a phenyl group to a benzyl group gives $\text{C}_6\text{H}_5\text{CH}_2\text{C}_6\text{H}_5$.
Step 5: Name that product. $\text{C}_6\text{H}_5\text{CH}_2\text{C}_6\text{H}_5$ is two benzene rings linked through a single $\text{CH}_2$ group. That molecule is diphenylmethane.
Step 6: Reject the others. Biphenyl would need two phenyls joined directly with no carbon between them, and the chloro or ester products do not arise from this coupling. So R is diphenylmethane.
