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(C) Step $1$: Cyclotrimerization of $15$ moles of acetylene $(HC \equiv CH)$ gives benzene $(A)$. Since $3$ moles of acetylene form $1$ mole of benzen

Vedclass · Accepted Answer

$136$

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(C) Step $1$: Cyclotrimerization of $15$ moles of acetylene $(HC \equiv CH)$ gives benzene $(A)$. Since $3$ moles of acetylene form $1$ mole of benzene,$15$ moles of acetylene would theoretically form $5$ moles of benzene. Given an $80\%$ yield,the actual amount of benzene $(A)$ formed is $5 	imes 0.8 = 4$ moles.Step $2$: Friedel-Crafts alkylation of $4$ moles of benzene with isopropyl chloride gives cumene $(B)$. Given a $50\%$ yield,the amount of cumene $(B)$ formed is $4 	imes 0.5 = 2$ moles.Step $3$: Cumene hydroperoxide rearrangement of $2$ moles of cumene gives phenol $(C)$. Given a $50\%$ yield,the amount of phenol $(C)$ formed is $2 	imes 0.5 = 1$ mole.Step $4$: Acetylation of $1$ mole of phenol with acetyl chloride $(CH_3COCl)$ in the presence of pyridine gives phenyl acetate $(D)$. Given a $100\%$ yield,the amount of phenyl acetate $(D)$ formed is $1$ mole.The molar mass of phenyl acetate $(C_8H_8O_2)$ is $(8 	imes 12) + (8 	imes 1) + (2 	imes 16) = 96 + 8 + 32 = 136 \ g/mol$.Therefore,the mass of $1$ mole of $D$ is $136 \ g$.

If the reaction sequence given below is carried out with $15$ moles of acetylene,the amount of the product $D$ formed (in $g$) is. . . . . The yields of $A, B, C$ and $D$ are given in parentheses. [Given: Atomic mass of $H=1, C=12, O=16, Cl=35$]

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