QJEEAdv21Ch55e5: JEE Advanced 2021 Chemistry - Class 11 JEEAdv21Ch55e5 Solution: For the following reaction scheme, percentage yields are given…

Mg₂C₃ reacts with water to give propyne.

$M{g_2}{C_3} + 4{H_2}O \to 2Mg{(OH)_2} + C{H_3} - \mathop {\mathop C\limits_{(P)} }\limits_{{\mathop{\rm Propyne}\nolimits} } \equiv CH$

Sodium amide (NaNH₂) takes proton from alkyne to form acetylide ion which, attacks on Me $-$ I to give an alkyne with one more carbon.


Mass of P formed = 4.0 g

Molar mass of P formed = 40 g mol ^{$-$ 1}

Number of moles of P = ${{Mass\,of\,P(W)} \over {Molar\,mass\,of\,P(M)}} = {4 \over {40}} = 0.1$ mol

1 mole of P forms 75% i.e., 0.75 = 0.075 mole of Q

Polymerisation of alkyne (Butyne) occurs in presence of red hot iron tube to form benzene derivative.


3 moles of alkyne forms 40% i.e., 0.40 mole of benzene derivative, R.

0.075 mole of Q forms ${{0.40} \over 3} \times 0.075 = 0.01$ mole of R.

Molar mass of R = 162 g/mol

Mass of R formed or value of x = Molar mass $\times$ Number of moles = 162 $\times$ 0.01 = 1.62 g