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JEE Advanced 2021 · Paper 1, Chemistry question 5

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Question Stem for Question Nos. 5 and 6

For the following reaction scheme, percentage yields are given along the arrow:

MgX2CX3→HX2OP\ce{Mg2C3 ->[H2O] P} (4.0 g); P→75 %NaNHX2,MeIQ→40 %red hot iron tube,873 KR\ce{P ->[NaNH2, MeI][75\%] Q ->[red hot iron tube, 873 K][40\%] R} (x\mathbf x g)

P→100 %HgX2+/HX+,333 KS→80 %Ba(OH)X2,heatT→80 %NaOClU\ce{P ->[Hg^{2+}/H+, 333 K][100\%] S ->[Ba(OH)2, heat][80\%] T ->[NaOCl][80\%] U} (y\mathbf y g; decolourises Baeyer's reagent)

x\mathbf x g and y\mathbf y g are mass of R\mathbf R and U\mathbf U, respectively.

(Use: Molar mass (in g mol−1^{-1}) of H, C and O as 1, 12 and 16, respectively)

Scheme: Mg2C3 + H2O gives P (4.0 g). P with NaNH2/MeI gives Q (75%); Q through a red hot iron tube at 873 K gives R (x g, 40%). P with Hg2+/H+ at 333 K gives S (100%); S with Ba(OH)2 and heat gives T (80%); T with NaOCl gives U (y g, 80%), which decolourises Baeyer's reagent.

The value of x\mathbf x is ____.

Answer and worked solution

Official answer: 1.62

Identify the compounds.

  • MgX2CX3+4 HX2O→2 Mg(OH)X2+CHX3C≡CH\ce{Mg2C3 + 4H2O -> 2Mg(OH)2 + CH3C#CH}, so P is propyne: 4.040=0.1\frac{4.0}{40}=0.1 mol.
  • NaNHX2\ce{NaNH2} then MeI methylates it to but-2-yne, Q: 0.1×0.75=0.0750.1\times0.75=0.075 mol.
  • A red hot iron tube trimerises the alkyne: 3 CHX3C≡CCHX3→CX6(CHX3)X6\ce{3 CH3C#CCH3 -> C6(CH3)6}. R is hexamethylbenzene, M=162M=162.

Mass of R.
nR=0.0753×0.40=0.010 mol,x=0.010×162=1.62 g.n_R=\frac{0.075}3\times0.40=0.010\text{ mol},\qquad x=0.010\times162=1.62\text{ g}.

Topics:AlkynesStoichiometryAromatic hydrocarbons and electrophilic substitution
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