JEE Advanced 2021 · Paper 1 · Chemistry Q6

ChemistryNumerical+2Question 25 of 57

Passage

Question Stem for Question Nos. 5 and 6

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

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

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

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

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

The value of $\mathbf y$ is ____.

Your answer

Answer and worked solution

Official answer: 3.2

Identify the compounds.

$\ce{Hg^{2+}/H+}$ hydrates propyne to acetone, S: 0.1 mol.
$\ce{Ba(OH)2}$ and heat bring about an aldol condensation, $\ce{2CH3COCH3 -> (CH3)2C=CHCOCH3 + H2O}$ . T is mesityl oxide: $\frac{0.1}2\times0.8=0.04$ mol.
NaOCl gives the haloform reaction on the methyl ketone. U is $\ce{(CH3)2C=CHCOOH}$ ( $M=100$ ), whose C=C decolourises Baeyer's reagent: $0.04\times0.8=0.032$ mol.

Mass of U. $y=0.032\times100=3.20$ g. (The key also accepts 3.90-3.91.)

JEE Advanced 2021 · Paper 1, Chemistry question 6