Carbonyl Compounds, Carboxylic Acids…

JEE Advanced 2023 Paper 2 Chemistry: Paragraph 2

Acetophenone (

\mathrm{C_6H_5COCH_3}

, molar mass

120\ \mathrm{g\,mol^{-1}}

) is warmed with excess iodine and aqueous sodium hydroxide:

\mathrm{C_6H_5COCH_3+3I_2+4NaOH\rightarrow C_6H_5COONa+CHI_3+3NaI+3H_2O}

. A sample of

12.0\ \mathrm{g}

of acetophenone is used and the reaction proceeds in

80\%

yield with respect to acetophenone for each of the two organic products. The yellow solid

\mathrm{CHI_3}

is filtered off, and the sodium benzoate formed is isolated in pure form and dissolved in water to give

500\ \mathrm{mL}

of solution at

298\ \mathrm{K}

. Take atomic masses

\mathrm{C}=12

\mathrm{H}=1

\mathrm{I}=127

\mathrm{g\,mol^{-1}}

;

K_a(\mathrm{C_6H_5COOH})=6.4\times10^{-5}

and

K_w=1.0\times10^{-14}

Question 1

Numerical answer

The mass of iodoform obtained (in g), correct to two decimal places, is

Show answer and solution

Answer: 31.52

Moles of acetophenone:

\dfrac{12.0}{120}=0.100\ \mathrm{mol}

. One mole of acetophenone gives one mole of

\mathrm{CHI_3}

(the haloform reaction cleaves the

\mathrm{COCH_3}

group,

\mathrm{CH_3}

becoming

\mathrm{CI_3}

and then

\mathrm{CHI_3}

) in the balanced equation. At

80\%

yield,

n(\mathrm{CHI_3})=0.100\times0.80=0.0800\ \mathrm{mol}

. Molar mass of

\mathrm{CHI_3}=12+1+3(127)=394\ \mathrm{g\,mol^{-1}}

. Mass

=0.0800\times394=31.52\ \mathrm{g}

. The

80\%

yield applies to each organic product, so the same

0.0800\ \mathrm{mol}

of sodium benzoate is also formed; this is used in the next question.

Numerical answer

The pH of the sodium benzoate solution, correct to two decimal places, is

Show answer and solution

Answer: 8.70

Sodium benzoate formed:

0.0800\ \mathrm{mol}

(same

80\%

yield as iodoform, one mole per mole of acetophenone). In

500\ \mathrm{mL}

the concentration is

c=\dfrac{0.0800}{0.500}=0.160\ \mathrm{M}

. The benzoate ion is the conjugate base of a weak acid and hydrolyses:

\mathrm{C_6H_5COO^-+H_2O\rightleftharpoons C_6H_5COOH+OH^-}

with

K_h=\dfrac{K_w}{K_a}=\dfrac{1.0\times10^{-14}}{6.4\times10^{-5}}=1.5625\times10^{-10}

. Since

K_h\ll c

[\mathrm{OH^-}]=\sqrt{K_hc}=\sqrt{1.5625\times10^{-10}\times0.160}=\sqrt{2.5\times10^{-11}}=5.0\times10^{-6}\ \mathrm{M}

. Then

\mathrm{pOH}=-\log(5.0\times10^{-6})=5.301

and

\mathrm{pH}=14-5.301=8.699\approx8.70

Original question written in the JEE Advanced 2023 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake