Step 1: Understand the chemical properties of the given compounds.
Chlorobenzene is a neutral, non-polar organic compound. It is insoluble in water but soluble in organic solvents like ethyl acetate. It does not react with NaOH.
Aniline is a weak organic base due to the lone pair of electrons on the nitrogen atom. It can react with strong acids to form salts. Its basicity is not strong enough to react with NaOH in a significant way to form a water-soluble salt under these conditions. Aniline is sparingly soluble in water but soluble in organic solvents like ethyl acetate.
Benzoic acid is a weak organic acid containing a carboxyl group (-COOH). It can react with strong bases like NaOH to form a water-soluble sodium benzoate salt.
Step 2: Analyze the reaction with NaOH.
When NaOH (an aqueous base) is added to the ethyl acetate solution containing chlorobenzene, aniline, and benzoic acid, a reaction will occur with the acidic component, benzoic acid: \[ \text{C}_6\text{H}_5\text{COOH} (in \, ethyl \, acetate) + \text{NaOH} (aq) \rightarrow \text{C}_6\text{H}_5\text{COONa} (aq) + \text{H}_2\text{O} \] The product, sodium benzoate, is an ionic salt and is soluble in the aqueous layer (NaOH solution) but insoluble in the organic layer (ethyl acetate).
Chlorobenzene, being neutral and non-polar, will not react with NaOH and will remain dissolved in the ethyl acetate layer. Aniline, being a weak base, will not react significantly with the strong base NaOH. It will mostly remain in its molecular form. Aniline has some solubility in water (though limited) and is also soluble in ethyl acetate. In a separating funnel, it will be distributed between the aqueous and organic layers based on its partition coefficient. However, since it did not undergo a chemical transformation into a highly water-soluble ionic species, a significant portion of aniline will likely remain in the ethyl acetate layer.
Step 3: Determine the contents of the ethyl acetate layer after separation.
After shaking the separating funnel and allowing the layers to separate, the aqueous layer (containing NaOH and sodium benzoate) will be at the bottom (due to higher density). The ethyl acetate layer (organic layer) will be at the top.
The compounds that would have preferentially dissolved in the ethyl acetate layer are those that are non-polar or have limited reactivity with NaOH and limited solubility in the aqueous phase under these conditions. These are chlorobenzene and aniline. Benzoic acid would have been converted to its sodium salt and moved into the aqueous layer. Therefore, the ethyl acetate layer will primarily contain chlorobenzene and aniline.
Choose the correct sets with respective observations:
(1) \( \text{CuSO}_4 \) (acidified with acetic acid) + \( K_2\text{Fe(CN)}_6 \) (neutralized with NaOH) → Blue precipitate
(2) \( 2\text{CuSO}_4 \) + \( K_2\text{Fe(CN)}_6 \) → Blue precipitate
(3) \( 4\text{FeCl}_3 \) + \( 3\text{K}_4\text{Fe(CN)}_6 \) → \( \frac{1}{2}K_4\text{Fe(CN)}_6 \)
(4) \( 37\text{Cl}_2 \) + \( 2\text{KFe(CN)}_6 \) → 6KC1
In the light of the above options, choose the correct set:
Identify the structure of the final product (D) in the following sequence of the reactions :
Total number of $ sp^2 $ hybridised carbon atoms in product D is _____.
A molecule with the formula $ \text{A} \text{X}_2 \text{Y}_2 $ has all it's elements from p-block. Element A is rarest, monotomic, non-radioactive from its group and has the lowest ionization energy value among X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
A transition metal (M) among Mn, Cr, Co, and Fe has the highest standard electrode potential $ M^{n}/M^{n+1} $. It forms a metal complex of the type $[M \text{CN}]^{n+}$. The number of electrons present in the $ e $-orbital of the complex is ... ...
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.
Consider the following equilibrium, $$ \text{CO(g)} + \text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} $$ 0.1 mol of CO along with a catalyst is present in a 2 dm$^3$ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH$_3$OH is formed. The $ K_p $ is ...... x $ 10^7 $ (nearest integer).
Given: $ R = 0.08 \, \text{dm}^3 \, \text{bar} \, \text{K}^{-1} \, \text{mol}^{-1} $
Assume only methanol is formed as the product and the system follows ideal gas behavior.