Practical Organic Chemistry

Acetate ions gives deep red colour ppt on reaction with netural ferric chloride solution.

$6 \mathrm{CH}_{3} \mathrm{COO}^{-}+3 \mathrm{Fe}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow {\left[\mathrm{Fe}_{3}(\mathrm{OH})_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{6}\right]^{+} + 2 \mathrm{H}^{+}}$ ${\left[\mathrm{Fe}_{3}(\mathrm{OH})_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{6}\right]^{+} 4 \mathrm{H}_{2} \mathrm{O} \longrightarrow} \left.\left.\underset{\text{Brown red ppt}}{3\left[\mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right)\right]}+3 \mathrm{CH}_{3} \mathrm{COOH}\right)_{2}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{6}\right]^{+}+2 \mathrm{H}^{+}$ Phenol reacts with freshly prepared ferric chloride solution and gives violet colour complex.

Sulphanilic acid is p-amino benzene sulphonic acid Since it contain both N and S so it give red colour in Lassaigne's test.

In dumas method for the estimation of N2, the sample is heated with copper oxide and the gas evolved is passed over copper gauze.

$\mathrm{CH}_{3} \mathrm{NH}_{2} \cdot \mathrm{HCl}$ will give positive Lassaigne's test for both nitrogen and halogen.

$\mathrm{X} \stackrel{\mathrm{Na}_2 \mathrm{O}_2}{\longrightarrow} \mathrm{Y} \stackrel{\mathrm{BaCl}_2}{\longrightarrow} \underset{\left[\mathrm{BaSO}_4\right]}{\mathrm{Z}}$ Methionine: $\mathrm{C}_5 \mathrm{H}_{11} \mathrm{NO}_2 \mathrm{S}$

Statement I: Potassium dichromate (K$_2$Cr$_2$O$_7$) is indeed commonly used as a primary standard in volumetric (titrimetric) analysis.

It is a powerful oxidizing agent and has a high equivalent weight.

Furthermore, it is readily available, it can be obtained in a high state of purity, it is not hygroscopic, and its solutions are stable.

However, the reason it is preferred over sodium dichromate (Na$_2$Cr$_2$O$_7$) is not due to its solubility, but rather its stability and molar mass.

Statement II: Typically, the solubility of salts increases with the size of the cation.

Sodium ions (Na$^+$) are smaller than potassium ions (K$^+$), so sodium salts tend to be more soluble in water than corresponding potassium salts.

Thus, we would expect sodium dichromate (Na$_2$Cr$_2$O$_7$) to be more soluble in water than potassium dichromate (K$_2$Cr$_2$O$_7$).

Therefore, the correct answer is Option A: Statement I is true but Statement II is false.

Phenolphthalein is a commonly used acid/base indicator that can be synthesized from phthalic anhydride and phenol (carbolic acid) in the presence of concentrated sulfuric acid (H₂SO₄).

Therefore, compound 'X' is phenol and compound 'Y' is phenolphthalein.

Statement I: Methyl orange is a weak base, not a weak acid.

This is because Methyl orange is actually a pH indicator commonly used in titrations.

In its deprotonated form, it appears yellow, and in its protonated form, it appears red.

Methyl orange has the ability to accept a proton ($H^+$) from water (a characteristic of a base) and is therefore considered a weak base.

Chemically, this can be represented as:

Here, $\text{MO}^-$ is the methyl orange anion, $\text{MOH}$ is the protonated form of methyl orange, and $\text{OH}^-$ is the hydroxide ion.

Statement II: The quinonoid form of methyl orange is more intensely colored than the benzenoid form.

Methyl orange exists in two resonance forms: a benzenoid form (yellow) and a quinonoid form (red).

In acidic solution, the equilibrium shifts towards the quinonoid form, which is red and more intensely colored than the yellow benzenoid form.

The equilibrium can be represented as follows: Yellow Benzenoid form $\leftrightharpoons$ Red Quinonoid form In conclusion, both statements are incorrect.

Methyl orange is a weak base and the quinonoid form is more intensely colored.

So, the correct answer is option A.

Both Statement I and Statement II are incorrect.

Organic compounds run slower than solvent in thin layer chromatography.

is not an integral value. It is the ratio of distance travelled by the organic compound to that of solvent.

value of a polar compound is smaller than that of non-polar compound as the polar compound is retained more by the adsorbent than non-polar compound.

The correct option here is : Option A : catalytic agent In the Kjeldahl method for the estimation of nitrogen, $\mathrm{CuSO}_4$ or copper sulfate acts as a catalytic agent.

The method consists of three main steps: digestion, neutralization, and titration.

During the digestion phase, the organic substance containing nitrogen is heated with concentrated sulfuric acid ($\mathrm{H_2SO_4}$), which digests or breaks down the organic matrix, releasing the nitrogen in the form of ammonium sulfate ($\mathrm{(NH_4)_2SO_4}$).

Organic compound $+\mathrm{H}_2 \mathrm{SO}_4 \xrightarrow{\mathrm{CuSO}_4}\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$ Catalysts such as $\mathrm{CuSO}_4$, along with other salts like potassium sulfate ($\mathrm{K_2SO_4}$), are added to the reaction mixture to raise the boiling point of the sulfuric acid and to accelerate the oxidation of the organic material.

This increases the rate of the digestion process, ensuring that the nitrogen within the sample is fully converted to ammonium ions ($\mathrm{NH_4}^+$).

The catalyst itself is not consumed in the reaction and does not become a part of the product; its role is to increase the reaction rate.

After digestion, the mixture is neutralized with a base, and then the liberated ammonia is distilled off and quantitatively measured, typically by titration with a standard acid solution.

Therefore, $\mathrm{CuSO}_4$ serves as a catalyst in this chemical analysis technique for quantitatively determining the nitrogen content of organic substances.