JEE Practical Organic Chemistry PYQs With Video Solutions PDF

In Lassaigne's test, Nitrogen present in an organic compound is converted into Sodium Cyanide ($$NaCN$$) during sodium fusion. On treatment with $$FeSO_4$$ followed by acidification, a Prussian blue complex, $$Fe_4[Fe(CN)_6]_3$$, is formed. Therefore, Nitrogen gives a Prussian blue coloration.

Hence:

$$I \rightarrow Q$$

Sulphur present in the organic compound is converted into Sodium Sulphide ($$Na_2S$$). When treated with Sodium nitroprusside, a violet-colored complex is formed.

Hence:

$$II \rightarrow S$$

When both Nitrogen and Sulphur are present together, sodium fusion produces Sodium Thiocyanate ($$NaSCN$$). On treatment with $$Fe^{3+}$$ ions, a blood red-colored ferric thiocyanate complex is formed.

Hence:

$$III \rightarrow P$$

Phosphorus is converted into phosphate ions during the test. The phosphate solution, when treated with Ammonium molybdate in the presence of nitric acid, produces a canary yellow precipitate of Ammonium phosphomolybdate.

Hence:

$$IV \rightarrow R$$

Therefore, the correct matching is $$I-Q,\ II-S,\ III-P,\ IV-R$$

Hence, the correct answer is $$(B)$$ $$I-Q,\ II-S,\ III-P,\ IV-R$$

The percentage of sulphur is calculated using the Carius method formula:

$$%\text{S} = \frac{32}{233} \times \frac{\text{Mass of }BaSO_4}{\text{Mass of organic compound}} \times 100$$

Substituting the given values, we get

=$$\frac{32}{233}\times\ \frac{0.4893}{0.160}\times\ 100\ \%\ $$

Which is $$42\%$$

Therefore, the percentage of sulphur present in the compound is, $$42$$

When an organic compound contains both nitrogen and sulfur, sodium fusion generally forms sodium thiocyanate.

$$Na + C + N + S \rightarrow NaSCN$$

On treatment with (FeCl_3), thiocyanate ions produce a blood-red complex.

$$Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+}$$

When excess sodium is used during fusion, the hot sodium decomposes sodium thiocyanate.

$$NaSCN + 2Na \xrightarrow{\Delta} NaCN + Na_2S$$

As a result, (SCN^-) ions are destroyed and the blood-red colour disappears. The cyanide ions formed give a positive nitrogen test, producing the characteristic Prussian blue colour.

$$Fe_4[Fe(CN)_6]_3$$

Hence, the correct answer is option (c).

Mass of carbon in $$1.76\text{ g}$$ of $$CO_2$$:

$$\text{Mass of C}=\frac{12}{44}\times1.76=0.48\text{ g}$$

$$\%C=48\%$$

Mass of bromine in $$0.94\text{ g}$$ of $$AgBr$$:

$$\text{Mass of Br}=\frac{80}{188}\times0.94=0.40\text{ g}$$

$$\%Br=\frac{0.40}{0.94}\times100\approx42.6\%$$

Therefore,

$$\%H=100-(48+42.6)=9.4\%$$

Nearest integer:

$$\boxed{9}$$

The initial amount of $$H_2SO_4$$ taken is:

$$\text{Moles of }H_2SO_4=0.5\times\frac{50}{1000}=0.025\text{ mol}$$

Since one mole of $$H_2SO_4$$ furnishes two moles of $$H^+$$, the total acid equivalents are:

$$0.025\times2=0.050\text{ mol of }H^+$$

The residual acid is neutralized by $$NaOH$$.

$$\text{Moles of }NaOH=0.5\times\frac{60}{1000}=0.030\text{ mol}$$

Thus, the remaining acid equivalents are:

$$0.030\text{ mol}$$

Therefore, acid consumed by ammonia is:

$$0.050-0.030=0.020\text{ mol}$$

Since:

$$NH_3+H^+\rightarrow NH_4^+$$

moles of $$NH_3$$ evolved are:

$$0.020\text{ mol}$$

Hence, moles of nitrogen present are also:

$$0.020\text{ mol}$$

Mass of nitrogen:

$$0.020\times14=0.28\text{ g}$$

Percentage of nitrogen:

$$\%N=\frac{0.28}{0.5}\times100=56\%$$

Hence, the percentage of nitrogen in the compound is:

$$\boxed{56\%}$$

Let us calculate the mass percentage of bromine using the quantitative principles of the Carius method:

Step 1: Identify Given Experimental Data

• Mass of the organic compound ($$w$$) = $$0.40\text{ g}$$
• Mass of the silver bromide precipitate ($$w_1$$) = $$0.376\text{ g}$$
• Molar mass of Bromine ($$\text{Br}$$) = $$80\text{ g mol}^{-1}$$
• Molar mass of Silver ($$\text{Ag}$$) = $$108\text{ g mol}^{-1}$$

Step 2: Calculate the Molar Mass of Silver Bromide ($$\text{AgBr}$$)
$$\text{Molar mass of AgBr} = \text{Atomic mass of Ag} + \text{Atomic mass of Br}$$
$$\text{Molar mass of AgBr} = 108 + 80 = 188\text{ g mol}^{-1}$$

Step 3: Establish Stoichiometric Relationship
From the molecular formula, $$1\text{ mole}$$ of $$\text{AgBr}$$ ($188\text{ g}$) contains exactly $$1\text{ mole}$$ of Bromine atoms ($80\text{ g}$).
Therefore, the mass of bromine contained inside $$0.376\text{ g}$$ of the collected $$\text{AgBr}$$ precipitate is:
$$\text{Mass of Bromine} = \frac{80}{188} \times 0.376\text{ g}$$
$$\text{Mass of Bromine} = 80 \times 0.002 = 0.16\text{ g}$$

Step 4: Calculate Mass Percentage of Bromine
The mass fraction of the element relative to the total mass of the starting organic sample is expressed as:
$$\text{Mass \% of Bromine} = \left( \frac{\text{Mass of Bromine}}{\text{Total mass of Organic Compound}} \right) \times 100$$
$$\text{Mass \% of Bromine} = \left( \frac{0.16\text{ g}}{0.40\text{ g}} \right) \times 100$$
$$\text{Mass \% of Bromine} = \frac{16}{40} \times 100 = 0.40 \times 100 = 40\%$$

Conclusion:
The mass percentage of bromine in the given organic compound is exactly 40%.

Answer:
The correct integer value is 40 .

Let us analyze the chemical principles behind Kjeldahl's method step by step:

Step 1: Digestion Phase (Conversion of Nitrogen)
When an organic compound containing nitrogen is heated strongly with concentrated sulfuric acid ($$\text{H}_2\text{SO}_4$$) in the presence of a catalyst (such as copper sulfate or mercury), the carbon and hydrogen are oxidized to $$\text{CO}_2$$ and $$\text{H}_2\text{O}$$. Concurrently, the nitrogen present in the organic framework is quantitatively converted into ammonium sulfate :
$$\text{Organic Compound Nitrogen} \xrightarrow{\text{Conc. H}_2\text{SO}_4} (\text{NH}_4)_2\text{SO}_4$$

Step 2: Subsequent Phases (Distillation and Titration)

• The digested mixture containing $$(\text{NH}_4)_2\text{SO}_4$$ is then treated with an excess of a strong alkali like sodium hydroxide ($$\text{NaOH}$$) to liberate ammonia gas ($$\text{NH}_3$$).
• The liberated $$\text{NH}_3$$ gas is distilled and absorbed into a known excess volume of a standard acid solution (like $$\text{HCl}$$ or $$\text{H}_3\text{BO}_3$$) to be measured via titration.

Distinction from other tests:
$$\text{NaCN}$$ is formed during the sodium fusion step of Lassaigne's test , not Kjeldahl's method.

Conclusion:
Since the question asks specifically for the chemical species formed during the digestion phase within the reaction mixture, the answer is ammonium sulfate.

Answer:
The correct choice is Option A .

sasasa

When both nitrogen and sulphur are present, sodium fusion forms sodium thiocyanate:

$$\mathrm{Na + C + N + S \xrightarrow{\Delta} NaSCN}$$

On treatment with $$\mathrm{FeCl_3}$$, thiocyanate ions form a blood-red ferrithiocyanate complex:

$$\mathrm{Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+}}$$

Hence, the species responsible for the blood-red colour is:

$$\mathrm{[Fe(SCN)]^{2+}}$$

Correct Option: $$\mathrm{(B)}$$

sasasa

1. Formation of Sodium Fusion Extract:
When an organic compound contains both Nitrogen ($$\text{N}$$) and Sulphur ($$\text{S}$$), fusing it with sodium metal converts these elements into sodium thiocyanate ($$\text{NaSCN}$$) instead of separate cyanide ($$\text{NaCN}$$) and sulphide ($$\text{Na}_2\text{S}$$) salts:
$$\text{Na} + \text{C} + \text{N} + \text{S} \xrightarrow{\Delta} \text{NaSCN}$$

2. Reaction with Ferric Chloride:
When ferric chloride ($$\text{FeCl}_3$$) is added to the acidified extract, the ferric ions ($$\text{Fe}^{3+}$$) react with the thiocyanate ions ($$\text{SCN}^-$$) to form a blood-red colored coordination complex ion:
$$\text{Fe}^{3+} + \text{SCN}^- \rightarrow [\text{Fe}(\text{SCN})]^{2+}$$

3. Note on Excess Sodium:
If excess sodium metal is used during fusion, the thiocyanate decomposes into separate cyanide and sulphide ions, which would instead yield a Prussian blue color ($$\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$$) upon testing.

Answer:
The complex ion responsible for the blood-red coloration is $$[\text{Fe}(\text{SCN})]^{2+}$$ . (Option C)