Practical Organic Chemistry

Mass of compound taken $= 0.314\ \text{g}$

Mass of $BaSO_4$ obtained $= 0.4813\ \text{g}$

In $BaSO_4$, sulphur present per mole $= 32\ \text{g}$

Molar mass of $BaSO_4 = 233\ \text{g mol}^{-1}$

So, mass of sulphur in $0.4813\ \text{g}$ of $BaSO_4$ is:

$ m(S)=0.4813\times\frac{32}{233} $

$ m(S)=0.4813\times\frac{32}{233}=0.06608\ \text{g (approx.)} $

Percentage of sulphur in the compound:

$ \%S=\frac{0.06608}{0.314}\times 100 \approx 21.05\% $

Correct option: B) $21.05\%$

Mass of organic compound taken $=0.2425\,\text{g}$

Mass of $\text{AgCl}$ obtained $=0.5253\,\text{g}$

In Carius method for estimation of halogens, the halogen present in the organic compound is converted into silver halide. Here chlorine is obtained as $\text{AgCl}$.

In $\text{AgCl}$, the fraction of chlorine is:

$\frac{\text{Atomic mass of Cl}}{\text{Molar mass of AgCl}}=\frac{35.5}{143.5}$

So, mass of chlorine in $0.5253\,\text{g}$ of $\text{AgCl}$ is:

$\frac{35.5}{143.5}\times 0.5253$

Therefore, percentage of chlorine in the compound is:

$ \begin{aligned} & \text { Percentage of } \mathrm{Cl}=\frac{35.5}{143.5} \times \frac{0.5253}{0.2425} \times 100 \\ & =53.58 \% \end{aligned} $

For each functional group, recall the standard qualitative test observation (NCERT):

A. Unsaturation (Baeyer’s test)

Baeyer’s reagent is cold, dilute alkaline $KMnO_4$ (pink/purple). With $C=C$ / $C\equiv C$, the pink colour gets discharged.

So, A $\rightarrow$ IV.

B. Alcoholic group (Ceric ammonium nitrate test)

Alcohols give red colour with ceric ammonium nitrate solution.

So, B $\rightarrow$ I.

C. Aldehyde group (Tollen’s reagent)

Aldehydes reduce Tollen’s reagent to metallic silver, giving a silver mirror.

So, C $\rightarrow$ II.

D. Phenolic group ($FeCl_3$ test)

Phenols give violet colour with neutral $FeCl_3$.

So, D $\rightarrow$ III.

Hence the correct matching is:

$A\text{-}IV,\; B\text{-}I,\; C\text{-}II,\; D\text{-}III$

Correct option: A

In Kjeldahl’s method, the ammonia ($\mathrm{NH_3}$) evolved neutralizes the acid taken. So,

$\text{equivalents of } \mathrm{H_2SO_4}=\text{equivalents of ammonia}$

Given: $15\,\text{mL}$ of $1\,\text{M } \mathrm{H_2SO_4}$ is neutralized.

$\mathrm{H_2SO_4}$ is a dibasic acid, so $1$ mole of $\mathrm{H_2SO_4}$ gives $2$ equivalents. Therefore, equivalents of $\mathrm{H_2SO_4}$ used are:

$\frac{15 \times 1 \times 2}{1000}$

Ammonia is monobasic (one $\mathrm{NH_3}$ neutralizes one equivalent of acid), so:

$\frac{15 \times 1 \times 2}{1000}=\text{moles of ammonia} \times 1$

So, moles of ammonia $=$ moles of nitrogen ($\mathrm{N}$), because each mole of $\mathrm{NH_3}$ contains $1$ mole of $\mathrm{N}$.

Now, mass of nitrogen:

$\text{Weight of nitrogen }=\frac{15 \times 1 \times 2}{1000} \times 14=0.42$

Percentage of nitrogen in $1\,\text{g}$ compound:

$\% \text{ weight of ' } \mathrm{N} \text { '}=\frac{0.42}{1} \times 100=42 \%$

Moles of $\text{BaSO}_4$ formed are

$n(\text{BaSO}_4)=\frac{1.2}{233}\ \text{mol}$

In $\text{BaSO}_4$, there is 1 atom of S per formula unit, so

$n(\text{S})=n(\text{BaSO}_4)=\frac{1.2}{233}\ \text{mol}$

Mass of sulphur in the compound:

$m(\text{S})=n(\text{S})\times 32=\frac{1.2}{233}\times 32=0.1648\ \text{g}$

Percentage of sulphur in the organic compound (mass of compound $=0.75\ \text{g}$):

$\%\,\text{S}=\frac{0.1648}{0.75}\times 100=21.97\%$

So, the correct option is D) $21.97\%$.

Organic layer in funnel are mixture of chloro benzene and aniline

(A) Aniline $-\mathrm{H}_2 \mathrm{O}$ : Steam Distillation

(B) Glycerol from spent-lye in soap industry

$-$ Distillation under reduced pressure

(C) Different fraction of crude oil in petroleum industry - Fractional distillation

(D) $\mathrm{CHCl}_3-$ Aniline - Simple distillation

To determine the percentage composition of nitrogen using Dumas' method, follow these calculations:

Calculate the Pressure of Nitrogen Gas:

$ \text{Pressure of } \mathrm{N}_2 = (715 \text{ mmHg} - 15 \text{ mmHg}) = 700 \text{ mmHg} $

Calculate the Moles of Nitrogen ($\mathrm{n}_{\mathrm{N}_2}$):

$ \mathrm{n}_{\mathrm{N}_2} = \frac{\mathrm{PV}}{\mathrm{RT}} $

Where:

$ P = 700 \text{ mmHg} $

$ V = 60 \times 10^{-3} \text{ L} $

$ R = 0.0821 \text{ L atm/mol K} $

$ T = 300 \text{ K} $

Plug the values into the equation:

$ \mathrm{n}_{\mathrm{N}_2} = \frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300} \approx 2.24 \times 10^{-3} \text{ mol} $

Calculate the Mass of Nitrogen Gas:

$ \text{Mass of } \mathrm{N}_2 = 2.24 \times 10^{-3} \text{ mol} \times 28 \text{ g/mol} = 0.06272 \text{ g} $

Determine the Percentage Composition of Nitrogen:

$ \% \mathrm{~N}_2 = \left(\frac{0.06272}{0.5}\right) \times 100 \approx 12.57\% $

Thus, the percentage composition of nitrogen in the compound is approximately 12.57%.

The correct matching is:

(A) Simple distillation → (III) Chloroform + Aniline

(B) Fractional distillation → (I) Diesel + Petrol

(C) Distillation under reduced pressure → (IV) Glycerol + Spent-lye

(D) Steam distillation → (II) Aniline + Water

So the answer is Option A.

Brief rationale:

Simple distillation (A):

• $bp(\text{CHCl}_3)=61^\circ\text{C},\;bp(\text{C}_6\text{H}_5\text{NH}_2)=184^\circ\text{C}$

• Δbp is large, so the lower-boiling chloroform distils off cleanly.

Fractional distillation (B):

• Petrol (40–200°C) and diesel (180–360°C) have overlapping but distinct boiling-range cuts.

• A fractionating column resolves them efficiently.

Vacuum distillation (C):

• Glycerol boils ≈290°C at 1 atm—too high for normal distillation.

• Lowering the pressure brings its boiling point down, letting you separate it from the spent-lye.

Steam distillation (D):

• Aniline (bp≈184°C) can be co-distilled with water around 98°C, avoiding thermal decomposition and separating it from the aqueous phase.

The sodium fusion extract of organic compound having N \& S gives blood red colour with $\mathrm{FeSO}_4$ and $\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

Let's match the given tests in List I with the correct observations in List II based on known chemical reactions and their typical outcomes:

List I with List II

• This test is used to detect unsaturation in organic compounds. If the compound has a double bond (alkenes) or a triple bond (alkynes), the $\mathrm{Br_2}$ in water (bromine water) will react and decolorize.


• Correct Observation: Reddish orange color of bromine water disappears (II).

• This test is used to detect alcohols. Ceric ammonium nitrate forms a red-colored complex with alcohols.


• Correct Observation: Red color appears (III).

• Commonly used to detect phenols, the ferric chloride test leads to the formation of a colored complex with phenols, typically varying shades like purple, green, blue, or even red, depending on the specific phenol.


• Correct Observation: Blue, Green, Violet, or Red color appear (IV).

• This test is used for the detection of carbonyl compounds (aldehydes and ketones). 2,4-Dinitrophenylhydrazine (2,4-DNP) reacts with carbonyl groups to form a yellow, orange, or red precipitate.


• Correct Observation: Yellow orange or orange red precipitate formed (I).

Conclusion:

Based on these observations, the correct matching of tests with outcomes should be:

• A with II


• B with III


• C with IV


• D with I

This makes the correct choice:

Let's evaluate each statement to identify the incorrect ones regarding the primary standard in titrimetric analysis:

(A) It should be purely available in dry form.

This statement is generally correct. A primary standard should be pure and available in a stable, dry form to ensure accurate weight measurements.

(B) It should not undergo chemical change in air.

This statement is also correct. A primary standard should be stable in air and not react with components of the atmosphere such as oxygen or carbon dioxide.

(C) It should be hygroscopic and should react with another chemical instantaneously and stoichiometrically.

This statement is incorrect. A primary standard should not be hygroscopic (it should not absorb moisture from the air) because hygroscopic substances cannot maintain a constant weight. The part about reacting stoichiometrically and instantaneously is generally correct, but being hygroscopic disqualifies a substance from being a primary standard.

(D) It should be readily soluble in water.

This statement is correct. A primary standard should be soluble in water to ensure that it can be dissolved to make a standard solution for titration.

(E) $\mathrm{KMnO}_4$ & $\mathrm{NaOH}$ can be used as primary standard.

This statement is incorrect. Both $\mathrm{KMnO}_4$ (potassium permanganate) and $\mathrm{NaOH}$ (sodium hydroxide) are not suitable as primary standards. $\mathrm{KMnO}_4$ is unstable and can decompose, while $\mathrm{NaOH}$ is hygroscopic, which makes precise weight measurements difficult.

Hence, the correct answer is:

Option A

(C) and (E) only

To match the tests in List I with the corresponding identifications in List II, let's analyze each test:

Bayer's test: This test is used to test for unsaturation in compounds, such as alkenes and alkynes. The presence of unsaturation is indicated by the decolorization of potassium permanganate solution.

Ceric ammonium nitrate test: This test is used to identify alcohols. A positive result is indicated by a color change.

Phthalein dye test: This test is used to identify phenols. Phenols react with phthalic anhydride to form a colored compound.

Schiff's test: This test is used for the detection of aldehydes. A positive result indicates the formation of a magenta or pink colored compound.

Based on the information above, the correct matching is:

• A. Bayer's test - IV. Unsaturation
• B. Ceric ammonium nitrate test - III. Alcoholic-OH group
• C. Phthalein dye test - I. Phenol
• D. Schiff's test - II. Aldehyde

Thus, the correct answer is:

Option D: (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Organic compounds run slower than solvent in thin layer chromatography. $\mathrm{R}_{\mathrm{f}}$ is not an integral value. It is the ratio of distance travelled by the organic compound to that of solvent. $\mathrm{R}_{\mathrm{f}}$ 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.

Statement A: Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point.

This statement is correct. Glycerol decomposes at its normal boiling point (290°C approximately), so vacuum distillation is used to lower the boiling point and prevent decomposition.

Statement B: Aniline can be purified by steam distillation as aniline is miscible in water.

This statement is incorrect. Aniline is not completely miscible in water. However, it can be purified by steam distillation due to its relatively high boiling point and water’s ability to carry it over as a vapor. The statement as given is not entirely correct.

Statement C: Ethanol can be separated from ethanol-water mixture by azeotropic distillation because it forms an azeotrope.

This statement is correct. Ethanol and water form a minimum boiling azeotrope, which has a constant boiling point lower than either pure component, enabling separation through azeotropic distillation.

Statement D: An organic compound is pure if the mixed melting point remains the same.

This statement is correct. If mixing a small quantity of an unknown compound with a known compound does not change the melting point, this suggests the unknown compound is the same as the known and thus pure.

Based on the analysis, the correct statements are A, C, and D.

Option A: A, C, D only.

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.

Let's examine each statement one by one to choose the most appropriate answer :

Statement (I) : Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution.

This statement is correct. Potassium hydrogen phthalate (often abbreviated as KHP, with the chemical formula $ C_8H_5KO_4 $) is commonly used as a primary standard in titrations because it is pure, stable, non-hygroscopic, and has a high molar mass. It reacts with sodium hydroxide (NaOH) in a stoichiometric reaction, where one mole of KHP reacts with one mole of NaOH. The equation for the reaction is as follows :

$ C_8H_5KO_4 + NaOH \rightarrow C_8H_4KO_4Na + H_2O $

Statement (II) : In this titration phenolphthalein can be used as indicator.

This statement is also correct. Phenolphthalein is a suitable indicator for the titration between KHP and NaOH. It changes color in the pH range of about 8.2 to 10, which is appropriate for the endpoint of a titration involving a weak acid (KHP) and a strong base (NaOH). In an acidic solution, phenolphthalein is colorless, but it turns pink in a basic solution, signaling the end of the titration when slight excess NaOH is present.

Given that both statements are correct, the most appropriate answer is :

• Option A : Both Statement I and Statement II are correct.

Steam distillation technique is applied to separate substances which are steam volatile and are immiscible with water.

To match List I (Technique) with List II (Application), we need to understand the relevance of each distillation technique:

1. Distillation is typically used for separating components of mixtures who have a significant difference in boiling points.

Application: Separation of glycerol from spent-lye.

2. Fractional distillation is used for separating mixtures of liquids with closer boiling points by using a fractionating column.

Application: Separation of crude oil fractions.

3. Steam distillation is generally used for compounds that decompose under high temperatures and are insoluble in water.

Application: Aniline - Water mixture.

4. Distillation under reduced pressure (vacuum distillation) is used for compounds with very high boiling points to avoid decomposition.

Application: Chloroform - Aniline.

Therefore, the correct match is:

Option D
A-IV (Distillation - Chloroform - Aniline)
B-III (Fractional distillation - Separation of crude oil fractions)
C-II (Steam distillation - Aniline - Water mixture)
D-I (Distillation under reduced pressure - Separation of glycerol from spent-lye)

Memory Based

The correct option for the purification of steam volatile water-immiscible substances is Option D, Steam distillation.

Steam distillation is a separation technique used to purify or isolate temperature-sensitive materials, such as natural aromatic compounds, that are volatile in steam at a temperature that is lower than their decomposition temperatures. It exploits the fact that immiscible liquids can be distilled at a lower temperature than the boiling points of the individual components.

Here is how steam distillation works:

• Water and the other immiscible substance are placed in the distillation apparatus.
• Steam is generated either within the same flask or introduced from an external source.
• The hot steam helps vaporize the other substance at a temperature below its normal boiling point.
• The vapors of water and the volatile substance both pass into the condenser, where they are cooled.
• Upon cooling, the vapors convert back into liquid, which is then collected in a separate container.
• Since the two substances are immiscible, they separate into two distinct layers, typically, with the organic compound floating on top of the water or sinking below it, depending on their densities.
• The layers can then be separated using a separating funnel or other means.

Steam distillation is different from fractional distillation (Option A), which is used for the separation of a mixture into its component parts, or fractions, of compounds that are miscible and have different boiling points. Fractional distillation often involves the use of a fractionating column, which brings more efficiency in separating substances with close boiling points.

Simple distillation (Option B) is suitable for separating a liquid from impurities or for separating liquids with significantly different boiling points (usually the difference in boiling points should be at least $25^\circ C$).

On the other hand, fractional distillation under reduced pressure (Option C) is used when a liquid has a very high boiling point or might decompose at high temperatures. By reducing the pressure, the boiling point of the liquid is lowered and it can be distilled at a temperature much lower than its normal boiling point, thus avoiding decomposition.

Therefore, for volatile substances that are not miscible with water, steam distillation remains the ideal choice.

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.

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:

$\text{MO}^- + \text{H}_2\text{O} \leftrightharpoons \text{MOH} + \text{OH}^-$

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.

Nitrogen Detection (Lassaigne's Test)

Nitrogen in organic compounds is detected by converting it into sodium cyanide, which further reacts with iron(II) sulfate to form sodium hexacyanoferrate(II). By reacting this with iron(III) ions, Prussian blue is formed:

$ \begin{aligned} &\mathrm{Na}+\mathrm{C}+\mathrm{N} \rightarrow \mathrm{NaCN} \\\\ &6 \mathrm{NaCN}+\mathrm{FeSO}_4 \rightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{Na}_2 \mathrm{SO}_4 \\\\ &\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \text{ (Prussian blue)} \end{aligned} $

Sulphur Detection

Sulphur can be detected using Sodium nitroprusside:

$ \mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]+\mathrm{Na}_2 \mathrm{S} \rightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right] \text{ [Purple]} $

Phosphorous Detection

Phosphorus detection involves reaction with ammonium molybdate to form a canary yellow compound:

$ \begin{aligned} &\mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{HNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_3 \\\\ &\mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+21 \mathrm{HNO}_3 \rightarrow\\\\& \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O} \text{ (canary yellow)} \end{aligned} $

Halogen Detection

Halogens react with silver nitrate to form specific colored precipitates:

$ \begin{aligned} &\mathrm{NaCl}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3 \text{ (White)} \\\\ &\mathrm{NaBr}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgBr}+\mathrm{NaNO}_3 \text{ (Pale yellow)} \\\\ &\mathrm{NaI}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgI}+\mathrm{NaNO}_3 \text{ (Yellow)} \end{aligned} $

Based on the above reactions, the correct matching is:

Option A

A-III (Nitrogen), B-I (Sulphur), C-IV (Phosphorous), D-II (Halogen).

Sulphanilic acid is p-amino benzene sulphonic acid

Since it contain both N and S so it give red colour in Lassaigne's test.

Hence at the end point methyl orange is present as quinonoid form.