Qualitative and Quantitative Analysis of Organic Compounds MCQ Quiz - Objective Question with Answer for Qualitative and Quantitative Analysis of Organic Compounds - Download Free PDF

CONCEPT:

Identification of Metal Cyanide Complexes

• When potassium ferrocyanide (K₄[Fe(CN)₆]) reacts with various metal salts, different colored precipitates are formed due to the formation of metal ferrocyanide complexes.
• Each metal salt reacts with potassium ferrocyanide to form characteristic precipitates, which can be used for identification purposes:
  • AgNO₃ + K₄[Fe(CN)₆] → Yellow precipitate (Formation of silver ferrocyanide)
  • Pb(NO₃)₂ + K₄[Fe(CN)₆] → White precipitate (Formation of lead(II) ferrocyanide)
  • CuSO₄ + K₄[Fe(CN)₆] (neutralized with NH₄OH) → Light blue precipitate (Formation of copper(II) ferrocyanide)
  • NiSO₄ + K₄[Fe(CN)₆] → Dark green precipitate (Formation of nickel(II) ferrocyanide)
  • CaCl₂ + K₄[Fe(CN)₆] (neutralized with NaOH) → White precipitate (Formation of calcium ferrocyanide)

EXPLANATION:

• Option A: AgNO₃ + K₄[Fe(CN)₆] → Yellow precipitate: This is correct, as silver ferrocyanide forms a yellow precipitate.
• Option B: Pb(NO₃)₂ + K₄[Fe(CN)₆] → White precipitate: This is correct, as lead(II) ferrocyanide forms a white precipitate.
  2Pb(NO₃)₂(aq) + K₄Fe(CN)₆](aq) → Pb₂[Fe(CN)₆ 
• Option C: CuSO₄ + K₄[Fe(CN)₆] (neutralized with NH₄OH) → Light blue precipitate: This is incorrect, as copper(II) ferrocyanide forms a light blue precipitate when neutralized with ammonium hydroxide.
  2CuSO₄(aq) + K₄[Fe(CN)₆](aq) → Cu₂[Fe(CN)₆
• Option D: NiSO₄ + K₄[Fe(CN)₆] → Dark green precipitate: This is correct, as nickel(II) ferrocyanide forms a dark green precipitate.
• Option E: CaCl₂ + K₄[Fe(CN)₆] (neutralized with NaOH) → White precipitate: This is incorrect, as calcium ferrocyanide forms a white precipitate when neutralized with NaOH.

Therefore, the correct answer is: A, B, and D only.

CONCEPT:

Conductometric Titration

• Conductometric titration involves the measurement of electrical conductivity during a titration process to monitor the progress of the reaction.
• The conductivity of the solution changes based on the nature of the ions present during the titration.
• In the case of a strong acid vs strong base titration, the reaction leads to the formation of neutral water molecules, which results in a distinct "V-shaped" graph.

EXPLANATION:

• In a titration between a strong acid and a strong base:
  • Initially, the solution contains excess H⁺ ions from the strong acid, which contribute to high conductivity.
  • As the strong base is added, OH^- ions react with H⁺ ions to form neutral water molecules (H₂O), reducing the number of free ions and decreasing conductivity.
  • At the equivalence point, all H⁺ ions are neutralized, and the conductivity reaches a minimum.
  • Beyond the equivalence point, excess OH^- ions from the strong base increase the conductivity, resulting in the "V-shaped" graph.
• This characteristic "V-shaped" graph is specific to the titration of a strong acid with a strong base.

Therefore, the correct answer is Strong acid vs strong base.

CONCEPT:

Dumas Method for Estimation of Nitrogen

• In this method, an organic compound is heated with CuO in a CO2 atmosphere, and nitrogen is collected over water.
• The partial pressure of nitrogen gas is calculated by subtracting the aqueous tension from the total pressure.
• The number of moles of nitrogen is calculated using the ideal gas equation:

  n = (P × V) / (R × T)

• The mass of nitrogen is then determined, and its percentage in the organic compound is found by:

  % N = (Mass of N2 / Mass of compound) × 100

EXPLANATION:

• Given data:
  • Volume of N2 = 60 mL = 60 × 10–3 L
  • Total pressure = 715 mmHg, aqueous tension = 15 mmHg
  • Pressure of N2 gas = 715 – 15 = 700 mmHg
  • Convert pressure to atm: 700 / 760 ≈ 0.921 atm
  • Temperature = 300 K, R = 0.0821 L·atm·K–1·mol–1

Pressure of N₂ gas = (715 – 15) = 700 mmHg

n_N2 = \(\frac{P V}{R T}\)

nN2 = \(\frac{700 × 60 × 10^{-3}}{760 × 0.0821 × 300}\)

= 2.24 × 10^–3 mol 

Mass of N₂ = 2.24 × 10^–3 × 28 g 

= 0.06272 g

% N₂ = \(\frac{0.06272}{0.5}\) × 100

= 12.544% \(\simeq\) 13%

​So, percentage of nitrogen in organic compound is 13%.

CONCEPT:

Sodium Nitroprusside Test

• The sodium nitroprusside test is a qualitative analysis method used to detect the presence of sulfide ions (S2−) in a sample.
• Sodium nitroprusside, Na2[Fe(CN)5NO], reacts specifically with sulfide ions to produce a characteristic violet-coloured complex.
• The appearance of a violet or purple colour confirms the presence of sulfide ions.

EXPLANATION:

\(\begin{array}{l}2 \mathrm{Na}+\mathrm{S} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}\\ \text{(from organic}\text{ compound)}\end{array}\)

\(\rm \mathrm{Na}_{2} \mathrm{~S}+\mathrm{Na}_{2}\left[\mathrm{Fe}^{\text {II }}(\mathrm{CN})_{5} \mathrm{NO}\right] \longrightarrow \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NOS}\right]\\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad violet \ colour\)

• When an organic compound containing sulfide ions reacts with sodium metal, sodium sulfide (Na₂S) is formed:
  • 2Na + S → Na₂S
• The Na₂S then reacts with sodium nitroprusside solution as follows:
  • Na₂S + Na₂[Fe(CN)₅NO] → Na₄[Fe(CN)₅NOS]
• This reaction produces a violet-coloured complex, confirming the presence of sulfide ions.
• Other species like SO₄²⁻, Na⁺, and PO₄³⁻ do not give a positive test with sodium nitroprusside.

The correct answer is S²⁻

CONCEPT:

Qualitative Analysis: Group Reagents for Cation Detection

• In salt analysis, cations are detected group-wise based on their precipitation with specific reagents under controlled conditions.
• Each group reagent selectively precipitates certain cations, helping classify them into analytical groups.

EXPLANATION:

• P: Pb²⁺, Cu²⁺ — Group II cations
  • Group reagent: H₂S gas in the presence of dilute HCl → Match: (i)
• Q: Al³⁺, Fe³⁺ — Group III cations
  • Group reagent: NH₄OH in the presence of NH₄Cl → Match: (iii)
• R: Co²⁺, Ni²⁺ — Group IV cations
  • Group reagent: H₂S in the presence of NH₄OH → Match: (iv)
• S: Ba²⁺, Ca²⁺ — Group V cations
  • Group reagent: (NH₄)₂CO₃ in the presence of NH₄OH → Match: (ii)

Correct Matching P → i, Q → iii, R → iv, S → ii.

Concept:

The thermal decomposition of potassium permanganate produces potassium manganate, manganese (IV) oxide, and oxygen. 

\(2{\rm{KMn}}{{\rm{O}}_4}{\rm{\;}}\left( {\rm{X}} \right)\mathop \to \limits^{513{\rm{\;K}}} {{\rm{K}}_2}{\rm{Mn}}{{\rm{O}}_4}{\rm{\;}}\left( {\rm{Y}} \right) + {\rm{Mn}}{{\rm{O}}_2} + {{\rm{O}}_2}{\rm{\;}}\left( {{\rm{gas}}} \right)\)

\({\rm{Mn}}{{\rm{O}}_2} + {\rm{NaCl}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{MnS}}{{\rm{O}}_4} + {\rm{C}}{{\rm{l}}_2}{\rm{\;}}\left( {\rm{Z}} \right) + {\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{O}}\)

The thermal decomposition of an ‘Mn’ compound (X) is KMnO₄ (potassium permanganate) at 513 K results in compound Y, K₂MnO₄ is (Potassium manganate) along with MnO₂ (manganese (IV) oxide) and gaseous product. Here the MnO₂ reacts with NaCl (sodium chloride) and concentrated H₂ SO₄ (sulfuric acid) to give a pungent gas Z is Cl₂ (chlorine).

X = KMnO₄

Y = K₂MnO₄

Z = Cl₂

Explanation:-

• Phenylhydrazine (C₆H₅NHNH₂):
  • This compound contains an NHNH₂ group, which should, in theory, generate sodium cyanide (NaCN) when fused with sodium, thereby giving a positive Lassaigne's test for nitrogen.
• Glycine (NH₂CH₂COOH):
  • Glycine, an amino acid, similarly should generate sodium cyanide (NaCN) when fused with sodium, giving a positive result in Lassaigne's test for nitrogen.
• Urea (NH₂CONH₂):
  • Urea consists of two NH₂ groups and a carbonyl group, which should generate sodium cyanide (NaCN) and hence give a positive Lassaigne's test for nitrogen.
• Hydrazine (NH₂NH₂):
  • Hydrazine has a unique structure (NH₂NH₂). When it is fused with sodium, it does not readily form sodium cyanide (NaCN). Due to this unique property, hydrazine does not produce a positive Lassaigne’s test for nitrogen.
  • Hydrazine (NH2–NH2) have no carbon so does not show Lassaigne’s test.
  • 

Hence, based on this re-evaluation, the correct answer indeed is  Hydrazine.

Concept:

The basic formula of Parts per Million is:

\({\rm{PPM}} = \frac{{{\rm{Weight}}}}{{{\rm{Volume}}}} \times {10^6}\)

The degree of hardness and the weight are interrelated. Thus,

\({\rm{Degree\;of\;hardness}} = \frac{{{\rm{\;Weight\;of\;hardness\;causing\;salt\;}}}}{{{\rm{Molecular\;weight}}}} \times 100\)

Calculation:

\(\Rightarrow {\rm{Degree\;of\;hardness}} = \left( {\frac{{0.81}}{{162}} + \frac{{0.73}}{{146}}} \right) \times 100\)

∴ Degree of hardness=1

Now, \({\rm{PPM}} = \frac{1}{{100}} \times {10^6}\)

∴ PPM = 10000 ppm

Concept:

(A) Chloroxylenol

Diagram

Chloroxylenol, also known as para-chloro-meta-xylenol, is an antiseptic and is infectant which is used for skin disinfection and cleaning surgical instruments. It is most effective and safe medicines needed in a health system. It is also commonly used in antibacterial soaps, wound-cleansing applications and household antiseptics.

Side effects of Chloroxylenol are generally few but can include skin irritation. It may be used mixed with water or alcohol Chloroxylenol is most effective against Gram-positive bacteria. It works by disruption of the cell wall and stopping the function of enzymes.

(B) Norethindrone

Diagram

Norithindrone also known as Northisterone, is a progestin medication used in birth control pills, menopausal hormone therapy, and for the treatment of gynocological disorders.

Northisterone has also been shown to be effective in inhibiting leutinizing hormone and follicle stimulating hormone.

(C) Sulphapyridine

Diagram

Sulfapyridine is a sulphonamide antibacterial medication. At one time, it was commonly reffered to as M&B 693. Sulfapyridine is no longer prescribed for treatment of infections in humans.

Sulfapyridine is a sulfa medicine. It is used to help control dermatitis herpetiformis (Duhring'sdisease). It is a chronic skin condition caused by a reaction to gluten ingestion. The vast majority of patients with DH also have an associated gluten sensitive enteropathy (celiac disease) a skin problem. It may also be used for other problems. However, this medicine will not work for any kind of infection as other sulfa medicines do.

(D) Pencillin

Diagram

Penicillin is a group of antibiotics which include penicillin G and penicillin V. Penicillin antibiotic were among the first medication to be effective against many bacterial infections caused by staphylococci and streptococci.

Procaine penicillin and benzathine penicillin have the same antibacterial activity as benzylpenicillin but act for a longer period of time. Phenoxymethylpenicillin is less active against gram-negative bacteria than benzylpenicillin.

Penicillin V is an antibiotic in the penicillin group of drugs. It fights against bacteria in your body. Penicillin V is used to treat many different types of infections caused by bacteria, such as ear infections.

Explanation:-

Role of Ammonium Chloride in Qualitative Analysis of Iron Group III

• During qualitative analysis, specifically in the precipitation of the iron group (Group III cations), ammonium chloride (NH4Cl) is added before adding ammonium hydroxide (NH4OH).
• Ammonium chloride provides a source of ammonium ions (NH4⁺), which play a crucial role in controlling the concentration of hydroxide ions (OH^-).
• This is important because a higher concentration of OH^- ions might cause the precipitation of Group IV cations (like Zn²⁺, Mn²⁺) in addition to the iron group cations.
• NH4⁺ ions from NH4Cl suppress the ionization of NH4OH, thereby maintaining a moderate concentration of OH^- ions just enough to precipitate the Group III hydroxides without causing the formation of unwanted hydroxides of subsequent groups.

NH4OH \(\rightleftharpoons\) NH4+ + OH-

NH4Cl → NH4+ + Cl

Due to common ion effect of NH4+,

[OH-] decreases in such extent that only group-III cation can be precipitated , due to their very low Ksp in the range of 10–38.

In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to decrease concentration of –OH ion

Concept:

The Graph 1 and 2 both represents the titration curve between strong acid and strong base, that is HCl (Hydrochloric acid) and NaOH (Sodium Hydroxide). Initially, the pH of NaOH is more than 7 but during the titration, it decreased. So, graph (1) is correct.

In a strong acid-strong base titration, the acid and base will react to form a neutral solution. At the equivalence point of the reaction, hydronium (H⁺) and hydroxide (OH^-) ions will react to form water, leading to a pH of 7. This is true for all strong acid and strong base titrations.

The titration of hydrochloric acid (strong acid) and sodium hydroxide (strong base) to form sodium chloride and water. The reaction as follows,

NaOH + HCl ⟶ NaCl + H₂O

In a weak base-strong acid titration, the acid and base will react to form an acidic solution. A conjugate acid will be produced during the titration, which then reacts with water to form hydronium ions. This results in a solution with a pH lower than 7.

A strong acid is an acid that is completely dissociated or ionized in an aqueous solution. It is a chemical species with a high capacity to lose a proton H⁺. In water, a strong acid loses one proton, which is captured by water to form the hydronium ion.

A strong base is a base that is completely dissociated in an aqueous solution. In contrast, a weak base only partially dissociates into its ions in water. Ammonia is a good example of a weak base. Strong bases react with strong acids to form stable compounds

Concept:

From the given data, we can infer that, when the compound reacts with Acid, it becomes insoluble.

When it reacts with a base, it becomes soluble. Hence, the compound is acidic in nature.

While reacts with Br₂/H₂O, it becomes an unsaturated compound and results in decolourization.

Phenol is acidic in nature while Aniline is basic in nature. Phenol is insoluble in dil. HCl but soluble in NaOH to form Sodium phenol.

Phenol decolourize Br₂ water to give 2, 4, 6 – tribromophenol

Here, option (c) and (d) cannot become an unsaturated compound due to their bond formation.

The organic compound that satisfies all the qualitative analysis given above is Phenol.

CONCEPT:

Lassaigne’s Test

• Lassaigne’s test is a qualitative test used to detect the presence of elements such as nitrogen (N), sulfur (S), and halogens (X) in an organic compound.
• In this test, the organic compound is fused with sodium metal, converting these elements into their respective sodium salts (e.g., NaCN, Na₂S, NaX), which can be detected by characteristic reactions.
• The test involves heating the mixture to allow the reaction of sodium with the element present in the organic compound.

• Nitrogen, sulphur, and halogens present in organic compounds are detected by Lassaigne’s test. Here, a small piece of Na metal is heated in a fusion tube with the organic compound. The principle is that, in doing so, Na converts all the elements present into ionic form.

  Na + C + N → NaCN
  2Na + S → Na2S
  Na + X → NaX ( X= Cl, Br, or I)
  The formed ionic salts are extracted from the fused mass by boiling it with distilled water. This is called sodium fusion extract.

EXPLANATION:

• Na + C + N → NaCN: This reaction forms sodium cyanide, used to detect nitrogen in the compound. This belongs to Lassaigne’s test.
• 2Na + S → Na₂S: This forms sodium sulfide, used to detect sulfur. This also belongs to Lassaigne’s test.
• Na + X → NaX: This forms sodium halide, used to detect halogens like Cl, Br, and I. This belongs to Lassaigne’s test.
• 2CuO + C → 2Cu + CO₂: This reaction involves the reduction of copper oxide by carbon and is unrelated to Lassaigne’s test for detecting N, S, or halogens.

Therefore, the reaction that does NOT belong to Lassaigne’s test is  2CuO + C → 2Cu + CO₂

CONCEPT:

Group Analysis of Cations

• The classification of cations into different groups is based on their solubility and behavior in qualitative analysis.
• Group I cations form insoluble chlorides when treated with dilute hydrochloric acid.
• Group II cations form insoluble sulfides when treated with hydrogen sulfide in an acidic medium.
• Group III cations form insoluble hydroxides when treated with ammonium hydroxide.
• Group IV cations form insoluble phosphates when treated with ammonium molybdate in an acidic medium.
• Group V cations do not form precipitates under normal conditions and are analyzed last.

Match the Group with their respective Cations and Group Reagents.

Group	Cations	Group Reagent
Group zero	NH4+	None
Group-I	Pb2+	Dilute HCl
Group-II	Pb2+, Cu2+, As3+	H2S gas in presence of dil. HCl
Group-III	Al3+, Fe3+	NH4OH in presence of NH4Cl
Group-IV	Co2+, Ni2+, Mn2+, Zn2+	H2S in presence of NH4OH
Group-V	Ba2+, Sr2+, Ca2+	(NH4)2CO3 in presence of NH4OH
Group-VI	Mg2+	None

 

EXPLANATION:

• Co²⁺ (Cobalt ion): This ion is typically classified as Group-IV in cation analysis because it forms hydroxides when treated with ammonium hydroxide.
• Mg²⁺ (Magnesium ion): This ion belongs to Group-VI as it forms sulfides in the presence of hydrogen sulfide in an acidic medium.
• Pb²⁺ (Lead ion): This ion is usually classified as Group-I because it forms phosphates when treated with ammonium molybdate in an acidic medium.
• Al³⁺ (Aluminum ion): This ion is classified as Group-III because it forms insoluble chlorides in the presence of dilute hydrochloric acid.

Therefore, the correct answer is A-III, B-IV, C-I, D-II

Concept:

The thermal decomposition of potassium permanganate produces potassium manganate, manganese (IV) oxide, and oxygen. 

\(2{\rm{KMn}}{{\rm{O}}_4}{\rm{\;}}\left( {\rm{X}} \right)\mathop \to \limits^{513{\rm{\;K}}} {{\rm{K}}_2}{\rm{Mn}}{{\rm{O}}_4}{\rm{\;}}\left( {\rm{Y}} \right) + {\rm{Mn}}{{\rm{O}}_2} + {{\rm{O}}_2}{\rm{\;}}\left( {{\rm{gas}}} \right)\)

\({\rm{Mn}}{{\rm{O}}_2} + {\rm{NaCl}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{MnS}}{{\rm{O}}_4} + {\rm{C}}{{\rm{l}}_2}{\rm{\;}}\left( {\rm{Z}} \right) + {\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{O}}\)

The thermal decomposition of an ‘Mn’ compound (X) is KMnO₄ (potassium permanganate) at 513 K results in compound Y, K₂MnO₄ is (Potassium manganate) along with MnO₂ (manganese (IV) oxide) and gaseous product. Here the MnO₂ reacts with NaCl (sodium chloride) and concentrated H₂ SO₄ (sulfuric acid) to give a pungent gas Z is Cl₂ (chlorine).

X = KMnO₄

Y = K₂MnO₄

Z = Cl₂