Organic Chemistry – Some Basic Principles and Techniques MCQ Quiz - Objective Question with Answer for Organic Chemistry – Some Basic Principles and Techniques - Download Free PDF

Acidity and Stabilization of Conjugate Bases

• Acidity: The acidity of a compound depends on how well the conjugate base (CB) can stabilize the negative charge. Factors such as the hybridization of the atom bearing the negative charge, resonance stabilization, and inductive effects play a key role in determining acidity.
• Hybridization: A conjugate base with a negative charge on a more electronegative atom (like oxygen) or with greater hybridization (like sp) tends to be less stable. The more localized the negative charge, the less acidic the compound.

EXPLANATION:

• Option A (Ethanol): The conjugate base (EtO⁻) is somewhat stabilized by oxygen, but it does not have resonance stabilization, making ethanol a relatively weak acid.
• Option B (Benzene sulfonic acid): The conjugate base (C₆H₅SO₃⁻) is resonance stabilized, as the negative charge is delocalized over two oxygen atoms. This makes it a stronger acid.
• Option C (Acetic acid): The conjugate base (CH₃COO⁻) is resonance stabilized, but it is not as stabilized as the conjugate base of sulfonic acid. Thus, acetic acid is moderately acidic.
• Option D (Ethynyl alcohol): The conjugate base (EtO₃C≡) has a negative charge on the sp-hybridized carbon, which is highly localized. This lack of resonance stabilization and the tightly held negative charge make ethynyl alcohol the least acidic compound in the list.

Therefore, the least acidic compound is Option D: Ethynyl alcohol (EtO₃C≡H).

CONCEPT:

Group Analysis in Qualitative Inorganic Chemistry

• Cations (basic radicals) are identified in a systematic manner using specific group reagents.
• Each group precipitates a specific set of metal ions as colored or white precipitates.
• Color and solubility of the precipitate with the group reagent help in identification.
• Matching Reagents and Precipitates:
  • Group III: NH₄OH in presence of NH₄Cl precipitates Fe³⁺ as Fe(OH)₃ (reddish brown).
  • Group II A: H₂S in presence of dil. HCl precipitates Cd²⁺ as CdS (yellow).
  • Group V: (NH₄)₂CO₃ in presence of NH₄OH precipitates Ca²⁺ as CaCO₃ (white).
  • Group II B: H₂S in presence of dil. HCl precipitates As³⁺ as As₂S₃ (yellow).

EXPLANATION:

• A - I: Fe³⁺ forms Fe(OH)₃ (reddish brown ppt) with NH₄OH in presence of excess NH₄Cl → Group III.
• B - II: Cd²⁺ forms CdS (yellow ppt) with H₂S in presence of dilute HCl → Group II A.
• C - III: Ca²⁺ forms CaCO₃ (white ppt) with (NH₄)₂CO₃ in presence of NH₄OH → Group V.
• D - IV: As³⁺ forms As₂S₃ (yellow ppt) with H₂S in presence of dilute HCl → Group II B.

Therefore, the correct answer is: Option 1) A - IV, B - II, C - III, D - V

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:

Separation of Mixtures

• Different mixtures require specific methods of separation based on their physical and chemical properties.
• Some common methods of separation include:
  • Simple Distillation: Used for separating liquids with significantly different boiling points.
  • Fractional Distillation: Used for separating mixtures of liquids with closer boiling points.
  • Distillation Under Reduced Pressure: Used for separating heat-sensitive liquids that may decompose at high temperatures.
  • Steam Distillation: Used for separating substances that are volatile in steam from non-volatile impurities.

EXPLANATION:

• A. CHCl₃ + C₆H₅NH₂: These are heat-sensitive liquids. Therefore, Distillation under reduced pressure is used.
• B. Crude oil in petroleum industry: Crude oil is a complex mixture of hydrocarbons that have close boiling points. Therefore, Fractional Distillation is used.
• C. Glycerol from spent-lye: Glycerol is separated using Simple Distillation.
• D. Aniline-water: Aniline and water are separated using Steam Distillation as aniline is volatile in steam.
• A - I (Distillation under reduced pressure)
• B - III (Fractional Distillation)
• C - IV (Simple Distillation)
• D - II (Steam Distillation)

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

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.

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

The correct answer is Option 1. 

 Key Points

• Arenes are a class of organic compounds that contain one or more aromatic rings, and they are also known as aromatic hydrocarbons.
• The term "arene" is often used to refer specifically to aromatic hydrocarbons that contain only one aromatic ring, whereas the term "polycyclic aromatic hydrocarbons" (PAHs) is used for compounds that contain two or more aromatic rings.
• Some common examples of arenes include benzene, toluene, and naphthalene.
• Aromatic hydrocarbons are widely used in industry and have many important applications, but some are also known to be harmful to human health and the environment.

Additional Information

•  Alkynes are a class of organic compounds that contain at least one carbon-carbon triple bond.
  • They are also known as acetylenes, which is a reference to the simplest alkyne, ethyne (also called acetylene).
  • Alkynes are characterized by their high reactivity due to the presence of the triple bond, which can undergo a variety of reactions such as addition, oxidation, and reduction.
  • Some common applications of alkynes include their use as starting materials for the synthesis of other organic compounds, as well as in the production of plastics, synthetic fibers, and pharmaceuticals.
  • Alkynes are also used in welding and cutting torches, as the high temperature produced by burning acetylene gas in the presence of oxygen can be used to melt or cut metals.

• Unsaturated hydrocarbons are a class of organic compounds that contain one or more carbon-carbon double bonds or triple bonds, which are referred to as "unsaturated" because they have fewer hydrogen atoms than their corresponding saturated hydrocarbons.

  • The most common types of unsaturated hydrocarbons are alkenes and alkynes, which contain carbon-carbon double bonds and triple bonds, respectively.

  • These double and triple bonds give unsaturated hydrocarbons their characteristic chemical reactivity, as they can undergo addition reactions to form new compounds.

• Saturated hydrocarbons are a class of organic compounds that contain only single bonds between carbon atoms, and therefore, each carbon atom is bonded to the maximum number of hydrogen atoms possible.

  • Because they have no carbon-carbon double or triple bonds, saturated hydrocarbons are referred to as "saturated" with hydrogen. The most common type of saturated hydrocarbon is the alkane, also known as a paraffin.

  • Alkanes have a general formula of CnH2n+2, where n is the number of carbon atoms in the molecule. For example, methane (CH4) is the simplest alkane and has one carbon atom, while ethane (C2H6) has two carbon atoms.

The correct answer is Option 3.

Key Points

• Calcium Sulphate Dihydrate (CaSO4.2H2O) is commonly known as “Gypsum”.
• When Gypsum is heated, it gets converted into Calcium Sulphate Hemihydrate (CaSO4.1/2H2O) or “Plaster of Paris (POP)”.
• This is prepared by heating the gypsum which contains calcium sulfate dihydrate (CaSO4.2H2O) to a temperature of about 150 deg C.
• When added water to the plaster of Paris (PoP), it will re-form into gypsum.​

Additional Information Some common chemical compounds with their common names are:-

Explanation:

The IUPAC name for the following compound is 3, 5-dimethyl-4-propylhept-1-en-6-yne.

In the case of the double bond and triple bond, the position is the same and the position of substituents is also the same, then the double bond is considered as senior so, the numbering starts from a double bond.

The correct answer is Furan.

Key Points

• Furan is a heterocyclic organic compound, consisting of a five-membered aromatic ring with four carbon atoms and one oxygen atom.
• Chemical compounds containing such rings are also referred to as furans.
• The chemical formula of Furan is C4H4O.
• Furan is a colorless, flammable, highly volatile liquid with a boiling point close to room temperature.

Additional Information

•  Acetic Acid:
  • Acetic acid, also known as ethanoic acid, is an organic compound with the formula CH3COOH.
  • It belongs to the family of carboxylic acids and is composed of a methyl group attached to a carboxyl functional group.
  • In pure form, acetic acid is often referred to as glacial acetic acid.
  • It is a colorless liquid that is corrosive and boils at 117.9 °C and melts at 16.6 °C. It is completely miscible with water.
  • Notably, acetic acid is the most important of the carboxylic acids, with a wide range of uses and applicability.
  • One common occurrence of acetic acid is in vinegar, where it is the primary component apart from water and trace elements.
  • In vinegar, the concentration of acetic acid is at least 4% by volume.
• Ethane:
  • Ethane is a colorless and odorless gas that belongs to the family of hydrocarbons, more specifically, the alkane series.
  • The chemical formula for ethane is C2H6, signifying that it contains two carbon atoms and six hydrogen atoms.
  • Saturated hydrocarbons like ethane are structurally the simplest hydrocarbon that contains a single carbon-carbon bond.
  • It can be found in abundant quantities in natural gas, making it a vital component for fuel generation.
  • In the laboratory, ethane can be prepared using sodium propionate.
  • The resulting compound is not only an important gaseous fuel but also holds a significant place in the study of organic chemistry.
• Methane:
  • ​Methane (CH4) is the simplest member of the paraffin series of hydrocarbons and is one of the most potent greenhouse gases.
  • This colorless, odorless gas can act as a significant heat-trapper in the atmosphere, thus contributing to global warming and climate change effects.
  • It's significant to note that although Methane is less prevalent in the atmosphere compared to Carbon Dioxide (CO2), it is a far more potent greenhouse gas.
  • In terms of its contribution to climate warming, it is considered the second-largest contributor after carbon dioxide.
  • In terms of its use, methane is a primary component of natural gas, a common source of energy for heating and electricity.
  • It is also a raw material in industrial chemical processes.

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

Concept:

Isomers:

• These are the compounds that have the same molecular formula but different structures or stereochemistries.
• There is a wide range of classification of organic molecules based on their structures.

​

• Metamerism: When isomers have the same molecular formula but differ in nature of alkyl groups attached to it.

• A class of compounds that have the same molecular formula but can have different functional groups exhibit functional isomerism.

• Positional isomers: Isomers having different positions of their functional groups.

(iii) C₃H₇Cl represents two position isomers

• Optical isomerism:

  • When two isomers rotate plane polarised light in different directions, then they are known as optical isomers.
  • The phenomenon is known as optical isomerism.

Explanation:

• The molecular formula of propyl alcohol is C₃H₈O.
• In the positional isomers, these things might differ:
  • The position of the substituent.
  • The position of the functional group.
  • Arrangement of carbon atoms in the chain.
• The structure of  n- propyl alcohol and isopropyl alcohol are:

• It is clearly seen that in n-propyl alcohol, the functional group -OH is in position 1 whereas in isopropyl alcohol it is in position two.

Hence, they are positional isomers.

The correct answer is -ide.Key Points

• The suffix "-ide" is added to the stem of the name of the element to form the name of negative ions consisting of a single atom.
• For example, chlorine forms chloride ion (Cl-) and oxygen forms oxide ion (O2-).

Additional Information

• The suffix "-ous" is used for naming cations with lower charge, such as Cu+ (cuprous ion).
• The suffix "-ic" is used for cations with higher charge, such as Cu2+ (cupric ion).
• The suffix "-onium" is used for naming positively charged ions composed of two or more atoms, such as ammonium ion (NH4+).
• The suffix "-ate" is used for naming oxyanions with higher oxidation state, such as sulfate ion (SO42-), while the suffix "-ite" is used for oxyanions with lower oxidation state, such as sulfite ion (SO32-).

The correct answer is Catalytic agent.

Key Points

Kjeldahl’s method:

• This method is used for the estimation of nitrogen.
• The nitrogen-containing compound is treated with concentrated H₂SO₄  to get ammonium sulfate which liberates ammonia on treatment with NaOH, ammonia is absorbed in known volume of standard acid.
• In this method CuSO₄ acts as a catalytic agent.
• Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

The correct answer is its chemical properties.Key Points

• The functional group in an organic compound determines its chemical properties.
• The properties of a functional group include its reactivity, polarity, and ability to participate in specific chemical reactions.
• A functional group is a specific arrangement of atoms within a molecule that imparts characteristic chemical properties to the compound.
• Examples of functional groups include alcohols, aldehydes, ketones, carboxylic acids, and amines.

Additional Information

• The nature of the carbon chain refers to whether it is straight or branched, saturated or unsaturated, or contains other substituents such as halogens or functional groups.
• These factors affect physical properties such as melting point, boiling point, and solubility.
• The length of the carbon chain can also affect physical properties such as boiling point and solubility.
• Molecular mass is a measure of the total mass of the atoms in a molecule, which affect physical properties such as density and boiling point.

Concept:

From question, the diagram given is:

\(\frac{{360^\circ }}{3} = 120^\circ \)

H’’ + H + H’