Organic chemistry is the study of carbon-containing compounds. It is a vast field with applications in various areas like medicine, pharmaceuticals, materials science, and more.
Key Concepts:
Tetravalency of Carbon: Carbon atoms have four valence electrons, allowing them to form stable bonds with other atoms, primarily hydrogen and other carbon atoms.
Homologous Series: A series of organic compounds that differ from each other by a constant -CH₂ unit.
Functional Groups: Specific groups of atoms or bonds that give organic compounds their characteristic properties. Examples include hydroxyl (-OH), carbonyl (C=O), and carboxyl (-COOH) groups.
Isomerism: Compounds with the same molecular formula but different structural arrangements are called isomers. Isomerism can be structural or stereoisomerism.
Nomenclature: The systematic naming of organic compounds using the International Union of Pure and Applied Chemistry (IUPAC) rules.
Purification Techniques: Methods used to isolate and purify organic compounds, such as distillation, crystallization, and chromatography.
Qualitative Analysis: Tests used to identify the functional groups present in organic compounds.
Quantitative Analysis: Methods used to determine the amount of a particular element or compound in a sample.
Important Topics:
Hydrocarbons: Alkanes, alkenes, alkynes, and aromatic hydrocarbons.
Haloalkanes and Haloarenes: Compounds containing halogen atoms.
Alcohols, Phenols, and Ethers: Compounds containing the -OH group.
Aldehydes, Ketones, and Carboxylic Acids: Compounds containing the carbonyl group.
Amines: Compounds containing the -NH₂ group.
Nitro Compounds: Compounds containing the -NO₂ group.
Cyanides and Isocyanides: Compounds containing the -CN and -NC groups.
Organic Compounds Containing Other Elements: Compounds containing elements like sulfur, phosphorus, and nitrogen.
Techniques Used in Organic Chemistry:
Spectroscopy: Techniques like IR, UV-Vis, NMR, and mass spectrometry to analyze the structure and composition of organic compounds.
Chromatography: Techniques like thin-layer chromatography (TLC) and column chromatography to separate and purify mixtures of organic compounds.
Titration: A quantitative analysis technique to determine the concentration of a solution.
By understanding these basic principles and techniques, you can delve deeper into the fascinating world of organic chemistry.
1. What are hybridisation states of each carbon atom in the following compounds ? CH2=C=O, CH3CH=CH2, (CH3) 2CO, CH2=CHCN, C6H6
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2. Indicate the σ and π bonds in the following molecules : C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3
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3. Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
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4. Give the IUPAC names of the following compounds :
Ans : (a) 1-Phenylbutane
(b) 2-Methyl-3-butanenitrile
(c) 2,5-Dimethylheptane
(d) 3-Bromo-3-chloropentane
(e) 2-Chloropropanal
(f) 2-Chloro-2-hydroxypropanal
These names follow the rules of IUPAC nomenclature for organic compounds, which specify the order of substituents, the numbering of the carbon chain, and the appropriate suffixes for different functional groups.
5. Which of the following represents the correct IUPAC name for the compounds concer ned ?
(a) 2,2-Dimethylpentane or 2-Dimethylpentane
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
(d) But-3-yn-1-ol or But-4-ol-1-yne.
Ans : (a) 2,2-Dimethylpentane
This is the correct name as the two methyl groups are attached to the same carbon atom, which is numbered 2 in the parent chain.
(b) 2,4,7-Trimethyloctane
This is the correct name as the three methyl groups are attached to carbons 2, 4, and 7 of the parent chain.
(c) 2-Chloro-4-methylpentane
This is the correct name as the chloro and methyl substituents are attached to carbons 2 and 4, respectively, of the parent chain.
(d) But-3-yn-1-ol
This is the correct name as the compound contains a triple bond (yne) at carbon 3, an alcohol group (ol) at carbon 1, and the parent chain is a four-carbon butane chain.
Therefore, all of the given options represent the correct IUPAC names for the corresponding compounds.
6. Draw formulas for the first five members of each homologous series beginning with the following compounds.
(a) H–COOH
(b) CH3COCH3
(c) H–CH=CH2
Ans : A homologous series is a series of organic compounds that differ from each other by a constant -CH₂ unit.
(a) H-COOH (Formic Acid)
Formic acid: HCOOH
Acetic acid: CH₃COOH
Propanoic acid: C₂H₅COOH
Butanoic acid: C₃H₇COOH
Pentanoic acid: C₄H₉COOH
(b) CH₃COCH₃ (Acetone)
Acetone: CH₃COCH₃
2-Butanone: CH₃CH₂COCH₃
3-Pentanone: CH₃CH₂CH₂COCH₃
2-Pentanone: CH₃COCH₂CH₂CH₃
3-Hexanone: CH₃CH₂CH₂COCH₂CH₃
(c) H-CH=CH₂ (Ethylene)
Ethylene: CH₂=CH₂
Propene: CH₃CH=CH₂
Butene: CH₃CH₂CH=CH₂
Pentene: CH₃CH₂CH₂CH=CH₂
Hexene: CH₃CH₂CH₂CH₂CH=CH₂
7. Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :
(a) 2,2,4-Trimethylpentane
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial
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8. Identify the functional groups in the following compounds
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Identifying Functional Groups
Functional groups are specific groups of atoms or bonds that give organic compounds their characteristic properties. Let’s identify the functional groups in the given compounds:
(a) CHO
This represents an aldehyde functional group. Aldehydes are characterized by a terminal carbonyl group.
(b) NH₂
This represents an amine functional group. Amines contain nitrogen atoms bonded to hydrogen atoms.
(c) OMe
This represents a methoxy group, which is an ether functional group. Ethers contain an oxygen atom bonded to two alkyl or aryl groups.
(d) OCH₂CH₂N(C₂H₅)₂
This represents an amide functional group. Amides feature a carbonyl group connected to a nitrogen atom.
(e) CH=CHNO₂
This represents a nitroalkene functional group. Nitroalkenes contain a nitro group (-NO₂) attached to a carbon atom that is part of a carbon-carbon double bond.
In summary, the functional groups present in the given compounds are:
(a) Aldehyde
(b) Amine
(c) Methoxy (ether)
(d) Amide
(e) Nitroalkene
9. Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?
Ans : O₂NCH₂CH₂O⁻ is expected to be more stable than CH₃CH₂O⁻.
Reasoning:
Electron-withdrawing effect: The nitro group (-NO₂) is a strong electron-withdrawing group. This means it pulls electron density away from the oxygen atom in the alkoxide ion.
Stabilization of negative charge: The electron-withdrawing effect of the nitro group helps to stabilize the negative charge on the oxygen atom, making the O₂NCH₂CH₂O⁻ ion more stable.
Inductive effect: The electron-withdrawing effect of the nitro group is transmitted through the sigma bonds of the carbon chain, leading to a decrease in electron density around the oxygen atom.
In contrast, the ethyl group in CH₃CH₂O⁻ is an electron-donating group. This increases electron density around the oxygen atom, making the ion less stable and more reactive.
10. Explain why alkyl groups act as electron donors when attached to a π system
Ans : Alkyl groups act as electron donors when attached to a π system due to their hyperconjugative effect.
Hyperconjugation is a type of delocalization of electrons involving the overlap of σ orbitals (from C-H bonds) with adjacent π orbitals. This delocalization results in a shift of electron density towards the π system.
Here’s a simplified explanation:
σ-π Overlap: The C-H bonds in an alkyl group have σ orbitals. These σ orbitals can overlap with the π orbitals of a nearby double bond or aromatic ring.
Electron Density Shift: This overlap allows for the delocalization of electrons from the C-H bonds towards the π system. This shift of electron density makes the alkyl group appear to be donating electrons to the π system.
Hyperconjugation is particularly effective in stabilizing carbocations and radicals. The increased electron density around the positive charge in a carbocation or the unpaired electron in a radical helps to disperse the charge or reduce the spin density, making the species more stable.
In summary, alkyl groups act as electron donors due to their hyperconjugative effect, which involves the delocalization of electron density from the C-H bonds to the adjacent π system.
11. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C6H5OH
(b) C6H5NO2
(c) CH3CH=CHCHO
(d) C6H5–CHO
(e) C6H5–C + H2
(f) CH3CH=CH C+ H2
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12. What are electrophiles and nucleophiles ? Explain with examples.
Ans : Electrophiles and nucleophiles are two fundamental concepts in organic chemistry that describe the reactivity of molecules and ions.
Electrophiles are electron-deficient species that seek electrons. Many of them carry a positive charge or have a partial positive character.. Examples of electrophiles include:
Carbocations: CH₃⁺, CH₃CH₂⁺
Carbonyl compounds: CH₃CHO, CH₃COCH₃
Acidic protons: H⁺ in HCl, H₂SO₄
Nucleophiles are electron-rich species that donate electrons. They often have a negative charge or a lone pair of electrons. Examples of nucleophiles include:
Anions: Cl⁻, OH⁻, CN⁻
Neutral molecules with lone pairs: NH₃, H₂O
Carbanions: CH₃⁻, CH₃CH₂⁻
Reactions between Electrophiles and Nucleophiles
Electrophiles and nucleophiles react with each other in a process known as nucleophilic addition or nucleophilic substitution. In these reactions, the nucleophile attacks the electrophilic center, forming a new bond and breaking an existing bond.
Examples:
Nucleophilic addition: CH₃CHO + CN⁻ → CH₃CH(CN)OH
Nucleophilic substitution: CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻
In brief, electrophiles are electron-deficient species that accept electrons, whereas nucleophiles are electron-rich species that donate electrons. These species interact in various chemical reactions, such as nucleophilic addition and substitution.
13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
(a) CH3COOH + HO– → CH3COO– +H2O 2024-25 organic chemistry – some basic principles and techniques 293
(b) CH3COCH3+ C – N → (CH3) 2C(CN)(OH)
(c) C6H6 + CH3C + O → C6H5COCH3
Ans : (a) and (b) are nucleophiles, while (c) is an electrophile.
Nucleophiles are electron-rich species that donate electrons, such as CN⁻ and OH⁻. Electrophiles are electron-deficient species that accept electrons, such as CH₃⁺.
14. Classify the following reactions in one of the reaction type studied in this unit. (a) CH3CH2Br + HS– → CH3CH2SH + Br–
(b) (CH3) 2C = CH2 + HCI → (CH3) 2CIC – CH3
(c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br–
(d) (CH3) 3C– CH2OH + HBr → (CH3) 2CBrCH2CH2CH3 + H2O
Ans : The given reactions can be classified as follows:
(a) CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻
This is a nucleophilic substitution reaction. The sulfur ion (HS⁻) acts as a nucleophile, attacking the electrophilic carbon atom attached to the bromine. The bromine atom is replaced by the sulfur atom.
(b) (CH₃)₂C=CH₂ + HCl → (CH₃)₂CClCH₃
This is an electrophilic addition reaction. The hydrogen ion (H⁺) from HCl acts as an electrophile, attacking the double bond in the alkene. The chloride ion (Cl⁻) then adds to the carbocation intermediate formed.
(c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻
This is an elimination reaction. The hydroxide ion (HO⁻) acts as a base, removing a proton from the beta-carbon (the carbon adjacent to the carbon attached to the bromine). This leads to the formation of a double bond and the elimination of the bromine atom.
(d) (CH₃)₃C–CH₂OH + HBr → (CH₃)₂CBrCH₂CH₂CH₃ + H₂O
This is a nucleophilic substitution reaction. The bromide ion (Br⁻) acts as a nucleophile, attacking the electrophilic carbon atom attached to the hydroxyl group. The hydroxyl group is replaced by the bromide atom.
15. What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors ?
Ans : The relationships between the members of the following pairs of structures are:
(a) Resonance contributors: The two structures have the same molecular formula and connectivity of atoms, but differ in the distribution of electrons. They represent different ways to draw the same molecule.
(b) Geometrical isomers: The two structures have the same molecular formula and connectivity of atoms, but differ in the spatial arrangement of their atoms due to restricted rotation around a double bond. They are cis and trans isomers.
(c) Resonance contributors: The two structures have the same molecular formula and connectivity of atoms, but differ in the placement of formal charges. They represent different ways to draw the same molecule.
In summary, the pairs (a) and (c) are resonance contributors, while the pair (b) represents geometrical isomers.
16. For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
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17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
(b) CH3CH2COOH > (CH3) 2CHCOOH > (CH3) 3C.COOH
Ans : The inductive effect is the gradual transmission of electrical charge through a sigma bond due to the electronegativity difference between atoms. This effect can be electron-withdrawing (e.g., halogen atoms) or electron-donating (e.g., alkyl groups).
The electromeric effect is a temporary polarization of a double bond caused by the approach of an electrophile or nucleophile. This effect is permanent only in conjugated systems.
The acidity of carboxylic acids is influenced by the electron-withdrawing or electron-donating effects of substituents on the alpha carbon.
In the series Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH, the acidity increases with the number of chlorine atoms. This is because chlorine is an electron-withdrawing group, and as the number of chlorine atoms increases, the electron-withdrawing effect on the carboxyl group also increases, making it easier to lose a proton. The inductive effect is the primary factor explaining this trend.
In the series CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C.COOH, the acidity decreases as the number of alkyl groups increases. This is because alkyl groups are electron-donating groups, and as the number of alkyl groups increases, the electron-donating effect on the carboxyl group also increases, making it more difficult to lose a proton. Again, the inductive effect is the primary factor explaining this trend.
In summary, the acidity of carboxylic acids is primarily influenced by the inductive effect of substituents on the alpha carbon. Electron-donating groups decrease acidity, whereas electron-withdrawing groups increase acidity.
(a)The decrease in the number of halogen atoms results in a weaker electron-withdrawing inductive effect, which, in turn, decreases the acidity.
(b) The acid strength decreases as the number of alkyl groups increases due to the enhanced electron-donating inductive effect.
18. Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation
(b) Distillation
(c) Chromatography
Ans : (a) Crystallization:
To purify an impure solid, it is dissolved in the minimum amount of a suitable solvent. This forms a solution that contains the soluble impurities while the insoluble ones remain as residue. The hot solution is then filtered to remove the insoluble impurities. The filtrate is then allowed to cool undisturbed until crystallization occurs. The crystals are separated from the remaining liquid (mother liquor) by filtration and dried.
(b) Distillation:
Distillation is a technique used to purify liquids from non-volatile impurities. The impure liquid is heated in a flask, and the resulting vapors are collected and condensed to obtain the pure liquid in a separate vessel. This method is suitable for purifying simple organic liquids like benzene, toluene, and xylene.
(c) Chromatography:
Chromatography is based on the selective distribution of the components of a mixture between two phases: a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or gas. When the stationary phase is solid, the separation is based on adsorption, and when it is a liquid, the separation is based on partition. Chromatography is commonly used to separate colored substances such as plant pigments or dyes
19. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Ans : Solvent Extraction is a suitable technique to separate two compounds with different solubilities in a solvent S.
Here’s how it works:
Dissolution: The mixture of two compounds is dissolved in a solvent S in which one compound is significantly more soluble than the other.
Immiscible Solvent Addition: A second solvent (immiscible with S) is added to the solution. This second solvent should be able to dissolve the less soluble compound from the mixture.
Separation: The two solvents form separate layers due to their immiscibility. The less soluble compound will preferentially dissolve in the second solvent, while the more soluble compound will remain in the first solvent.
Extraction: The two layers are separated, and the desired compound can be recovered from its respective solvent using techniques like evaporation or distillation.
Key Factors:
Solvent Choice: The choice of solvents is crucial. The first solvent should dissolve the desired compound well, while the second solvent should selectively dissolve the other compound.
Distribution Coefficient: The distribution coefficient (K) is a measure of how much a compound prefers one solvent over the other. A higher K value indicates a greater preference for the second solvent.
Multiple Extractions: To achieve complete separation, multiple extractions with the second solvent may be necessary.
Solvent extraction is a versatile technique used in various fields, including chemistry, biochemistry, and environmental science, to isolate and purify compounds based on their solubility differences.
20. What is the difference between distillation, distillation under reduced pressure and steam distillation ?
Ans : Distillation, Distillation Under Reduced Pressure, and Steam Distillation
These three techniques are used to separate components of a mixture based on their boiling points.
Distillation
Principle: The process involves heating a liquid mixture to its boiling point, causing the component with the lower boiling point to vaporize. The vapor is transformed into a liquid state and collected.
Application: Suitable for separating components with significantly different boiling points.
Distillation Under Reduced Pressure
Principle: The pressure above the liquid mixture is reduced, lowering the boiling points of the components. This allows for the distillation of compounds that would decompose at their atmospheric boiling point.
Application: Used for compounds that are heat-sensitive or have high boiling points.
Steam Distillation
Principle: The mixture is heated with steam. The steam carries the organic compound with a high vapor pressure, allowing it to distill at a temperature lower than its normal boiling point.
Application: Suitable for separating organic compounds that are insoluble in water but have a relatively high vapor pressure.
21. Discuss the chemistry of Lassaigne’s test.
Ans : Lassaigne’s Test: A Qualitative Analysis for Nitrogen, Sulfur, and Halogens
Lassaigne’s test is a qualitative organic analysis technique used to detect the presence of nitrogen, sulfur, and halogens in organic compounds. It involves fusing the organic compound with sodium metal (Na) under anhydrous conditions to convert the elements of interest into their corresponding sodium salts.
Steps Involved:
Fusion with Sodium Metal: A small amount of the organic compound is fused with metallic sodium in a dry test tube. This reaction converts the nitrogen, sulfur, and halogen atoms in the organic compound into their respective sodium salts (sodium cyanide, sodium sulfide, and sodium halide).
Testing for Nitrogen:
The fused mass is dissolved in water and a few drops of ferrous sulfate (FeSO₄) solution are added.
Then, a few drops of ferric chloride (FeCl₃) solution are added.
The formation of a Prussian blue color (Fe₄[Fe(CN)₆]₃) indicates the presence of nitrogen.
Testing for Sulfur:
The fused mass is dissolved in water and a few drops of sodium nitroprusside (Na₂[Fe(CN)₅NO]) solution are added.
The presence of sulfur is signaled by a violet color..
Testing for Halogens:
The fused mass is dissolved in water and acidified with dilute nitric acid (HNO₃).
Silver nitrate (AgNO₃) solution is added.
The formation of a precipitate indicates the presence of a halogen:
White precipitate: Chloride (Cl⁻)
Cream-colored precipitate: Bromide (Br⁻)
Yellow precipitate: Iodide (I⁻)
Chemistry Behind the Test:
Nitrogen: The nitrogen in the organic compound is converted to sodium cyanide (NaCN) upon fusion with sodium. NaCN reacts with ferrous sulfate to form sodium ferrocyanide (Na₄[Fe(CN)₆]). Sodium ferrocyanide then reacts with ferric chloride to form Prussian blue.
Sulfur: The sulfur in the organic compound is converted to sodium sulfide (Na₂S) upon fusion with sodium. Na₂S reacts with sodium nitroprusside to form a violet-colored complex.
Halogens: The halogens in the organic compound are converted to their respective sodium halides (NaCl, NaBr, NaI) upon fusion with sodium. These halides react with silver nitrate to form insoluble silver halide precipitates.
In conclusion, Lassaigne’s test is a simple and effective method for detecting the presence of nitrogen, sulfur, and halogens in organic compounds.
22. Differentiate between the principle of estimation of nitrogen in an organic compound by
(i) Dumas method and
(ii) Kjeldahl’s method.
Ans : Comparison of Dumas and Kjeldahl Methods for Nitrogen Estimation
Dumas Method
Principle: The organic compound is heated with copper oxide (CuO) in a sealed tube. The nitrogen in the compound is converted to nitrogen gas (N₂). The volume of nitrogen gas is measured and used to calculate the percentage of nitrogen.
Process:
The organic compound is mixed with copper oxide and heated in a combustion tube.
The nitrogen from the compound is transformed into nitrogen gas.
The nitrogen gas is collected over water and its volume is measured.
The volume of nitrogen gas is used to calculate the mass of nitrogen in the compound.
Kjeldahl Method
Principle: The organic compound is heated with concentrated sulfuric acid and a catalyst (usually copper sulfate or potassium sulfate) to convert the nitrogen into ammonium sulfate (NH₄)₂SO₄. The ammonium sulfate is then treated with sodium hydroxide (NaOH) to release ammonia gas (NH₃). The ammonia gas is absorbed in a standard acid solution, and the amount of acid neutralized is used to calculate the amount of nitrogen.
Process:
The organic compound is heated with concentrated sulfuric acid and a catalyst.
The compound’s nitrogen undergoes a transformation into ammonium sulfate.
The ammonium sulfate is treated with sodium hydroxide to release ammonia gas.
The ammonia gas is absorbed in a standard acid solution.
The amount of acid neutralized is used to calculate the amount of ammonia, which is then used to calculate the amount of nitrogen in the compound.
Key Differences:
Dumas Method: Directly measures the volume of nitrogen gas produced. Suitable for compounds with high nitrogen content.
Kjeldahl Method: Indirectly measures nitrogen by determining the amount of ammonia produced. Suitable for a wide range of nitrogen-containing compounds.
Choice of Method:
The choice of method depends on the nature of the organic compound and the desired accuracy. For compounds with high nitrogen content, the Dumas method might be more suitable. For a wider range of compounds, the Kjeldahl method is often preferred due to its simplicity and versatility.
23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Ans : The estimation of halogens, sulfur, and phosphorus in organic compounds involves oxidizing the organic compound to convert the elements of interest into inorganic forms that can be easily detected and quantified.
Halogens:
Principle: The organic compound is heated with fuming nitric acid in the presence of silver nitrate. The halogen is converted to its corresponding silver halide salt, which is insoluble and can be precipitated, filtered, and weighed.
Example: In the case of chlorine, the reaction can be represented as:
R-Cl + AgNO₃ + HNO₃ → AgCl(s) + R-NO₂ + H₂O
The weight of silver chloride (AgCl) precipitate gives the amount of chlorine present in the organic compound.
Sulfur:
Principle: The organic compound is fused with sodium peroxide (Na₂O₂) or sodium hydroxide (NaOH). This converts sulfur into sulfate ions (SO₄²⁻). The sulfate ions can then be precipitated as barium sulfate (BaSO₄) by adding barium chloride (BaCl₂).
Example:
C₂H₅SH + 3Na₂O₂ → Na₂SO₄ + 2NaOH + 2H₂O
BaCl₂ + Na₂SO₄ → BaSO₄(s) + 2NaCl
The weight of barium sulfate (BaSO₄) precipitate gives the amount of sulfur present in the organic compound.
Phosphorus:
Principle: The organic compound is fused with sodium peroxide (Na₂O₂) or sodium hydroxide (NaOH). This converts phosphorus into phosphate ions (PO₄³⁻). The phosphate ions can then be precipitated as magnesium ammonium phosphate (MgNH₄PO₄·6H₂O) by adding magnesium chloride (MgCl₂) and ammonium hydroxide (NH₄OH).
Example:
C₆H₅PO₃H₂ + 3Na₂O₂ → Na₃PO₄ + 2NaOH + H₂O
MgCl₂ + NH₄OH + Na₃PO₄ → MgNH₄PO₄·6H₂O(s) + 2NaCl + 2NH₄Cl
The weight of magnesium ammonium phosphate precipitate gives the amount of phosphorus present in the organic compound.
In all these methods, the weight of the precipitate formed is used to calculate the amount of the respective element present in the organic compound.
24. Explain the principle of paper chromatography.
Ans : The principle of paper chromatography relies on the varying adsorption of components onto paper and their solubility in a solvent.
Here’s how it works:
Preparation: A small amount of the mixture is spotted onto a piece of filter paper.
Development: The paper is placed in a container with a solvent (mobile phase). The solvent moves up the paper by capillary force.
Separation: As the solvent moves, the components of the mixture interact differently with the paper and the solvent. Components that are more soluble in the solvent will travel further up the paper, while those that are more strongly adsorbed to the paper will move more slowly.
Visualization: The separated components can be visualized using various techniques, such as UV light, iodine vapor, or specific reagents.
The principle behind paper chromatography is the partitioning of the components between the stationary phase (paper) and the mobile phase (solvent). Components with a higher affinity for the mobile phase will travel further, while those with a higher affinity for the stationary phase will remain closer to the starting point.
Factors affecting the separation:
Nature of the solvent: The polarity of the solvent affects the solubility of the components.
Nature of the paper: The type of paper and its surface properties influence the adsorption of the components.
Temperature: The temperature can affect the solubility of the components and the rate of solvent movement.
Paper chromatography is a simple and effective technique for separating mixtures of small molecules, such as dyes, pigments, and amino acids.
25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Ans : Nitric acid is added to the sodium extract before adding silver nitrate to test for halogens to prevent the formation of interfering precipitates.
Here’s a breakdown of why:
Removal of Cyanide and Sulfide: Nitric acid reacts with cyanide (CN⁻) and sulfide (S²⁻) ions, which may be present in the sodium extract if the organic compound contains nitrogen or sulfur. This reaction converts these ions into harmless gases, preventing them from interfering with the test for halogens.
Acidic Medium: The addition of nitric acid creates an acidic environment, which is necessary for the reaction between silver nitrate and halide ions (Cl⁻, Br⁻, I⁻) to occur.
Formation of Silver Halides: In an acidic medium, silver nitrate reacts with halide ions to form insoluble silver halide precipitates, which can be used to identify the presence of halogens.
Therefore, nitric acid is added to the sodium extract before adding silver nitrate to ensure a reliable test for halogens by preventing interference from cyanide and sulfide ions and providing an acidic environment for the reaction to occur.
NaCN + HNO3 ——-> NaNO3 + HCN
Na2S + 2HNO3 ——> 2NaNO3 + H2S
26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Ans : The fusion of an organic compound with metallic sodium in the Lassaigne’s test is a crucial step to convert the elements of interest (nitrogen, sulfur, and halogens) into more reactive forms that can be easily detected.
Here’s a breakdown of why this process is necessary:
Conversion of Elements:
Nitrogen: Organic compounds containing nitrogen are converted into sodium cyanide (NaCN) upon fusion with sodium metal.
Sulfur: Organic compounds containing sulfur are converted into sodium sulfide (Na₂S).
Halogens: Organic compounds containing halogens are converted into sodium halides (e.g., NaCl, NaBr, NaI).
Increased Reactivity: The resulting sodium salts are more reactive than the original organic compounds. This makes it easier to detect the presence of these elements using specific tests.
Solubility: The sodium salts are generally soluble in water, allowing for easy testing and analysis.
In summary, the fusion of an organic compound with metallic sodium in the Lassaigne’s test is essential to convert the elements of interest into a more reactive and soluble form, making them easier to detect.
27. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Ans : Sublimation is a suitable technique for separating calcium sulfate and camphor.
Here’s a breakdown of why:
Camphor: This compound has a unique property of directly converting from a solid to a gas state without passing through the liquid state. This process is called sublimation.
Calcium Sulfate: This compound is non-sublimable, meaning it does not undergo sublimation.
Therefore, when the mixture of calcium sulfate and camphor is heated, the camphor will sublime, leaving behind the calcium sulfate.
The sublimed camphor can then be condensed back into a solid form, separating it from the calcium sulfate
28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ?
Ans : In steam distillation, an organic liquid vaporizes at a temperature below its boiling point due to the presence of steam.
Here’s a breakdown of the process:
Mixture Formation: The organic compound and water are mixed together in a distillation flask.
Heating: The mixture is heated.
Vaporization: The water in the mixture vaporizes, creating steam.
Partial Pressure Reduction: The presence of steam lowers the total vapor pressure of the mixture. According to Dalton’s Law of Partial Pressures, the total pressure of a gas mixture is the sum of the individual gas pressures.
Boiling Point Depression: The lowered total vapor pressure means that the mixture can boil at a temperature lower than the boiling point of either component alone.
Vaporization of Organic Compound: As the mixture boils, both the water vapor and the organic compound vaporize.
In essence, the addition of water to the organic compound lowers its effective vapor pressure, allowing it to vaporize at a temperature below its normal boiling point. This is the principle behind steam distillation.
29. Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Ans : No, CCl₄ will not give a white precipitate of AgCl when heated with silver nitrate.
Reason:
The carbon-chlorine bonds in CCl₄ are very strong due to the high electronegativity of chlorine. This makes it difficult for the silver ions (Ag⁺) to displace the chloride ions (Cl⁻) from CCl₄. As a result, no significant reaction occurs between CCl₄ and silver nitrate, and no white precipitate of AgCl is formed
In contrast, organic compounds containing a more reactive carbon-halogen bond, such as alkyl halides, can react with silver nitrate to form a white precipitate of AgCl. This is because the carbon-halogen bond in these compounds is weaker, making it easier for the silver ions to displace the halide ions.
30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Ans : Potassium hydroxide (KOH) is used to absorb carbon dioxide (CO₂) evolved during the estimation of carbon present in an organic compound because it forms a stable, water-soluble carbonate salt.
When carbon dioxide is passed through a solution of potassium hydroxide, it reacts to form potassium carbonate (K₂CO₃):
CO₂(g) + 2KOH(aq) → K₂CO₃(aq) + H₂O(l)
The formation of potassium carbonate removes the carbon dioxide from the gas mixture, allowing for accurate measurement of the remaining gases. This ensures that the carbon dioxide produced from the combustion of the organic compound is completely absorbed, leading to a precise determination of the carbon content.
31. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Ans : The use of acetic acid instead of sulfuric acid for acidification in the lead acetate test for sulfur is essential to avoid interference from sulfate ions.
Here’s a breakdown of why:
Sulfate Ion Interference: Sulfuric acid contains sulfate ions (SO₄²⁻), which can react with lead acetate to form lead sulfate (PbSO₄), a white precipitate. This precipitate can mask the formation of lead sulfide (PbS), the black precipitate that indicates the presence of sulfur.
Acetic Acid’s Role: Acetic acid, being a weak acid, provides a suitable acidic environment for the reaction without introducing additional sulfate ions. It ensures that the lead acetate reacts specifically with sulfide ions (S²⁻) to form the characteristic black precipitate of lead sulfide (PbS).
In summary, acetic acid is used for acidification in the lead acetate test to prevent interference from sulfate ions and ensure a reliable detection of sulfur.
32. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Ans : The given image shows the calculation of the masses of CO₂ and H₂O produced from the combustion of an organic compound containing 69% carbon and 4.8% hydrogen.
Step I: Calculation of mass of CO₂ produced
The mass of the organic compound is 0.20 g.
% of carbon in the compound = 69%.
The molar mass of carbon is 12 g/mol, and the molar mass of CO₂ is 44 g/mol.
Using the formula:
(Percentage of carbon / 12) x (Mass of CO₂ formed) / (44) x 100 = Mass of compound
find mass of CO₂ formed:
Mass of CO₂ formed
= (69 x 44 x 0.20 g) / (12 x 100)
= 0.506 g
Step II: Calculation of mass of H₂O produced
Using a similar approach, we can calculate the mass of H₂O formed:
Mass of H₂O formed
= (4.8 x 18 x 0.20 g) / (2 x 100)
= 0.0864 g
Therefore, the combustion of 0.20 g of the organic compound produces 0.506 g of CO₂ and 0.0864 g of H₂O.
33. A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Ans : The given image shows the calculation of the percentage of nitrogen in an organic compound using the Kjeldahl method. The Kjeldahl method involves converting the nitrogen in the organic compound to ammonium sulfate, which is then reacted with sodium hydroxide to release ammonia gas. The ammonia gas is absorbed in a standard acid solution, and the amount of acid consumed is used to determine the amount of nitrogen in the original compound.
Here’s a breakdown of the calculations:
Step I: Calculation of volume of unused acid
The volume of NaOH solution required to neutralize the excess acid is 60 cm³.
The normality of NaOH solution is 1/2 N.
The normality of H₂SO₄ solution is 1/N.
Using the normality equation, we can calculate the volume of unused acid:
(Normality of acid) x (Volume of acid) = (Normality of base) x (Volume of base)
Substituting the values, we get:
(1/N) x V = (1/2 N) x 60 cm³
Solving for V, we get:
V = 30 cm³
Step II: Calculation of volume of acid used
The total volume of acid added is 50 cm³.
The volume of unused acid is 30 cm³.
Therefore, the volume of acid used is 50 cm³ – 30 cm³ = 20 cm³.
Step III: Calculation of percentage of nitrogen
The mass of the compound is 0.50 g.
The volume of acid used is 20 cm³.
The normality of acid used is 1 N.
The percentage of nitrogen can be calculated using the formula:
Percentage of N = (1.4 x Volume of acid used x Normality of acid used) / Mass of the compound
Substituting the values, we get:
Percentage of N = (1.4 x 20 x 1) / 0.50
= 56%
Therefore, the percentage of nitrogen in the organic compound is 56%
34. 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Ans : To calculate the percentage of chlorine in the organic compound, we need to determine the mass of chlorine present in the silver chloride formed.
Calculate the moles of silver chloride:
Moles of AgCl =
mass of AgCl
——————————————-
molar mass of AgCl
Moles of AgCl = 0.5740 g / 143.32 g/mol ≈ 0.004 mol
Determine the moles of chlorine:
One mole of AgCl contains one mole of chlorine.
Moles of Cl =
Moles of AgCl
= 0.004 mol
Calculate the mass of chlorine:
Mass of Cl =
Moles of Cl * molar mass of Cl
Mass of Cl =
0.004 mol * 35.45 g/mol ≈ 0.1418 g
Calculate the percentage of chlorine in the organic compound:
Percentage of Cl = (mass of Cl / mass of organic compound) * 100%
Percentage of Cl = (0.1418 g / 0.3780 g) * 100% ≈ 37.57%
Therefore, the percentage of chlorine present in the organic compound is approximately 37.57%.
35. In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Ans : The Carius method involves heating an organic compound with fuming nitric acid and silver nitrate in a sealed tube. The sulfur in the compound is converted to sulfuric acid, which then reacts with barium chloride to form a precipitate of barium sulfate. The percentage of sulfur in the compound can be calculated by measuring the mass of barium sulfate formed.
Calculations:
Calculate the moles of barium sulfate:
Moles of BaSO₄ = mass of BaSO₄ / molar mass of BaSO₄
Moles of BaSO₄ = 0.668 g / 233.39 g/mol ≈ 0.00286 mol
Calculate the moles of sulfur:
Each mole of barium sulfate includes one mole of sulfur.
Moles of S = Moles of BaSO₄ = 0.00286 mol
Calculate the mass of sulfur:
Mass of S = Moles of S * molar mass of S
Mass of S = 0.00286 mol * 32.07 g/mol ≈ 0.0917 g
Ascertain the percentage of sulfur present in the compound.
Percentage of S = (mass of S / mass of organic compound) * 100%
Percentage of S = (0.0917 g / 0.468 g) * 100% ≈ 19.6%
Therefore, the percentage of sulfur in the given organic compound is approximately 19.6%.
36. In the organic compound CH2 = CH – CH2 – CH2 – C ≡ CH, the pair of hydridised orbitals involved in the formation of: C2 – C3 bond is:
(a) sp – sp2
(b) sp – sp3
(c) sp2 – sp3
(d) sp3 – sp3
Ans : The C2-C3 bond in the molecule CH2=CH-CH2-CH2-C≡CH involves sp2 – sp3 hybridization.
Here’s the breakdown:
C2: The carbon at position 2 is involved in a double bond with C3. Double bonds involve one σ bond (formed by sp² hybrid orbitals) and one π bond (formed by overlapping p orbitals).
C3: This carbon is also involved in a double bond with C2. Therefore, it is also sp² hybridized.
So, the orbitals involved in the C2-C3 bond are sp² hybrid orbitals.
37. In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
(a) Na4[Fe(CN)6]
(b) Fe4[Fe(CN)6] 3
(c) Fe2[Fe(CN)6]
(d) Fe3[Fe(CN)6] 4
Ans : The Prussian blue color obtained in the Lassaigne’s test for nitrogen is due to the formation of Fe₄[Fe(CN)₆]₃.
Explanation:
In the Lassaigne’s test, the organic compound is fused with sodium metal to convert the nitrogen into sodium cyanide (NaCN). Then, ferrous sulfate (FeSO₄) is added followed by ferric chloride (FeCl₃).
The reaction between ferrous sulfate and sodium cyanide produces sodium ferrocyanide (Na₄[Fe(CN)₆]).
Na₄[Fe(CN)₆] + 2FeCl₃ → Fe₄[Fe(CN)₆]₃ + 4NaCl
The Fe₄[Fe(CN)₆]₃ compound is Prussian blue in color, indicating the presence of nitrogen in the original organic compound.
38. Which of the following carbocation is most stable ?
(a) (CH3) 3C. C+ H2
(b) (CH3) 3C +
(c) CH3CH2C + H2
(d) CH3C + H CH2CH3
Ans : The most stable carbocation is (b) (CH₃)₃C⁺.
Explanation:
The stability of a carbocation is directly correlated to the number of alkyl groups bonded to its positive carbon. More alkyl groups increase the stability of the carbocation due to the electron-donating inductive effect of these groups, which helps to disperse the positive charge.
In the given options:
(a) (CH₃)₃C. C⁺H₂: This is a primary carbocation, with only one alkyl group attached to the positive carbon.
(b) (CH₃)₃C⁺: This is a tertiary carbocation, with three alkyl groups attached to the positive carbon.
(c) CH₃CH₂C⁺H₂: This is a primary carbocation, with only one alkyl group attached to the positive carbon.
(d) CH₃C⁺HCH₂CH₃: This is a secondary carbocation, with two alkyl groups attached to the positive carbon.
Therefore, (b) (CH₃)₃C⁺, being a tertiary carbocation, is the most stable due to the electron-donating effect of the three methyl groups.
39. The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation
(b) Distillation
(c) Sublimation
(d) Chromatography
Ans : Chromatography is the most advanced method for isolating, purifying, and separating organic substances.
Here’s a brief explanation of why chromatography is superior to the other options:
Versatility: Chromatography can be used to separate a wide range of organic compounds, from simple molecules to complex mixtures.
Sensitivity: Chromatography can detect and separate even small amounts of compounds.
Efficiency: Chromatography is often more efficient than other techniques, requiring less solvent and time.
Specificity: Chromatography can be tailored to separate specific types of compounds based on their properties.
While crystallization, distillation, and sublimation are also valuable techniques in organic chemistry, chromatography offers a more comprehensive and versatile approach for the isolation, purification, and separation of organic compounds.
40. The reaction: CH3CH2I + KOH(aq) → CH3CH2OH + KI is classified as :
(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition
Ans : The reaction CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI is classified as nucleophilic substitution.
Here’s why:
Nucleophile: The hydroxide ion (OH⁻) from KOH acts as a nucleophile, a species that is attracted to positive centers.
Electrophile: The carbon atom attached to the iodine in CH₃CH₂I is electron-deficient (electrophilic) due to the electronegativity of iodine.
Substitution: The nucleophile (OH⁻) attacks the electrophilic carbon, leading to the substitution of the iodide ion (I⁻) with the hydroxide ion.
Therefore, the reaction is a nucleophilic substitution reaction.
