GSEB Class 11 Chemistry Solutions Chapter 12 Organic Chemistry Some Basic Principles and Techniques

 

Question 1. What are hybridisation states of each carbon atom in the following compounds? CH2=C=O, CH3-CH=CH2, (CH3)2C=O, CH2=CHCN, C6H6
Answer: The hybridisation states for each carbon atom in the listed compounds are as follows:

• \( \text{CH}_2=\text{C}=\text{O} \): \( \text{sp}^2 \), \( \text{sp} \), \( \text{sp}^2 \)
• \( \text{CH}_3-\text{CH}=\text{CH}_2 \): \( \text{sp}^3 \), \( \text{sp}^2 \), \( \text{sp}^2 \)
• \( (\text{CH}_3)_2\text{C}=\text{O} \): \( \text{sp}^3 \), \( \text{sp}^3 \), \( \text{sp}^2 \)
• \( \text{CH}_2=\text{CHCN} \): \( \text{sp}^2 \), \( \text{sp}^2 \), \( \text{sp} \)
• \( \text{C}_6\text{H}_6 \): \( \text{sp}^2 \)

In simple words: We look at the bonds around each carbon atom to figure out its hybridisation. Single bonds often mean sp³ , double bonds mean sp², and triple bonds mean sp. In a benzene ring, all carbons are sp².

Exam Tip: Remember to count the number of sigma bonds and lone pairs (steric number) around each carbon atom to determine its hybridisation state. Double bonds count as one sigma and one pi bond, and triple bonds as one sigma and two pi bonds.

 

Question 1. Indicate the σ and π bonds in the following molecules: C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3.
Answer: The σ (sigma) and π (pi) bonds for each of the given molecules are indicated below:

• \( \text{C}_6\text{H}_6 \) (Benzene): This molecule has 6 C-C sigma bonds, 6 C-H sigma bonds, and 3 C-C pi bonds.
• \( \text{C}_6\text{H}_{12} \) (Cyclohexane): This compound possesses 6 C-C sigma bonds and 12 C-H sigma bonds.
• \( \text{CH}_2\text{Cl}_2 \) (Dichloromethane): It contains 2 C-H sigma bonds and 2 C-Cl sigma bonds.
• \( \text{CH}_2=\text{C}=\text{CH}_2 \) (Allene): There are two C-C sigma bonds, four C-H sigma bonds, and two C-C pi bonds.
• \( \text{CH}_3\text{NO}_2 \) (Nitromethane): This molecule has three C-H sigma bonds, one C-N sigma bond, two N-O sigma bonds, and one N=O pi bond.
• \( \text{HCONHCH}_3 \) (N-Methylformamide): This structure contains one C=O pi bond, one C-N sigma bond, three C-H sigma bonds (from CH3), one N-H sigma bond, and one C-H sigma bond (from CHO group). (Note: The OCR text for this molecule was corrupt; the bond count is based on the standard structure of N-Methylformamide.)

In simple words: Sigma bonds are the first connection between two atoms, like a single thread. Pi bonds are extra connections, like a second or third thread, found in double or triple bonds. We count them to understand how atoms link together in a molecule.

Exam Tip: A single bond is always a sigma bond. A double bond consists of one sigma and one pi bond. A triple bond has one sigma and two pi bonds. Drawing the Lewis structure can help you count correctly.

 

Question 3. Write bond line formulas for: isopropyl alcohol, 2, 3-Dimethyl butanal, Heptan-4-one.
Answer: The bond line formulas for the given compounds are standard representations in organic chemistry.

• **Isopropyl alcohol**: This is a three-carbon chain with a hydroxyl (-OH) group attached to the middle carbon. Its formula is \( (\text{CH}_3)_2\text{CHOH} \).
• **2, 3-Dimethyl butanal**: This is a four-carbon aldehyde chain (butanal) with two methyl groups attached at the 2nd and 3rd carbon positions. Its formula is \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}(\text{CH}_3)\text{CHO} \).
• **Heptan-4-one**: This is a seven-carbon chain (heptane) with a ketone (C=O) group at the 4th carbon position. Its formula is \( \text{CH}_3\text{CH}_2\text{CH}_2\text{C}(=\text{O})\text{CH}_2\text{CH}_2\text{CH}_3 \).

In simple words: Bond line formulas show molecules simply, using lines for carbon bonds and assuming carbons and hydrogens are at line ends and bends. We then add other atoms like oxygen or nitrogen.

Exam Tip: When drawing bond line formulas, remember that each vertex or end of a line represents a carbon atom, and hydrogen atoms attached to carbons are usually not shown but are implied to complete carbon's valency (4 bonds).

 

Question 4. Give the I.U.P. A.C. names of the following compounds:
Answer: The IUPAC names for the compounds shown in the figures are:
(a) Propylbenzene
(b) 3-Methylpentane nitrile
(c) 2, 5-Dimethylheptane
(d) 3-Bromo-3-Chloroheptane
(e) 3-Chloropropan-1-al
(f) 2, 2-Dichloroethanol.In simple words: IUPAC names are like official names for chemicals, based on rules. They tell us exactly what atoms are in a molecule and how they are joined together, making it simple to understand its structure.

Exam Tip: Always identify the longest continuous carbon chain, then number it to give the lowest possible numbers to functional groups and substituents. Finally, arrange substituents alphabetically.

 

Question 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.
Answer: The formulas for the first five members of each homologous series are as follows:
(a) For compounds beginning with \( \text{H-COOH} \) (carboxylic acids):

• \( \text{H-COOH} \) (Methanoic acid)
• \( \text{CH}_3\text{COOH} \) (Ethanoic acid)
• \( \text{CH}_3\text{CH}_2\text{COOH} \) (Propanoic acid)
• \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \) (Butanoic acid)
• \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH} \) (Pentanoic acid)

(b) For compounds beginning with \( \text{CH}_3\text{COCH}_3 \) (ketones):

• \( \text{CH}_3\text{COCH}_3 \) (Propan-2-one)
• \( \text{CH}_3\text{COCH}_2\text{CH}_3 \) (Butan-2-one)
• \( \text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3 \) (Pentan-3-one)
• \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 \) (Pentan-2-one)
• \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_2\text{CH}_2\text{CH}_3 \) (Heptan-4-one)

(c) For compounds beginning with \( \text{H-CH}=\text{CH}_2 \) (alkenes):

• \( \text{H}_2\text{C}=\text{CH}_2 \) (Ethene)
• \( \text{CH}_3-\text{CH}=\text{CH}_2 \) (Propene)
• \( \text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}_2 \) (But-1-ene)
• \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}=\text{CH}_2 \) (Pent-1-ene)
• \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}=\text{CH}_2 \) (Hex-1-ene)

In simple words: A homologous series is a group of compounds with the same basic structure, where each member differs from the next by a simple \( \text{CH}_2 \) unit. We are listing the first five compounds in these groups.

Exam Tip: To generate members of a homologous series, simply add a \( -\text{CH}_2- \) unit to the carbon chain each time, ensuring the functional group remains in its characteristic position.

 

Question 7. Give condensed and bond line structural formulas and identify the functional group(s) present if any, for: (a) 2, 2, 4-Trimethypentane (b) 2-Hydroxy-I, 2, 3-propanetricarboxylic acid (c) Hexanedial.
Answer: The condensed structural formulas and functional groups for the given compounds are:
(a) **2, 2, 4-Trimethylpentane**

• Condensed formula: \( (\text{CH}_3)_3\text{CCH}_2\text{CH}(\text{CH}_3)_2 \)
• Functional group(s): Alkane (only C-C and C-H single bonds)

(b) **2-Hydroxy-1, 2, 3-propanetricarboxylic acid**

• Condensed formula: \( \text{HOOC-CH}_2\text{C(OH)(COOH)CH}_2\text{COOH} \)
• Functional group(s): Hydroxyl (alcohol) group, Carboxylic acid groups

(c) **Hexanedial**

• Condensed formula: \( \text{OHC(CH}_2)_4\text{CHO} \)
• Functional group(s): Aldehyde groups

In simple words: We write a short version of the molecule's structure (condensed formula) and then name the special groups of atoms (functional groups) that give the molecule its main properties.

Exam Tip: When writing condensed formulas, show branching by enclosing the branched group in parentheses. For functional groups, look for characteristic atoms like oxygen (alcohols, ethers, aldehydes, ketones, acids), nitrogen (amines, nitriles), or halogens.

 

Question 8. Identify the functional groups in the following compounds.
Answer: The functional groups in the given compounds are:
(a) The compound contains a -CHO (aldehyde) group, an -OH (hydroxy) group, and an -OMe (methoxy) group (which is an ether linkage).
(b) The compound contains an -NH\( _2 \) (amino) group (specifically a primary amine) and a -CH\( _2 \text{N(C}_2\text{H}_5)_2 \) group, which is a tertiary amino group. It also features a -COOCH\( _3 \) (ester) group.
(c) The compound contains a -CH=CH- (alkene) group (an ethylenic double bond) and an -NO\( _2 \) (nitro) group.In simple words: Functional groups are specific sets of atoms within a molecule that determine how it reacts. We just need to point out these special parts.

Exam Tip: Familiarize yourself with common functional groups and their characteristic bonding patterns (e.g., C=O for carbonyls, -OH for alcohols, -NH2 for amines) to quickly identify them in a given structure.

 

Question 9. Which of the two: O2N-CH2-CH2O or CH3-CH2-O¯ is expected to be more stable and why?
Answer: \( \text{O}_2\text{N-CH}_2\text{CH}_2\text{O}^- \) is expected to be more stable than \( \text{CH}_3-\text{CH}_2\text{O}^- \).
In \( \text{O}_2\text{N} \leftarrow \text{CH}_2\text{O-CH}_2\text{O}^- \), the \( \text{-NO}_2 \) group applies a -I (inductive) effect. This effect helps to spread out the negative charge, making it less concentrated. However, in \( \text{CH}_3 \rightarrow \text{CH}_2\text{O}^- \), the \( \text{-CH}_3 \) group exhibits a +I effect, which intensifies the negative charge on the oxygen atom. The dispersal of a charge generally results in greater stability for the ion.In simple words: The first molecule is more stable because the \( \text{NO}_2 \) part pulls electrons away from the negatively charged oxygen, spreading out the charge and making it steadier. The second molecule has a \( \text{CH}_3 \) part that pushes electrons towards the oxygen, making the negative charge more intense and less stable.

Exam Tip: Electron-withdrawing groups stabilize negative charges by dispersing them (negative inductive effect), while electron-donating groups destabilize negative charges by intensifying them (positive inductive effect).

 

Question 10. Explain why alkyl groups act as electron donors when attached to a π system.
Answer: Alkyl groups act as electron donors when attached to a \( \pi \) system primarily due to the +I (positive inductive) effect. Because of this effect, electrons are released towards the carbon atom holding the \( \pi \) bond next to it. This electron donation can be seen in reactions such as the addition of an attacking reagent to an alkene.
When an attacking reagent is present, the \( \pi \) electron-pair in a multiple bond will shift. This shifting of the \( \pi \) electron-pair in a multiple bond at the call of the attacking reagent is termed the Electromeric effect. \[ \text{R}-\text{CH}=\text{CH}_2 \quad \xrightarrow{\text{attacking reagent}} \quad \text{R}-\text{CH}^+-\text{CH}_2^- \]In simple words: Alkyl groups give electrons to a double or triple bond. This happens because they have a slight push of electrons (called the inductive effect). When another molecule tries to react with the double bond, these electrons can shift easily (called the electromeric effect), helping the reaction to occur.

Exam Tip: Remember that the positive inductive effect (+I effect) is a permanent electron displacement through sigma bonds, while the electromeric effect (E-effect) is a temporary, reagent-induced electron transfer in pi systems.

 

Question 11. Draw the resonance structures for the following compounds. Show the electron shift rrow notation, (a) Phenol (b) C6H5NO2 (c) CH3CH = CHCHO (d) C6H5-CHO (e) C6H5-C+H2 (f) CH3CH = CHCH+2
Answer: The resonance structures illustrate the delocalisation of electrons within a molecule, which contributes to its stability. Electron shifts are typically shown using curved arrows. Due to the complexity of drawing these intricate structures in plain text, a conceptual explanation is provided for each compound.
(a) **Resonance structures of Phenol**: Phenol has five canonical (contributing) resonance structures. The lone pair of electrons on the oxygen atom is delocalised into the benzene ring, creating partial negative charges at the ortho and para positions and a partial positive charge on the oxygen. Phenol is a resonance hybrid of these five structures.
(b) **Resonance structures of nitrobenzene (\( \text{C}_6\text{H}_5\text{NO}_2 \))**: Nitrobenzene also shows resonance, primarily involving the nitro group. The \( \text{-NO}_2 \) group is an electron-withdrawing group, so it pulls electrons from the benzene ring, leading to positive charges at the ortho and para positions and delocalisation within the nitro group itself. It typically has four resonating structures.
(c) **Resonance structures of \( \text{CH}_3\text{CH}=\text{CHCHO} \)** (Crotonaldehyde): This conjugated system involves electron delocalization from the carbon-carbon double bond towards the more electronegative oxygen of the aldehyde group. The \( \pi \) electrons shift across the conjugated system, creating partial positive and negative charges at different atoms.
(d) **Resonance structures of \( \text{C}_6\text{H}_5\text{CHO} \)** (Benzaldehyde): In benzaldehyde, the \( \pi \) electrons of the carbonyl group and the benzene ring are conjugated. The oxygen atom of the carbonyl group draws electron density, causing electron delocalization from the ring, resulting in a positive charge at the ortho and para positions of the benzene ring and a negative charge on the oxygen.
(e) **Resonance structures of \( \text{C}_6\text{H}_5\text{C}^+\text{H}_2 \)** (Benzyl carbocation): In this carbocation, the positive charge on the carbon adjacent to the benzene ring is delocalized into the ring. This delocalization stabilizes the carbocation by spreading the positive charge over the ortho and para positions of the ring.
(f) **Resonance structures of \( \text{CH}_3\text{CH}=\text{CHCH}^+_2 \)** (But-2-enyl carbocation): This is an allylic carbocation, where the positive charge is conjugated with the carbon-carbon double bond. The \( \pi \) electrons of the double bond shift to stabilize the positive charge, creating another resonating structure where the positive charge is located at the other end of the double bond.In simple words: Resonance structures are different ways to draw a molecule where electrons can be in more than one place. They show how electrons move around, especially in double bonds or lone pairs, which makes the molecule more stable. We use curved arrows to show this electron movement.

Exam Tip: When drawing resonance structures, remember that only electrons (pi electrons and lone pairs) move, not atoms. The overall charge of the molecule must be conserved in all contributing structures. Electron-donating groups direct to ortho/para, while electron-withdrawing groups direct to meta and deactivate the ring towards electrophilic attack.

 

Question 12. What are electrophiles and nucleophiles? Explain with examples.
Answer:
**Electrophiles**: These are reagents that are looking for electrons. They are called electron-seeking reagents because they can accept a pair of electrons. Electrophiles are typically positively charged or have an incomplete octet, making them Lewis acids.

• **Neutral electrophiles**: Examples include \( \text{BF}_3 \), carbene (\( :\text{CR}_2 \)), \( \text{AlCl}_3 \), \( \text{FeCl}_3 \). These have an empty orbital to accept electrons.
• **Positive electrophiles**: Examples include hydrogen ion (\( \text{H}^+ \)), hydronium ion (\( \text{H}_3\text{O}^+ \)), chlorine cation (\( \text{Cl}^+ \)), bromine cation (\( \text{Br}^+ \)), iodine cation (\( \text{I}^+ \)), nitronium ion (\( \text{NO}_2^+ \)), nitrosonium ion (\( \text{NO}^+ \)), and alkyl carbocations (\( \text{R}^+ \)).

**Nucleophiles**: These are reagents that are looking for a relatively positive center (a nucleus-loving site). They are electron-rich species that possess at least one lone pair of electrons or a \( \pi \) bond, making them Lewis bases.

• **Neutral nucleophiles**: Examples include water (\( \text{H}_2\ddot{\text{O}} \)), ammonia (\( \text{N}\text{H}_3 \)), primary amines (\( \text{R}\ddot{\text{N}}\text{H}_2 \)), alcohols (\( \text{ROH} \)), thiols (\( \text{RSH} \)), and ethers (\( \text{ROR} \)).
• **Negative nucleophiles**: Examples include hydride ion (\( \text{H}^- \)), chloride ion (\( \text{Cl}^- \)), bromide ion (\( \text{Br}^- \)), iodide ion (\( \text{I}^- \)), alkyl anions (\( \text{R}^- \)), hydroxide ion (\( \text{OH}^- \)), alkoxide ion (\( \text{OR}^- \)), thiolate ion (\( \text{SR}^- \)), amide ion (\( \text{NH}_2^- \)), cyanide ion (\( \text{CN}^- \)), and carboxylate ion (\( \text{RCOO}^- \)).

In simple words: Electrophiles are like electron magnets; they are attracted to electrons because they need them. Nucleophiles are like electron donors; they have extra electrons and like to find places with fewer electrons to share them with.

Exam Tip: Remember that electrophiles are Lewis acids (electron pair acceptors), and nucleophiles are Lewis bases (electron pair donors). The presence of a positive charge or an empty orbital signifies an electrophile, while a lone pair or a negative charge indicates a nucleophile.

 

Question 13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles. (a) CH3COOH+ HO¯ → CH3COO + H2O (b) CH3COCH3 + CN¯ → (CH3)2C(CN) (OH) (c) C6H5 + CH3C+O → C6H5COCH3.
Answer: We need to identify whether the bolded reagents are nucleophiles or electrophiles in each reaction:
(a) In the reaction \( \text{CH}_3\text{COOH} + \text{HO}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \), \( \text{HO}^- \) is a **nucleophile**. It donates its electron pair to the acidic proton of acetic acid.
(b) In the reaction \( \text{CH}_3\text{COCH}_3 + \text{CN}^- \rightarrow (\text{CH}_3)_2\text{C}(\text{CN})(\text{OH}) \), \( \text{CN}^- \) is a **nucleophile**. It attacks the electrophilic carbon of the ketone.
(c) In the reaction \( \text{C}_6\text{H}_5 + \text{CH}_3\text{C}^+\text{O} \rightarrow \text{C}_6\text{H}_5\text{COCH}_3 \), \( \text{CH}_3\text{C}^+\text{O} \) (acetyl cation) is an **electrophile**. It accepts electrons from the benzene ring.In simple words: We check if the bolded molecule is giving away electrons (nucleophile) or wanting electrons (electrophile) in the reaction. If it has a negative charge or a lone pair and attacks something positive, it's a nucleophile. If it has a positive charge or needs electrons and attacks something electron-rich, it's an electrophile.

Exam Tip: Look at the electron flow. If the reagent has a negative charge or lone pair and initiates the attack, it's a nucleophile. If it has a positive charge or an electron deficiency and is being attacked, it's an electrophile.

 

Question 14. Classify the following reactions in one of the reaction type studied in this unit: (a) CH3CH2Br + HS¯ → CH3CH2SH (b) (CH3)2C = CH2 + HCI → (CH3)2 CIC (c) CH3CH2Br + HO¯ → CH2 = CH2 + H2O (d) (CH3)3 C - CH2OH + HBr → (CH3)2 CBrCH2OH
Answer: Here are the classifications for each of the provided reactions:
(a) \( \text{CH}_3\text{CH}_2\text{Br} + \text{HS}^- \rightarrow \text{CH}_3\text{CH}_2\text{SH} \): This is a **nucleophilic substitution reaction**. The \( \text{HS}^- \) nucleophile replaces the \( \text{Br}^- \) leaving group.
(b) \( (\text{CH}_3)_2\text{C} = \text{CH}_2 + \text{HCl} \rightarrow (\text{CH}_3)_2\text{CClCH}_3 \): This is an **electrophilic addition reaction**. The \( \pi \) bond of the alkene attacks the \( \text{H}^+ \) electrophile, followed by \( \text{Cl}^- \) addition.
(c) \( \text{CH}_3\text{CH}_2\text{Br} + \text{HO}^- \rightarrow \text{CH}_2 = \text{CH}_2 + \text{H}_2\text{O} + \text{Br}^- \): This is a **beta-elimination reaction** (or simply elimination reaction). The hydroxide acts as a base, removing a proton and leading to the formation of an alkene.
(d) \( (\text{CH}_3)_3\text{C}-\text{CH}_2\text{OH} + \text{HBr} \rightarrow (\text{CH}_3)_2\text{CBrCH}_2\text{OH} \): This is a **substitution reaction and rearrangement reaction**. The hydroxyl group is substituted by bromine, likely involving a carbocation intermediate that undergoes rearrangement to a more stable structure before the bromine attaches.In simple words: We are sorting chemical reactions into groups based on how atoms change places. Substitution means one part is swapped, addition means parts are added to a double bond, elimination means parts are removed to make a double bond, and rearrangement means atoms move within the molecule.

Exam Tip: Distinguish between substitution and elimination by observing if a new group replaces an old one (substitution) or if a double bond forms by removing two groups (elimination). Addition reactions break pi bonds to form new sigma bonds.

 

Question 15. What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Answer: We need to determine the relationship between the members of each pair of structures:
(a) The structures are position isomers. They differ in the position of the functional group. Since their connectivity of atoms is different, they are also considered **structural isomers**.
(b) These structures are **geometrical isomers** (specifically, trans-isomer and cis-isomer). They have the same connectivity of atoms but differ in the spatial arrangement of groups around a double bond.
(c) The structures are **resonance contributors**. They represent different possible electron distributions within the same molecule, illustrating electron delocalization.In simple words: We're checking how molecules that look a bit different are related. Structural isomers have different atom connections. Geometrical isomers have atoms arranged differently in space around a double bond. Resonance contributors are just different ways to draw the same molecule, showing how electrons move around.

Exam Tip: Structural isomers (constitutional isomers) have different atom-to-atom bonding sequences. Geometrical isomers (cis/trans or E/Z) arise from restricted rotation, typically around double bonds. Resonance contributors show electron delocalization, meaning only electrons move, not atoms.

 

Question 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. (a) CH3O-OCH3 → CH3O+ OCH₃ (b) =O + OH- → =O- + H₂O (c) Br → + Br- (d) + E → + E
Answer: Here's the classification of each bond cleavage and the reactive intermediate formed:
(a) \( \text{CH}_3\text{O-OCH}_3 \rightarrow \text{CH}_3\text{O} \cdot + \cdot \text{OCH}_3 \)

• This is a case of **homolytical fission**.
• The reactive intermediates formed are **free-radicals**.

(b) \( \text{C=O} + \text{OH}^- \rightarrow \text{C=O}^- + \text{H}_2\text{O} \) (Representing a generic carbonyl reacting)

• This is a case of **heterolysis**.
• The reaction intermediate formed is a **carbanion**. (In the source, it describes \( \text{C=O} + \text{OH}^- \rightarrow \text{C=O}^- + \text{H}_2\text{O} \), which implies the oxygen of the carbonyl becomes negatively charged, resembling a carbanion for the carbon center in the larger molecule where \( \text{O}^- \) is one of the groups).

(c) \( \text{R-Br} \rightarrow \text{R}^+ + \text{Br}^- \) (Representing an alkyl halide)

• This is a case of **heterolysis** (also called heterolytic bond fission).
• The reaction intermediate formed is a **carbocation**.

(d) \( \text{ArH} + \text{E}^+ \rightarrow \text{Ar-E} + \text{H}^+ \) (Representing electrophilic aromatic substitution)

• This is a case of **heterolysis** (heterolytic bond fission).
• The reaction intermediate formed is a **carbocation** (specifically, a sigma complex in aromatic substitution).

In simple words: When a chemical bond breaks, it can do so in two ways. Homolysis is when each atom gets one electron, forming neutral, highly reactive "free radicals." Heterolysis is when one atom takes both electrons, forming charged particles called "carbocations" (positive) or "carbanions" (negative).

Exam Tip: Homolytic cleavage generates free radicals, each with an unpaired electron. Heterolytic cleavage generates ions (a carbocation and an anion), where one atom gains both bonding electrons and the other loses them.

 

Question 17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? (a) CI3CCOOH > CI2CHCOOH > CICH2COOH (b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH
Answer:
**Inductive effect**: This refers to the displacement of an electron cloud along a saturated carbon chain. It happens whenever an electron-withdrawing or electron-donating group is present at the end of the chain. This effect weakens steadily as the distance from the substituent increases, usually dying down after three carbon atoms.
There are two main types of Inductive effects:
(i) **Negative Inductive effect (-I effect)**: This happens if the substituent attached to the end of the chain is electron-withdrawing. It pulls electron density away from the chain.
\( \delta\delta\delta^+ \leftarrow \text{C} \leftarrow \delta\delta^+ \leftarrow \text{C} \leftarrow \delta^+ \leftarrow \text{C} \leftarrow \text{X} \) (\( \text{X} \) is electron-withdrawing)
The -I effect decreases in the order:
\( \text{-NO}_2 > \text{-CN} > \text{-COOH} > \text{-F} > \text{-Cl} > \text{-Br} > \text{-I} \)
(ii) **Positive Inductive effect (+I effect)**: This happens if the substituent attached to the end of the carbon chain is electron-donating. It pushes electron density into the chain.
\( \delta^- \rightarrow \text{C} \rightarrow \delta\delta^- \rightarrow \text{C} \rightarrow \delta\delta\delta^- \rightarrow \text{CH}_3 \)
The +I effect of some atoms or groups in decreasing order is:
\( (\text{CH}_3)_3\text{C} \rightarrow (\text{CH}_3)_2\text{CH} \rightarrow \text{CH}_3\text{CH}_2 \rightarrow \text{CH}_3 \rightarrow \text{D} > \text{H} \)
(t-Butyl > iso-propyl > Ethyl > Methyl > Deuterium > Hydrogen)
The inductive effect is a permanent effect that operates in the ground state of organic molecules. It helps explain properties like high melting points, boiling points, and dipole moments of polar compounds.

**Electromeric Effect**: This involves the complete transfer of electrons from a multiple bond (double or triple bond) to one of the bonded atoms (usually the more electronegative one) in the presence of an attacking reagent.
\[ \text{C=O} \xrightarrow{\text{reagent added}} \text{C}^+-\text{O}^- \] \[ \text{C}^+-\text{O}^- \xleftarrow{\text{reagent removed}} \text{C=O} \]
This effect is temporary and only occurs when a reagent is present. Once the reagent is removed, the molecule returns to its original position. The electromeric effect also has two types:
**+E effect**: If the electrons of the \( \pi \) bond transfer to the atom of the double bond where the reagent ultimately attaches. For example, in the addition of acids to alkenes:
\( \text{C=C} + \text{H}^+ \rightarrow \text{C}^+-\text{C} \text{H} \)
**-E effect**: If the electrons of the double bond transfer to an atom of the double bond other than the one where the reagent finally attaches.

**Explanation of acidity order for carboxylic acids**:
The acidity order of carboxylic acids is explained by the **Inductive effect (-I effect and +I effect)**.
(a) \( \text{Cl}_3\text{CCOOH} > \text{Cl}_2\text{CHCOOH} > \text{ClCH}_2\text{COOH} \)
This order is due to the **-I effect** of chlorine atoms. Chlorine is an electron-withdrawing group. More chlorine atoms present mean a stronger -I effect, which pulls electron density away from the carboxylate anion ( \( \text{COO}^- \) ), stabilizing it. Greater stability of the conjugate base translates to stronger acidity.
(b) \( \text{CH}_3\text{CH}_2\text{COOH} > (\text{CH}_3)_2\text{CHCOOH} > (\text{CH}_3)_3\text{C.COOH} \)
This order is due to the **+I effect** of alkyl groups. Alkyl groups are electron-donating. As the number and branching of alkyl groups increase, the +I effect intensifies, pushing more electron density towards the carboxylate anion. This destabilizes the conjugate base, making the acid weaker.In simple words: The Inductive effect is a tiny push or pull of electrons through single bonds that makes molecules a bit positive or negative in certain spots. The Electromeric effect is a big, temporary shift of electrons in double or triple bonds when a reaction is happening. For acids, electron-pulling groups make them stronger by making their leftover negative part more stable, while electron-pushing groups make them weaker.

Exam Tip: For acidity, always consider the stability of the conjugate base. Electron-withdrawing groups (like halogens, -NO2) stabilize conjugate bases via the -I effect, increasing acidity. Electron-donating groups (like alkyl groups) destabilize conjugate bases via the +I effect, decreasing acidity.

 

Question 18. Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography.
Answer: Here are the brief descriptions of the principles of each technique:
(a) **Crystallisation**:
**Principle**: Crystallisation is a purification process that converts an impure compound into its pure crystalline form. Pure substances often form crystals with definite geometrical shapes. The method relies on differences in solubility: the impure substance is dissolved in a suitable solvent where it is sparingly soluble at room temperature but significantly soluble at higher temperatures.
**Process**: The solution is concentrated to form a nearly saturated solution. Upon cooling, the pure substance separates out as crystals, leaving the impurities in the solution (or outside the crystal lattice).
**Example**: Impure sugar can be purified using this method. The impure sugar is dissolved in hot water to make a concentrated solution, then cooled slowly. Pure sugar crystals will form, which can then be collected.

(b) **Distillation**:
**Principle**: Distillation involves converting a liquid into its vapor phase by heating, followed by condensing the vapors back into liquid by cooling. This technique is used for liquids that are stable at their boiling points and contain non-volatile impurities, or for separating two liquids with different boiling points.
**Process**: If two liquids with different boiling points are to be separated, the liquid with the lower boiling point (or higher vapor pressure) will vaporize first and be collected in the receiver.
**Example**: A mixture of toluene and hexane can be separated by distillation because they have different boiling points. Hexane, having a lower boiling point, will distill first.

(c) **Chromatography**:
**Principle**: Chromatography is a separation technique where components of a mixture are separated based on their differential movement (adsorption or partition) through a stationary phase under the influence of a mobile phase.
**Process**: The stationary phase can be a solid or a liquid tightly bound to a solid support, while the mobile phase can be either a liquid or a gas. Components of the mixture interact differently with the stationary and mobile phases, causing them to move at different speeds and thus separate.
**Example**: Colored components from a mixture, such as pigments in ink, can be separated by column chromatography or thin-layer chromatography. The dyes, having different affinities for the stationary phase (e.g., silica gel) and mobile phase (e.g., solvent), will travel at different rates, leading to their separation.In simple words: These are ways to clean up mixed substances. Crystallisation cleans by letting a pure solid form from a hot solution. Distillation separates liquids by heating them to turn into steam and then cooling the steam. Chromatography separates things based on how well they stick to a surface or dissolve in a liquid as they move along.

Exam Tip: For each technique, focus on the core principle (e.g., solubility differences for crystallisation, boiling point differences for distillation, differential adsorption/partition for chromatography) and provide a common example for illustration.

 

Question 19. Describe the method which can be used to separate two compounds with different solubilities in a solvent S.
Answer: Two compounds that possess different solubilities in a solvent S can be separated from each other through **fractional crystallisation**. This method involves a series of repeated crystallisations.
**Process**:
1. The mixture of the two compounds is first heated in solvent S to create a saturated solution.
2. When this hot solution is allowed to cool gradually, the substance that is less soluble will crystallise out first.
3. The more soluble substance will remain dissolved in the solution (the mother liquor).
4. The crystals of the first compound are then separated from the mother liquor by filtration.
5. The mother liquor is concentrated again (e.g., by evaporation) and allowed to cool, which then yields crystals of the second compound. This process can be repeated to enhance purity.In simple words: To separate two compounds that dissolve differently in a liquid, we use a trick called fractional crystallisation. We dissolve them in hot liquid, then cool it slowly. The compound that doesn't dissolve as well comes out as crystals first. We take those out, then cool the rest of the liquid to get the second compound.

Exam Tip: Fractional crystallisation is effective when the two compounds have significantly different solubilities in the chosen solvent, especially across a temperature range. Remember it involves a series of repeated steps for optimal separation.

 

Question 20. What is the difference between distillation, distillation under reduced pressure and steam
Answer: Here are the key differences between standard distillation, distillation under reduced pressure, and steam distillation:
**1. Standard Distillation (Simple Distillation)**:

• **Principle**: Separates liquids with significantly different boiling points or separates a liquid from non-volatile impurities. The liquid is heated to its boiling point, vaporizes, and the vapor is then condensed and collected.
• **Application**: Used for purifying simple organic liquids like benzene, ethanol, or acetone from non-volatile impurities.

**2. Distillation under Reduced Pressure (Vacuum Distillation)**:

• **Principle**: This technique is used for liquids that have very high boiling points or tend to decompose at or below their normal boiling points. Reducing the external pressure lowers the boiling point of the liquid, allowing it to distill at a lower, safer temperature.
• **Application**: Ideal for purifying substances like glycerol or for liquids that are thermally unstable.

**3. Steam Distillation**:

• **Principle**: Separates volatile organic compounds that are immiscible with water and possess a high vapor pressure from non-volatile organic or inorganic impurities. The mixture is heated with steam; the organic compound co-distills with water at a temperature below the boiling point of either component.
• **Application**: Commonly used for extracting essential oils from plants or purifying aniline.

In simple words: Distillation is about turning a liquid into vapor and back again to clean it. Normal distillation works for simple liquids. Distillation under reduced pressure helps when liquids boil at very high temperatures or break down easily, by lowering the pressure to boil them at a cooler temperature. Steam distillation is for liquids that don't mix with water but can be carried along by steam at lower temperatures.

Exam Tip: Focus on the conditions and types of compounds each distillation method is best suited for. Simple distillation is for stable compounds with distinct boiling points. Vacuum distillation is for high-boiling or heat-sensitive compounds. Steam distillation is for water-immiscible, volatile compounds.

 

Question 21. Discuss the chemistry of Lassaigne's test.
Answer:Preparation of Lassaigne's filtrate: A small, dried piece of sodium, about the size of a pea, is put into a small ignition tube and heated until it melts. Then, a small amount of the given organic compound (or a drop if it's a liquid) is added to the molten sodium and heated again until it glows red. The contents of the fusion tube are then moved to a china dish containing 10-15 ml of water. The tube breaks, and its contents are boiled in the china dish. The resulting solution, after filtering, is known as Lassaigne's filtrate. This filtrate is used for testing elements like nitrogen (N), sulfur (S), both N and S, and halogens in the organic compound provided.
In simple words: Lassaigne's test starts by melting sodium metal with a tiny bit of the organic substance. This mixture is then boiled in water and filtered to make a liquid called Lassaigne's filtrate, which helps us check for nitrogen, sulfur, or halogens in the original compound.

Exam Tip: Remember to always use a fresh, pea-sized piece of sodium for the fusion, and ensure thorough heating to red hot to ensure complete reaction for accurate results.

 

Question 21. (Continued) Discuss the chemistry of Lassaigne's test.
Answer:Test for Nitrogen:
\( \text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN} \)
From organic compound
Next, freshly made \( \text{FeSO}_4 \) is added to a portion of the Lassaigne's filtrate, and a small amount of \( \text{H}_2\text{SO}_4 \) is included. The following reactions then take place:
\( 2\text{NaCN} + \text{FeSO}_4 \rightarrow \text{Fe(CN)}_2 + \text{Na}_2\text{SO}_4 \)
\( \text{Fe(CN)}_2 + 4\text{NaCN} \rightarrow \text{Na}_4[\text{Fe(CN)}_6] \)
Sodium hexacyano ferrate (II)
Some \( \text{Fe}^{2+} \) ions become oxidized to \( \text{Fe}^{3+} \) ions during heating:
\( 3\text{Na}_4[\text{Fe(CN)}_6] + 4\text{Fe}^{3+} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 + 12\text{Na}^{+} \)
Iron (III) hexacyanoferrate (II) (Prussian blue)
These \( \text{Fe}^{3+} \) ions produce a Prussian blue color, which confirms the presence of nitrogen in the compound.
Test for N & S if present, together:
\( \text{Na} + \text{C} + \text{N} + \text{S} \xrightarrow{\Delta} \text{NaSCN} \)
Sodium thiocyanate
\( \text{Fe}^{3+} + 3\text{NaSCN} \rightarrow \text{Fe(SCN)}_3 + 3\text{Na}^{+} \)
(Blood red color)
If a blood red color appears after treating with \( \text{FeSO}_4 \) and heating, the organic compound contains both N and S.
If S is present alone:
\( 2\text{Na} + \text{S} \xrightarrow{\Delta} \text{Na}_2\text{S} \)
When lead acetate \( \text{Pb(CH}_3\text{COO)}_2 \) is added in a small amount to the Lassaigne's filtrate and acidified with acetic acid, a black precipitate of \( \text{PbS} \) is obtained; this means S alone is present in the organic compound.
\( \text{Na}_2\text{S} + \text{Pb(CH}_3\text{COO)}_2 \rightarrow \text{PbS}\downarrow + 2\text{CH}_3\text{COONa} \)
Test for Halogens:
\( \text{Na} + \text{X} \rightarrow \text{NaX} \) [\( \text{X} = \text{Cl, Br, I} \)]
From the organic compound
The Lassaigne's filtrate is boiled with dilute \( \text{HNO}_3 \) and then cooled. A few drops of \( \text{AgNO}_3 \) are then added.
(a) \( \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl}\downarrow + \text{NaNO}_3 \)
If a white precipitate is formed that dissolves in \( \text{NH}_3 \) but not in \( \text{HNO}_3 \), the organic compound contains chlorine.
(b) \( \text{NaBr} + \text{AgNO}_3 \rightarrow \text{AgBr}\downarrow + \text{NaNO}_3 \)
Formation of a pale yellow precipitate that is partially soluble in ammonia indicates the presence of bromine.
(c) \( \text{NaI} + \text{AgNO}_3 \rightarrow \text{AgI}\downarrow + \text{NaNO}_3 \)
yellow ppt.
A yellow precipitate that is insoluble in ammonia indicates the presence of iodine in the organic compound.
In simple words: Lassaigne's test involves fusing the organic sample with sodium to create ionic compounds. Then, these compounds are tested with specific reagents. For nitrogen, a Prussian blue color shows up. For sulfur, a black precipitate with lead acetate, or a blood red color if N and S are both there. For halogens, silver nitrate forms precipitates of different colors and solubilities, helping identify chlorine, bromine, or iodine.

Exam Tip: For the halogen test, carefully observe the color and solubility of the silver halide precipitate in ammonia to differentiate between chlorine, bromine, and iodine.

 

Question 22. Differentiate between the principle of estimation of nitrogen in an organic compound by
(i) Duma's method
(ii) Kjeldahl's method.
Answer:Estimation of N in the organic compound by DUMA's method: This method is suitable for all organic compounds that contain nitrogen. A known weight of the organic compound is heated with excess copper oxide in a \( \text{CO}_2 \) atmosphere. Carbon, hydrogen, and sulfur get oxidized to \( \text{CO}_2 \), \( \text{H}_2\text{O} \), and \( \text{SO}_2 \), while nitrogen gas \( (\text{N}_2) \) is set free. Any nitrogen oxides formed are reduced back to free \( \text{N}_2 \) by passing them over hot reduced copper gauze.
\( \text{C} + 2\text{CuO} \xrightarrow{\Delta} \text{CO}_2 + 2\text{Cu} \)
\( 2\text{H} + \text{CuO} \xrightarrow{\Delta} \text{H}_2\text{O} + \text{Cu} \)
\( \text{Nitrogen} + \text{CuO} \xrightarrow{\Delta} \text{N}_2 + \text{a small amount of oxides of N} \)
\( \text{Oxides of N} + \text{Cu} \xrightarrow{\Delta} \text{CuO} + \text{N}_2 \)
The volume of \( \text{N}_2 \) collected over \( \text{KOH} \) is measured in a Schiff's nitrometer. This volume is then converted to NTP (Normal Temperature and Pressure). If V is the volume in liters and 22.4 liters of \( \text{N}_2 \) at STP/NTP weigh 28g, the weight of N present in the organic compound can be calculated using this formula:
\( \text{% of N in the organic compound} = \frac{\text{V}}{22.4} \times \frac{28}{\text{W}} \times 100 \)
Where W = weight of the organic compound.
Estimation of N in the organic compound by KJELDAHL's method: This is a simpler method for estimating nitrogen and is widely used for nitrogen in foodstuffs, drugs, etc. This method is not suitable for compounds containing N directly linked to \( \text{N} = \text{O} \) or another N atom (as in diazo compounds \( -\text{N} = \text{N}- \)). It is also not suitable for organic compounds that have nitrogen in a ring structure (like pyridine \( \text{C}_5\text{H}_5\text{N} \) or quinoline).
A known weight of the organic compound is heated with concentrated \( \text{H}_2\text{SO}_4 \) in the presence of \( \text{K}_2\text{SO}_4 \) and a small amount of \( \text{CuSO}_4 \) or \( \text{Hg} \) in a long-necked flask called a Kjeldahl's flask. \( \text{K}_2\text{SO}_4 \) increases the boiling point of \( \text{H}_2\text{SO}_4 \), and \( \text{CuSO}_4 \) acts as a catalyst.
Nitrogen present in the organic compound is completely converted into ammonium sulfate \( (\text{NH}_4)_2\text{SO}_4 \). This \( (\text{NH}_4)_2\text{SO}_4 \) is then boiled with excess \( \text{NaOH} \), and the released ammonia is absorbed in a known excess of a standard acid like \( \text{H}_2\text{SO}_4 \) or \( \text{HCl} \).
\( \text{N (of organic compd)} + \text{Conc. H}_2\text{SO}_4 \xrightarrow{\Delta} (\text{NH}_4)_2\text{SO}_4 \)
\( (\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{NH}_3 \)
\( 2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 \)
The volume of the unused acid is determined by titration against a standard alkali.
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
Using the normality equation, the normality and strength of \( \text{NH}_3 \) or N can be determined. From there, the percentage of N can be estimated:
\( \text{%N} = \frac{1.4 \times \text{Molarity of H}_2\text{SO}_4 \times \text{twice the vol. of H}_2\text{SO}_4 \text{ used}}{\text{Mass of substance taken}} \)
In simple words: Duma's method burns the sample to turn all nitrogen into \( \text{N}_2 \) gas, which is then measured. Kjeldahl's method converts nitrogen into ammonium sulfate, which is then reacted with a base to release ammonia, which is measured by titration. Duma's works for all nitrogen compounds, while Kjeldahl's is simpler but not for all types of nitrogen-containing compounds.

Exam Tip: Remember that Kjeldahl's method is not applicable to all nitrogen-containing compounds, especially those with nitrogen in ring structures or directly bonded to oxygen or other nitrogen atoms.

 

Question 23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Answer:Halogens, sulfur, and phosphorus found in organic compounds are estimated using CARIUS METHOD, as explained below.
Carius Method for Estimation of Halogens:
A known weight of the organic compound is heated with fuming nitric acid \( (\text{HNO}_3) \) and silver nitrate \( (\text{AgNO}_3) \) in a sealed hard glass tube. This tube is kept in a furnace under specific conditions. In these conditions, carbon (C) and hydrogen (H) in the organic compound are converted to \( \text{CO}_2 \) and \( \text{H}_2\text{O} \), while the halogen changes into a silver halide. The precipitates of silver halide are filtered, washed, dried, and weighed. Knowing the weight of the substance taken, the percentage of halogen is calculated as follows:
Let the weight of the organic compound taken = \( \text{W g} \).
Mass of the silver halide formed = \( \text{x} \).
Percentage of Halogen in the compound:
\( \text{% of Halogen} = \frac{\text{Atomic mass of X}}{108 + \text{Atomic mass of X}} \times \frac{\text{x}}{\text{W}} \times 100 \)
\( \text{% of Chlorine} = \frac{35.5}{143.5} \times \frac{\text{x}}{\text{W}} \times 100 \)
\( \text{% of Bromine} = \frac{80}{188} \times \frac{\text{x}}{\text{W}} \times 100 \)
\( \text{% of Iodine} = \frac{127}{235} \times \frac{\text{Mass of AgI}}{\text{Mass of substance taken}} \times 100 \)
(ii) Estimation of Sulfur: In the Carius method, to estimate sulfur, which is completely converted to \( \text{H}_2\text{SO}_4 \), it is precipitated by adding excess barium chloride solution.
\( \text{C} + 2\text{O (from HNO}_3) \xrightarrow{\Delta} \text{CO}_2 \)
\( 2\text{H} + \text{O (from HNO}_3) \xrightarrow{\Delta} \text{H}_2\text{O} \)
\( \text{S} + \text{H}_2\text{O} + 3\text{O (from HNO}_3) \xrightarrow{\Delta} \text{H}_2\text{SO}_4 \)
\( \text{H}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4\downarrow + 2\text{HCl} \)
white ppt.
The \( \text{BaSO}_4 \) precipitate is filtered, washed, dried, and weighed. Knowing the mass of the substance taken (W) and the mass of \( \text{BaSO}_4 \), the percentage of sulfur can be calculated.
\( \text{% of S} = \frac{\text{Atomic mass of S}}{\text{Molecular mass of BaSO}_4} \times \frac{\text{x}}{\text{W}} \times 100 \)
\( = \frac{32}{137 + 32 + 4 \times 16} \times \frac{\text{x}}{\text{W}} \times 100 \)
\( = \frac{32}{233} \times \frac{\text{x}}{\text{W}} \times 100 \) [At Mass of Ba = 137]
(iii) Estimation of Phosphorus: Phosphorus present in the organic compound in the Carius tube is oxidized by concentrated \( \text{HNO}_3 \) to phosphoric acid \( (\text{H}_3\text{PO}_4) \). It is then precipitated as magnesium ammonium phosphate by adding magnesia mixture (a mix of \( \text{MgCl}_2 \), \( \text{NH}_4\text{Cl} \), and \( \text{NH}_3 \)).
\( 2\text{P} + 5\text{O [from HNO}_3] \xrightarrow{\Delta} \text{P}_2\text{O}_5 \)
\( \text{P}_2\text{O}_5 + 3\text{H}_2\text{O} \xrightarrow{\Delta} 2\text{H}_3\text{PO}_4 \)
\( \text{MgCl}_2 + \text{NH}_4\text{Cl} + \text{H}_3\text{PO}_4 \xrightarrow{\Delta} \text{Mg(NH}_4)\text{PO}_4 \)
The precipitate of magnesium ammonium phosphate is filtered, washed, dried, and then ignited to give magnesium pyrophosphate.
\( 2\text{Mg NH}_4\text{PO}_4 \xrightarrow{\Delta} \text{Mg}_2\text{P}_2\text{O}_7 + 2\text{NH}_3 + \text{H}_2\text{O} \)
The percentage of P in the organic compound is calculated as:
\( \text{% of P in the organic compound} = \frac{2 \times \text{At. Mass of P}}{\text{Molecular mass of Mg}_2\text{P}_2\text{O}_7} \times \frac{\text{Wt. of Mg}_2\text{P}_2\text{O}_7}{\text{Wt. of compound}} \times 100 \)
\( = \frac{2 \times 31}{2 \times 24 + 2 \times 31 + 7 \times 16} \times \frac{\text{x}}{\text{W}} \times 100 \)
\( = \frac{62}{222} \times \frac{\text{x}}{\text{W}} \times 100 \)
In simple words: The Carius method checks for halogens, sulfur, and phosphorus in organic compounds. For halogens, the sample is heated with nitric acid and silver nitrate, forming a silver halide precipitate which is weighed. For sulfur, it's converted to sulfuric acid, then precipitated as barium sulfate and weighed. For phosphorus, it's oxidized to phosphoric acid, precipitated as magnesium ammonium phosphate, then ignited to magnesium pyrophosphate and weighed. All these measurements help figure out the percentage of each element.

Exam Tip: For Carius method calculations, ensure you use the correct molecular weights for the precipitates (silver halide, barium sulfate, magnesium pyrophosphate) and the atomic weights of the elements being estimated.

 

Question 24. Explain the principle of paper chromatography.
Answer:In paper chromatography, a specific type of paper called chromatographic paper is used. Even though paper is mostly cellulose, the stationary phase in paper chromatography is not the cellulose itself but the water that is adsorbed or chemically bound to it. The mobile phase is a different liquid, often a mixture of two or three solvents, with water as one component. This technique is a kind of partition chromatography, where both the mobile and stationary phases are liquids.
Paper chromatography works on the idea of partition, which means it relies on the continuous differential partitioning or distribution of various components of a mixture between the stationary and mobile phases. The retention factor \( (\text{R}_f) \) is remembered here, which is:
\( \text{R}_f = \frac{\text{Distance travelled by the compound (x)}}{\text{Distance travelled by the solvent (y)}} \)
The paper strip that is developed is called the CHROMATOGRAM. If a mixture of compounds is used, the spots of the separated colored compounds are visible at different heights from the starting line and are identified by their \( \text{R}_f \) values. However, the spots of colorless compounds can be seen under UV light or by using an appropriate spray reagent.
In simple words: Paper chromatography separates mixtures based on how much each part prefers to stay in the paper's water (stationary phase) versus moving with the solvent liquid (mobile phase). Different compounds travel different distances, and we use the \( \text{R}_f \) value to identify them.

Exam Tip: Always make sure the starting line for applying the sample is above the solvent level in the chromatography chamber to prevent the sample from dissolving directly into the solvent pool instead of being carried up the paper.

 

Question 25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer:Function of Nitric acid: When testing halogens present in an organic compound, we add nitric acid \( (\text{HNO}_3) \) to the Lassaigne's filtrate or sodium extract to boil off any \( \text{H}_2\text{S} \) and \( \text{HCN} \) gases that might be present from sulfur and nitrogen in the organic compound:
\( \text{NaCN} + \text{HNO}_3 \rightarrow \text{NaNO}_3 + \text{HCN} \)
\( \text{Na}_2\text{S} + 2\text{HNO}_3 \rightarrow 2\text{NaNO}_3 + \text{H}_2\text{S} \)
If cyanide and sulfide are not removed, they will react with \( \text{AgNO}_3 \), leading to incorrect results:
\( \text{NaCN} + \text{AgNO}_3 \rightarrow \text{AgCN (white ppt.)} + \text{NaNO}_3 \)
\( \text{Na}_2\text{S} + 2\text{AgNO}_3 \rightarrow \text{Ag}_2\text{S (black ppt.)} + 2\text{NaNO}_3 \)
In simple words: Nitric acid is added before silver nitrate to remove any cyanide or sulfide from the sample. If these are not removed, they would react with the silver nitrate and give false positive results for halogens.

Exam Tip: Boiling with nitric acid is a critical step to eliminate interfering ions; incomplete removal of cyanide or sulfide will lead to erroneous results in the halogen test.

 

Question 26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:Before proceeding to test for nitrogen (N), sulfur (S), or phosphorus (P) in a given organic compound, the compound is fused with metallic sodium. These elements are present in the organic compound as covalent bonds, such as nitrogen in urea \( (\text{NH}_2\text{-CO-NH}_2) \) or sulfur in thiourea \( (\text{NH}_2\text{CS NH}_2) \). By fusing the organic compound with sodium, these elements convert into ionic compounds, like \( \text{NaCN} \), \( \text{Na}_2\text{S} \), or \( \text{NaX} \) (where X is a halogen). Ionic compounds are much easier to detect using simple ionic reactions than covalent ones.
In simple words: We fuse the organic compound with sodium metal to change covalently bonded nitrogen, sulfur, or halogens into ionic compounds. Ionic compounds are easier to detect in tests.

Exam Tip: Fusion with sodium helps break strong covalent bonds in organic compounds, converting elements like N, S, and halogens into water-soluble ionic forms that are readily detectable by standard qualitative tests.

 

Question 27. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Answer:Calcium sulfate \( (\text{CaSO}_4) \), an inorganic compound that does not sublime, can be separated from camphor, which is an organic compound and sublimes, using the simple technique of SUBLIMATION. Sublimation involves the direct change of a solid into a gaseous state when heated, bypassing the liquid state, and vice-versa upon cooling. Camphor sublimes and then deposits on the walls of the cooler part of the funnel, while \( \text{CaSO}_4 \) remains as it is. Thus, the two components can be separated.

(Image description: A sublimation apparatus showing a mixture of camphor and calcium sulfate being heated. A funnel inverted over the mixture, with a cotton plug at the top, is cooled. Camphor sublimes and deposits on the inner walls of the funnel, while calcium sulfate remains in the dish.)
In simple words: The best way to separate calcium sulfate and camphor is sublimation. Camphor turns into a gas and then back to a solid, leaving the calcium sulfate behind.

Exam Tip: Remember that sublimation is only effective for separating a volatile solid from a non-volatile solid, or from another volatile solid with a significantly different sublimation temperature.

 

Question 28. Explain why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Answer:In steam distillation, the mixture boils when the combined vapor pressures of water and the organic liquid become equal to the external atmospheric pressure. Since the vapor pressure of water is notably higher than that of the organic liquid at a given temperature, the organic liquid will vaporize at a much lower temperature than its usual boiling point. This process allows the organic liquid to distill without reaching its normal, often higher, boiling point, preventing decomposition of heat-sensitive compounds.
In simple words: In steam distillation, the organic liquid boils at a lower temperature because its vapor pressure adds to water's vapor pressure, reaching the total atmospheric pressure sooner.

Exam Tip: This principle is crucial for purifying compounds that are immiscible with water and decompose at or near their normal boiling points.

 

Question 29. Will \( \text{CCl}_4 \) give white ppt. of AgCl on heating it with silver nitrate ? Give reason for your answer.
Answer:\( \text{CCl}_4 \) will not produce a white precipitate of \( \text{AgCl} \) when it is treated with a silver nitrate solution. This is because \( \text{CCl}_4 \) is a covalent compound and therefore does not ionize to produce \( \text{Cl}^- \) ions, which are a necessary requirement for the \( \text{AgNO}_3 \) test to form \( \text{AgCl} \). For the silver nitrate test to work, free chloride ions must be present in the solution to react with silver ions.
In simple words: No, \( \text{CCl}_4 \) won't form a white precipitate with silver nitrate because it's a covalent compound and doesn't release \( \text{Cl}^- \) ions needed for the reaction.

Exam Tip: The silver nitrate test for halides specifically detects free halide ions; covalent halides do not yield a positive test unless they undergo hydrolysis to release the halide ions.

 

Question 30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:Carbon dioxide \( (\text{CO}_2) \) is slightly acidic by nature. Therefore, \( \text{KOH} \), which is a strong alkali, dissolves the \( \text{CO}_2 \) evolved from the compound containing carbon and reacts with it to form \( \text{K}_2\text{CO}_3 \). The increase in the weight of the potash bulbs, where the absorption happens, will give the weight of \( \text{CO}_2 \) evolved, allowing for the calculation of carbon content. The reaction is:
\( 2\text{KOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O(l)} \)
In simple words: Potassium hydroxide solution absorbs carbon dioxide because \( \text{CO}_2 \) is acidic and \( \text{KOH} \) is a strong base. They react to form potassium carbonate, and the weight gain helps measure the carbon content.

Exam Tip: Remember that \( \text{KOH} \) is used because of its strong basic nature, which efficiently traps the acidic \( \text{CO}_2 \) gas, making it suitable for quantitative estimation of carbon.

 

Question 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?
Answer:For testing sulfur in an organic compound using the lead acetate test, the Lassaigne's extract must be acidified with acetic acid \( (\text{CH}_3\text{COOH}) \) and not sulfuric acid. This is important because lead acetate is soluble in acetic acid and therefore does not interfere with the test. However, if sulfuric acid \( (\text{H}_2\text{SO}_4) \) is used, lead acetate will react with \( \text{H}_2\text{SO}_4 \) itself to form a white precipitate of \( \text{PbSO}_4 \), as shown below:
\( \text{Pb(CH}_3\text{COO)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{PbSO}_4\downarrow + 2\text{CH}_3\text{COOH} \)
white ppt.
This white precipitate of \( \text{PbSO}_4 \) formed would interfere with the sulfur test, leading to false results. The actual reaction for sulfur detection is:
\( \text{Pb(CH}_3\text{COO)}_2 + \text{Na}_2\text{S} \rightarrow \text{PbS}\downarrow + 2\text{CH}_3\text{COONa} \)
(sulfur from the org. compd.) black ppt.
In simple words: Acetic acid is used because lead acetate dissolves in it and doesn't interfere with the sulfur test. Sulfuric acid, however, would react with lead acetate to form a white precipitate, confusing the test results.

Exam Tip: Always select an acid for acidification that does not form a precipitate with the test reagent itself, ensuring the accuracy of qualitative analysis.

 

Question 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.
Answer:The percentages are given as: Carbon = 69%, Hydrogen = 4.8%. The remainder is oxygen.
Percentage of Oxygen = \( 100\% - (69\% + 4.8\%) = 100\% - 73.8\% = 26.2\% \).
Let the mass of the organic compound taken be \( \text{W} = 0.20 \text{ g} \).
We use the formulas for percentage of carbon and hydrogen:
\( \text{%C} = \frac{12}{44} \times \frac{\text{Mass of CO}_2 \text{ formed}}{\text{Mass of substance taken}} \times 100 \)
\( 69 = \frac{12}{44} \times \frac{\text{Mass of CO}_2 \text{ formed}}{0.20} \times 100 \)
Therefore, Mass of \( \text{CO}_2 \) formed \( = \frac{69 \times 44 \times 0.20}{12 \times 100} = 0.506 \text{ g} \)
\( \text{%H} = \frac{2}{18} \times \frac{\text{Mass of H}_2\text{O} \text{ formed}}{\text{Mass of substance taken}} \times 100 \)
\( 4.8 = \frac{2}{18} \times \frac{\text{Mass of H}_2\text{O} \text{ formed}}{0.20} \times 100 \)
Therefore, Mass of water formed \( = \frac{4.8 \times 18 \times 0.20}{2 \times 100} = 0.0864 \text{ g} \)
In simple words: For a 0.20 g sample, given the percentages of carbon and hydrogen, we calculate that 0.506 g of carbon dioxide and 0.0864 g of water will be produced during complete combustion.

Exam Tip: When calculating masses from percentage composition, always ensure you use the correct stoichiometric ratios (e.g., 12/44 for carbon in \( \text{CO}_2 \) and 2/18 for hydrogen in \( \text{H}_2\text{O} \)).

 

Question 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 \( \text{H}_2\text{SO}_4 \). The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Answer:Step-I: Determine the volume of \( \text{H}_2\text{SO}_4 \) used.
Volume of acid taken = 50 ml of 0.5 M \( \text{H}_2\text{SO}_4 \).
This is equivalent to \( 25 \text{ ml} \) of 1.0 M \( \text{H}_2\text{SO}_4 \).
Volume of alkali used for neutralization of excess acid:
= 60 ml of 0.5 M \( \text{NaOH} \).
This is equivalent to 30 ml of 1.0 M \( \text{NaOH} \).
According to the reaction, 1 mole of \( \text{H}_2\text{SO}_4 \) neutralizes 2 moles of \( \text{NaOH} \):
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
So, 30 ml of 1 M \( \text{NaOH} \) would be neutralized by 15 ml of 1 M \( \text{H}_2\text{SO}_4 \).
Volume of acid used by ammonia:
= \( 25 \text{ ml} - 15 \text{ ml} = 10 \text{ ml} \).
Step-II: To determine the percentage of nitrogen in the compound.
1 mole of \( \text{H}_2\text{SO}_4 \) neutralizes 2 moles of \( \text{NH}_3 \).
Therefore, 10 ml of 1 M \( \text{H}_2\text{SO}_4 \) is equivalent to 20 ml of 1 M \( \text{NH}_3 \).
Since 1000 ml of 1 M \( \text{NH}_3 \) contains 14g of Nitrogen,
10 ml of 1 M \( \text{NH}_3 \) will contain N \( = \frac{14}{1000} \times 20 = 0.28 \text{ g} \).
This is the amount of N present in 0.50 g of the compound.
\( \text{% N} = \frac{\text{Mass of N}}{\text{Mass of compound}} \times 100 = \frac{0.28}{0.50} \times 100 = 56.0\% \).
In simple words: First, we figure out how much acid was used by the ammonia. Then, using that amount, we calculate the total nitrogen present. Finally, we express this as a percentage of the initial compound's mass. The calculations show 56.0% nitrogen.

Exam Tip: In Kjeldahl's method calculations, carefully track the stoichiometry of the acid-base reactions to correctly determine the amount of ammonia, and thus nitrogen, absorbed.

 

Question 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.
Answer:Mass of organic compound (W) = 0.3780 g
Mass of silver chloride \( (\text{AgCl}) \) formed (x) = 0.5740 g
Atomic mass of Chlorine \( (\text{Cl}) \) = 35.5
Molecular mass of \( \text{AgCl} \) = Atomic mass of Ag + Atomic mass of Cl = 108 + 35.5 = 143.5
The percentage of chlorine is calculated using the formula:
\( \text{% of chlorine} = \frac{\text{Atomic mass of Cl}}{\text{Molecular mass of AgCl}} \times \frac{\text{Wt. of AgCl}}{\text{Wt. of org. compound}} \times 100 \)
\( = \frac{35.5}{143.5} \times \frac{0.5740}{0.3780} \times 100 \)
\( = 37.57\% \)
In simple words: By comparing the weight of silver chloride produced to the original compound's weight and using the atomic and molecular masses, we calculate that the organic compound contains 37.57% chlorine.

Exam Tip: Always use the precise atomic and molecular weights to ensure accuracy in Carius method calculations for halogens.

 

Question 35. In the estimation of sulphur by Carius method, 0.468g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Answer:Mass of organic sulfur compound (W) = 0.468 g
Mass of barium sulfate \( (\text{BaSO}_4) \) formed (x) = 0.668 g
Atomic mass of Sulfur \( (\text{S}) \) = 32
Molecular mass of \( \text{BaSO}_4 \) = Atomic mass of Ba + Atomic mass of S + \( 4 \times \) Atomic mass of O
= \( 137 + 32 + 4 \times 16 = 137 + 32 + 64 = 233 \)
The percentage of sulfur is calculated using the formula:
\( \text{% of S} = \frac{\text{Atomic mass of S}}{\text{Molecular mass of BaSO}_4} \times \frac{\text{Wt. of BaSO}_4}{\text{Wt. of org. comp.}} \times 100 \)
\( = \frac{32}{233} \times \frac{0.668}{0.468} \times 100 \)
\( = 19.66\% \)
In simple words: Given the weights of the organic compound and the barium sulfate produced, and knowing the atomic weights, we calculated that the compound contains 19.66% sulfur.

Exam Tip: It's important to know the molecular weight of \( \text{BaSO}_4 \) (233) and the atomic weight of S (32) for accurate sulfur estimation in Carius method problems.