General Introduction

Organic chemistry is the study of carbon-containing compounds. Carbon is unique due to its ability to form strong covalent bonds with other carbon atoms (catenation) and with various other elements like hydrogen, oxygen, nitrogen, sulphur, phosphorus, and halogens. Organic compounds are fundamental to life (DNA, proteins) and everyday materials (fuels, plastics, medicines).

The field of organic chemistry developed historically from studying compounds derived from living organisms. Initially, it was believed that a 'vital force' was necessary for their formation. However, this idea was disproven by scientists like F. Wohler (synthesis of urea from inorganic ammonium cyanate in 1828), Kolbe (acetic acid synthesis in 1845), and Berthelot (methane synthesis in 1856), who showed that organic compounds could be synthesized in the laboratory from inorganic precursors.

The development of electronic theory of covalent bonding in the early 20th century provided a scientific foundation for organic chemistry.

Tetravalence Of Carbon: Shapes Of Organic Compounds

Carbon is tetravalent, forming four covalent bonds. The shapes of simple organic molecules like methane (CH$_4$), ethene (C$_2$H$_4$), and ethyne (C$_2$H$_2$) are explained by the hybridization of carbon's atomic orbitals (sp$^3$, sp$^2$, and sp, respectively).

The Shapes Of Carbon Compounds

• In **methane (CH$_4$)**, the carbon is sp$^3$ hybridized and forms four sigma ($\sigma$) bonds with hydrogen atoms. The molecule has a **tetrahedral** shape with bond angles of 109.5$^\circ$.
• In **ethene (C$_2$H$_4$)**, each carbon is sp$^2$ hybridized, forming three $\sigma$ bonds (one C-C, two C-H) and one pi ($\pi$) bond (C=C). The atoms around each carbon lie in a single plane, resulting in a **trigonal planar** geometry around each carbon and a **planar** molecule overall. The bond angles around each carbon are approximately 120$^\circ$.
• In **ethyne (C$_2$H$_2$)**, each carbon is sp hybridized, forming two $\sigma$ bonds (one C-C, one C-H) and two $\pi$ bonds (C$\equiv$C). The molecule is **linear** with bond angles of 180$^\circ$.

Hybridization affects bond properties: sp hybrid orbitals (50% s character) are closer to the nucleus, forming shorter and stronger bonds than sp$^2$ (33% s character) or sp$^3$ (25% s character) hybrid orbitals. Electronegativity of carbon also increases with s character (sp > sp$^2$ > sp$^3$).

Some Characteristic Features Of P Bonds

A $\pi$ bond is formed by the sideways overlap of two parallel p-orbitals on adjacent atoms. For maximum overlap, the atoms involved in the $\pi$ bond and the atoms directly attached to them must lie in the same plane. Rotation around a double bond is restricted because it would break the $\pi$ overlap.

The electron cloud of a $\pi$ bond is located above and below the plane of the sigma bond. $\pi$ bonds are typically weaker than $\sigma$ bonds but are more exposed, making them reactive sites in molecules with multiple bonds.

Answer:

Structural Representations Of Organic Compounds

The structures of organic compounds can be represented in various ways to show connectivity and arrangement of atoms.

Complete, Condensed And Bond-line Structural Formulas

• **Complete Structural Formula:** Shows all atoms and all covalent bonds using dashes (single bond = one dash, double bond = two dashes, triple bond = three dashes). Lone pairs on heteroatoms may or may not be shown.
• **Condensed Structural Formula:** Omits some or all dashes and indicates the number of identical groups attached to an atom using subscripts. Chains of CH$_2$ groups are often condensed further (e.g., (CH$_2$)$_n$).
• **Bond-line Structural Formula (Skeletal Structure):** A simplified representation where carbon and hydrogen atoms bonded to carbon are not explicitly shown. Carbon atoms are implied at the junctions of lines and at the ends of lines. Lines represent carbon-carbon bonds, drawn in a zig-zag manner. Hydrogen atoms attached to carbon are assumed to be present in sufficient number to satisfy carbon's valency (4). Heteroatoms (like O, N, S, halogens) and hydrogens bonded to heteroatoms are shown explicitly.

Examples:

Ethane (C$_2$H$_6$)

• Complete: H–H$_3$C – CH$_3$
• Condensed: CH$_3$CH$_3$
• Bond-line: \\

Ethene (C$_2$H$_4$)

• Complete: H$_2$C=CH$_2$
• Condensed: CH$_2$=CH$_2$
• Bond-line: \

Ethyne (C$_2$H$_2$)

• Complete: HC$\equiv$CH
• Condensed: HC$\equiv$CH
• Bond-line: —

Cyclic compounds in bond-line formula are represented by polygons.

Answer:

Three-Dimensional Representation Of Organic Molecules

To visualise the 3D structure of molecules on a 2D surface like paper, specific conventions are used in structural formulas:

• **Solid Wedge ( ):** Represents a bond projecting **out of the plane** of the paper towards the observer.
• **Dashed Wedge ( ):** Represents a bond projecting **into the plane** of the paper away from the observer.
• **Normal Line (—):** Represents a bond lying **in the plane** of the paper.

These are used in wedge-and-dash formulas (Fig. 12.1).

Additionally, physical **molecular models** (Framework, Ball-and-stick, Space-filling) and **computer graphics** are used for better 3D visualisation (Fig. 12.2).

Answer:

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Classification Of Organic Compounds

Organic compounds are classified based on their structure:

• **Acyclic (Open chain) compounds:** Straight or branched chains of carbon atoms. Also called aliphatic compounds. Examples: Ethane, isobutane, acetaldehyde, acetic acid.
• **Cyclic (Closed chain or Ring) compounds:** Carbon atoms form rings. * **Alicyclic compounds:** Cyclic compounds resembling aliphatic compounds. Rings contain only carbon atoms (homocyclic) or heteroatoms (heterocyclic). Examples: Cyclopropane, cyclohexane, tetrahydrofuran. * **Aromatic compounds:** Special cyclic compounds with conjugated $\pi$ systems, exhibiting aromaticity (benzene and related compounds - benzenoid; or rings with heteroatoms - heterocyclic aromatic). Examples: Benzene, aniline, furan, pyridine. Some non-benzenoid aromatic compounds exist like tropone.

Organic compounds are also classified by functional groups into **homologous series**.

Functional Group

A **functional group** is an atom or group of atoms within a molecule that is primarily responsible for its characteristic chemical properties (e.g., -OH, -CHO, -COOH).

Homologous Series

A **homologous series** is a family of organic compounds with the same functional group and similar chemical properties. Successive members (homologues) in a series differ by a -CH$_2$ unit. Examples: Alkanes (CH$_4$, C$_2$H$_6$, C$_3$H$_8$, etc.), alkenes, alkanols, alkanoic acids.

Compounds can have multiple identical or different functional groups (polyfunctional compounds).

Nomenclature Of Organic Compounds

With millions of organic compounds, a systematic naming system is essential for clear communication. The **IUPAC (International Union of Pure and Applied Chemistry) system of nomenclature** provides rules for naming organic compounds based on their structure.

Before IUPAC, compounds were named based on origin or properties (trivial/common names). While many common names are still used (e.g., formic acid, toluene), systematic IUPAC names are preferred.

The Iupac System Of Nomenclature

An IUPAC name consists of a **root name** (based on the parent hydrocarbon chain or ring), **suffix(es)** (indicating functional group(s)), and **prefix(es)** (indicating substituents other than those given by the suffix). Numbers (locants) are used to indicate the positions of functional groups and substituents.

Iupac Nomenclature Of Alkanes

• **Straight chain alkanes:** Named using a prefix indicating the number of carbons (meth-, eth-, prop-, but-, pent-, hex-, etc.) followed by the suffix '-ane'. Examples: Methane (CH$_4$), Ethane (C$_2$H$_6$), Propane (C$_3$H$_8$), Butane (C$_4$H$_{10}$), Pentane (C$_5$H$_{12}$), Hexane (C$_6$H$_{14}$).

Name	Molecular formula	Structure
Methane	CH$_4$	CH$_4$
Ethane	C$_2$H$_6$	CH$_3$CH$_3$
Propane	C$_3$H$_8$	CH$_3$CH$_2$CH$_3$
Butane	C$_4$H$_{10}$	CH$_3$CH$_2$CH$_2$CH$_3$
Pentane	C$_5$H$_{12}$	CH$_3$(CH$_2$)$_3$CH$_3$
Hexane	C$_6$H$_{14}$	CH$_3$(CH$_2$)$_4$CH$_3$
Heptane	C$_7$H$_{16}$	CH$_3$(CH$_2$)$_5$CH$_3$
Octane	C$_8$H$_{18}$	CH$_3$(CH$_2$)$_6$CH$_3$
Nonane	C$_9$H$_{20}$	CH$_3$(CH$_2$)$_7$CH$_3$
Decane	C$_{10}$H$_{22}$	CH$_3$(CH$_2$)$_8$CH$_3$

• **Branched chain alkanes:** Named by identifying the longest continuous carbon chain as the parent alkane. Branches are alkyl groups (alkane minus one H, e.g., methyl -CH$_3$, ethyl -CH$_2$CH$_3$). Alkyl groups are named by replacing '-ane' with '-yl'. Positions of branches are indicated by numbering the parent chain such that the branches get the lowest possible numbers. Multiple identical branches use prefixes (di-, tri-). Different branches are listed alphabetically.

Alkane	Molecular formula	Alkyl group	Formula
Methane	CH$_4$	Methyl	-CH$_3$
Ethane	C$_2$H$_6$	Ethyl	-CH$_2$CH$_3$
Propane	C$_3$H$_8$	Propyl	-CH$_2$CH$_2$CH$_3$
Butane	C$_4$H$_{10}$	Butyl	-CH$_2$CH$_2$CH$_2$CH$_3$

Common branched alkyl groups (isopropyl, sec-butyl, isobutyl, tert-butyl, neopentyl) are also used (Table 12.3). The carbon atom attaching to the main chain in a branched alkyl group is numbered 1 within that group.

Problem 12.7. Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect. (a) (2,5,6-Trimethyloctane) (3,4,7-Trimethyloctane) (b) (3-Ethyl-5-methylheptane) (5-Ethyl-3-methylheptane) (c) (3-Ethyl-1,1-dimethylcyclohexane) (1-Ethyl-3,3-dimethylcyclohexane)

Answer:

(a)

Structure: A chain of 8 carbons (octane) with three methyl branches.

H$_3$C—CH—CH$_2$—CH$_2$—CH—CH—CH$_2$—CH$_3$

|

CH$_3$

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CH$_3$ CH$_3$

Correct parent chain is 8 carbons (octane).

If numbering from left to right: Branches are on carbons 2, 5, 6. Name: 2,5,6-Trimethyloctane.

If numbering from right to left: Branches are on carbons 3, 4, 7. Name: 3,4,7-Trimethyloctane.

Rule: Numbering should give the lowest possible set of locants (numbers) to the substituents. Compare the sets of locants: (2, 5, 6) vs (3, 4, 7). Since 2 < 3, the set (2, 5, 6) is the lowest set.

The name given in parenthesis (3,4,7-Trimethyloctane) is incorrect because the numbering results in a higher set of locants (3,4,7) compared to the correct numbering from the left (2,5,6).

(b)

Structure: A chain of 7 carbons (heptane) with an ethyl branch and a methyl branch.

H$_3$C—CH$_2$—CH—CH$_2$—CH—CH$_2$—CH$_3$

| |

CH$_2$CH$_3$ CH$_3$

Correct parent chain is 7 carbons (heptane).

If numbering from left to right: Ethyl on C3, Methyl on C5. Name: 3-Ethyl-5-methylheptane.

If numbering from right to left: Methyl on C3, Ethyl on C5. Name: 5-Ethyl-3-methylheptane.

Rule: If the two substituents are found in equivalent positions when numbering from either end (sum of locants is the same, or the first points of difference are the same when comparing sets), the lower number is given to the substituent that comes first in the alphabetical listing. In this case, Ethyl (E) comes before Methyl (M) alphabetically. The set of locants is (3, 5) in both numbering directions. So, alphabetical order decides which group gets the lower number. Numbering from left to right gives Ethyl (on C3) a lower number than Methyl (on C5). Numbering from right to left gives Methyl (on C3) a lower number than Ethyl (on C5).

Since Ethyl (E) comes before Methyl (M) alphabetically, the numbering should be from left to right so that Ethyl gets the lower number (3). The correct name is 3-Ethyl-5-methylheptane.

The name given in parenthesis (5-Ethyl-3-methylheptane) is incorrect because the numbering is done from the right, giving a lower number to the methyl group (3) which comes later in alphabetical order than the ethyl group (on C5).

(c)

Structure: A 6-membered ring (cyclohexane) with a methyl group and a chlorine atom attached.

This is a cyclic compound with two different substituents. The parent name is cyclohexane. The substituents are Methyl (-CH$_3$) and Chloro (-Cl). The rule is to number the ring carbons to give the lowest possible locants to the substituents. If there are different substituents, numbering starts at the substituent that comes first in alphabetical order (unless a specific functional group is present). Chloro (C) comes before Methyl (M) alphabetically.

Let's look at the names given:

3-Ethyl-1,1-dimethylcyclohexane (incorrect parenthesis name: 1-ethyl-3,3-dimethylcyclohexane). Let's analyse the given name: 3-Ethyl-1,1-dimethylcyclohexane. Parent is cyclohexane. Substituents are one ethyl and two methyls. Locants are 1, 1, 3. The correct name is 1,1-Dimethyl-3-ethylcyclohexane (alphabetical order). So the name given is correct, but the parenthesised one is incorrect.

Let's analyse the parenthesised name: 1-ethyl-3,3-dimethylcyclohexane. Parent is cyclohexane. Substituents are one ethyl and two methyls. Locants are 1, 3, 3. Comparing the set of locants (1, 1, 3) vs (1, 3, 3). The first set is lower. So 1,1-dimethyl-3-ethylcyclohexane is the correct name.

However, the original structure shown in the text for (c) is methylchlorocyclohexane, not dimethyl ethylcyclohexane. The text seems to have used the example names incorrectly for the problem. Let's assume the structure shown for (c) (a cyclohexane ring with a Cl and a CH$_3$) corresponds to the names 3-chloro-1-methylcyclohexane or similar.

Let's assume the intended structure for (c) is 1-chloro-3-methylcyclohexane. Parent is cyclohexane. Substituents are Chloro (-Cl) and Methyl (-CH$_3$). Chloro comes before Methyl alphabetically. So, numbering should start at the carbon with Cl and proceed to give the lowest number to the next substituent (Methyl). If Cl is on C1, Methyl is on C3 (1,3). If Methyl is on C1, Cl is on C3 (1,3). Locant set is (1, 3) in both cases. Alphabetical order (Chloro before Methyl) dictates that Cl gets the lower number (1). The correct name is 1-chloro-3-methylcyclohexane.

Let's assume the names given in (c) (3-Ethyl-1,1-dimethylcyclohexane) refer to the structure of a cyclohexane ring with two methyl groups on one carbon and one ethyl group on another carbon.

CH$_3$

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C—CH$_3$

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Ring—C

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CH$_2$CH$_3$

Correct numbering: Ring carbons get lowest locants. Alphabetical order of substituents (Ethyl before Methyl). The two methyls are on the same carbon. Let's say Ethyl is on C1. Then Methyls are on C3. Name: 1-ethyl-3,3-dimethylcyclohexane. Locants (1, 3, 3). What if Methyls are on C1? Then Ethyl is on C3. Name: 1,1-dimethyl-3-ethylcyclohexane. Locants (1, 1, 3). This set is lower. So 1,1-dimethyl-3-ethylcyclohexane is the correct parent name structure.

Given name: 3-Ethyl-1,1-dimethylcyclohexane. This means Ethyl is on C3 and two Methyls are on C1. Name: 1,1-dimethyl-3-ethylcyclohexane. Locants (1,1,3). The given name uses 3-Ethyl and 1,1-dimethyl. This indicates a disconnect. Let's assume the numbers refer to positions relative to some starting point, and the alphabetical order is applied to the full name. If we number such that the dimethyl group is on C1, the ethyl group is on C3. The name is 1,1-dimethyl-3-ethylcyclohexane. Alphabetical order is ethyl before dimethyl. So ethyl should get the lower number if there's a choice. Let's assume the given names in (c) are meant to show the application of the alphabetical rule when locants are equally low. If the substituents are on carbons 1 and 3 (which is the lowest possible combination for two distinct monosubstituted carbons), then alphabetical order determines which gets number 1. Chloro before Methyl. But here it is dimethyl and ethyl.

Let's focus on the rule explanation provided by the text in point 5: "If the two substituents are found in equivalent positions, the lower number is given to the one coming first in the alphabetical listing. Thus, the following compound is 3-ethyl-6-methyloctane and not 6-ethyl-3-methyloctane." This example confirms that if the set of locants is the same from either direction (3,6 vs 3,6 in the example), then alphabetical order determines priority for the lower locant. Ethyl (E) before Methyl (M), so Ethyl gets the lower locant (3). My earlier analysis of 3-ethyl-5-methylheptane vs 5-ethyl-3-methylheptane was correct based on this rule.

Let's revisit (a) and (b) in Problem 12.7 in the context of the explanation provided.

(a) 2,5,6-Trimethyloctane (correct) vs 3,4,7-Trimethyloctane (incorrect). The set of locants (2,5,6) is lower than (3,4,7). This follows rule 2/3. Not rule 5.

(b) 3-Ethyl-5-methylheptane (correct) vs 5-Ethyl-3-methylheptane (incorrect). Parent chain 7 carbons. Substituents at 3 and 5. Locants (3,5) in both directions. Ethyl at 3, Methyl at 5 (from left). Methyl at 3, Ethyl at 5 (from right). Alphabetical order: Ethyl before Methyl. Ethyl gets the lower number (3). Name 3-ethyl-5-methylheptane is correct. 5-ethyl-3-methylheptane is incorrect because Methyl (on C3) comes later in alphabetical order than Ethyl (on C5). This illustrates rule 5.

(c) 3-Ethyl-1,1-dimethylcyclohexane (incorrect parenthesis name: 1-ethyl-3,3-dimethylcyclohexane). Let's analyse the given names assuming they refer to the structure with one ethyl and two methyl groups on a cyclohexane ring. Name: 3-Ethyl-1,1-dimethylcyclohexane. Parent is cyclohexane. Substituents: Ethyl and two Methyls. Locants: 1,1,3. Alphabetical order: Ethyl (E) before Dimethyl (M). Let's try numbering: If the carbons with substituents are 1 and 3. If Ethyl is on C1 and two Methyls on C3, name is 1-ethyl-3,3-dimethylcyclohexane. Locants (1, 3, 3). If two Methyls are on C1 and Ethyl is on C3, name is 1,1-dimethyl-3-ethylcyclohexane. Locants (1, 1, 3). The set (1, 1, 3) is lower than (1, 3, 3). So the correct numbering starts at the carbon with the two methyls. The name is 1,1-dimethyl-3-ethylcyclohexane.

The given name is 3-Ethyl-1,1-dimethylcyclohexane. It seems the text is trying to show that alphabetical order is used *after* the lowest locant set is determined. The lowest locant set is (1, 1, 3). The substituents are on positions 1 (two methyls) and 3 (one ethyl). The name is constructed as [locant]-[substituent name]... Parent. Substituents are listed alphabetically: Ethyl, Dimethyl. So Ethyl comes first. The correct name should be 3-ethyl-1,1-dimethylcyclohexane.

Wait, the text marks the name (3-Ethyl-1,1-dimethylcyclohexane) as correct, and (1-ethyl-3,3-dimethylcyclohexane) as incorrect. This is opposite to the rules established! Let's reread Rule 5 and apply it rigorously to the original example in the text: "3-ethyl-6-methyloctane and not 6-ethyl-3-methyloctane". Octane, substituents at 3 and 6. Set (3,6). From left: Ethyl at 3, Methyl at 6. From right: Methyl at 3, Ethyl at 6. Alphabetical: Ethyl (E) before Methyl (M). So Ethyl gets the lower locant if there's a tie like (3,6) vs (3,6) if counted from opposite ends. This rule seems to be about choosing between numbering directions when the locant *sets* are identical. Here the set is (3,6) from both directions. So, the one that comes first alphabetically (Ethyl) gets the lower number (3). So 3-ethyl-6-methyloctane is correct. My previous analysis was correct. Rule 5 applies when the set of locants is the same regardless of numbering direction (e.g., 3,6 vs 3,6). The alphabet rule applies to which group gets the lower number in that set.

Let's re-examine the case in (c): 1,1-dimethyl-3-ethylcyclohexane vs 1-ethyl-3,3-dimethylcyclohexane. The set of locants from the first name is (1, 1, 3). The set of locants from the second name is (1, 3, 3). The set (1, 1, 3) is the lower set of locants. So the numbering starts at the carbon with the two methyl groups, and the ethyl group is on carbon 3. The correct name must use the lowest locant set (1,1,3). Within this set, the substituents are two methyls at 1 and one ethyl at 3. The name construction is [locant]-[substituent name]... followed by the parent. Substituents are listed alphabetically. Ethyl (E) comes before Dimethyl (M). So the name is 3-ethyl-1,1-dimethylcyclohexane.

The text marks (3-Ethyl-1,1-dimethylcyclohexane) as the correct name and (1-ethyl-3,3-dimethylcyclohexane) as incorrect. My derivation based on lowest locant set followed by alphabetical ordering matches this. The confusion was in applying rule 5 to this case, which is not about equivalent positions, but about which set of locants is lower.

Final analysis of Problem 12.7:

(a) 2,5,6-Trimethyloctane (correct) vs 3,4,7-Trimethyloctane (incorrect): Incorrect because (3,4,7) is not the lowest set of locants; (2,5,6) is lower.

(b) 3-Ethyl-5-methylheptane (correct) vs 5-Ethyl-3-methylheptane (incorrect): The set of locants is (3,5) from either direction. Alphabetical order (Ethyl before Methyl) determines that Ethyl gets the lower locant (3).

(c) 3-Ethyl-1,1-dimethylcyclohexane (correct) vs 1-ethyl-3,3-dimethylcyclohexane (incorrect): The lowest set of locants for the substituents is (1,1,3), achieved by numbering the carbon with the two methyls as C1. The substituents are two methyls at C1 and one ethyl at C3. Substituents are named alphabetically: Ethyl before Dimethyl. The name is 3-ethyl-1,1-dimethylcyclohexane. The parenthesised name uses the higher locant set (1,3,3) by starting numbering at the carbon with the ethyl group and proceeding to the carbon with the two methyls. This is incorrect.

So the explanations provided in the text for why the parenthesised names are incorrect are based on applying the IUPAC rules correctly (lowest locants first, then alphabetical order for ties in locants). My initial detailed analysis aligns with this. The confusion was in the interpretation of Rule 5 and applying it to cases it didn't directly apply to, and a potential misunderstanding of the example structures in the problem statement versus the names provided.

Final answer explanation for Problem 12.7 will state why the parenthesised name violates the specific IUPAC rule.

• **Cyclic Alkanes:** Named by prefixing 'cyclo' to the name of the corresponding straight-chain alkane. Substituent naming follows similar rules as branched alkanes, with numbering starting to give the lowest locants to substituents (alphabetical order for ties).

Nomenclature Of Organic Compounds Having Functional Group(s)

Functional groups determine the chemical properties of organic compounds. The IUPAC name of a compound with a functional group is derived from the parent hydrocarbon, with the functional group indicated by a suffix or prefix (Table 12.4).

Steps for naming compounds with functional groups:

1. Identify the principal functional group (highest priority).
2. Identify the longest carbon chain containing the principal functional group; this is the parent chain.
3. Number the parent chain to give the principal functional group the lowest possible locant.
4. Other functional groups (subordinate) and substituents are named as prefixes.
5. Name the compound using prefixes, locants, parent chain name, and suffix(es) for the functional group(s).

Order of priority for common functional groups (decreasing): -COOH > -SO$_3$H > -COOR > -COCl > -CONH$_2$ > -CN > -CHO > >C=O > -OH > -NH$_2$ > >C=C< > -C$\equiv$C-.

Substituents like alkyl groups, halogens, -NO$_2$, -OR are always named as prefixes.

Answer:

Answer:

Nomenclature Of Substituted Benzene Compounds

Substituted benzene compounds are named by placing the substituent as a prefix to 'benzene'. Common names for some substituted benzenes are widely used (e.g., Toluene for methylbenzene, Aniline for aminobenzene, Anisole for methoxybenzene).

For disubstituted benzene, positions are indicated by numbering the ring carbons to give the lowest possible locants (e.g., 1,2-, 1,3-, 1,4-). Trivial prefixes ortho (o-) for 1,2-, meta (m-) for 1,3-, and para (p-) for 1,4- are also used.

For tri- or higher substituted benzenes, lowest locant rule applies, and substituents are named alphabetically. If a common name derivative is used as the base, its principal functional group carbon is C1, and numbering follows to give the next substituent lowest locant.

When the benzene ring acts as a substituent on an alkane chain containing a functional group, the benzene ring is named as a phenyl group (C$_6$H$_5$-, abbreviated Ph).

Answer:

Isomerism

Isomerism is the phenomenon where two or more compounds have the same molecular formula but different structures and properties. Such compounds are called **isomers**. Isomerism is broadly classified into structural isomerism and stereoisomerism.

Structural Isomerism

Structural isomers have the **same molecular formula** but differ in the **connectivity** of atoms (how they are linked). Types of structural isomerism:

• **Chain isomerism:** Different carbon skeletons (straight vs branched). Example: C$_5$H$_{12}$ exists as pentane (straight chain), isopentane (2-methylbutane, branched), neopentane (2,2-dimethylpropane, branched).
• **Position isomerism:** Same carbon skeleton and functional group, but the position of the substituent or functional group on the chain or ring is different. Example: C$_3$H$_8$O can be Propan-1-ol (OH on C1) or Propan-2-ol (OH on C2).
• **Functional group isomerism:** Same molecular formula but different functional groups. Example: C$_3$H$_6$O can be Propanal (aldehyde, -CHO) or Propanone (ketone, >C=O).
• **Metamerism:** Arises in compounds with a functional group containing a heteroatom (like ether, ketone, ester) due to different alkyl chains on either side of the functional group. Example: C$_4$H$_{10}$O can be Methoxypropane (CH$_3$OC$_3$H$_7$) or Ethoxyethane (C$_2$H$_5$OC$_2$H$_5$).

Stereoisomerism

Stereoisomers have the **same molecular formula** and the **same connectivity** of atoms, but they differ in the **spatial arrangement** of their atoms or groups. Types of stereoisomerism:

• **Geometrical isomerism:** Occurs in compounds with restricted rotation (like double bonds or rings) where substituents on each side of the restricted rotation are arranged differently in space (cis vs trans).
• **Optical isomerism:** Occurs in compounds with chiral centers (usually a carbon bonded to four different groups), resulting in non-superimposable mirror images (enantiomers) that rotate plane-polarized light in opposite directions.

Fundamental Concepts In Organic Reaction Mechanism

An organic reaction involves a chemical transformation where an organic molecule (the **substrate**) reacts with an **attacking reagent** to form new product(s), often through intermediate species. The step-by-step description of how bonds break and form during a reaction is called the **reaction mechanism**. Understanding mechanisms helps predict reactivity and design synthetic routes.

General reaction: Substrate + Attacking Reagent $\rightarrow$ [Intermediate(s)] $\rightarrow$ Product(s) (+ Byproducts)

Fission Of A Covalent Bond

Covalent bonds can break in two main ways:

• **Heterolytic Cleavage (Heterolysis):** The shared electron pair goes entirely with one of the bonded atoms, forming ions. The atom that takes the electron pair gains a negative charge and has a complete valence shell (octet). The atom that loses the electron pair gains a positive charge and has an incomplete valence shell (sextet).

  Example: $\text{CH}_3\text{–Cl} \rightarrow \text{CH}_3^+ \text{ (carbocation)} + \text{Cl}^- \text{ (chloride ion)}$

  Species with a positively charged carbon with a sextet are called **carbocations** (primary, secondary, tertiary). They are electron-deficient and reactive. Stability increases with increasing alkyl substitution (methyl < primary < secondary < tertiary) due to inductive and hyperconjugation effects. Carbocations are sp$^2$ hybridised and have a trigonal planar shape. Species with a negatively charged carbon having a valence octet and a lone pair are called **carbanions** (primary, secondary, tertiary). Carbon is typically sp$^3$ hybridised, resulting in a pyramidal shape. They are also reactive. Electron-donating alkyl groups destabilise carbanions (opposite to carbocations). Reactions proceeding via heterolysis are called **ionic, heteropolar, or polar reactions**.
• **Homolytic Cleavage (Homolysis):** Each atom involved in the bond takes one electron from the shared pair, forming neutral species with unpaired electrons called **free radicals**. Single electron movement is shown by 'fish hook' curved arrows (half-headed arrows).

  Example: $\text{Cl–Cl} \rightarrow \text{Cl} \cdot + \text{Cl} \cdot \text{ (chlorine free radicals)}$

  $\text{CH}_3\text{–Br} \rightarrow \text{CH}_3 \cdot \text{ (methyl free radical)} + \text{Br} \cdot \text{ (bromine free radical)}$

  Alkyl free radical stability increases with increasing alkyl substitution (methyl < primary < secondary < tertiary) similar to carbocations. Reactions proceeding via homolysis are called **free radical, homopolar, or nonpolar reactions**.

Substrate And Reagent

In an organic reaction, it's often convenient to distinguish between the two reacting species: the **substrate** and the **reagent**. The substrate is typically the organic molecule that provides the carbon skeleton where the reaction occurs. The reagent is the species that attacks the substrate. If both molecules contribute carbon atoms to new bonds, the distinction is arbitrary, and the molecule of focus is called the substrate.

Electron Movement In Organic Reactions

Curved arrows are used to represent the movement of electron pairs or single electrons in organic reaction mechanisms. A full-headed curved arrow ($\curvearrowright$) shows the movement of an electron pair, starting from the source of the electron pair (e.g., a lone pair, a $\pi$ bond, or a $\sigma$ bond) and ending at the atom or bond where the electron pair moves. A half-headed curved arrow ( ) shows the movement of a single electron (as in free radical reactions).

Examples of electron movement with curved arrows:

• Formation of a bond by donation of a lone pair: $\text{Nu:}^- + \text{E}^+ \curvearrowright \text{Nu–E}$
• Breaking a $\pi$ bond: $ \curvearrowright $
• Movement of a lone pair to form a $\pi$ bond: $ \curvearrowright $

Answer:

Heterolytic cleavage (heterolysis) involves the bond breaking such that the shared electron pair stays with one of the fragments, forming ions. The arrow starts from the bond and points towards the atom that receives the electron pair.

Electron Displacement Effects In Covalent Bonds

Electron displacement effects in covalent bonds influence the polarity and reactivity of organic molecules. These can be permanent or temporary.

• **Permanent effects:** Caused by the influence of nearby atoms or groups, even in the absence of a reagent. * **Inductive effect:** Polarization of a $\sigma$ bond due to the presence of an electronegative or electropositive atom or group in the chain. The effect is transmitted along the $\sigma$ bonds but weakens rapidly with distance. Electron-withdrawing groups (-I effect) pull electron density; electron-donating groups (+I effect) push electron density. Halogens, -NO$_2$, -CN, -COOH are typical electron-withdrawing groups. Alkyl groups are considered electron-donating.
* **Resonance effect (Mesomeric effect):** Delocalisation of $\pi$ electrons or lone pairs in conjugated systems. It causes a redistribution of electron density and can create partial charges. (+R effect: electron donation away from the substituent into the conjugated system; -R effect: electron withdrawal towards the substituent from the conjugated system). Occurs in systems with alternating single and double bonds, or adjacent lone pairs/unoccupied p orbitals.
• **Temporary effects:** Occur only in the presence of an attacking reagent. * **Electromeric effect (E effect):** Complete transfer of a shared pair of $\pi$ electrons to one of the atoms joined by a multiple bond, triggered by an attacking reagent. It is temporary and is annulled when the reagent is removed. (+E effect: $\pi$ electrons move towards the atom to which the reagent attaches; -E effect: $\pi$ electrons move away from the atom to which the reagent attaches).
* **Polarisability effect:** Temporary dipole induced in a molecule by the approach of an attacking reagent.

Inductive Effect

When a $\sigma$ bond connects two atoms of different electronegativity, the electron density shifts towards the more electronegative atom, creating partial positive and negative charges ( bond polarity). This polarity in one $\sigma$ bond can induce polarity in adjacent $\sigma$ bonds, albeit to a lesser extent. This sequential polarization along a carbon chain is the inductive effect.

Example: Chloroethane (CH$_3$CH$_2$Cl). The C–Cl bond is polar ($\text{C}^\delta{^+}–\text{Cl}^\delta{^-}$). The carbon attached to Cl ($\text{C}_1$) becomes partially positive. This $\text{C}_1$ then attracts electron density from the adjacent C–C bond, causing $\text{C}_2$ to become partially positive ($\text{C}_2^{\delta\delta+}$). The inductive effect weakens rapidly with distance.

Substituents are classified by their inductive effect relative to hydrogen:

• **Electron-withdrawing groups (-I effect):** Pull electron density through $\sigma$ bonds (e.g., halogens, -NO$_2$, -CN, -COOH).
• **Electron-donating groups (+I effect):** Push electron density through $\sigma$ bonds (e.g., alkyl groups like -CH$_3$, -CH$_2$CH$_3$).

Answer:

Bond polarity is determined by the difference in electronegativity between the bonded atoms. A larger electronegativity difference leads to a more polar bond.

Answer:

Resonance Structure

Many organic molecules cannot be adequately represented by a single Lewis structure because of the delocalisation of $\pi$ electrons or lone pairs. The actual structure is a **resonance hybrid** (a weighted average) of two or more contributing Lewis structures called **resonance structures** (or canonical structures). Resonance structures are hypothetical and differ only in the arrangement of electrons, not the positions of nuclei or number of unpaired electrons.

Example: Benzene (C$_6$H$_6$). Its structure with alternating single and double bonds doesn't explain its uniform bond lengths. Benzene is a resonance hybrid of two equivalent Kekulé structures.

The resonance hybrid is more stable than any single contributing structure; the energy difference is the resonance energy. More stable contributing structures contribute more to the hybrid.

Rules for writing resonance structures:

1. Same connectivity of atoms.
2. Same number of valence electrons and unpaired electrons.
3. Arrows show electron movement.
4. More stable structures (more covalent bonds, complete octets, less charge separation, negative charge on more electronegative atom) contribute more.

Answer:

The acetate ion (CH$_3$COO$^-$) is the conjugate base of acetic acid. The negative charge is delocalised over the two oxygen atoms. The structure is CH$_3$—C(═O)—O$^-$ with a lone pair on the negatively charged oxygen and lone pairs on the double-bonded oxygen.

Answer:

The compound is propenal (acrolein). It has a conjugated system: a double bond (CH$_2$=CH) conjugated with a carbonyl double bond (CH=O). Delocalisation of $\pi$ electrons can occur.

Answer:

The molecule is methyl acetate (an ester). Its standard Lewis structure (major contributor) is CH$_3$—C(═O)—O—CH$_3$. Oxygen atoms have lone pairs. Resonance can occur involving the lone pair on the oxygen adjacent to the carbonyl carbon. Let's draw the major contributor (Structure III) and the resonance structure involving electron donation from the ester oxygen (Structure IV).

Resonance Effect

The resonance effect (also called the mesomeric effect) is the polarity induced in a conjugated system by the delocalisation of $\pi$ electrons or lone pairs. It's a permanent effect. It's classified as:

• **Positive Resonance Effect (+R or +M):** Electron-donating effect. Groups with lone pairs adjacent to a $\pi$ system donate electrons into the system (e.g., -OH, -OR, -NH$_2$, halogens). Electron density increases at certain positions in the conjugated system.
• **Negative Resonance Effect (-R or -M):** Electron-withdrawing effect. Groups with multiple bonds or positive charges adjacent to a $\pi$ system withdraw electrons from the system (e.g., -NO$_2$, -CN, -CHO, -COOH). Electron density decreases at certain positions.

Conjugated systems (alternating single and double bonds) are key to resonance. Resonance/Mesomeric effect is important in explaining reactivity and stability of conjugated compounds.

Electromeric Effect (E effect)

The electromeric effect is a **temporary** effect that occurs in compounds with multiple bonds when attacked by a reagent. It involves the complete transfer of a shared pair of $\pi$ electrons to one of the atoms involved in the multiple bond.

• **+E effect:** The $\pi$ electrons move towards the atom bonded to the attacking reagent. Typically occurs in addition reactions of alkenes/alkynes with electrophiles. Example: Addition of H$^+$ to ethene: $\text{CH}_2=\text{CH}_2 + \text{H}^+ \rightarrow \overset{+}{\text{CH}}_2-\overset{\curvearrowright}{\text{CH}_2}$ (Incorrect arrow, should be $\pi$ electrons moving to carbon bonded to H$^+$). Correct: $\text{CH}_2=\text{CH}_2 + \text{H}^+ \curvearrowright \overset{+}{\text{CH}}_2-\text{CH}_3$.
• **-E effect:** The $\pi$ electrons move away from the atom bonded to the attacking reagent. Typically occurs in addition reactions of carbonyl compounds with nucleophiles. Example: Addition of CN$^-$ to carbonyl: $ >\overset{\curvearrowright}{\text{C}}=\text{O} + \text{Nu}^- \curvearrowright >\overset{-}{\text{C}}-\text{O}^-$.

Electromeric effect predominates over the inductive effect when they oppose each other.

Hyperconjugation

Hyperconjugation is a **permanent** stabilising effect involving the delocalisation of $\sigma$ electrons of a C–H bond (in an alkyl group) into an adjacent empty p orbital or a $\pi$ system (double bond, benzene ring, positive charge). It is also called "no-bond resonance".

Example: Ethyl cation ($\text{CH}_3\overset{+}{\text{C}}\text{H}_2$). A C–H $\sigma$ bond of the methyl group can align with the empty p orbital on the positively charged carbon. The $\sigma$ electrons delocalise into the empty p orbital, helping to disperse the positive charge and stabilise the carbocation.

More alkyl groups attached to the positively charged carbon mean more C–H bonds available for hyperconjugation, leading to greater stabilisation. Stability of carbocations increases with increasing hyperconjugation (methyl < primary < secondary < tertiary).

Hyperconjugation also stabilises alkenes (delocalisation of $\sigma$ electrons into the adjacent $\pi$ system) and alkylarenes (delocalisation of $\sigma$ electrons into the adjacent aromatic $\pi$ system).

In alkenes, more alkyl substituents on the double bond lead to greater hyperconjugation and higher stability.

Answer:

Types Of Organic Reactions And Mechanisms

Organic reactions are broadly classified based on how the connectivity of atoms changes. The main types are:

• **Substitution reactions:** An atom or group in a molecule is replaced by another atom or group.
• **Addition reactions:** Atoms or groups are added across a multiple bond (double or triple). The multiple bond is converted to a single bond or a double bond.
• **Elimination reactions:** Atoms or groups are removed from adjacent carbons, typically resulting in the formation of a multiple bond.
• **Rearrangement reactions:** Atoms or groups within a molecule rearrange to form a structural isomer.

Each type of reaction proceeds through a specific mechanism involving intermediates and electron movements.

Methods Of Purification Of Organic Compounds

Organic compounds obtained from natural sources or synthesized in the laboratory often contain impurities. Various purification techniques are used to obtain pure compounds, based on differences in physical properties between the compound and the impurities.

Common methods:

• **Sublimation:** Used for solids that change directly from solid to vapour upon heating (sublimable solids) to separate them from non-sublimable impurities.
• **Crystallisation:** Most common method for purifying solid organic compounds. Based on differences in solubility of the compound and impurities in a suitable solvent. Impure solid is dissolved in a hot solvent, filtered to remove insoluble impurities, and the solution is cooled to crystallise the pure compound. Coloured impurities are removed by activated charcoal.
• **Distillation:** Used to separate volatile liquids from non-volatile impurities or liquids with significant differences in boiling points. Liquid is heated to vaporise, vapour is condensed and collected.
• **Fractional Distillation:** Used when boiling points are close. Vapours pass through a fractionating column, allowing repeated vaporisation and condensation steps, enriching the vapour in the lower boiling component at each stage.
• **Distillation under reduced pressure:** Used for liquids with very high boiling points or those that decompose upon heating at atmospheric pressure. Pressure is reduced, lowering the boiling point.
• **Steam Distillation:** Used for steam-volatile compounds that are immiscible with water. Steam is passed through the liquid, the mixture of steam and compound vapours is condensed, and the organic compound is separated from water (often using a separating funnel). The liquid boils below its normal boiling point because the total vapour pressure (liquid + water) equals atmospheric pressure.
• **Differential Extraction:** Used to separate an organic compound from an aqueous solution by shaking with an immiscible organic solvent where the compound is more soluble. The organic layer is separated and the solvent removed. Continuous extraction is used for less soluble compounds.
• **Chromatography:** Powerful technique for separation, purification, and purity testing. Based on differential distribution of components between a stationary phase and a mobile phase.

Types of Chromatography:

• **Adsorption Chromatography:** Separation based on differential adsorption on a solid stationary phase (adsorbent like silica gel or alumina). * **Column Chromatography:** Separation in a column packed with adsorbent. Different components are eluted at different rates. * **Thin Layer Chromatography (TLC):** Separation on a thin layer of adsorbent coated on a plate. Components separate as spots based on their retardation factor (Rf value).
• **Partition Chromatography:** Separation based on differential partitioning between a stationary liquid phase (often water trapped on paper) and a mobile solvent phase. * **Paper Chromatography:** A common type of partition chromatography using special paper.

Purity is checked by melting/boiling points or chromatographic/spectroscopic methods.

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Qualitative Analysis Of Organic Compounds

Qualitative analysis determines the elements present in an organic compound. Organic compounds always contain carbon and hydrogen. They may also contain oxygen, nitrogen, sulphur, halogens, and phosphorus.

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Detection Of Carbon And Hydrogen

Detected by heating the organic compound with copper(II) oxide. Carbon is oxidised to CO$_2$ (tested with lime water, which turns turbid: $\text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O}$). Hydrogen is oxidised to H$_2$O (tested with anhydrous copper sulfate, which turns blue: $\text{5H}_2\text{O} + \text{CuSO}_4 \rightarrow \text{CuSO}_4 \cdot \text{5H}_2\text{O}$).

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Detection Of Other Elements

Nitrogen, sulphur, halogens, and phosphorus are detected by converting them to ionic forms by fusing the organic compound with metallic sodium ("Lassaigne's test").

Na fusion reactions:

• Nitrogen: $\text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN}$ (Sodium cyanide)
• Sulphur: $\text{2Na} + \text{S} \rightarrow \text{Na}_2\text{S}$ (Sodium sulfide)
• Halogen (X=Cl, Br, I): $\text{Na} + \text{X} \rightarrow \text{NaX}$ (Sodium halide)
• Nitrogen and Sulphur: $\text{Na} + \text{C} + \text{N} + \text{S} \rightarrow \text{NaSCN}$ (Sodium thiocyanate)

The fused mass is extracted with water to get the **sodium fusion extract** (containing NaCN, Na$_2$S, NaX, NaSCN etc., depending on the elements present). Tests are performed on this extract.

• Test for Nitrogen: Sodium fusion extract + $\text{FeSO}_4$ solution + Acidification with concentrated $\text{H}_2\text{SO}_4$. Formation of **Prussian blue** colour confirms nitrogen. ($\text{CN}^- + \text{Fe}^{2+} \rightarrow [\text{Fe(CN)}_6]^{4-}$, then $\text{Fe}^{3+} + [\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3$, Prussian blue).
• Test for Sulphur: * Extract + Acetic acid + Lead acetate solution. **Black precipitate** (PbS) confirms sulfur. ($\text{S}^{2-} + \text{Pb}^{2+} \rightarrow \text{PbS} \downarrow$) * Extract + Sodium nitroprusside solution. **Violet colouration** confirms sulfur. ($\text{S}^{2-} + [\text{Fe(CN)}_5\text{NO}]^{2-} \rightarrow [\text{Fe(CN)}_5\text{NOS}]^{4-}$) If N and S are both present, NaSCN forms. Tests are done to avoid interference (e.g., boil extract with $\text{HNO}_3$ before halogen test).
• Test for Halogens: Extract + Nitric acid + Silver nitrate solution. **White precipitate** (AgCl) soluble in $\text{NH}_4\text{OH}$ (aq) $\implies$ Chlorine. **Yellowish precipitate** (AgBr) sparingly soluble in $\text{NH}_4\text{OH}$ (aq) $\implies$ Bromine. **Yellow precipitate** (AgI) insoluble in $\text{NH}_4\text{OH}$ (aq) $\implies$ Iodine. ($\text{X}^- + \text{Ag}^+ \rightarrow \text{AgX} \downarrow$). Extract must be boiled with concentrated HNO$_3$ before test if N or S are present to decompose NaCN or Na$_2$S.
• Test for Phosphorus: Heat compound with an oxidising agent (like $\text{Na}_2\text{O}_2$) to oxidise P to phosphate ($\text{PO}_4^{3-}$). Boil with $\text{HNO}_3$. Add ammonium molybdate solution. **Yellow colouration or precipitate** indicates phosphorus. ($\text{PO}_4^{3-} + \text{Ammonium Molybdate} \rightarrow \text{Ammonium phosphomolybdate}$)

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Quantitative Analysis

Quantitative analysis determines the mass percentage of each element in an organic compound, essential for empirical and molecular formula determination.

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Carbon And Hydrogen

Estimated simultaneously by combustion (Liebig's method). Known mass of compound burnt in excess oxygen and CuO. C $\rightarrow$ CO$_2$, H $\rightarrow$ H$_2$O. Produced H$_2$O is absorbed in weighed $\text{CaCl}_2$ tube, CO$_2$ in weighed KOH solution tube. Increase in mass of tubes corresponds to mass of H$_2$O and CO$_2$ formed.

Percentage of C = $\frac{\text{Mass of } \text{CO}_2 \times 12.011}{\text{Molar mass of } \text{CO}_2 \times \text{Mass of compound}} \times 100$. Molar mass CO$_2$ $\approx 44.011$. Using atomic mass 12 for C, Molar mass CO$_2$ = 44.

Percentage of C = $\frac{\text{Mass of } \text{CO}_2 \times 12}{\text{44} \times \text{Mass of compound}} \times 100$.

Percentage of H = $\frac{\text{Mass of } \text{H}_2\text{O} \times 2.016}{\text{Molar mass of } \text{H}_2\text{O} \times \text{Mass of compound}} \times 100$. Molar mass H$_2$O $\approx 18.015$. Using atomic mass 1 for H, Molar mass H$_2$O = 18.

Percentage of H = $\frac{\text{Mass of } \text{H}_2\text{O} \times 2}{\text{18} \times \text{Mass of compound}} \times 100$.

Problem 12.20. On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. Determine the percentage composition of carbon and hydrogen in the compound.

Answer:

Given: Mass of organic compound = 0.246 g. Mass of CO$_2$ produced = 0.198 g. Mass of H$_2$O produced = 0.1014 g.

Percentage of carbon = $\frac{\text{Mass of C in } \text{CO}_2}{\text{Mass of compound}} \times 100\%$.

Mass of C in 0.198 g CO$_2$: Molar mass of C = 12.011 g/mol. Molar mass of CO$_2$ = 44.009 g/mol. (Using 12 and 44 as in formula derivation).

Mass of C = $\frac{12}{44} \times \text{Mass of } \text{CO}_2 = \frac{12}{44} \times 0.198 \text{ g} \approx 0.054 \text{ g}$.

Percentage of carbon = $\frac{0.054 \text{ g}}{0.246 \text{ g}} \times 100\% \approx 21.95\%$. (Text matches)

Percentage of hydrogen = $\frac{\text{Mass of H in } \text{H}_2\text{O}}{\text{Mass of compound}} \times 100\%$.

Mass of H in 0.1014 g H$_2$O: Molar mass of H = 1.008 g/mol. Molar mass of H$_2$O = 18.015 g/mol. (Using 2 and 18 as in formula derivation).

Mass of H = $\frac{2}{18} \times \text{Mass of } \text{H}_2\text{O} = \frac{2}{18} \times 0.1014 \text{ g} \approx 0.011267 \text{ g}$.

Percentage of hydrogen = $\frac{0.011267 \text{ g}}{0.246 \text{ g}} \times 100\% \approx 4.58\%$. (Text matches)

The percentage composition is approximately **21.95% carbon** and **4.58% hydrogen**.

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Nitrogen

Two methods: Dumas method and Kjeldahl's method.

• **Dumas method:** Compound heated with CuO in $\text{CO}_2$ atmosphere $\rightarrow$ free N$_2$. $\text{N}_2$ collected over KOH solution (absorbs $\text{CO}_2$). Volume of $\text{N}_2$ measured at STP. Percentage calculated from $\text{N}_2$ volume. Not applicable for compounds with N in nitro, azo, or ring systems.
• **Kjeldahl's method:** Compound heated with concentrated $\text{H}_2\text{SO}_4 \rightarrow$ ammonium sulfate. Heated with excess NaOH $\rightarrow$ $\text{NH}_3$ gas. $\text{NH}_3$ absorbed in known volume of standard acid (e.g., $\text{H}_2\text{SO}_4$). Unreacted acid titrated with standard alkali. Amount of acid consumed $\rightarrow$ amount of $\text{NH}_3 \rightarrow$ amount of N. Not applicable for compounds with N in nitro, azo, or ring systems.

Problem 12.21. In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)

Answer:

Given: Mass of organic compound = 0.3 g. Volume of $\text{N}_2$ collected = 50 mL. Temperature $T = 300$ K. Pressure $p_{\text{Total}} = 715$ mm Hg. Aqueous tension at 300K = 15 mm Hg.

First, calculate the pressure of dry nitrogen gas: $p_{\text{N}_2} = p_{\text{Total}} - \text{Aqueous tension} = 715 \text{ mm Hg} - 15 \text{ mm Hg} = 700$ mm Hg.

Now, convert the volume of $\text{N}_2$ collected under these conditions to volume at STP (273.15 K and 760 mm Hg).

Using the combined gas law: $\frac{p_1V_1}{T_1} = \frac{p_{STP}V_{STP}}{T_{STP}}$.

$\frac{(700 \text{ mm Hg})(50 \text{ mL})}{300 \text{ K}} = \frac{(760 \text{ mm Hg})V_{STP}}{273.15 \text{ K}}$.

$V_{STP} = \frac{(700)(50)(273.15)}{(300)(760)} \text{ mL}$. Using 273 K for STP temperature as in text example:

$V_{STP} = \frac{(700)(50)(273)}{(300)(760)} \text{ mL} = \frac{9555000}{228000} \text{ mL} \approx 41.907$ mL. (Text gets 41.9 mL)

At STP, 1 mole of $\text{N}_2$ gas (28.014 g) occupies 22.414 L (or 22414 mL). (Using 28 g and 22400 mL as in text example).

22400 mL of $\text{N}_2$ at STP weighs 28 g.

So, 41.9 mL of $\text{N}_2$ at STP weighs $\frac{28 \text{ g}}{22400 \text{ mL}} \times 41.9 \text{ mL}$.

Mass of nitrogen = $\frac{28 \times 41.9}{22400} \text{ g} \approx 0.052375$ g.

Percentage of nitrogen = $\frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100\%$.

Percentage of nitrogen = $\frac{0.052375 \text{ g}}{0.3 \text{ g}} \times 100\% \approx 17.458\%$.

Rounding to two decimal places:

Percentage of nitrogen $\approx 17.46\%$. (Text matches)

The percentage composition of nitrogen in the compound is approximately **17.46%**.

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Halogens

Estimated by **Carius method**. Known mass of compound heated in Carius tube with fuming $\text{HNO}_3$ and $\text{AgNO}_3$. Halogen $\rightarrow$ AgX precipitate (AgCl white, AgBr pale yellow, AgI yellow). Precipitate filtered, dried, weighed. Percentage calculated from mass of AgX.

Percentage of halogen = $\frac{\text{Mass of AgX}}{\text{Molar mass of AgX}} \times \frac{\text{Atomic mass of halogen}}{\text{Mass of organic compound}} \times 100$.

Problem 12.23. In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

Answer:

Given: Mass of organic compound = 0.15 g. Mass of AgBr obtained = 0.12 g.

We need to find the percentage of bromine (Br) in the compound.

Molar mass of AgBr = Atomic mass of Ag + Atomic mass of Br. (Using approximate atomic masses: Ag = 107.87, Br = 79.904. Using 108 and 80 as in text example).

Molar mass of AgBr = 108 + 80 = 188 g/mol.

188 g of AgBr contains 80 g of Br.

Mass of Br in 0.12 g of AgBr = $\frac{80 \text{ g Br}}{188 \text{ g AgBr}} \times 0.12 \text{ g AgBr}$.

Mass of Br = $\frac{80 \times 0.12}{188} \text{ g} = \frac{9.6}{188} \text{ g} \approx 0.051064$ g.

Percentage of bromine = $\frac{\text{Mass of bromine}}{\text{Mass of organic compound}} \times 100\%$.

Percentage of bromine = $\frac{0.051064 \text{ g}}{0.15 \text{ g}} \times 100\% \approx 34.042\%$.

Rounding to two decimal places:

Percentage of bromine $\approx 34.04\%$. (Text matches)

The percentage of bromine in the compound is approximately **34.04%**.

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Sulphur

Estimated by **Carius method**. Compound heated with $\text{Na}_2\text{O}_2$ or fuming $\text{HNO}_3 \rightarrow$ S oxidised to $\text{H}_2\text{SO}_4$. $\text{H}_2\text{SO}_4$ precipitated as $\text{BaSO}_4$ by adding $\text{BaCl}_2$ solution. $\text{BaSO}_4$ filtered, dried, weighed. Percentage calculated from mass of $\text{BaSO}_4$.

Percentage of S = $\frac{\text{Mass of } \text{BaSO}_4}{\text{Molar mass of } \text{BaSO}_4} \times \frac{\text{Atomic mass of S}}{\text{Mass of organic compound}} \times 100$. Molar mass BaSO$_4 \approx 233.38$ (137.33 + 32.06 + 4*16.00). Atomic mass S $\approx 32.06$. Using 233 and 32 as in text example).

Percentage of S = $\frac{\text{Mass of } \text{BaSO}_4 \times 32}{\text{233} \times \text{Mass of organic compound}} \times 100$.

Problem 12.24. In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?

Answer:

Given: Mass of organic compound = 0.157 g. Mass of barium sulphate (BaSO$_4$) obtained = 0.4813 g.

We need to find the percentage of sulphur (S) in the compound.

Molar mass of BaSO$_4$ = Atomic mass of Ba + Atomic mass of S + 4 $\times$ Atomic mass of O. (Using approximate atomic masses: Ba = 137, S = 32, O = 16).

Molar mass of BaSO$_4$ = 137 + 32 + 4(16) = 137 + 32 + 64 = 233 g/mol. (Text uses 233)

233 g of BaSO$_4$ contains 32 g of S.

Mass of S in 0.4813 g of BaSO$_4$ = $\frac{32 \text{ g S}}{233 \text{ g BaSO}_4} \times 0.4813 \text{ g BaSO}_4$.

Mass of S = $\frac{32 \times 0.4813}{233} \text{ g} \approx \frac{15.3996}{233} \text{ g} \approx 0.06609$ g.

Percentage of sulphur = $\frac{\text{Mass of sulphur}}{\text{Mass of organic compound}} \times 100\%$.

Percentage of sulphur = $\frac{0.06609 \text{ g}}{0.157 \text{ g}} \times 100\% \approx 42.0955\%$.

Rounding to two decimal places:

Percentage of sulphur $\approx 42.10\%$. (Text matches)

The percentage of sulphur in the given compound is approximately **42.10%**.

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Phosphorus

Estimated by oxidising P to phosphoric acid ($\text{H}_3\text{PO}_4$). $\text{H}_3\text{PO}_4$ precipitated as ammonium phosphomolybdate or $\text{MgNH}_4\text{PO}_4$. $\text{MgNH}_4\text{PO}_4$ on ignition gives $\text{Mg}_2\text{P}_2\text{O}_7$. Percentage calculated from mass of precipitate.

Percentage of P = $\frac{\text{Mass of precipitate}}{\text{Molar mass of precipitate}} \times \frac{\text{Mass of P (or P2) in precipitate}}{\text{Mass of organic compound}} \times 100$.

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Oxygen

Percentage of oxygen is usually determined by difference: 100% - (sum of percentages of all other elements).

Alternatively, oxygen can be estimated directly: Compound decomposed in $\text{N}_2$ stream $\rightarrow$ oxygen converted to CO over hot coke $\rightarrow$ CO oxidised to $\text{CO}_2$ over $\text{I}_2\text{O}_5$ $\rightarrow$ $\text{CO}_2$ measured. Mass of $\text{CO}_2$ $\rightarrow$ mass of CO $\rightarrow$ mass of $\text{O}_2$.

Modern CHN elemental analysers determine C, H, and N simultaneously using combustion followed by detection of $\text{CO}_2$, H$_2$O, and $\text{N}_2$. Oxygen is usually determined by difference or sometimes directly by pyrolysis. These methods use very small sample quantities.

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Question 12.4 Give the IUPAC names of the following compounds :

(a)

(b)

(c)

(d)

(e)

(f) $Cl_2CHCH_2OH$

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Question 12.8 Identify the functional groups in the following compounds

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(b)

(c)

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Question 12.15 What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors ?

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(b)

(c)

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Question 12.16 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)

(b)

(c)

(d)

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