Organic Chemistry Q&A

a) Heterocyclic compound: A cyclic compound that contains atoms of at least two different elements as members of its ring(s).

Example: Pyridine (a six-membered ring containing five carbon atoms and one nitrogen atom).

b) Acyclic compound: An organic compound that does not contain a ring structure.

Example: Butane (CH₃CH₂CH₂CH₃).

Position Isomerism: Isomers that differ in the position of the functional group or a substituent atom on the carbon chain.

Example: But-1-ene (CH₂=CH-CH₂-CH₃) and But-2-ene (CH₃-CH=CH-CH₃). The double bond is at a different position.

Functional Isomerism: Isomers that have the same molecular formula but different functional groups.

Example: Ethanol (CH₃CH₂OH, an alcohol) and Dimethyl ether (CH₃OCH₃, an ether). Both have the molecular formula C₂H₆O.

Chain Isomerism (or Skeletal Isomerism): Isomers that differ in the arrangement of the carbon chain (straight chain vs. branched chain).

Example: n-Butane (CH₃CH₂CH₂CH₃, a straight chain) and Isobutane (2-methylpropane, (CH₃)₃CH, a branched chain). Both have the molecular formula C₄H₁₀.

Metamerism: A type of isomerism in which different alkyl chains are attached to the same functional group. It is observed in compounds like ethers, ketones, and secondary amines.

Example: Diethyl ether (CH₃CH₂-O-CH₂CH₃) and Methyl propyl ether (CH₃-O-CH₂CH₂CH₃). Both have the molecular formula C₄H₁₀O.

Nucleophiles: Electron-rich species that are attracted to positively charged or electron-deficient centers (nuclei). They are Lewis bases.

Example: Hydroxide ion (OH^-), Ammonia (NH₃).

Electrophiles: Electron-deficient species that are attracted to negatively charged or electron-rich centers. They are Lewis acids.

Example: Carbocations (CH₃⁺), Boron trifluoride (BF₃).

• SO₃: Electrophile (Sulfur has an incomplete octet and can accept electrons).
• C₂H₅O^-: Nucleophile (Has a negative charge and lone pair on oxygen).
• OH^-: Nucleophile (Has a negative charge and lone pair on oxygen).
• BF₃: Electrophile (Boron has an incomplete octet, acts as a Lewis acid).
• NH₃: Nucleophile (Nitrogen has a lone pair of electrons).
• C₆H₅⁺: Electrophile (Carbocation, electron deficient).

Propanone and propanal exhibit functional isomerism.

• Propanone (Acetone): CH₃-CO-CH₃ (Functional group: Ketone, -CO-)
• Propanal (Propionaldehyde): CH₃CH₂-CHO (Functional group: Aldehyde, -CHO)

Both have the molecular formula C₃H₆O.

Homologous Series: A series of organic compounds in which all members have the same general formula, similar chemical properties, and successive members differ by a CH₂ group.

Characteristics:

1. All members can be represented by a general formula (e.g., alkanes C_nH_2n+2).
2. Successive members differ by a -CH₂ group in their molecular formula.
3. They show similar chemical properties due to the same functional group.
4. There is a gradual change in physical properties (e.g., melting point, boiling point) as the molecular mass increases.
5. They can be prepared by general methods.

Carbocation: A species in which a carbon atom carries a positive charge and has only six electrons in its valence shell. It is an electron-deficient species.

Stability of alkyl carbocations: Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (CH₃⁺).

This is due to the electron-donating inductive effect (+I effect) of alkyl groups, which helps to disperse the positive charge, and hyperconjugation.

Free Radicals: Atoms or groups of atoms that have one or more unpaired electrons. They are highly reactive and electrically neutral.

Example: Methyl radical (CH₃•), Chlorine radical (Cl•).

Heterolytic Cleavage (Heterolysis): The breaking of a covalent bond in such a way that both the shared electrons are retained by one of the bonded atoms. This results in the formation of a cation and an anion.

Example: R-X → R⁺ + X^- (where X is more electronegative than R).

Homolytic Cleavage (Homolysis): The breaking of a covalent bond in such a way that each of the bonded atoms retains one of the shared electrons. This results in the formation of two free radicals.

Example: A-B → A• + B•

Preparation of Sodium Fusion Extract (Lassaigne's Extract):

1. A small piece of dry sodium metal is heated strongly in a fusion tube with the organic compound.
2. The tube is then plunged into distilled water in a porcelain dish.
3. The contents are boiled, cooled, and then filtered. The filtrate is called sodium fusion extract or Lassaigne's extract.

This process converts elements like N, S, and halogens present in the organic compound into their ionic forms (NaCN, Na₂S, NaX), which are soluble in water.

Detection of Sulphur using Sodium Fusion Extract:

• Sodium Nitroprusside Test: To a portion of the Lassaigne's extract, add a few drops of sodium nitroprusside solution. A violet or purple coloration indicates the presence of sulphur (due to the formation of sodium thionitroprusside, Na₄[Fe(CN)₅NOS]).
• Lead Acetate Test: Acidify a portion of the extract with acetic acid and add lead acetate solution. A black precipitate of lead sulphide (PbS) indicates sulphur.

Detection of Nitrogen using Sodium Fusion Extract:

1. To a portion of the Lassaigne's extract, add freshly prepared ferrous sulphate solution and heat gently.
2. Then, add a few drops of ferric chloride solution and acidify with concentrated sulphuric acid.

A Prussian blue or green precipitate/coloration indicates the presence of nitrogen (due to the formation of ferric ferrocyanide, Fe₄[Fe(CN)₆]₃).

Detection of Halogens using Sodium Fusion Extract:

1. Acidify a portion of the Lassaigne's extract with dilute nitric acid and boil it (to decompose any NaCN or Na₂S which would interfere with the test).
2. Cool the solution and add silver nitrate solution.

• Chloride (Cl): A white precipitate soluble in ammonium hydroxide indicates chlorine (AgCl).
• Bromide (Br): A pale yellow precipitate sparingly soluble in ammonium hydroxide indicates bromine (AgBr).
• Iodide (I): A yellow precipitate insoluble in ammonium hydroxide indicates iodine (AgI).

Estimation of Carbon and Hydrogen in an Organic Compound using Liebig's Method:

1. A known mass of the organic compound is heated strongly in a combustion tube in the presence of excess dry copper(II) oxide.
2. Carbon is oxidized to carbon dioxide (CO₂), and hydrogen is oxidized to water (H₂O).
3. The H₂O produced is absorbed by a weighed U-tube containing anhydrous calcium chloride (CaCl₂).
4. The CO₂ produced is absorbed by a weighed U-tube containing concentrated KOH solution.

The increase in mass of the CaCl₂ tube gives the mass of water, and the increase in mass of the KOH tube gives the mass of carbon dioxide. From these masses, the percentages of carbon and hydrogen in the original compound can be calculated.

Carius Method of Estimation of Halogens:

1. A known mass of the organic compound is heated in a sealed Carius tube with fuming nitric acid and a few crystals of silver nitrate.
2. The halogen present in the compound is converted into its silver halide (AgX).
3. The silver halide precipitate is filtered, washed, dried, and weighed.

From the mass of silver halide, the percentage of the halogen in the organic compound can be calculated.

Dumas Method of Estimation of Nitrogen:

1. A known mass of the organic compound is heated with copper(II) oxide in an atmosphere of carbon dioxide.
2. Nitrogen present in the compound is converted into free nitrogen gas (N₂). Carbon and hydrogen are oxidized to CO₂ and H₂O.
3. The gaseous mixture (N₂, CO₂, H₂O) is passed over a heated copper gauze (to reduce any oxides of nitrogen to N₂) and then through a nitrometer containing KOH solution.

The KOH solution absorbs CO₂, and the nitrogen gas collects in the upper part of the nitrometer. The volume of nitrogen gas is measured, and from this, its mass is calculated using the gas equation. Finally, the percentage of nitrogen in the organic compound is determined.

Differences between Resonance Effect (Mesomeric Effect) and Inductive Effect:

Differences between Inductive Effect and Electromeric Effect:

One use of Chromatography:

Separation and purification of mixtures of organic compounds (e.g., separating pigments from plant extracts, separating amino acids).

a) Propan-1-ol (CH₃CH₂CH₂OH) and Propan-2-ol (CH₃CH(OH)CH₃): Position Isomerism (The -OH functional group is at different positions on the carbon chain).

b) n-Pentane (CH₃CH₂CH₂CH₂CH₃) and Isopentane (2-methylbutane, CH₃CH(CH₃)CH₂CH₃): Chain Isomerism (Different arrangements of the carbon skeleton).

Free radicals are formed by the homolytic cleavage of a covalent bond. This typically occurs under conditions like:

• High temperatures: Thermal decomposition.
• UV light or visible light: Photolysis.
• Presence of initiators: Peroxides, azo compounds, etc.
• Redox reactions: Electron transfer reactions.

The element estimated by Kjeldahl method is Nitrogen.

The lower homologue of CH₃COOH (Acetic Acid) is HCOOH (Formic Acid).

An example for a non-benzenoid aromatic compound is Azulene.

(Note: The structure provided for part (a) is ambiguous. Assuming the structure refers to 3-methylpentane (CH₃-CH₂-CH(CH₃)-CH₂-CH₃) for part (a), and parts (b) and (c) refer to general concepts or a related compound like 2-bromobutane for hybridization.)

a) Bond line formula for 3-methylpentane:

      CH₃
      │
CH₃─CH₂─CH─CH₂─CH₃
                

In bond-line notation, this would be represented as:

    /
---C---
    \
                

b) Hybridization of carbon attached to bromine:

In an organic compound, a carbon atom directly attached to bromine (e.g., in bromoalkanes like CH₃CH₂Br or CH₃CHBrCH₃) is typically sp³ hybridized, as it forms four single bonds.

c) How many sigma bonds are present in it (assuming 3-methylpentane, C₆H₁₄):

For an acyclic alkane with molecular formula C_nH_2n+2, the number of sigma bonds is (3n + 1).

For 3-methylpentane (C₆H₁₄):

Number of sigma bonds = (3 × 6) + 1 = 18 + 1 = 19 sigma bonds.