Question 1

Q1. What are hybridisation states of each carbon atom in the following compounds? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6.

Accepted Answer

To determine the hybridisation state of each carbon atom, we count the number of sigma (σ) bonds and lone pairs around it. For carbon, the hybridisation is sp3 if it forms four σ bonds, sp2 if it forms three σ bonds (and one π bond), and sp if it forms two σ bonds (and two π bonds).

**1. CH2=C=O (Ketene)**
   *   **C1 (CH2=)**: Forms two σ bonds with H atoms and one σ bond with C2. Total 3 σ bonds. Therefore, **sp2 hybridised**.
   *   **C2 (=C=)**: Forms one σ bond with C1 and one σ bond with O. Total 2 σ bonds. Therefore, **sp hybridised**.
   *   **C3 (=O)**: Not a carbon atom.

**2. CH3CH=CH2 (Propene)**
   *   **C1 (CH3-)**: Forms three σ bonds with H atoms and one σ bond with C2. Total 4 σ bonds. Therefore, **sp3 hybridised**.
   *   **C2 (-CH=)**: Forms one σ bond with C1, one σ bond with H, and one σ bond with C3. Total 3 σ bonds. Therefore, **sp2 hybridised**.
   *   **C3 (=CH2)**: Forms one σ bond with C2 and two σ bonds with H atoms. Total 3 σ bonds. Therefore, **sp2 hybridised**.

**3. (CH3)2CO (Acetone or Propan-2-one)**
   *   **C1 (CH3-)**: Forms three σ bonds with H atoms and one σ bond with C2. Total 4 σ bonds. Therefore, **sp3 hybridised**.
   *   **C2 (-CO-)**: Forms one σ bond with C1, one σ bond with C3, and one σ bond with O. Total 3 σ bonds. Therefore, **sp2 hybridised**.
   *   **C3 (-CH3)**: Forms three σ bonds with H atoms and one σ bond with C2. Total 4 σ bonds. Therefore, **sp3 hybridised**.

**4. CH2=CHCN (Acrylonitrile or Propenenitrile)**
   *   **C1 (CH2=)**: Forms two σ bonds with H atoms and one σ bond with C2. Total 3 σ bonds. Therefore, **sp2 hybridised**.
   *   **C2 (-CH=)**: Forms one σ bond with C1, one σ bond with H, and one σ bond with C3. Total 3 σ bonds. Therefore, **sp2 hybridised**.
   *   **C3 (-CN)**: Forms one σ bond with C2 and one σ bond with N (triple bond consists of one σ and two π bonds). Total 2 σ bonds. Therefore, **sp hybridised**.

**5. C6H6 (Benzene)**
   *   Each carbon atom in benzene is bonded to two other carbon atoms and one hydrogen atom. It forms three σ bonds (two C-C and one C-H) and participates in one π bond (part of the delocalised π system). Total 3 σ bonds. Therefore, **all six carbon atoms are sp2 hybridised**.

Detailed structures showing bonds would clarify the hybridisation for each carbon atom.

Question 2

Q2. Indicate the σ and π bonds in the following molecules: C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCN, C2H4.

Accepted Answer

To indicate the number of sigma (σ) and pi (π) bonds, we first draw the Lewis structure or expanded structural formula for each molecule. Remember:
*   A single bond consists of one σ bond.
*   A double bond consists of one σ bond and one π bond.
*   A triple bond consists of one σ bond and two π bonds.

**1. C6H6 (Benzene)**
   *   Structure: A cyclic six-membered ring with alternating single and double bonds (or a delocalised π system). Each carbon is bonded to one hydrogen.
   *   C-C single bonds: 3
   *   C=C double bonds: 3
   *   C-H single bonds: 6
   *   Total σ bonds = (3 C-C single) + (3 C-C double, each contributing 1σ) + (6 C-H) = 3 + 3 + 6 = **12 σ bonds**.
   *   Total π bonds = (3 C=C double, each contributing 1π) = **3 π bonds**.

**2. C6H12 (Cyclohexane)**
   *   Structure: A cyclic six-membered ring with only single bonds. Each carbon is bonded to two hydrogens.
   *   C-C single bonds: 6
   *   C-H single bonds: 12
   *   Total σ bonds = (6 C-C) + (12 C-H) = **18 σ bonds**.
   *   Total π bonds = **0 π bonds**.

**3. CH2Cl2 (Dichloromethane)**
   *   Structure: Central carbon bonded to two hydrogens and two chlorines, all single bonds.
   *   C-H single bonds: 2
   *   C-Cl single bonds: 2
   *   Total σ bonds = (2 C-H) + (2 C-Cl) = **4 σ bonds**.
   *   Total π bonds = **0 π bonds**.

**4. CH2=C=CH2 (Allene)**
   *   Structure: Two double bonds adjacent to a central carbon.
   *   C-H single bonds: 4 (two on each terminal carbon)
   *   C=C double bonds: 2
   *   Total σ bonds = (4 C-H) + (2 C=C, each contributing 1σ) = 4 + 2 = **6 σ bonds**.
   *   Total π bonds = (2 C=C, each contributing 1π) = **2 π bonds**.

**5. CH3NO2 (Nitromethane)**
   *   Structure: Methyl group attached to a nitro group. The nitro group has one N=O double bond and one N-O single bond, with formal charges and resonance. For counting bonds, we consider the resonance hybrid or one of the canonical forms. Let's consider the form with one N=O and one N-O.
   *   C-H single bonds: 3
   *   C-N single bond: 1
   *   N=O double bond: 1 (1σ, 1π)
   *   N-O single bond: 1 (1σ)
   *   Total σ bonds = (3 C-H) + (1 C-N) + (1 N-O from double) + (1 N-O single) = 3 + 1 + 1 + 1 = **6 σ bonds**.
   *   Total π bonds = (1 N=O from double) = **1 π bond**.

**6. HCN (Hydrogen Cyanide)**
   *   Structure: Hydrogen bonded to carbon, which is triple-bonded to nitrogen.
   *   C-H single bond: 1
   *   C≡N triple bond: 1 (1σ, 2π)
   *   Total σ bonds = (1 C-H) + (1 C≡N, contributing 1σ) = **2 σ bonds**.
   *   Total π bonds = (1 C≡N, contributing 2π) = **2 π bonds**.

**7. C2H4 (Ethene)**
   *   Structure: Two carbons double-bonded to each other, each carbon bonded to two hydrogens.
   *   C-H single bonds: 4
   *   C=C double bond: 1
   *   Total σ bonds = (4 C-H) + (1 C=C, contributing 1σ) = 4 + 1 = **5 σ bonds**.
   *   Total π bonds = (1 C=C, contributing 1π) = **1 π bond**.

Question 3

Q3. Write the bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Accepted Answer

Bond-line formulas (or skeletal structures) represent carbon-carbon bonds as lines and omit carbon and hydrogen atoms attached to carbon. Heteroatoms (like O, N, Cl) and hydrogens attached to them are shown explicitly. Each vertex or end of a line represents a carbon atom, and sufficient hydrogens are assumed to complete carbon's valency (4).

**1. Isopropyl alcohol**
   *   IUPAC name: Propan-2-ol
   *   Molecular formula: CH3CH(OH)CH3
   *   Description: A three-carbon chain with an -OH group on the middle carbon. The bond-line formula will show a 'V' shape with an -OH group attached to the central carbon.

**2. 2,3-Dimethylbutanal**
   *   Molecular formula: CH3CH(CH3)CH(CH3)CHO
   *   Description: A four-carbon main chain (butanal) with an aldehyde group (-CHO) at C1. Methyl groups are attached at C2 and C3. The bond-line formula will show a zig-zag four-carbon chain with a double-bonded oxygen at one end (aldehyde) and two short lines representing methyl groups at the second and third carbons from the aldehyde end.

**3. Heptan-4-one**
   *   Molecular formula: CH3CH2CH2COCH2CH2CH3
   *   Description: A seven-carbon main chain (heptane) with a ketone group (-CO-) at C4. The bond-line formula will show a zig-zag seven-carbon chain with a double-bonded oxygen at the fourth carbon from either end.

Question 4

Q4. What are hybridisation states of each carbon atom in the following compounds? CH₂=C=O, CH₃CH=CH₂, (CH₃)₂C=O, CH₂=CH-CN, C₆H₆.

Accepted Answer

The hybridization state of a carbon atom is determined by the number of sigma (σ) bonds it forms and the number of lone pairs it possesses. For carbon, which typically forms four bonds and has no lone pairs, the hybridization is determined solely by the number of sigma bonds:

*   **sp³ hybridization**: 4 sigma bonds (e.g., in alkanes, -CH₃, -CH₂-)
*   **sp² hybridization**: 3 sigma bonds (e.g., in alkenes, carbonyl carbons, aromatic carbons)
*   **sp hybridization**: 2 sigma bonds (e.g., in alkynes, nitriles, allenes' central carbon)

Let's analyze each compound:

**1. CH₂=C=O (Ketene)**
    *   **Carbon 1 (CH₂=)**: Forms two C-H sigma bonds and one C=C double bond (which contains one sigma and one pi bond). Total sigma bonds = 2 (C-H) + 1 (C=C) = 3 sigma bonds. Therefore, Carbon 1 is **sp² hybridized**.
    *   **Carbon 2 (=C=O)**: Forms one C=C double bond (one sigma, one pi) and one C=O double bond (one sigma, one pi). Total sigma bonds = 1 (C=C) + 1 (C=O) = 2 sigma bonds. Therefore, Carbon 2 is **sp hybridized**.

**2. CH₃CH=CH₂ (Propene)**
    *   **Carbon 1 (CH₃-)**: Forms three C-H sigma bonds and one C-C sigma bond. Total sigma bonds = 3 (C-H) + 1 (C-C) = 4 sigma bonds. Therefore, Carbon 1 is **sp³ hybridized**.
    *   **Carbon 2 (-CH=)**: Forms one C-C sigma bond, one C-H sigma bond, and one C=C double bond (one sigma, one pi). Total sigma bonds = 1 (C-C) + 1 (C-H) + 1 (C=C) = 3 sigma bonds. Therefore, Carbon 2 is **sp² hybridized**.
    *   **Carbon 3 (=CH₂)**: Forms one C=C double bond (one sigma, one pi) and two C-H sigma bonds. Total sigma bonds = 1 (C=C) + 2 (C-H) = 3 sigma bonds. Therefore, Carbon 3 is **sp² hybridized**.

**3. (CH₃)₂C=O (Acetone / Propan-2-one)**
    *   **Carbon 1 & 3 (CH₃-)**: Each methyl carbon forms three C-H sigma bonds and one C-C sigma bond. Total sigma bonds = 3 (C-H) + 1 (C-C) = 4 sigma bonds. Therefore, both methyl carbons are **sp³ hybridized**.
    *   **Carbon 2 (=C=O)**: Forms two C-C sigma bonds and one C=O double bond (one sigma, one pi). Total sigma bonds = 2 (C-C) + 1 (C=O) = 3 sigma bonds. Therefore, the carbonyl carbon is **sp² hybridized**.

**4. CH₂=CH-CN (Acrylonitrile / Propenenitrile)**
    *   **Carbon 1 (CH₂=)**: Forms two C-H sigma bonds and one C=C double bond (one sigma, one pi). Total sigma bonds = 2 (C-H) + 1 (C=C) = 3 sigma bonds. Therefore, Carbon 1 is **sp² hybridized**.
    *   **Carbon 2 (-CH=)**: Forms one C=C double bond (one sigma, one pi), one C-H sigma bond, and one C-C sigma bond. Total sigma bonds = 1 (C=C) + 1 (C-H) + 1 (C-C) = 3 sigma bonds. Therefore, Carbon 2 is **sp² hybridized**.
    *   **Carbon 3 (-CN)**: Forms one C-C sigma bond and one C≡N triple bond (one sigma, two pi bonds). Total sigma bonds = 1 (C-C) + 1 (C≡N) = 2 sigma bonds. Therefore, Carbon 3 is **sp hybridized**.

**5. C₆H₆ (Benzene)**
    *   All six carbon atoms in benzene are identical due to resonance. Each carbon atom forms one C-H sigma bond and two C-C sigma bonds (one to each adjacent carbon). Total sigma bonds = 1 (C-H) + 2 (C-C) = 3 sigma bonds. Therefore, all six carbon atoms are **sp² hybridized**.

Question 5

Q5. Indicate the σ and π bonds in the following molecules: C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃.

Accepted Answer

To determine the number of sigma (σ) and pi (π) bonds, we need to draw the Lewis structure for each molecule and apply the following rules:
*   **Single bond**: Consists of 1 σ bond.
*   **Double bond**: Consists of 1 σ bond and 1 π bond.
*   **Triple bond**: Consists of 1 σ bond and 2 π bonds.

Let's analyze each molecule:

**1. C₆H₆ (Benzene)**
    *   **Structure**: Benzene is a cyclic molecule with 6 carbon atoms and 6 hydrogen atoms. Its structure can be represented by Kekule structures with alternating single and double bonds, or as a resonance hybrid. For counting σ and π bonds, we can consider one Kekule structure.
    *   **Bonds**: There are 6 C-H single bonds, and 6 C-C bonds (3 single, 3 double in a Kekule structure).
    *   **σ bonds**: 6 (C-H) + 6 (C-C) = 12 σ bonds.
    *   **π bonds**: 3 (from 3 C=C double bonds) = 3 π bonds.
    *   **Total**: 12 σ bonds, 3 π bonds.

**2. C₆H₁₂ (Cyclohexane)**
    *   **Structure**: Cyclohexane is a cyclic alkane with 6 carbon atoms and 12 hydrogen atoms, all connected by single bonds.
    *   **Bonds**: There are 6 C-C single bonds and 12 C-H single bonds.
    *   **σ bonds**: 6 (C-C) + 12 (C-H) = 18 σ bonds.
    *   **π bonds**: 0 π bonds (all single bonds).
    *   **Total**: 18 σ bonds, 0 π bonds.

**3. CH₂Cl₂ (Dichloromethane)**
    *   **Structure**: A central carbon atom bonded to two hydrogen atoms and two chlorine atoms, all by single bonds.
    *   **Bonds**: There are 2 C-H single bonds and 2 C-Cl single bonds.
    *   **σ bonds**: 2 (C-H) + 2 (C-Cl) = 4 σ bonds.
    *   **π bonds**: 0 π bonds (all single bonds).
    *   **Total**: 4 σ bonds, 0 π bonds.

**4. CH₂=C=CH₂ (Allene)**
    *   **Structure**: A molecule with two adjacent double bonds.
    *   **Bonds**: There are 4 C-H single bonds and 2 C=C double bonds.
    *   **σ bonds**: 4 (C-H) + 2 (C=C, each contributing 1 σ) = 6 σ bonds.
    *   **π bonds**: 2 (from 2 C=C double bonds, each contributing 1 π) = 2 π bonds.
    *   **Total**: 6 σ bonds, 2 π bonds.

**5. CH₃NO₂ (Nitromethane)**
    *   **Structure**: A methyl group attached to a nitro group. The nitro group (NO₂) has resonance structures, but for counting, we can consider one canonical form with a C-N single bond, one N=O double bond, and one N-O single bond (with formal charges).
    *   **Bonds**: There are 3 C-H single bonds, 1 C-N single bond, 1 N=O double bond, and 1 N-O single bond.
    *   **σ bonds**: 3 (C-H) + 1 (C-N) + 1 (N=O) + 1 (N-O) = 6 σ bonds.
    *   **π bonds**: 1 (from N=O double bond) = 1 π bond.
    *   **Total**: 6 σ bonds, 1 π bond.

**6. HCONHCH₃ (N-Methylformamide)**
    *   **Structure**: An amide functional group where the nitrogen is bonded to a methyl group.
    *   **Bonds**: There is 1 C-H single bond (formyl group), 1 C=O double bond, 1 C-N single bond, 1 N-H single bond, and 3 C-H single bonds (methyl group).
    *   **σ bonds**: 1 (C-H) + 1 (C=O) + 1 (C-N) + 1 (N-H) + 3 (C-H) = 8 σ bonds.
    *   **π bonds**: 1 (from C=O double bond) = 1 π bond.
    *   **Total**: 8 σ bonds, 1 π bond.

Question 6

Q6. What are the essential conditions for a molecule to be chiral?

Accepted Answer

For a molecule to be chiral, it must satisfy the following essential conditions:

1.  **Presence of a Chiral Centre (Stereocentre):** The most common condition is the presence of a carbon atom bonded to four different atoms or groups. This carbon atom is called a chiral carbon or an asymmetric carbon. While a chiral carbon is a strong indicator, it's important to note that molecules without chiral carbons can also be chiral (e.g., allenes, biphenyls with restricted rotation), but this is less common at the introductory level.

2.  **Non-superimposable Mirror Image:** The molecule must not be superimposable on its mirror image. If a molecule and its mirror image are identical (superimposable), the molecule is achiral. If they are different (non-superimposable), the molecule is chiral.

3.  **Absence of Elements of Symmetry:** A chiral molecule must not possess any elements of symmetry, specifically:
    *   **Plane of Symmetry:** An imaginary plane that divides the molecule into two halves that are mirror images of each other. If a molecule has a plane of symmetry, it is achiral.
    *   **Centre of Symmetry:** A point within the molecule from which, if lines are drawn to any atom and extended an equal distance in the opposite direction, they encounter an identical atom. If a molecule has a centre of symmetry, it is achiral.
    *   **Alternating Axis of Symmetry (Sn):** This is a more complex element of symmetry. If a molecule possesses an Sn axis, it is achiral.

In summary, the primary condition is that a molecule and its mirror image are non-superimposable. The presence of a chiral carbon and the absence of a plane or center of symmetry are practical indicators to determine chirality.

Question 7

Q7. Write the structural formulae of the compounds whose IUPAC names are given below:
(i) 2-Methylbutane
(ii) Pent-2-ene
(iii) 3,4-Dimethylhex-3-ene
(iv) 2-Iodopropan-2-ol

Accepted Answer

The structural formulae for the given IUPAC names are as follows:

(i) **2-Methylbutane**
   *   Parent chain: Butane (4 carbons in a straight chain).
   *   Methyl group at position 2.
   *   Structure: CH₃-CH(CH₃)-CH₂-CH₃

(ii) **Pent-2-ene**
   *   Parent chain: Pentene (5 carbons).
   *   Double bond starts at position 2.
   *   Structure: CH₃-CH=CH-CH₂-CH₃

(iii) **3,4-Dimethylhex-3-ene**
   *   Parent chain: Hexene (6 carbons).
   *   Double bond starts at position 3.
   *   Two methyl groups: one at position 3 and one at position 4.
   *   Structure: CH₃-CH₂-C(CH₃)=C(CH₃)-CH₂-CH₃

(iv) **2-Iodopropan-2-ol**
   *   Parent chain: Propanol (3 carbons with an -OH group).
   *   -OH group at position 2.
   *   Iodo group at position 2.
   *   Structure: CH₃-C(I)(OH)-CH₃

Question 8

Q8. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C6H5OH
(b) CH3CH=CHCHO
(c) CH2=CH-CN
(d) C6H5NO2
(e) CH3CH=

Accepted Answer

Resonance structures show the delocalization of electrons within a molecule. Curved arrows indicate the movement of electron pairs (either lone pairs or pi electrons).

(a) **C₆H₅OH (Phenol)**
    *   The lone pair on the oxygen atom of the -OH group is in conjugation with the pi electrons of the benzene ring. Oxygen donates its lone pair into the ring, activating it towards electrophilic substitution and creating negative charge at ortho and para positions.

(b) **CH₃CH=CHCHO (But-2-enal)**
    *   The pi electrons of the C=C double bond are in conjugation with the pi electrons of the C=O double bond. The oxygen atom is more electronegative, drawing electron density towards itself, leading to a positive charge on the carbon of the C=O group and subsequent electron shifts.

(c) **CH₂=CH-CN (Acrylonitrile)**
    *   The pi electrons of the C=C double bond are in conjugation with the pi electrons of the C≡N triple bond. Nitrogen is more electronegative than carbon, drawing electron density towards itself from the triple bond, which in turn pulls electrons from the double bond.

(d) **C₆H₅NO₂ (Nitrobenzene)**
    *   The nitro group (-NO₂) is an electron-withdrawing group. The nitrogen atom of the nitro group has a positive charge and is directly attached to the benzene ring. It withdraws electron density from the ring, creating positive charges at the ortho and para positions.

(e) **CH₃CH=CHCH₂⁺ (But-2-en-1-yl carbocation)**
    *   This is an allylic carbocation where the positive charge is in conjugation with a C=C double bond. The pi electrons of the double bond can shift to delocalize the positive charge, creating another resonance structure where the positive charge is on the other end of the conjugated system.

Question 9

Q9. What are electrophiles and nucleophiles? Explain with examples.

Accepted Answer

### Introduction
In organic chemistry, reactions often involve the interaction between electron-rich and electron-deficient species. These species are broadly classified as nucleophiles and electrophiles, respectively. Understanding these terms is fundamental to predicting and explaining reaction mechanisms.

### 1. Electrophiles (Electron-Loving Species)

**Definition:** Electrophiles are electron-deficient species that are attracted to electron-rich centers. They are typically positively charged or have an incomplete octet, making them capable of accepting an electron pair. Due to their ability to accept electron pairs, electrophiles are also known as **Lewis acids**.

**Characteristics of Electrophiles:**
*   They are electron-deficient.
*   They have a vacant orbital to accept an electron pair.
*   They are positively charged or neutral molecules with an electron-deficient atom.
*   They seek out and react with nucleophiles.

**Types and Examples of Electrophiles:**
*   **Positively Charged Species:** These are strong electrophiles due to their inherent electron deficiency.
    *   **H⁺** (proton, from acids like HCl, H₂SO₄)
    *   **NO₂⁺** (nitronium ion, used in nitration reactions)
    *   **R₃C⁺** (carbocations, e.g., CH₃⁺, (CH₃)₃C⁺)
    *   **X⁺** (halonium ions, e.g., Cl⁺, Br⁺, generated in situ)
    *   **Acylium ions (RCO⁺)**
*   **Neutral Molecules with Electron-Deficient Atoms:** These molecules have atoms with incomplete octets or atoms bonded to highly electronegative atoms, creating a partial positive charge.
    *   **BF₃, AlCl₃** (Lewis acids with incomplete octets)
    *   **SO₃** (sulfur trioxide, used in sulfonation)
    *   **Carbonyl compounds (R₂C=O):** The carbon atom of the carbonyl group (C=O) is electrophilic due to the electronegativity of oxygen, which pulls electron density away from carbon, creating a partial positive charge (δ⁺C=Oδ⁻).
    *   **Alkyl halides (R-X):** The carbon atom bonded to the halogen (X) is electrophilic because the halogen is more electronegative, creating a partial positive charge on carbon (Rδ⁺-Xδ⁻).
    *   **Carbene (R₂C: )**

### 2. Nucleophiles (Nucleus-Loving Species)

**Definition:** Nucleophiles are electron-rich species that are attracted to positively charged or electron-deficient centers (nuclei). They possess at least one lone pair of electrons or a pi bond that can be donated to form a new covalent bond. Due to their ability to donate electron pairs, nucleophiles are also known as **Lewis bases**.

**Characteristics of Nucleophiles:**
*   They are electron-rich.
*   They have at least one lone pair of electrons or a pi electron system.
*   They are negatively charged or neutral molecules with available electron pairs.
*   They seek out and react with electrophiles.

**Types and Examples of Nucleophiles:**
*   **Negatively Charged Species (Anions):** These are typically strong nucleophiles due to their excess electron density.
    *   **OH⁻** (hydroxide ion)
    *   **RO⁻** (alkoxide ion, e.g., CH₃O⁻)
    *   **CN⁻** (cyanide ion)
    *   **X⁻** (halide ions, e.g., Cl⁻, Br⁻, I⁻)
    *   **RCOO⁻** (carboxylate ion)
    *   **Carbanions (R₃C⁻)**
*   **Neutral Molecules with Lone Pairs:** These molecules have unshared electron pairs that can be donated.
    *   **H₂O** (water, oxygen has two lone pairs)
    *   **ROH** (alcohols, oxygen has two lone pairs)
    *   **RNH₂** (amines, nitrogen has one lone pair)
    *   **R₂S** (sulfides, sulfur has two lone pairs)
    *   **Ethers (R-O-R)**
*   **Molecules with Pi Bonds:** The pi electrons in double or triple bonds can act as nucleophiles, especially in reactions like electrophilic addition.
    *   **Alkenes (C=C)**
    *   **Alkynes (C≡C)**
    *   **Aromatic rings (e.g., Benzene)**

### Conclusion
Electrophiles and nucleophiles are complementary species. In a typical organic reaction, a nucleophile attacks an electrophile, leading to the formation of new bonds and the transformation of molecules. Understanding their nature and reactivity is crucial for comprehending the mechanisms of various organic reactions, such as substitution, addition, and elimination reactions.

Question 10

Q10. Identify the electrophilic centre in the following:
CH₃CHO, CH₃CN, CH₃I, CH₂=CH₂, C₆H₆

Accepted Answer

An electrophilic centre is an atom or a region within a molecule that is electron-deficient and can accept an electron pair from a nucleophile. This deficiency often arises from a partial positive charge (δ⁺) due to bonding with more electronegative atoms or the presence of a positive charge.

Let's identify the electrophilic centre in each compound:

1.  **CH₃CHO (Acetaldehyde)**
    *   The carbon atom of the carbonyl group (C=O) is bonded to a highly electronegative oxygen atom. This causes the oxygen to pull electron density away from the carbon, creating a significant partial positive charge on the carbonyl carbon.
    *   **Electrophilic Centre:** Carbonyl carbon (C in C=O)

2.  **CH₃CN (Acetonitrile)**
    *   The carbon atom of the cyano group (C≡N) is bonded to a highly electronegative nitrogen atom. The nitrogen pulls electron density from the carbon, making the carbon partially positive.
    *   **Electrophilic Centre:** Carbon atom of the cyano group (C in C≡N)

3.  **CH₃I (Iodomethane)**
    *   The carbon atom is bonded to iodine, which is more electronegative than carbon. Although iodine is less electronegative than other halogens like Cl or Br, it still creates a partial positive charge on the carbon atom.
    *   **Electrophilic Centre:** Carbon atom bonded to iodine (C in C-I)

4.  **CH₂=CH₂ (Ethene)**
    *   Alkenes are generally considered nucleophilic due to their pi electron cloud. However, under certain conditions (e.g., in the presence of very strong electrophiles or in reactions involving radical intermediates), the carbons of the double bond can become electrophilic. But in the context of typical electrophilic attack, the pi bond itself acts as a nucleophile. If forced to identify an electrophilic *centre* in a neutral alkene, it's less straightforward than in polar molecules. However, if an electrophile attacks one carbon, the other carbon temporarily becomes electron-deficient (a carbocation).
    *   **Electrophilic Centre:** No inherent electrophilic centre in the neutral molecule under normal conditions. The pi bond acts as a nucleophile. If an electrophile attacks, one of the carbons of the double bond can become an electrophilic site (carbocation).

5.  **C₆H₆ (Benzene)**
    *   Benzene is an electron-rich aromatic system due to its delocalized pi electrons, making it primarily a nucleophile that undergoes electrophilic aromatic substitution. It does not have an inherent electrophilic carbon atom in its neutral state. The entire ring acts as an electron-rich system.
    *   **Electrophilic Centre:** No inherent electrophilic centre in the neutral molecule. The pi electron cloud acts as a nucleophile.

Question 11

Q11. Write the complete structural formulae of the following compounds:
(a) 2,2-Dimethylpentane
(b) Cyclohexylmethanol
(c) But-1-en-3-yne
(d) Bicyclo[2.2.1]heptane
(e) 4-Ethyl-1-is

Accepted Answer

The complete structural formulae for the given compounds are described below. For a clear visual representation, refer to the image prompt provided.

**(a) 2,2-Dimethylpentane**
*   **Parent chain:** Pentane (5 carbon atoms in a straight chain).
*   **Substituents:** Two methyl groups (-CH3) attached to the second carbon atom of the pentane chain.
*   **Structure description:** A five-carbon chain (CH3-CH2-CH2-CH2-CH3) where the second carbon atom (from either end, but typically numbered from the end closer to the substituent for lowest locant) has two methyl groups attached to it, replacing two hydrogen atoms. The structure would be CH3-C(CH3)2-CH2-CH2-CH3.

**(b) Cyclohexylmethanol**
*   **Parent group:** Cyclohexane ring (a six-membered saturated carbon ring).
*   **Substituent:** A methanol group (-CH2OH) attached to one of the carbon atoms of the cyclohexane ring.
*   **Structure description:** A six-membered carbon ring, with each carbon saturated with hydrogens (CH2 groups). One of these CH2 groups is replaced by a -CH2OH group. The structure is a cyclohexane ring with a -CH2OH group branching off one of its carbons.

**(c) But-1-en-3-yne**
*   **Parent chain:** Butane (4 carbon atoms).
*   **Functional groups:** A double bond (ene) starting at carbon 1 and a triple bond (yne) starting at carbon 3.
*   **Structure description:** A four-carbon chain. Carbon 1 and Carbon 2 are connected by a double bond (CH2=CH-). Carbon 3 and Carbon 4 are connected by a triple bond (-C≡CH). The complete structure is CH2=CH-C≡CH.

**(d) Bicyclo[2.2.1]heptane**
*   **Parent system:** Bicyclic system with 7 carbon atoms (heptane).
*   **Bridgehead carbons:** Two carbons that are part of all three bridges.
*   **Bridges:** The numbers [2.2.1] indicate the number of carbon atoms in each of the three bridges connecting the two bridgehead carbons. So, there are two bridges with 2 carbon atoms each, and one bridge with 1 carbon atom.
*   **Structure description:** This is a complex cage-like structure. Imagine a cyclohexane ring. Two carbons on opposite sides of this ring are the bridgehead carbons. One bridge (with 1 carbon) connects these two bridgehead carbons from 'above' the ring. The remaining 4 carbons of the original cyclohexane ring form two bridges of 2 carbons each between the bridgehead carbons. All carbon atoms are sp3 hybridized. This structure is also known as norbornane.

**(e) 4-Ethyl-1-isopropyl-2-methylbenzene**
*   **Parent ring:** Benzene ring (C6H6).
*   **Substituents:** An ethyl group (-CH2CH3) at position 4, an isopropyl group (-CH(CH3)2) at position 1, and a methyl group (-CH3) at position 2.
*   **Structure description:** A six-membered aromatic benzene ring. If the isopropyl group is considered at position 1, then moving clockwise or counter-clockwise, the methyl group is at position 2, and the ethyl group is at position 4. The benzene ring carbons are sp2 hybridized, and the carbons in the alkyl substituents are sp3 hybridized.

Question 12

Q12. An organic compound has an empirical formula CH2O. Its vapour density is 45. Find the molecular formula of the compound.

Accepted Answer

The molecular formula of the compound is C3H6O3.

Question 13

Q13. Determine the empirical formula and the molecular formula of an organic compound containing carbon, hydrogen and oxygen. The percentage composition is 40.68% carbon, 5.08% hydrogen

Accepted Answer

The empirical formula of the compound is C2H3O2 and the molecular formula is C4H6O4.

Question 14

Q14. What are the types of hybridisation of each carbon in the following compounds?
(a) CH3Cl
(b) (CH3)2CO
(c) CH2=C=CH2
(d) C6H6
(e) CH3CN
(f) CH3NC

Accepted Answer

The type of hybridisation of a carbon atom depends on the number of sigma bonds it forms and the presence of pi bonds. Generally:
*   **sp3 hybridisation:** Carbon forms four single (sigma) bonds.
*   **sp2 hybridisation:** Carbon forms one double bond and two single (sigma) bonds (total three sigma bonds).
*   **sp hybridisation:** Carbon forms one triple bond and one single (sigma) bond, OR two double bonds (total two sigma bonds).

Here are the hybridisations for each carbon in the given compounds:

**(a) CH3Cl (Chloromethane)**
*   The carbon atom is bonded to three hydrogen atoms and one chlorine atom, all by single bonds. It forms 4 sigma bonds.
*   **Hybridisation: sp3**

**(b) (CH3)2CO (Propanone or Acetone)**
*   **Carbonyl carbon (C=O):** This carbon is bonded to two methyl groups (single bonds) and one oxygen atom by a double bond. It forms 3 sigma bonds and 1 pi bond.
    *   **Hybridisation: sp2**
*   **Methyl carbons (CH3):** Each methyl carbon is bonded to three hydrogen atoms and one central carbon atom, all by single bonds. Each forms 4 sigma bonds.
    *   **Hybridisation: sp3**

**(c) CH2=C=CH2 (Allene)**
*   **Terminal carbons (CH2):** Each terminal carbon is bonded to two hydrogen atoms (single bonds) and one central carbon atom by a double bond. It forms 3 sigma bonds and 1 pi bond.
    *   **Hybridisation: sp2**
*   **Central carbon (=C=):** This carbon is bonded to two other carbon atoms by two double bonds. It forms 2 sigma bonds and 2 pi bonds.
    *   **Hybridisation: sp**

**(d) C6H6 (Benzene)**
*   Each carbon atom in the benzene ring is bonded to two other carbon atoms and one hydrogen atom. It forms 3 sigma bonds and is involved in the delocalised pi system (effectively one pi bond per carbon).
*   **Hybridisation: sp2** (for all 6 carbon atoms)

**(e) CH3CN (Acetonitrile)**
*   **Methyl carbon (CH3):** This carbon is bonded to three hydrogen atoms and one nitrile carbon atom, all by single bonds. It forms 4 sigma bonds.
    *   **Hybridisation: sp3**
*   **Nitrile carbon (C≡N):** This carbon is bonded to the methyl carbon by a single bond and to the nitrogen atom by a triple bond. It forms 2 sigma bonds and 2 pi bonds.
    *   **Hybridisation: sp**

**(f) CH3NC (Methyl isocyanide)**
*   **Methyl carbon (CH3):** This carbon is bonded to three hydrogen atoms and one nitrogen atom, all by single bonds. It forms 4 sigma bonds.
    *   **Hybridisation: sp3**
*   **Isocyanide carbon (N≡C):** This carbon is bonded to the nitrogen atom by a triple bond (and the nitrogen is bonded to the methyl group). It forms 2 sigma bonds and 2 pi bonds.
    *   **Hybridisation: sp**

Question 15

Q15. Write the state of hybridisation of carbon atoms in the following compounds and shape of each of the molecules.
(a) H2C=O
(b) CH3F
(c) HC≡N

Accepted Answer

The state of hybridisation of a carbon atom determines the electron geometry around it, which in turn influences the molecular shape. The VSEPR theory helps predict the molecular geometry based on the number of electron domains (sigma bonds and lone pairs) around the central atom.

**(a) H2C=O (Formaldehyde)**
*   **Carbon Hybridisation:** The carbon atom is bonded to two hydrogen atoms by single bonds and to one oxygen atom by a double bond. It forms 3 sigma bonds and 1 pi bond. Therefore, the carbon atom is **sp2 hybridized**.
*   **Molecular Shape:** With three electron domains (three sigma bonds) and no lone pairs on the central carbon, the electron geometry and molecular geometry are both **trigonal planar**. The bond angles around the carbon atom are approximately 120°.

**(b) CH3F (Fluoromethane)**
*   **Carbon Hybridisation:** The carbon atom is bonded to three hydrogen atoms and one fluorine atom, all by single bonds. It forms 4 sigma bonds. Therefore, the carbon atom is **sp3 hybridized**.
*   **Molecular Shape:** With four electron domains (four sigma bonds) and no lone pairs on the central carbon, the electron geometry and molecular geometry are both **tetrahedral**. The bond angles around the carbon atom are approximately 109.5°.

**(c) HC≡N (Hydrogen Cyanide)**
*   **Carbon Hybridisation:** The carbon atom is bonded to one hydrogen atom by a single bond and to one nitrogen atom by a triple bond. It forms 2 sigma bonds and 2 pi bonds. Therefore, the carbon atom is **sp hybridized**.
*   **Molecular Shape:** With two electron domains (two sigma bonds) and no lone pairs on the central carbon, the electron geometry and molecular geometry are both **linear**. The bond angle around the carbon atom is 180°.

Question 16

Q16. What is the hybridisation of each carbon atom in the following compounds?
(a) CH₂=C=O
(b) CH₃CH=CH₂
(c) (CH₃)₂CO
(d) CH₂=CH-CN
(e) C₆H₆

Accepted Answer

The hybridisation of each carbon atom is determined by the number of sigma bonds and lone pairs (if any) around it. For carbon, it's primarily based on the number of sigma bonds (single, double, or triple bonds count as one sigma bond each, plus bonds to hydrogen or other atoms).

(a) **CH₂=C=O (Ketene)**
    *   **C1 (CH₂)**: This carbon is bonded to two hydrogen atoms and one carbon atom via a double bond. It forms 3 sigma bonds (2 C-H, 1 C=C). Therefore, it is **sp² hybridised**.
    *   **C2 (=C=O)**: This carbon is bonded to one carbon atom via a double bond and one oxygen atom via a double bond. It forms 2 sigma bonds (1 C=C, 1 C=O). Therefore, it is **sp hybridised**.

(b) **CH₃CH=CH₂ (Propene)**
    *   **C1 (CH₃)**: This carbon is bonded to three hydrogen atoms and one carbon atom via a single bond. It forms 4 sigma bonds (3 C-H, 1 C-C). Therefore, it is **sp³ hybridised**.
    *   **C2 (CH=)**: This carbon is bonded to one hydrogen atom, one carbon atom via a single bond, and one carbon atom via a double bond. It forms 3 sigma bonds (1 C-H, 1 C-C, 1 C=C). Therefore, it is **sp² hybridised**.
    *   **C3 (=CH₂)**: This carbon is bonded to two hydrogen atoms and one carbon atom via a double bond. It forms 3 sigma bonds (2 C-H, 1 C=C). Therefore, it is **sp² hybridised**.

(c) **(CH₃)₂CO (Acetone)**
    *   **C1 (CH₃)**: This carbon is bonded to three hydrogen atoms and one carbon atom via a single bond. It forms 4 sigma bonds. Therefore, it is **sp³ hybridised**.
    *   **C2 (C=O)**: This carbon (the carbonyl carbon) is bonded to two methyl carbons via single bonds and one oxygen atom via a double bond. It forms 3 sigma bonds (2 C-C, 1 C=O). Therefore, it is **sp² hybridised**.
    *   **C3 (CH₃)**: This carbon is bonded to three hydrogen atoms and one carbon atom via a single bond. It forms 4 sigma bonds. Therefore, it is **sp³ hybridised**.

(d) **CH₂=CH-CN (Acrylonitrile)**
    *   **C1 (CH₂=)**: This carbon is bonded to two hydrogen atoms and one carbon atom via a double bond. It forms 3 sigma bonds. Therefore, it is **sp² hybridised**.
    *   **C2 (=CH-)**: This carbon is bonded to one hydrogen atom, one carbon atom via a double bond, and one carbon atom via a single bond. It forms 3 sigma bonds. Therefore, it is **sp² hybridised**.
    *   **C3 (-CN)**: This carbon is bonded to one carbon atom via a single bond and one nitrogen atom via a triple bond. It forms 2 sigma bonds (1 C-C, 1 C≡N). Therefore, it is **sp hybridised**.

(e) **C₆H₆ (Benzene)**
    *   Each carbon atom in benzene is bonded to two other carbon atoms and one hydrogen atom. It forms 3 sigma bonds (2 C-C, 1 C-H) and participates in a delocalized pi system. Therefore, each carbon atom is **sp² hybridised**.

Question 17

Q17. Explain the effect of the following on the acidity of carboxylic acids:
(a) Presence of an electron withdrawing group on the α-carbon atom.
(b) Presence of an electron donating g

Accepted Answer

The acidity of carboxylic acids is primarily determined by the stability of their conjugate base, the carboxylate anion (R-COO⁻). A more stable carboxylate anion corresponds to a stronger acid. The stability of the carboxylate anion is influenced by factors that can delocalize or disperse its negative charge. The inductive effect of groups attached to the α-carbon plays a significant role.

### (a) Presence of an electron withdrawing group on the α-carbon atom

**Effect:** Electron-withdrawing groups (EWGs) increase the acidity of carboxylic acids.

**Explanation:**
1.  **Inductive Effect (-I effect):** Electron-withdrawing groups (e.g., -F, -Cl, -Br, -I, -NO₂, -CN, -COOH) exert a negative inductive effect (-I effect). This means they pull electron density away from the carbon chain through sigma bonds.
2.  **Stabilization of Carboxylate Anion:** When an EWG is present on the α-carbon (the carbon atom adjacent to the carboxyl group), it pulls electron density away from the carboxylate group. This effectively disperses or delocalizes the negative charge on the oxygen atoms of the carboxylate anion.
3.  **Increased Stability:** The dispersal of negative charge stabilizes the carboxylate anion. A more stable conjugate base is more readily formed, which means the corresponding carboxylic acid is stronger (more acidic).
4.  **Distance Dependence:** The -I effect decreases rapidly with distance. Therefore, an EWG at the α-carbon has a greater acid-strengthening effect than the same group at the β-carbon, and so on.

**Example:** Chloroacetic acid (ClCH₂COOH) is more acidic than acetic acid (CH₃COOH) because the chlorine atom is an electron-withdrawing group that stabilizes the chloroacetate anion.

### (b) Presence of an electron donating group on the α-carbon atom

**Effect:** Electron-donating groups (EDGs) decrease the acidity of carboxylic acids.

**Explanation:**
1.  **Inductive Effect (+I effect):** Electron-donating groups (e.g., alkyl groups like -CH₃, -C₂H₅) exert a positive inductive effect (+I effect). This means they push electron density towards the carbon chain through sigma bonds.
2.  **Destabilization of Carboxylate Anion:** When an EDG is present on the α-carbon, it pushes electron density towards the carboxylate group. This intensifies the negative charge on the oxygen atoms of the carboxylate anion.
3.  **Decreased Stability:** The intensification of negative charge destabilizes the carboxylate anion. A less stable conjugate base is less readily formed, which means the corresponding carboxylic acid is weaker (less acidic).
4.  **Number of EDGs:** The acid-weakening effect increases with the number of electron-donating groups present on the α-carbon.

**Example:** Acetic acid (CH₃COOH) is less acidic than formic acid (HCOOH) because the methyl group (-CH₃) is an electron-donating group that destabilizes the acetate anion. Propanoic acid (CH₃CH₂COOH) is even weaker than acetic acid due to the larger ethyl group's stronger +I effect.

Question 18

Q18. Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation
(b) Distillation
(c) Chromatography

Accepted Answer

Organic compounds are often obtained from natural sources or synthesized in the laboratory as impure mixtures. Various purification techniques are employed to separate and purify these compounds. Here's a description of three common techniques:

### (a) Crystallisation

**Principle:** Crystallisation is based on the difference in solubilities of the compound and its impurities in a suitable solvent. The compound to be purified is usually more soluble in the chosen solvent at higher temperatures than at lower temperatures, while the impurities are either sparingly soluble or remain soluble in the mother liquor (the solvent remaining after crystallisation).

**Description:**
1.  **Selection of Solvent:** A suitable solvent is chosen in which the compound is sparingly soluble at room temperature but appreciably soluble at higher temperatures, and impurities are either very soluble or insoluble.
2.  **Dissolution:** The impure solid is dissolved in the minimum amount of the chosen solvent at or near its boiling point.
3.  **Hot Filtration:** The hot solution is filtered to remove any insoluble impurities.
4.  **Cooling:** The hot, clear filtrate is then allowed to cool slowly. As the solution cools, the solubility of the desired compound decreases, and it separates out as pure crystals.
5.  **Isolation:** The crystals are then separated from the mother liquor by filtration, washed with a small amount of cold solvent, and dried.

**Example:** Purification of benzoic acid from water. Benzoic acid is sparingly soluble in cold water but highly soluble in hot water. Impurities can be removed by hot filtration, and pure benzoic acid crystals form upon cooling.

### (b) Distillation

**Principle:** Distillation is a technique used to separate liquids based on differences in their boiling points. When a liquid mixture is heated, the component with the lower boiling point (more volatile) vaporizes preferentially. These vapors are then condensed back into liquid form and collected separately.

**Description:**
1.  **Heating:** The liquid mixture is heated in a distillation flask.
2.  **Vaporisation:** The component with the lower boiling point vaporizes first, forming a vapor phase enriched in this component.
3.  **Condensation:** The vapors rise, pass into a condenser, where they are cooled by circulating water, and condense back into liquid.
4.  **Collection:** The condensed liquid (distillate) is collected in a receiver. The component with the higher boiling point remains in the distillation flask.

**Types of Distillation:**
*   **Simple Distillation:** Used for separating liquids with a significant difference in boiling points (typically > 25°C).
*   **Fractional Distillation:** Used for separating liquids with close boiling points. It involves using a fractionating column that provides a large surface area for repeated vaporization and condensation cycles, leading to better separation.
*   **Distillation under Reduced Pressure:** Used for liquids that decompose at or below their normal boiling points. The boiling point is lowered by reducing the pressure.
*   **Steam Distillation:** Used for separating substances that are steam volatile and immiscible with water.

**Example:** Separation of chloroform (b.p. 334 K) and aniline (b.p. 457 K) by simple distillation. Chloroform will distill over first, followed by aniline.

### (c) Chromatography

**Principle:** Chromatography is a powerful technique used for separating, identifying, and purifying components of a mixture. Its principle is based on the differential distribution (adsorption or partition) of the components between two phases: a stationary phase and a mobile phase.

**Description:**
1.  **Stationary Phase:** This is a fixed phase, which can be a solid (e.g., silica gel, alumina in column chromatography) or a liquid supported on a solid (e.g., water on paper in paper chromatography).
2.  **Mobile Phase:** This is a moving phase, which can be a liquid (solvent) or a gas, that flows over or through the stationary phase.
3.  **Separation:** The mixture is dissolved in the mobile phase and introduced to the stationary phase. As the mobile phase moves, the components of the mixture interact differently with the stationary phase and the mobile phase. Components that have a greater affinity for the stationary phase (e.g., stronger adsorption) will move slower, while those with a greater affinity for the mobile phase (e.g., more soluble) will move faster.
4.  **Differential Movement:** This differential movement causes the components to separate and emerge from the stationary phase at different times or positions.

**Types of Chromatography:** Common types include column chromatography, thin-layer chromatography (TLC), paper chromatography, gas chromatography (GC), and high-performance liquid chromatography (HPLC).

**Example:** Separation of plant pigments (like chlorophylls and carotenoids) using paper chromatography or column chromatography. Different pigments will travel at different rates up the paper or down the column, leading to their separation into distinct bands or spots.

Question 19

Q19. Explain the effect of the following on the acidity of carboxylic acids:
(i) Electron withdrawing groups
(ii) Electron donating groups

Accepted Answer

The acidity of carboxylic acids is determined by the stability of the carboxylate anion (R-COO⁻) formed after the donation of a proton (H⁺). A more stable carboxylate anion corresponds to a stronger acid.

(i) **Effect of Electron Withdrawing Groups (EWGs):**
Electron withdrawing groups (like -NO₂, -CN, -F, -Cl, -Br, -I, -COOH, -CHO, -COR) attached to the carbon chain of a carboxylic acid increase its acidity. This is primarily due to the **inductive effect (-I effect)**. EWGs pull electron density away from the carboxylate group through the sigma bonds. This delocalization of negative charge stabilizes the carboxylate anion by dispersing its charge. The more effectively the negative charge is dispersed, the more stable the anion, and thus, the stronger the acid. The strength of the EWG and its proximity to the carboxyl group also influence the extent of acidity increase.

*Example:* Trichloroacetic acid (CCl₃COOH) is much stronger than acetic acid (CH₃COOH) because the three chlorine atoms are strong EWGs that stabilize the trichloroacetate anion.

(ii) **Effect of Electron Donating Groups (EDGs):**
Electron donating groups (like -CH₃, -CH₂CH₃, -OCH₃, -NH₂, -OH) attached to the carbon chain of a carboxylic acid decrease its acidity. This is primarily due to the **inductive effect (+I effect)**. EDGs push electron density towards the carboxylate group. This intensifies the negative charge on the carboxylate anion, making it less stable. A less stable anion means the acid is weaker and less likely to donate a proton.

*Example:* Acetic acid (CH₃COOH) is a weaker acid than formic acid (HCOOH) because the methyl group (-CH₃) is an EDG that destabilizes the acetate anion compared to the formate anion.

Question 20

Q20. Give a brief description of the principles of the following techniques taking an example in each case:
(i) Crystallisation
(ii) Distillation
(iii) Chromatography

Accepted Answer

Purification of organic compounds is a crucial step in organic chemistry. Here are descriptions of three common techniques:

(i) **Crystallisation**

*   **Principle:** Crystallisation is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The compound is either sparingly soluble at room temperature but appreciably soluble at a higher temperature, or it is more soluble in one solvent than in another. Impurities are either much more soluble or much less soluble in the chosen solvent than the compound itself.
*   **Method:** The impure solid organic compound is dissolved in a minimum amount of a suitable solvent at or near its boiling point to form a saturated solution. The hot solution is then filtered to remove insoluble impurities. The filtrate is allowed to cool slowly, causing the pure compound to crystallise out, as its solubility decreases with temperature. The impurities, being either more soluble or present in smaller quantities, remain in the solution (mother liquor). The crystals are then separated by filtration, washed with a small amount of cold solvent, and dried.
*   **Example:** Purification of benzoic acid from a mixture containing salicylic acid. Benzoic acid is sparingly soluble in cold water but highly soluble in hot water. Salicylic acid is more soluble in water than benzoic acid. By dissolving the mixture in hot water and then cooling, pure benzoic acid crystals will form, while salicylic acid remains dissolved in the mother liquor.

(ii) **Distillation**

*   **Principle:** Distillation is used to separate volatile liquids from non-volatile impurities or to separate two or more liquids having different boiling points. The principle is that liquids vaporise at different temperatures, and the vapours can then be condensed and collected separately.
*   **Method:** The liquid mixture is heated in a distillation flask. The component with the lower boiling point vaporises first. The vapours rise, pass into a condenser, where they are cooled by circulating water, and condense back into liquid. This pure liquid (distillate) is collected in a receiver. The component with the higher boiling point remains in the distillation flask.
*   **Types and Examples:**
    *   **Simple Distillation:** Used for separating liquids boiling below 150°C at atmospheric pressure and for liquids that are non-volatile impurities. *Example:* Separation of chloroform (b.p. 61°C) and aniline (b.p. 184°C).
    *   **Fractional Distillation:** Used to separate two or more liquids with boiling points close to each other (difference less than 25°C). A fractionating column is fitted between the distillation flask and the condenser. The column provides a large surface area for repeated vaporisation and condensation cycles, enriching the vapour with the more volatile component as it moves up the column. *Example:* Separation of ethanol (b.p. 78°C) and water (b.p. 100°C).
    *   **Distillation under Reduced Pressure (Vacuum Distillation):** Used for liquids that decompose at or below their normal boiling points. By reducing the pressure, the liquid boils at a lower temperature, preventing decomposition. *Example:* Purification of glycerol (b.p. 290°C, decomposes) can be done at 180°C under reduced pressure.
    *   **Steam Distillation:** Used for separating substances that are steam volatile (volatile in steam) and immiscible with water. The substance distils along with steam at a temperature lower than its normal boiling point. *Example:* Purification of aniline, nitrobenzene, essential oils.

(iii) **Chromatography**

*   **Principle:** Chromatography is a highly versatile technique used for the separation, purification, and testing of purity of organic compounds. It is based on the differential adsorption or partition of the components of a mixture between two phases: a stationary phase and a mobile phase.
    *   **Stationary Phase:** A solid or a liquid supported on a solid. It is fixed.
    *   **Mobile Phase:** A liquid or a gas that moves over the stationary phase.
    The components of the mixture travel at different rates through the stationary phase depending on their relative affinities for the stationary and mobile phases. Components that are more strongly adsorbed by the stationary phase or more soluble in the stationary phase move slower, while those with less affinity for the stationary phase or more soluble in the mobile phase move faster.
*   **Types and Examples:**
    *   **Adsorption Chromatography:** Based on the principle of differential adsorption of components of a mixture on a suitable adsorbent (e.g., silica gel, alumina). *Examples:*
        *   **Column Chromatography:** A mixture is loaded onto a column packed with adsorbent. The mobile phase (eluent) is passed through the column. Components separate into bands and are collected as fractions. *Example:* Separation of plant pigments like chlorophylls and carotenoids.
        *   **Thin Layer Chromatography (TLC):** A thin layer of adsorbent (e.g., silica gel) is spread on a glass plate. A spot of the mixture is applied near one edge. The plate is placed in a developing chamber with the mobile phase. Components separate as spots at different heights, characterized by their Rf values (retardation factor).
    *   **Partition Chromatography:** Based on the differential partitioning of components between two immiscible liquid phases (one stationary, one mobile). *Examples:*
        *   **Paper Chromatography:** A strip of special paper (cellulose acts as stationary phase, holding water) is used. The mixture is spotted, and a suitable solvent (mobile phase) moves up the paper by capillary action, separating components.
        *   **Gas-Liquid Chromatography (GLC):** The mobile phase is an inert gas, and the stationary phase is a non-volatile liquid coated on a solid support. Used for separating volatile organic compounds.

Question 21

Q21. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S but both are equally soluble in another solvent S'.

Accepted Answer

The method that can be used to separate two compounds with different solubilities in a solvent S, but both are equally soluble in another solvent S', is **Crystallisation**, specifically **Fractional Crystallisation** if both compounds are solids.

**Principle:**
This method relies on the difference in solubility of the two compounds in solvent S at different temperatures. Although both compounds might be equally soluble in solvent S', this solvent is not suitable for separation by crystallisation. Solvent S is the key, as it shows differential solubility.

**Method (Fractional Crystallisation):**
1.  **Choose Solvent S:** Select solvent S in which the two compounds have significantly different solubilities, ideally one being much more soluble than the other, especially at higher temperatures.
2.  **Dissolution:** Dissolve the mixture of the two compounds in a minimum amount of hot solvent S to form a saturated solution. Both compounds will dissolve.
3.  **Cooling:** Allow the hot solution to cool slowly. As the temperature drops, the solubility of both compounds decreases. However, the compound that is less soluble in solvent S will start to crystallise out first, leaving the more soluble compound dissolved in the mother liquor.
4.  **Separation:** Filter the solution to separate the crystals of the less soluble compound. These crystals can then be washed with a small amount of cold solvent S and dried.
5.  **Further Purification (Optional):** If the mother liquor still contains a significant amount of the more soluble compound, it can be concentrated by evaporating some of the solvent S and then cooled again to obtain crystals of the second compound. This process can be repeated to achieve higher purity for both compounds.

**Example:**
Consider separating a mixture of compound A and compound B. Compound A is sparingly soluble in cold ethanol (solvent S) but highly soluble in hot ethanol. Compound B is highly soluble in both cold and hot ethanol. Both are equally soluble in benzene (solvent S').

1.  Dissolve the mixture of A and B in a minimum amount of hot ethanol.
2.  Cool the solution slowly. Compound A, being less soluble in cold ethanol, will crystallise out.
3.  Filter off the crystals of pure A. Compound B will remain dissolved in the ethanol mother liquor.
4.  The mother liquor can then be evaporated to recover compound B.

Question 22

Q22. Explain the terms:
(i) Electrophile
(ii) Nucleophile

Accepted Answer

(i) **Electrophile (Electron-loving species):**

*   **Definition:** An electrophile is a reagent (atom, ion, or molecule) that is deficient in electrons and seeks an electron-rich center (a nucleophile) to form a new bond. They are 'electron-loving' species.
*   **Characteristics:**
    *   They are typically positively charged or have an electron-deficient atom.     *   They act as Lewis acids (electron pair acceptors).     *   They attack regions of high electron density.
*   **Examples:**
    *   **Cations:** H⁺, NO₂⁺ (nitronium ion), Br⁺, CH₃⁺ (carbocation).
    *   **Neutral molecules with incomplete octet:** BF₃, AlCl₃, SO₃.
    *   **Neutral molecules with polar multiple bonds:** Carbonyl carbon in R₂C=O, carbon in CO₂.
    *   **Halogens:** Cl₂, Br₂ (due to polarizability, one halogen atom can act as an electrophile).

(ii) **Nucleophile (Nucleus-loving species):**

*   **Definition:** A nucleophile is a reagent (atom, ion, or molecule) that is rich in electrons and seeks an electron-deficient center (an electrophile) to form a new bond. They are 'nucleus-loving' species.
*   **Characteristics:**
    *   They typically have a lone pair of electrons or a negative charge.     *   They act as Lewis bases (electron pair donors).     *   They attack regions of low electron density.
*   **Examples:**
    *   **Anions:** OH⁻ (hydroxide ion), CN⁻ (cyanide ion), Cl⁻, Br⁻, R-O⁻ (alkoxide ion), R-COO⁻ (carboxylate ion).
    *   **Neutral molecules with lone pairs:** H₂O (water), NH₃ (ammonia), R-OH (alcohols), R-NH₂ (amines), R₂S (sulfides).
    *   **Molecules with pi (π) bonds:** Alkenes (C=C), Alkynes (C≡C), Benzene (electron-rich π-systems can donate electrons to strong electrophiles).

Question 23

Q23. Write the IUPAC names of the following compounds:
(i) CH3CH=CHCH2CH(CH3)CH3
(ii) CH3CH2CH2CH(CH2CH3)CH2CH2CH3
(iii) CH3CH2C(CH3)2CH2CH(CH3)COOH
(iv) CH3CH(OH)CH2COCH3
(v) CH3C

Accepted Answer

The IUPAC names for the given compounds are as follows:

(i) **CH3CH=CHCH2CH(CH3)CH3**
   *   Longest carbon chain containing the double bond has 6 carbons. Numbering starts from the end closer to the double bond.
   *   Parent chain: Hex-2-ene
   *   Substituent: Methyl group at position 5.
   *   IUPAC Name: **5-Methylhex-2-ene**

(ii) **CH3CH2CH2CH(CH2CH3)CH2CH2CH3**
   *   Longest carbon chain is 8 carbons long (octane). Numbering starts from the end closer to the substituent.
   *   Parent chain: Octane
   *   Substituent: Ethyl group at position 4.
   *   IUPAC Name: **4-Ethyloctane**

(iii) **CH3CH2C(CH3)2CH2CH(CH3)COOH**
   *   The functional group is carboxylic acid (-COOH), which defines the parent chain and numbering (C1 is the carboxyl carbon).
   *   Longest carbon chain containing -COOH has 6 carbons.
   *   Parent chain: Hexanoic acid
   *   Substituents: Methyl groups at positions 2 and 4.
   *   IUPAC Name: **2,4,4-Trimethylhexanoic acid**

(iv) **CH3CH(OH)CH2COCH3**
   *   This compound contains two functional groups: a ketone (-CO-) and an alcohol (-OH). Ketone has higher priority than alcohol.
   *   Longest carbon chain containing both functional groups has 5 carbons. Numbering starts from the end closer to the ketone.
   *   Parent chain: Pentan-2-one
   *   Substituent: Hydroxy group at position 4.
   *   IUPAC Name: **4-Hydroxypentan-2-one**

(v) **CH3CH2CH(CH3)CH(C2H5)CH2OH**
   *   The functional group is alcohol (-OH), which defines the parent chain and numbering (C1 is the carbon bearing -OH).
   *   Longest carbon chain containing -OH has 6 carbons.
   *   Parent chain: Hexan-1-ol
   *   Substituents: Methyl group at position 3, Ethyl group at position 4.
   *   IUPAC Name: **4-Ethyl-3-methylhexan-1-ol**

(vi) **CH3CH2C≡CCH2CH3**
   *   Longest carbon chain containing the triple bond has 6 carbons. Numbering starts from the end closer to the triple bond (symmetrical in this case).
   *   Parent chain: Hex-3-yne
   *   No substituents.
   *   IUPAC Name: **Hex-3-yne**

(vii) **CH3CH(Cl)CH2CH2CH(OH)CH3**
   *   This compound contains two functional groups: an alcohol (-OH) and a chloro group (-Cl). Alcohol has higher priority.
   *   Longest carbon chain containing -OH has 6 carbons. Numbering starts from the end closer to the -OH group.
   *   Parent chain: Hexan-2-ol
   *   Substituent: Chloro group at position 5.
   *   IUPAC Name: **5-Chlorohexan-2-ol**

(viii) **CH3CH2CH=CHCH2C≡CH**
   *   This compound contains both a double bond and a triple bond. When both are present, the chain is numbered to give the lowest locants to the multiple bonds. If there's a tie, the double bond gets priority.
   *   Longest carbon chain containing both multiple bonds has 7 carbons.
   *   Numbering starts from the right end (closer to the triple bond) to give locants 1 (for triple bond) and 3 (for double bond). If numbered from left, locants would be 2 (for double bond) and 5 (for triple bond). So, 1,3 is preferred over 2,5.
   *   Parent chain: Hept-3-en-1-yne
   *   IUPAC Name: **Hept-3-en-1-yne**

Question 24

Q24. Draw the structures of the following compounds:
(i) 3-Ethyl-4,4-dimethylheptane
(ii) 4-Hydroxy-4-methylpentan-2-one
(iii) 2-Ethoxy-3-methylbutane
(iv) 3-Chloro-4-methylhexan-2-

Accepted Answer

The structures of the given compounds are as follows:

(i) **3-Ethyl-4,4-dimethylheptane**
    *   Parent chain: Heptane (7 carbons, all single bonds).
    *   Substituents: Ethyl group at C3, two methyl groups at C4.
    *   Structure: CH3-CH2-CH(CH2CH3)-C(CH3)2-CH2-CH2-CH3

(ii) **4-Hydroxy-4-methylpentan-2-one**
    *   Parent chain: Pentan-2-one (5 carbons, ketone at C2).
    *   Substituents: Hydroxy group at C4, methyl group at C4.
    *   Structure: CH3-CO-CH2-C(CH3)(OH)-CH3

(iii) **2-Ethoxy-3-methylbutane**
    *   Parent chain: Butane (4 carbons).
    *   Functional group: Ether (ethoxy group, -OCH2CH3) at C2.
    *   Substituent: Methyl group at C3.
    *   Structure: CH3-CH(OCH2CH3)-CH(CH3)-CH3

(iv) **3-Chloro-4-methylhexan-2-ol**
    *   Parent chain: Hexan-2-ol (6 carbons, alcohol at C2).
    *   Substituents: Chloro group at C3, methyl group at C4.
    *   Structure: CH3-CH(OH)-CH(Cl)-CH(CH3)-CH2-CH3

(v) **Pent-4-en-2-ol**
    *   Parent chain: Pentane with a double bond (pent-4-ene) and an alcohol group (pentan-2-ol).
    *   Functional groups: Double bond at C4-C5, alcohol at C2.
    *   Structure: CH2=CH-CH2-CH(OH)-CH3

(vi) **3-Cyclohexylpentan-3-ol**
    *   Parent chain: Pentan-3-ol (5 carbons, alcohol at C3).
    *   Substituent: Cyclohexyl group at C3.
    *   Structure: CH3-CH2-C(OH)(C6H11)-CH2-CH3 (where C6H11 represents the cyclohexyl ring)

Question 25

Q25. Which of the following represents the correct IUPAC name for the compounds concerned?
(a) 2,2-Dimethylpentane or 2,2-Dimethylpentane
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethylo

Accepted Answer

Let's analyze each option to determine the correct IUPAC name:

**(a) 2,2-Dimethylpentane or 2,2-Dimethylpentane**
   *   The question seems to have a typo, presenting the same name twice. Assuming the structure is: CH3-C(CH3)2-CH2-CH2-CH3
   *   Longest chain: Pentane (5 carbons).
   *   Substituents: Two methyl groups at C2.
   *   Correct IUPAC name: **2,2-Dimethylpentane**.
   *   This option is correctly named.

**(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane**
   *   Let's consider the structure: CH3-CH(CH3)-CH2-CH(CH3)-CH2-CH(CH3)-CH2-CH3
   *   Longest chain: Octane (8 carbons).
   *   Numbering from left: Methyl groups at C2, C4, C7. Locants: (2,4,7).
   *   Numbering from right: Methyl groups at C2, C5, C7. Locants: (2,5,7).
   *   According to the lowest set of locants rule, (2,4,7) is preferred over (2,5,7) because the first point of difference (4 vs 5) gives a lower number.
   *   Correct IUPAC name: **2,4,7-Trimethyloctane**.
   *   Therefore, '2,4,7-Trimethyloctane' is the correct name.

**(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane**
   *   Let's consider the structure: CH3-CH(CH3)-CH2-CH(Cl)-CH3
   *   Longest chain: Pentane (5 carbons).
   *   Substituents: Chloro and Methyl.
   *   Numbering from left: Methyl at C2, Chloro at C4. Locants: (2,4).
   *   Numbering from right: Chloro at C2, Methyl at C4. Locants: (2,4).
   *   Since both numbering directions give the same set of locants (2,4), we apply alphabetical order for substituents. 'Chloro' comes before 'Methyl'. So, the carbon bearing 'chloro' should get the lower number if possible, but here the locants are the same. In such a tie-breaking situation, the substituent that comes first alphabetically should be written first in the name, but the numbering is decided by the lowest set of locants. If the locants are identical, then alphabetical priority is used to assign the lowest number to the alphabetically preferred substituent. Here, numbering from the right gives Cl at C2 and Me at C4. Numbering from the left gives Me at C2 and Cl at C4. The set of locants (2,4) is the same. Therefore, the substituent that comes first alphabetically (chloro) should be given the lower number. So, numbering from the right is preferred.
   *   Correct IUPAC name: **2-Chloro-4-methylpentane**.
   *   Therefore, '2-Chloro-4-methylpentane' is the correct name.

**(d) But-3-yn-1-ol or But-4-ol-1-yne**
   *   Let's consider the structure: HC≡C-CH2-CH2-OH
   *   Parent chain: Butane.
   *   Functional groups: Alcohol (-OH) and Alkyne (triple bond).
   *   Priority: Alcohol has higher priority than alkyne. So, the carbon bearing -OH is C1.
   *   Numbering: From the right, C1 has -OH, C3 has the triple bond.
   *   Correct IUPAC name: **But-3-yn-1-ol**.
   *   'But-4-ol-1-yne' is incorrect because the alcohol group is at C1, not C4, and the triple bond is at C3, not C1 in relation to the alcohol. Also, the numbering for 'yne' should be based on the lowest possible number for the triple bond after prioritizing the -OH group.
   *   Therefore, 'But-3-yn-1-ol' is the correct name.

**Conclusion:** All the first names in each pair are the correct IUPAC names based on the standard rules.

Question 26

Q26. Explain the effect of the following factors on the acidity of carboxylic acids: (a) electron withdrawing groups, (b) electron donating groups.

Accepted Answer

The acidity of carboxylic acids is determined by the stability of the carboxylate anion (R-COO⁻) formed after the donation of a proton (H⁺). The more stable the carboxylate anion, the stronger the acid. Factors that stabilize the carboxylate anion increase acidity, while factors that destabilize it decrease acidity. This stability is primarily influenced by the inductive effect of substituents attached to the carbon atom bearing the carboxyl group.

Question 27

Q27. Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography (d) Differential extr

Accepted Answer

Organic compounds obtained from natural sources or synthesized in the laboratory are often impure. Various techniques are used to purify them based on their physical properties.

Question 28

Q28. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S but having almost same boiling points.

Accepted Answer

The method that can be used to separate two compounds with different solubilities in a solvent S but having almost the same boiling points is **Crystallisation**.

**Principle:** Crystallisation relies on the difference in solubility of the components of a mixture in a given solvent at different temperatures. If two compounds have almost the same boiling points, distillation would not be an effective separation method. However, if their solubilities in a particular solvent 'S' differ significantly, especially at varying temperatures, crystallisation can be employed.

**Process:**
1.  The mixture of the two compounds is dissolved in a minimum amount of hot solvent 'S' to form a saturated solution. 2.  The solution is then slowly cooled. 3.  The compound that is less soluble at lower temperatures will crystallize out first, leaving the more soluble compound and impurities in the solution (mother liquor). 4.  The crystals are then filtered, washed with a small amount of cold solvent, and dried to obtain the pure compound.

This method effectively separates compounds based on their solubility differences, even if their boiling points are very close.

Question 29

Q29. Write the IUPAC names of the following compounds:
(a) CH3CH=C(CH2CH3)CH2CH3
(b) (CH3)3CCH2C(CH3)2CH=CH2
(c) CH3CH(CH3)CH2CH(CH3)CH(CH3)2
(d) (CH3)3CCH2CH(CH3)2
(e) CH3C(CH3)2C

Accepted Answer

The IUPAC names of the given compounds are as follows:

**(a) CH3CH=C(CH2CH3)CH2CH3**
*   **Structure:** CH3–CH=C(CH2CH3)–CH2–CH3
*   **Parent Chain:** The longest carbon chain containing the double bond is 5 carbons long (pent-2-ene). The ethyl group is a substituent.
*   **Numbering:** Number from the end closer to the double bond, giving it the lowest possible locant.
*   **IUPAC Name:** 3-Ethylpent-2-ene

**(b) (CH3)3CCH2C(CH3)2CH=CH2**
*   **Structure:** (CH3)3C–CH2–C(CH3)2–CH=CH2
*   **Parent Chain:** The longest carbon chain containing the double bond is 5 carbons long (pent-1-ene). There are three methyl groups and one tert-butyl group as substituents.
*   **Numbering:** Number from the end closer to the double bond.
*   **IUPAC Name:** 2,2,4,4-Tetramethylpent-1-ene

**(c) CH3CH(CH3)CH2CH(CH3)CH(CH3)2**
*   **Structure:** CH3–CH(CH3)–CH2–CH(CH3)–CH(CH3)2
*   **Parent Chain:** The longest continuous carbon chain is 6 carbons long (hexane).
*   **Substituents:** There are four methyl groups.
*   **Numbering:** Number from the end that gives the lowest sum of locants for the substituents.
*   **IUPAC Name:** 2,3,4,5-Tetramethylhexane

**(d) (CH3)3CCH2CH(CH3)2**
*   **Structure:** (CH3)3C–CH2–CH(CH3)2
*   **Parent Chain:** The longest continuous carbon chain is 5 carbons long (pentane).
*   **Substituents:** There are four methyl groups.
*   **Numbering:** Number from the end that gives the lowest sum of locants for the substituents.
*   **IUPAC Name:** 2,2,4-Trimethylpentane

**(e) CH3C(CH3)2CH2C(CH3)2CH3**
*   **Structure:** CH3–C(CH3)2–CH2–C(CH3)2–CH3
*   **Parent Chain:** The longest continuous carbon chain is 5 carbons long (pentane).
*   **Substituents:** There are four methyl groups.
*   **Numbering:** Number from either end, as it's symmetrical.
*   **IUPAC Name:** 2,2,4,4-Tetramethylpentane

**(f) CH3CH2C(CH3)2CH2CH(CH3)CH2CH3**
*   **Structure:** CH3–CH2–C(CH3)2–CH2–CH(CH3)–CH2–CH3
*   **Parent Chain:** The longest continuous carbon chain is 7 carbons long (heptane).
*   **Substituents:** There are three methyl groups.
*   **Numbering:** Number from the end that gives the lowest sum of locants for the substituents.
*   **IUPAC Name:** 3,3,5-Trimethylheptane

**(g) CH3CH(CH3)CH2CH(CH3)CH2CH3**
*   **Structure:** CH3–CH(CH3)–CH2–CH(CH3)–CH2–CH3
*   **Parent Chain:** The longest continuous carbon chain is 6 carbons long (hexane).
*   **Substituents:** There are two methyl groups.
*   **Numbering:** Number from the end that gives the lowest sum of locants for the substituents.
*   **IUPAC Name:** 2,4-Dimethylhexane

**(h) CH3CH2CH2CH(CH3)CH2CH2CH3**
*   **Structure:** CH3–CH2–CH2–CH(CH3)–CH2–CH2–CH3
*   **Parent Chain:** The longest continuous carbon chain is 7 carbons long (heptane).
*   **Substituent:** One methyl group.
*   **Numbering:** Number from the end that gives the lowest locant for the substituent.
*   **IUPAC Name:** 4-Methylheptane

**(i) CH3CH(CH3)CH2CH2CH(CH3)CH2CH2CH3**
*   **Structure:** CH3–CH(CH3)–CH2–CH2–CH(CH3)–CH2–CH2–CH3
*   **Parent Chain:** The longest continuous carbon chain is 8 carbons long (octane).
*   **Substituents:** Two methyl groups.
*   **Numbering:** Number from the end that gives the lowest sum of locants for the substituents.
*   **IUPAC Name:** 2,5-Dimethyloctane

**(j) CH3CH2CH(CH3)CH2CH(CH3)CH2CH2CH3**
*   **Structure:** CH3–CH2–CH(CH3)–CH2–CH(CH3)–CH2–CH2–CH3
*   **Parent Chain:** The longest continuous carbon chain is 8 carbons long (octane).
*   **Substituents:** Two methyl groups.
*   **Numbering:** Number from the end that gives the lowest sum of locants for the substituents.
*   **IUPAC Name:** 3,5-Dimethyloctane

Question 30

Q30. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C6H5OH
(b) C6H5NO2
(c) CH3CH=CHCHO
(d) C6H5CHO
(e) C6H5–CH2

Accepted Answer

Resonance structures are drawn by delocalizing π-electrons or lone pairs using curved arrows to show the movement of electrons. The stability of resonance structures depends on factors like the number of covalent bonds, charge separation, and electronegativity of atoms bearing charges.

**(a) C6H5OH (Phenol)**
Phenol has a hydroxyl group (-OH) attached to a benzene ring. The lone pair of electrons on the oxygen atom can be delocalized into the benzene ring, activating the ortho and para positions for electrophilic substitution. This results in negative charges at these positions in the resonance structures.

**(b) C6H5NO2 (Nitrobenzene)**
Nitrobenzene has a nitro group (-NO2) attached to a benzene ring. The nitro group is an electron-withdrawing group, which pulls electron density from the benzene ring, deactivating it, especially at the ortho and para positions. This results in positive charges at these positions in the resonance structures.

**(c) CH3CH=CHCHO (But-2-enal)**
But-2-enal is an α,β-unsaturated aldehyde. The π-electrons of the C=C double bond and the C=O double bond are conjugated. The electron-withdrawing nature of the carbonyl group pulls electron density from the C=C bond.

**(d) C6H5CHO (Benzaldehyde)**
Benzaldehyde has an aldehyde group (-CHO) attached to a benzene ring. The aldehyde group is an electron-withdrawing group, similar to the nitro group, which deactivates the ortho and para positions of the benzene ring by creating positive charges at these positions.

**(e) C6H5–CH2+ (Benzyl carbocation)**
In a benzyl carbocation, the positive charge on the carbon atom adjacent to the benzene ring is stabilized by resonance with the π-electrons of the benzene ring. The positive charge is delocalized to the ortho and para positions of the ring.

**(f) CH3CH=CH–CH2+ (But-2-enyl carbocation)**
This is an allylic carbocation. The positive charge is adjacent to a C=C double bond, allowing for resonance stabilization through the delocalization of the π-electrons. The positive charge can shift between the terminal carbons.

Chapter 12: Organic Chemistry: Basic Principles