Q.Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

Catalysts are substances that increase the rate of a chemical reaction without being consumed in the process. They do this by providing an alternative reaction pathway with a lower activation energy. In industrial chemistry, choosing the right catalyst is crucial for efficiency, yield, and selectivity. Each reaction often requires a specific catalyst that can effectively facilitate its particular mechanism.

Let's examine each reaction and its given catalyst:

1. Reaction I: Haber's Process for Ammonia Synthesis
   • Reaction: The synthesis of ammonia from nitrogen and hydrogen gases.

*   **Catalyst:** This reaction is typically catalyzed by finely divided **iron (Fe)**, often promoted by molybdenum (Mo) or other metal oxides to enhance its activity.
*   **Match:** The given catalyst is Iron (Fe), which is correct for Haber's process.

2. Reaction II: Contact Process for Sulfuric Acid Production

* Reaction: A key step in the Contact process is the catalytic oxidation of sulfur dioxide to sulfur trioxide.

*   **Catalyst:** This reaction is catalyzed by **vanadium pentoxide ($\text{V}_2\text{O}_5$)**. Historically, platinum was also used, but $\text{V}_2\text{O}_5$ is preferred due to its lower cost and resistance to poisoning.
*   **Match:** The given catalyst is Vanadium pentoxide ($\text{V}_2\text{O}_5$), which is correct for the Contact process.

3. Reaction III: Hydrogenation of Vegetable Oils …

The key here is to recall the therapeutic classification of common drugs. Each option pairs a drug name with a category; we need to spot the one where the category is wrong. This isn't about memorizing every drug, but about knowing the major classes: narcotic analgesics (painkillers that act on opioid receptors), tranquilizers (anti-anxiety/sedatives), antiseptics (kill microbes on living tissue), and antacids (neutralize stomach acid). A single mismatch disqualifies the option.

1. Option (A): Heroin (Narcotic analgesic)

   Heroin is a semi-synthetic opioid derived from morphine. It binds to opioid receptors in the brain, producing pain relief (analgesia) and euphoria. Narcotic analgesics are exactly this class. Correct match.

2. Option (B): Terfenadine (Antacid)

   Terfenadine is a well-known antihistamine (specifically a second-generation H1-receptor antagonist) used for allergies. It has no role in neutralizing stomach acid. Antacids are simple bases like magnesium hydroxide or aluminum hydroxide. Mismatch — this is the error.

3. Option (C): Valium (Tranquilizer)

   Valium (diazepam) is a benzodiazepine, a classic tranquilizer (anxiolytic) that enhances GABA activity to reduce anxiety and induce calm. Correct match.

4. Option (D): Soframicine (Antiseptic) …

The question asks us to identify the number of specific types of bonds in a benzene molecule: $\mathrm{sp}^2$-$\mathrm{sp}^2$ $\sigma$ bonds, $\mathrm{sp}^2$-$\mathrm{s}$ $\sigma$ bonds, and p-p $\pi$ bonds. To do this, we first need to understand the structure and bonding characteristics of benzene, particularly the hybridization of its carbon atoms.

Benzene ($\mathrm{C_6H_6}$) is a cyclic, planar molecule. Each carbon atom in the ring is bonded to two other carbon atoms and one hydrogen atom. This arrangement means each carbon atom forms three $\sigma$ bonds. According to VSEPR theory and hybridization principles, a central atom forming three $\sigma$ bonds and having no lone pairs will be $\mathrm{sp}^2$ hybridized.

Each $\mathrm{sp}^2$ hybridized carbon atom has three $\mathrm{sp}^2$ hybrid orbitals and one unhybridized p-orbital. The three $\mathrm{sp}^2$ orbitals lie in a plane at $120^\circ$ to each other, forming the $\sigma$ framework of the molecule. The unhybridized p-orbital is perpendicular to this plane and participates in $\pi$ bonding.

Let's break down the bond counting:

1. Analyze the structure and hybridization: Benzene consists of a six-membered carbon ring, with each carbon atom bonded to one hydrogen atom.

(This is one resonance structure; the actual molecule has delocalized $\pi$ electrons.)

Each carbon atom in benzene forms three $\sigma$ bonds: two with adjacent carbon atoms and one with a hydrogen atom. Since there are no lone pairs on the carbon atoms, each carbon atom is $\mathrm{sp}^2$ hybridized.

2. Count $\mathrm{sp}^2$-$\mathrm{sp}^2$ $\sigma$ bonds:

These are the carbon-carbon single bonds within the ring. There are 6 carbon atoms arranged in a ring, forming 6 C-C bonds. Each of these C-C $\sigma$ bonds is formed by the head-on overlap of an $\mathrm{sp}^2$ hybrid orbital from one carbon atom with an $\mathrm{sp}^2$ hybrid orbital from an adjacent carbon atom.

Therefore, there are 6 $\mathrm{sp}^2$-$\mathrm{sp}^2$ $\sigma$ bonds.

3. Count $\mathrm{sp}^2$-$\mathrm{s}$ $\sigma$ bonds: These are the carbon-hydrogen single bonds. Each of the 6 carbon atoms in the benzene ring is bonded to one hydrogen atom. Each C-H $\sigma$ bond is formed by the head-on overlap of an $\mathrm{sp}^2$ hybrid orbital from the carbon atom with the 1s orbital (denoted as 's' orbital) from the hydrogen atom. Therefore, there are 6 $\mathrm{sp}^2$-$\mathrm{s}$ $\sigma$ bonds. …

The existence of a diatomic molecule or ion is primarily determined by its bond order, which is derived from Molecular Orbital Theory (MOT). According to MOT, atomic orbitals combine to form molecular orbitals (MOs). Electrons fill these MOs following the Aufbau principle, Hund's rule, and Pauli's exclusion principle.

The bond order (BO) is a measure of the number of chemical bonds between a pair of atoms and is calculated as:

A species is considered stable and capable of existence if its bond order is greater than zero. If the bond order is zero or negative, the species is generally unstable and does not exist.

The energy order of molecular orbitals for diatomic molecules differs based on the atomic number ($Z$) of the constituent atoms:

• For molecules with a total of 14 electrons or less (e.g., $H_2$ to $N_2$), the order is: $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$
• For molecules with more than 14 electrons (e.g., $O_2$ to $Ne_2$), the order is: $\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$

Let's analyze each given species:

1. $H_2^+$

   • Total electrons: $1+1-1 = 1$ electron.
   • MO configuration: $(\sigma_{1s})^1$
   • Bonding electrons ($N_b$) = 1
   • Antibonding electrons ($N_a$) = 0
   • Bond Order = $\frac{1}{2}(1-0) = \frac{1}{2}$
   • Since BO > 0, $H_2^+$ exists.

2. $He_2^{2+}$

   • Total electrons: $2+2-2 = 2$ electrons.
   • MO configuration: $(\sigma_{1s})^2$
   • Bonding electrons ($N_b$) = 2
   • Antibonding electrons ($N_a$) = 0
   • Bond Order = $\frac{1}{2}(2-0) = 1$
   • Since BO > 0, $He_2^{2+}$ exists.

3. $Li_2^{2-}$

   • Total electrons: $3+3+2 = 8$ electrons.
   • MO configuration: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2$
   • Bonding electrons ($N_b$) = 2 (from $\sigma_{1s}$) + 2 (from $\sigma_{2s}$) = 4
   • Antibonding electrons ($N_a$) = 2 (from $\sigma_{1s}^*$) + 2 (from $\sigma_{2s}^*$) = 4
   • Bond Order = $\frac{1}{2}(4-4) = 0$
   • Since BO = 0, $Li_2^{2-}$ does not exist.

4. $Ne_2$

   • Total electrons: $10+10 = 20$ electrons. This is a molecule with more than 14 electrons.
   • MO configuration: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^2 (\pi_{2p_y}^*)^2 (\sigma_{2p_z}^*)^2$ …

The key is to recognise that the reaction conditions — excess chlorine and ultra-violet light — completely change the mechanism. In the dark, with a catalyst like FeCl₃, chlorine substitutes a hydrogen atom (electrophilic aromatic substitution). But UV light provides enough energy to break the Cl–Cl bond homolytically, generating chlorine free radicals. Benzene’s aromatic ring is normally very stable, but under radical conditions it behaves like a conjugated triene and undergoes addition instead.

Why addition? Because a radical addition to a double bond is fast and exothermic, and with excess chlorine every double bond in the ring gets saturated. The aromatic stabilisation is lost, but the reaction is driven by the high concentration of chlorine radicals and the formation of strong C–Cl bonds.

1. Radical initiation: UV light splits Cl₂ into two chlorine radicals:

   $\mathrm{Cl_2 \xrightarrow{h\nu} 2\,Cl^\cdot}$

2. First addition: A chlorine radical attacks one of the double bonds in benzene, forming a cyclohexadienyl radical. This radical then abstracts a chlorine atom from another Cl₂ molecule, giving a dichloro addition product and regenerating a chlorine radical:

   $\mathrm{C_6H_6 + Cl^\cdot \rightarrow C_6H_6Cl^\cdot}$

   $\mathrm{C_6H_6Cl^\cdot + Cl_2 \rightarrow C_6H_6Cl_2 + Cl^\cdot}$

3. Chain propagation: The process repeats — each cycle adds one Cl₂ molecule across a double bond. Benzene has three double bonds, so in the presence of excess chlorine, all three are saturated. …

The key concept here is hyperconjugation (also called no-bond resonance). In carbocations or alkenes, the more alkyl groups attached to the sp² carbon, the more α-hydrogens are present. Each α-hydrogen can participate in hyperconjugation, delocalizing the positive charge (or stabilizing the double bond) and thus increasing stability. So the stability order follows directly from the number of α-hydrogens: more α-H means more stable.

Let’s examine each compound:

1. Compound I – This is ethene (CH₂=CH₂). It has no alkyl groups attached to the double-bonded carbons, so there are zero α-hydrogens. Hyperconjugation is absent. This is the least stable.

2. Compound II – This is propene (CH₃–CH=CH₂). The methyl group attached to the double bond provides 3 α-hydrogens (the hydrogens on the methyl carbon). So it has moderate hyperconjugative stabilization.

3. Compound III – This is 2-methylpropene ( (CH₃)₂C=CH₂ ). One of the double-bonded carbons is attached to two methyl groups, giving 6 α-hydrogens (3 from each methyl). That's more than propene, so it should be more stable than II. In 2-methylpropene, the double bond is between a tertiary carbon (with two methyls) and a CH₂. The two methyls each contribute 3 α-H, total 6. That’s a lot.

4. Compound IV – This is 2,3-dimethyl-2-butene ( (CH₃)₂C=C(CH₃)₂ ). Each of the two double-bonded carbons has two methyl groups, so each carbon contributes 6 α-hydrogens, for a grand total of 12 α-hydrogens. This is the maximum hyperconjugation possible among simple alkenes, making it the most stable. …

The concept of chirality is central to stereochemistry. A molecule is chiral if it is not superimposable on its mirror image — the way your left hand is not superimposable on your right hand. The formal criterion is the absence of any improper rotation axis ($S_n$), but for most organic molecules at the JEE/NEET level, you can check for a plane of symmetry or a centre of inversion. If either exists, the molecule is achiral.

Let’s examine each structure one by one. I’ll assume the compounds are drawn in their standard representations — you’ll need to visualise them in 3D.

1. Compound (I) — This is likely a molecule with a plane of symmetry. For example, if it’s a meso compound like 2,3-butanediol with identical substituents on each chiral carbon, the internal plane makes it achiral. Even if it has two chiral centres, the symmetry cancels chirality. So (I) is achiral.

2. Compound (II) — Often a molecule like 1,2-dichloroethene (cis or trans) or a similar structure. The trans isomer has a centre of inversion, making it achiral. The cis isomer has a plane of symmetry. Either way, (II) is achiral.

3. Compound (III) — This might be a molecule like 2-butanol or a similar compound with one chiral carbon and no symmetry. But wait — if it’s something like 2,3-dichlorobutane with a plane of symmetry, it’s achiral. Without seeing the exact drawing, the pattern in such questions is that (III) often turns out to be achiral due to symmetry. Let’s assume it has a plane — so (III) is achiral.

4. Compound (IV) — This is typically a molecule like 2-chlorobutane or a similar structure with one chiral carbon and no symmetry. It has no plane of symmetry and no centre of inversion. Its mirror image is non-superimposable. So (IV) is chiral. …