CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 2 - 56/2/2)

Concept:
The structure of glucose was initially proposed as an open-chain polyhydroxy aldehyde. While this model successfully accounts for many properties, it fails to explain several experimental observations:

Aldehyde tests: Despite having an aldehyde group, glucose does not give Schiff's test and does not form the hydrogen sulphite addition product with \(NaHSO_3\).
Pentaacetate reactivity: The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free \(-CHO\) group.
Anomerism: Glucose exists in two distinct crystalline forms (\(\alpha\) and \(\beta\)) which exhibit mutarotation.



Step 1: Analyzing the validity of the open-chain model for each option.

(A) Prolonged heating with \(HI\) reduces all carbons to a straight chain of six carbons (n-hexane). This confirms the 6-carbon straight chain skeleton, which is consistent with an open chain.


(B) Reaction with \(NH_2OH\) forms an oxime. This is a standard reaction for carbonyl groups (\(C=O\)) and is explained by the open-chain aldehyde structure.


(C) Bromine water is a mild oxidizing agent that converts the aldehyde group (\(-CHO\)) to a carboxylic acid group (\(-COOH\)), forming gluconic acid. This also confirms the presence of an aldehyde group.


(D) The existence of \(\alpha\)- and \(\beta\)-anomers implies that the \(-OH\) group at \(C-5\) adds to the \(-CHO\) group to form a cyclic hemiacetal. This creates a new chiral center at \(C-1\). The open-chain structure cannot explain this because it only shows one form of the aldehyde. Quick Tip: The cyclic structure of glucose is a six-membered ring called a pyranose ring. The two forms (\(\alpha\) and \(\beta\)) differ only in the configuration of the hydroxyl group at the hemiacetal carbon (\(C-1\)).

Concept:
Proteins are biological macromolecules. They are polymers where the repeating monomeric units are \(\alpha\)-amino acids. Each amino acid contains:

An amino group (\(-NH_2\)).
A carboxyl group (\(-COOH\)).
A hydrogen atom and a variable side chain (\(R\)) attached to the same \(\alpha\)-carbon.



Step 1: Understanding the polymerization process.

When two amino acids join, the carboxyl group (\(-COOH\)) of one amino acid reacts with the amino group (\(-NH_2\)) of the next amino acid. This is a condensation reaction that involves the loss of a water molecule (\(H_2O\)).


Step 2: Defining the resulting linkage.

The resulting \(-CO-NH-\) linkage is specifically termed a peptide bond (or peptide linkage). When many such bonds are formed, the resulting chain is called a polypeptide. A protein consists of one or more such polypeptide chains.


Step 3: Eliminating other options.


Covalent bond: While a peptide bond is a type of covalent bond, the term is too general.
Glycosidic bond: These link monosaccharide units in carbohydrates.
Coordinate bond: These involve the sharing of an electron pair from a single atom, typically found in complex compounds. Quick Tip: To remember biological linkages: 1. Proteins \(\rightarrow\) Peptide. 2. Polysaccharides \(\rightarrow\) Glycosidic. 3. Nucleic Acids \(\rightarrow\) Phosphodiester.

Concept:
Nucleic acids are polymers of nucleotides. Each nucleotide is composed of a sugar, a phosphate group, and a nitrogenous base. Nitrogenous bases are categorized into:

Purines: Adenine (A) and Guanine (G).
Pyrimidines: Cytosine (C), Thymine (T), and Uracil (U).



Step 1: Comparing the base composition of DNA and RNA.

DNA (Deoxyribonucleic acid) contains four bases:

Adenine (A)
Guanine (G)
Cytosine (C)
Thymine (T)



RNA (Ribonucleic acid) also contains four bases, but with one crucial difference:

Adenine (A)
Guanine (G)
Cytosine (C)
Uracil (U)



Step 2: Identifying the unique base.

From the comparison above, it is clear that Thymine is present exclusively in DNA. In RNA, Uracil is present instead of Thymine. Quick Tip: Mnemonic: DNA uses "T" for "Tall" (double helix), RNA uses "U" for "Understudy" (often single-stranded).

Concept:

Nitriles (or cyanides) contain the –C≡N functional group. They can be reduced to primary amines (–NH₂) using strong reducing agents such as Lithium aluminium hydride (LiAlH₄) or by catalytic hydrogenation.

Step 1: Determining the structure of the starting material

The compound is Ethanenitrile. In IUPAC nomenclature, "ethane" indicates a two-carbon chain. Therefore, its structure is:

\(CH_3-C\equiv N\)

Step 2: Writing the reduction reaction

When CH₃CN reacts with LiAlH₄, four hydrogen atoms are added across the triple bond:

Step 3: Naming the product

The product formed is:

\(CH_3CH_2NH_2\)

This consists of an ethyl group attached to an amino group. Hence, the product is Ethylamine (Ethanamine).

Step 4: Verification of carbon count

The starting material contains two carbon atoms (including the carbon of the nitrile group). Reduction of nitriles preserves the carbon count.

• Methylamine has 1 carbon — Incorrect.
• Propylamine has 3 carbons — Incorrect.
• Ethylamine has 2 carbons — Correct.

Quick Tip: Always count the total number of carbon atoms in the nitrile (including the carbon in the –CN group) to determine the name of the resulting amine.

Concept:
The boiling point of organic compounds like alcohols is influenced by two main factors:

Molecular Weight/Mass: As the number of carbon atoms increases, the magnitude of Van der Waals forces (London dispersion forces) increases, leading to a higher boiling point.
Branching: For isomeric compounds (same molecular formula), increased branching results in a more spherical shape. This reduces the surface area in contact with other molecules, thereby weakening the Van der Waals forces and lowering the boiling point.



Step 1: Ranking by carbon chain length.

The given compounds are:

Propan-1-ol (\(3 carbons\))
Butan-1-ol (\(4 carbons\))
Butan-2-ol (\(4 carbons\))
Pentan-1-ol (\(5 carbons\))

By mass: \(Propan-1-ol < Butanols < Pentan-1-ol\).


Step 2: Comparing the isomers (Butan-1-ol vs Butan-2-ol).

Both have 4 carbons.

Butan-1-ol is a primary alcohol with a straight chain.
Butan-2-ol has the hydroxyl group on the second carbon, which effectively creates a "branch" in terms of molecular surface area.

Because Butan-2-ol is more branched than the straight-chain Butan-1-ol, its boiling point is lower.
Thus: \(Butan-2-ol < Butan-1-ol\).


Step 3: Final Assembly.

Combining the observations:
Propan-1-ol (\(lowest\)) < Butan-2-ol < Butan-1-ol < Pentan-1-ol (\(highest\)). Quick Tip: Boiling Point \(\propto\) Molecular Mass.
Boiling Point \(\propto \frac{1}{Branching}\).
Straight chains always have higher boiling points than their branched isomers.

Concept:
Chirality is a fundamental property of three-dimensional objects. In organic chemistry, a molecule is said to be chiral if it cannot be superimposed on its mirror image.

The most common cause of chirality is the presence of an asymmetric carbon atom (also known as a chiral center or stereocenter).
An asymmetric carbon is a \(sp^3\) hybridized carbon atom that is covalently bonded to four different atoms or groups of atoms.
If a molecule possesses a plane of symmetry or a center of symmetry, it is achiral.



Step 1: Detailed structural analysis of Propan-2-ol.

The structure of Propan-2-ol is \( CH_3-CH(OH)-CH_3 \).
Let us examine the central carbon (\(C-2\)):

It is bonded to a hydrogen atom (\(-H\)).
It is bonded to a hydroxyl group (\(-OH\)).
It is bonded to a methyl group (\(-CH_3\)) on the left.
It is bonded to a methyl group (\(-CH_3\)) on the right.

Since two of the four groups (the two methyl groups) are identical, the carbon atom is not asymmetric. Consequently, the molecule has a plane of symmetry passing through the \( C-H \) and \( C-OH \) bonds, making it achiral.


Step 2: Detailed structural analysis of Butan-2-ol.

The structure of Butan-2-ol is \( CH_3-CH(OH)-CH_2-CH_3 \).
Let us examine the carbon at position 2 (\(C-2\)):

Group 1: \(-H\) (Hydrogen atom)
Group 2: \(-OH\) (Hydroxyl group)
Group 3: \(-CH_3\) (Methyl group)
Group 4: \(-CH_2CH_3\) (Ethyl group)

All four groups attached to \(C-2\) are distinctly different from one another. Therefore, \(C-2\) is a chiral center. This makes the molecule non-superimposable on its mirror image, confirming that Butan-2-ol is chiral.


Step 3: Analysis of 1-Bromobutane and 2-Bromopropane.


1-Bromobutane (\(CH_2Br-CH_2-CH_2-CH_3\)): Every carbon atom in this chain is bonded to at least two hydrogen atoms. Since two bonds lead to identical atoms (H), none of these carbons are chiral.
2-Bromopropane (\(CH_3-CHBr-CH_3\)): The central carbon is bonded to one H, one Br, and two identical methyl groups. Just like Propan-2-ol, the presence of two identical groups makes it achiral. Quick Tip: To identify chirality quickly in exams, look for a "fork" in the carbon chain where the groups on either side of the functional group are of different lengths or compositions (like Methyl vs Ethyl).

Concept:
Coordination compounds are formed when ligands (donor atoms/molecules) donate electron pairs to a central metal ion. Depending on the variety of ligands present in the coordination sphere, complexes are categorized into two types:

Homoleptic Complexes: These are complexes in which the central metal atom or ion is bound to only one kind of donor group/ligand. For example, a metal surrounded only by ammonia molecules.
Heteroleptic Complexes: These are complexes in which the central metal atom or ion is bound to more than one kind of donor group/ligand. These complexes often exhibit various types of isomerism, such as geometrical or optical isomerism.



Step 1: Analysis of Homoleptic options (A, B, and C).

(A) \( [Co(NH_3)_6]^{3+} \): Here, the Cobalt(III) ion is surrounded by six ammine (\(NH_3\)) ligands. Since all six ligands are identical, it is a homoleptic complex.


(B) \( [Cr(NH_3)_6]^{3+} \): Similarly, the Chromium(III) ion is coordinated with six identical ammine ligands. This is also a homoleptic complex.


(C) \( [Ni(H_2O)_6]^{2+} \): In this case, the Nickel(II) ion is coordinated with six aqua (\(H_2O\)) molecules. Since only one type of ligand (water) is present, it is homoleptic.


Step 2: Detailed analysis of option (D).

In the complex \( [Co(NH_3)_4Cl_2]^+ \), the Cobalt ion is the central metal. Looking at the coordination sphere (inside the square brackets), we find:

Four ammine (\(NH_3\)) ligands.
Two chlorido (\(Cl^-\)) ligands.

Because there are two different species (ammine and chlorido) acting as ligands to the same metal center, this complex perfectly fits the definition of a heteroleptic complex.


Step 3: Conclusion.

The presence of multiple types of ligands in the coordination sphere of \( [Co(NH_3)_4Cl_2]^+ \) distinguishes it from the other options which contain only a single type of ligand. Quick Tip: Heteroleptic complexes are important because they are the ones that typically show Geometrical Isomerism (cis/trans forms). Homoleptic complexes like \( [Co(NH_3)_6]^{3+} \) cannot show cis-trans isomerism.

Concept:
IUPAC rules for naming coordination compounds:

Ligands are named first in alphabetical order.
Use prefixes like di, tri, etc., to indicate the number of ligands.
The name of the metal is followed by its oxidation state in Roman numerals in parentheses.
For neutral complexes, the metal name remains unchanged (Platinum).



Step 1: Determining the oxidation state of Platinum.

Let the oxidation state of \(Pt\) be \(x\).
Ligands:

Ammine (\(NH_3\)) is neutral (charge = \(0\)).
Chlorido (\(Cl\)) has a charge of \(-1\).

The complex is neutral, so the sum of charges is zero: \[ x + 2(0) + 2(-1) = 0 \quad \Rightarrow \quad x - 2 = 0 \quad \Rightarrow \quad x = +2 \]
The oxidation state is (II).


Step 2: Naming the ligands in alphabetical order.


Ammine (starts with 'a')
Chlorido (starts with 'c')

Alphabetically, ammine comes before chlorido. Since there are two of each, we use "diammine" and "dichlorido".


Step 3: Combining the name.

Ligands: diammine + dichlorido
Metal: platinum
Oxidation state: (II)
Full Name: diamminedichloridoplatinum (II) Quick Tip: Note the spelling of "ammine" (with double 'm'). In IUPAC, "amine" refers to organic compounds, while "ammine" refers to the \(NH_3\) ligand.

Concept:
The electronic configuration follows the Aufbau principle, Pauli's exclusion principle, and Hund's rule. However, certain elements show anomalies due to the extra stability associated with half-filled or fully-filled subshells.


Step 1: Determining the atomic number.

Chromium (\(Cr\)) has an atomic number of \(Z = 24\).


Step 2: Applying the Aufbau principle.

Expected configuration based on energy levels: \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^4 \]
Using the noble gas Argon (\(Z=18\)) as a core:
Expected: \( [Ar] \, 4s^2 \, 3d^4 \).


Step 3: Accounting for stability exceptions.

In the case of Chromium, an electron from the \(4s\) orbital shifts to the \(3d\) orbital. This results in a \(3d^5 \, 4s^1\) configuration.

Why? The \(3d^5\) configuration is exactly half-filled.
Result: Half-filled subshells have increased stability due to the symmetrical distribution of electrons and high exchange energy.

Therefore, the correct configuration is \( [Ar] \, 3d^5 \, 4s^1 \). Quick Tip: The two most common exceptions in the 3d series are: 1. Chromium (Z=24): \(3d^5 \, 4s^1\) (Half-filled) 2. Copper (Z=29): \(3d^{10} \, 4s^1\) (Fully-filled)

Concept:
The Order of Reaction is an empirical (experimentally determined) value that describes the dependency of the reaction rate on the concentration of the reactants.

It is defined as the sum of the exponents of the concentration terms of the reactants in the rate law expression.
If Rate = \( k[X]^a [Y]^b \), then the order with respect to X is \(a\), the order with respect to Y is \(b\), and the overall order is \( (a+b) \).
Order can be an integer (0, 1, 2), a fraction, or even negative.



Step 1: Identifying the exponents from the given Rate Law.

The problem provides the specific rate equation determined by experiment: \[ Rate = k[A]^{1/2} [B]^1 \]
From this expression, we can identify:

The exponent of reactant A is \( \frac{1}{2} \) (which is \( 0.5 \)). This means the reaction is half-order with respect to A.
The exponent of reactant B is \( 1 \). This means the reaction is first-order with respect to B.



Step 2: Mathematical calculation of the overall order.

To find the total or overall order of the chemical reaction, we simply need to calculate the arithmetic sum of these exponents: \[ Overall Order (n) = Order w.r.t. A + Order w.r.t. B \] \[ n = \frac{1}{2} + 1 \]
Converting to decimals for clarity: \[ n = 0.5 + 1.0 = 1.5 \]


Step 3: Significance of the result.

The calculated overall order is 1.5. A fractional order such as 1.5 indicates that the reaction is not an elementary (single-step) reaction. Instead, it follows a complex mechanism involving multiple elementary steps, where the rate-determining step involves these specific dependencies. Quick Tip: Never use the coefficients from the balanced equation (like the '2' in 2B) to determine the order unless you are explicitly told the reaction is "elementary". Always use the exponents given in the Rate Law.

Concept:
Chemical reactions are classified into elementary (single-step) and complex (multi-step) reactions. To understand their kinetics, we use two terms: Molecularity and Order.

Molecularity: The number of reacting species (atoms, ions, or molecules) that must collide simultaneously to bring about a chemical reaction in an elementary step.
Complex Reaction: A reaction that proceeds through a sequence of elementary steps (a mechanism).



Step 1: Distinguishing between theoretical and experimental quantities.

Molecularity is a theoretical concept derived from the reaction mechanism/elementary step. It is not something measured in a lab. In contrast, the Order of Reaction is an experimental quantity. Therefore, statement (A) is incorrect.


Step 2: Analyzing Molecularity in the context of complex reactions.

For a complex reaction, the overall reaction doesn't occur in a single collision. Each individual step in the mechanism has its own molecularity. However, the term "molecularity" as applied to the overall balanced equation of a complex reaction is physically meaningless. We only discuss the "order" of a complex reaction (usually determined by the slowest step). Thus, statement (B) is the correct choice.


Step 3: Evaluating possible values for Molecularity.

Molecularity represents a count of colliding particles. You cannot have zero molecules colliding to form a product, nor can you have half a molecule participating in a collision. Therefore, molecularity is always a positive integer (1, 2, or 3). Statement (C) is incorrect.


Step 4: Considering high molecularity reactions.

The probability of four or more molecules colliding at the exact same moment with the correct orientation is statistically negligible. Consequently, molecularity greater than three is extremely rare. Statement (D) is incorrect.


Final Answer: The correct statement is (B). Quick Tip: Molecularity is only defined for elementary reactions. For complex reactions, the order of the reaction is determined by the "Rate Determining Step" (the slowest step).

Concept:
The standard Gibbs energy change (\( \Delta_r G^\circ \)) is the maximum work that can be obtained from a chemical reaction. In electrochemistry, it is calculated using the formula: \[ \Delta_r G^\circ = -nFE^\circ_{cell} \]
where \( n \) is the number of electrons transferred, \( F \) is Faraday's constant, and \( E^\circ_{cell} \) is the standard cell potential.


Step 1: Determining the standard cell potential (\( E^\circ_{cell} \)).

First, identify the half-cells:

Anode (Oxidation): \( Zn \rightarrow Zn^{2+} + 2e^- \); \( E^\circ_{anode} = -0.76 \, V \)
Cathode (Reduction): Silver ions are reduced; \( E^\circ_{cathode} = 0.80 \, V \)
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ E^\circ_{cell} = 0.80 \, V - (-0.76 \, V) = 1.56 \, V \]


Step 2: Identifying the value of \( n \).

The balanced equation shows that Zinc loses 2 electrons and two Silver atoms are formed (each Ag\(^+\) in the oxide gaining 1 electron, total 2 electrons). Thus, \( n = 2 \).


Step 3: Calculating \( \Delta_r G^\circ \) and converting units.

Using the formula: \[ \Delta_r G^\circ = -2 \times 96500 \, C/mol \times 1.56 \, V \] \[ \Delta_r G^\circ = -301080 \, J/mol \]
Convert Joules to kiloJoules (kJ) by dividing by 1000: \[ \Delta_r G^\circ = -301.080 \, kJ/mol \]


Final Answer: \( \Delta_r G^\circ \) for the reaction is -301.080 kJ mol\(^{-1}\). Quick Tip: The unit of \( \Delta G \) calculated from \( nFE \) is Joules (\(J\)). Always remember to convert it to kiloJoules (\(kJ\)) as most multiple-choice options are provided in kJ.

Concept:
This question tests knowledge of the functional groups in glucose and their chemical tests. Glucose (\(C_6H_{12}O_6\)) is an aldohexose, containing an aldehyde group and five hydroxyl groups in its open-chain form.


Step 1: Evaluating the Assertion.

Bromine water (\( Br_2/H_2O \)) is a mild oxidizing agent. It is strong enough to oxidize the aldehyde group (\(-CHO\)) of glucose to a carboxylic acid group (\(-COOH\)), but not strong enough to oxidize the alcohol groups. The resulting 6-carbon mono-carboxylic acid is called gluconic acid. Thus, the Assertion is True.


Step 2: Evaluating the Reason.

The open-chain structure of glucose does contain a carbonyl group (specifically an aldehyde group at \(C-1\)). This is proven by various reactions such as the formation of an oxime with hydroxylamine and a cyanohydrin with hydrogen cyanide. Therefore, the statement that the carbonyl group is "absent" is scientifically False.


Step 3: Selecting the correct code.

Since the Assertion is a factually correct statement and the Reason is factually incorrect, the correct option is (C).


Final Answer: (C) Assertion (A) is true, but Reason (R) is false. Quick Tip: To distinguish between an aldehyde and a ketone, bromine water is used. It will oxidize Glucose (aldehyde) to Gluconic acid, but it will not react with Fructose (ketone).

Concept:
Gabriel Phthalimide synthesis is used to prepare primary amines. The mechanism involve nucleophilic substitution (\(S_N2\)) of an alkyl halide by the phthalimide anion.


Step 1: Analyzing the scope of Gabriel Phthalimide synthesis (Assertion).

Gabriel synthesis works well for primary aliphatic amines (like methylamine, ethylamine). However, it cannot be used to prepare primary aromatic amines like aniline (\(C_6H_5NH_2\)). Therefore, the Assertion is False.


Step 2: Analyzing the chemical reason (Reason).

To make aniline, we would need to react potassium phthalimide with an aryl halide (e.g., chlorobenzene). Aryl halides are extremely unreactive toward nucleophilic substitution because:

The \( C-X \) bond has partial double bond character due to resonance.
The carbon attached to the halogen is \(sp^2\) hybridized (more electronegative).
Electronic repulsion between the nucleophile and the electron-rich benzene ring.

Consequently, the phthalimide anion cannot displace the halogen from the benzene ring. Thus, the Reason is True.


Step 3: Conclusion.

Assertion is False and Reason is True. This corresponds to option (D).


Final Answer: (D) Assertion (A) is false, but Reason (R) is true. Quick Tip: Gabriel Phthalimide synthesis is the preferred method for preparing pure primary aliphatic amines, as it prevents the formation of secondary or tertiary amines.

Concept:
The definition of a transition element according to IUPAC is an element whose atom has a partially filled d-subshell, or which can give rise to cations with an incomplete d-subshell.


Step 1: Checking the electronic configurations (Reason).

Let's look at Zinc (\(Zn\)), Cadmium (\(Cd\)), and Mercury (\(Hg\)):

Zinc (\(Z=30\)): Ground state = \([Ar] 3d^{10} 4s^2 \); Ion (\(Zn^{2+}\)) = \([Ar] 3d^{10} \).
Cadmium (\(Z=48\)): Ground state = \([Kr] 4d^{10} 5s^2 \); Ion (\(Cd^{2+}\)) = \([Kr] 4d^{10} \).
Mercury (\(Z=80\)): Ground state = \([Xe] 5d^{10} 6s^2 \); Ion (\(Hg^{2+}\)) = \([Xe] 5d^{10} \).

In both the neutral ground state and their most common oxidation states (+2), these elements have a completely filled (\(d^{10}\)) subshell.


Step 2: Linking to the definition (Assertion).

Since transition elements must have an incomplete (partially filled) d-orbital in at least one of their common forms, Zn, Cd, and Hg fail to meet this criterion. Although they are part of the d-block, they are not transition elements. Thus, the Assertion is True.


Step 3: Conclusion.

The Reason provides the exact scientific basis for the Assertion. Hence, both are true and the Reason explains the Assertion correctly.


Final Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation. Quick Tip: A common exam trick: All transition elements are d-block elements, but not all d-block elements are transition elements.

Concept:
The molecularity of a reaction refers to the number of reacting species (atoms, ions, or molecules) that must collide simultaneously in an elementary step to bring about a chemical change.

Unimolecular: One molecule decomposes or rearranges.
Bimolecular: Two species collide simultaneously.
Termolecular: Three species collide simultaneously (rare).



Step 1: Analyzing the chemical equation provided.

The reaction given is the decomposition of hydrogen iodide: \[ 2HI \longrightarrow H_2 + I_2 \]
In this elementary process, two molecules of hydrogen iodide (\(HI\)) must come together and collide with sufficient energy and proper orientation to break the existing \(H-I\) bonds and form \(H-H\) and \(I-I\) bonds.


Step 2: Linking stoichiometry to molecularity.

Since the stoichiometric coefficient of \(HI\) in the elementary step is 2, and no other reactants are involved, exactly two molecules are required for the collision. This makes the molecularity of the reaction equal to 2 (bimolecular). Therefore, the Assertion (A) is True.


Step 3: Evaluating the Reason.

The Reason states that two molecules are involved in a simultaneous collision. This is the fundamental definition of why a reaction is termed "bimolecular" or has a molecularity of 2. Thus, the Reason is True and it successfully explains the Assertion.


Final Answer: The correct choice is (A). Quick Tip: Molecularity is only applicable to elementary reactions. For complex reactions, molecularity has no overall meaning; instead, we look at the molecularity of the rate-determining (slowest) step.

Step 1: Identify the parent chain

The parent hydrocarbon is pent-2-ene, which consists of five carbon atoms with a double bond between carbon-2 and carbon-3.



Step 2: Locate the substituents


A methyl group (\(-CH_3\)) is attached to carbon-3.
A bromo atom (Br) is attached to carbon-4.




Step 3: Draw the structure






Structure of 4--Bromo--3--methylpent--2--ene




Condensed structural formula:
\[ \boxed{\mathrm{CH_3-CH=C(CH_3)-CH(Br)-CH_3}} \]



Quick Tip: The longest chain contains 5 carbon atoms \(\Rightarrow\) Pent. The double bond starts from carbon-2 \(\Rightarrow\) Pent-2-ene. Number the chain so that the double bond gets the lowest possible number. Attach the methyl group at carbon-3 and bromine at carbon-4.

Concept:
Haloalkanes undergo nucleophilic substitution reactions (\(S_N1\) or \(S_N2\)) when reacted with reagents containing strong nucleophiles.

Aqueous \(KOH\) dissociates completely to provide free hydroxide ions (\(OH^-\)).
Hydroxide ions are strong nucleophiles that replace the halogen atom.



Step 1: Identifying the reactants and mechanism.


Substrate: Chloroethane (\(CH_3CH_2Cl\)) - a primary alkyl halide.
Reagent: Aqueous \(KOH\) (provides \(OH^-\) in a polar medium).
Mechanism: Primary alkyl halides undergo bimolecular nucleophilic substitution (\(S_N2\)). The \(OH^-\) ion attacks the electrophilic carbon from the backside, displacing the chloride ion (\(Cl^-\)).



Step 2: Writing the chemical reaction.
\[ CH_3CH_2Cl + KOH_{(aq)} \xrightarrow{\Delta} CH_3CH_2OH + KCl \]


Observation:
Chloroethane undergoes nucleophilic substitution to yield Ethanol (an alcohol) along with potassium chloride as a byproduct.


Final Answer: Chloroethane is converted into Ethanol (\(CH_3CH_2OH\)). Quick Tip: Aqueous \(KOH\) leads to Substitution (Alcohol formation), whereas Alcoholic \(KOH\) leads to Elimination (Alkene formation).

Concept:
According to the Collision Theory of chemical kinetics, a reaction occurs when reactant molecules physically collide with each other with sufficient kinetic energy (activation energy) and proper orientation.


Step 1: Formal Definition.

Collision Frequency (\(Z\)) is defined as the number of collisions that take place per second per unit volume of the reaction mixture.


Step 2: Mathematical representation in Rate Theory.

For a bimolecular elementary reaction \( A + B \rightarrow Products \), the rate of reaction is expressed as: \[ Rate = Z_{AB} \cdot e^{-E_a/RT} \]
Where:

\( Z_{AB} \) = Collision frequency of reactants A and B.
\( e^{-E_a/RT} \) = Fraction of molecules having energy equal to or greater than Activation Energy (\(E_a\)).



Final Answer: Collision frequency is the total number of collisions occurring per second per unit volume of the reaction mixture. Quick Tip: Higher collision frequency generally increases the reaction rate, provided the orientation factor (\(P\)) and energy criteria are met.

Concept:
The unit of the rate constant (\(k\)) depends on the overall order of the reaction (\(n\)). It can be derived using the general rate law expression: \[ Rate = k [Concentration]^n \quad \Rightarrow \quad k = \frac{Rate}{[Concentration]^n} \]


Step 1: Deriving the general unit formula.

Substitute the units of Rate (\(mol L^{-1}s^{-1}\)) and Concentration (\(mol L^{-1}\)): \[ Unit of k = \frac{mol L^{-1}s^{-1}}{(mol L^{-1})^n} = (mol L^{-1})^{1-n} s^{-1} \]


Step 2: Evaluating for First Order Reaction (\(n = 1\)).

Substitute \(n = 1\) into the formula: \[ Unit = (mol L^{-1})^{1-1} s^{-1} = (mol L^{-1})^0 s^{-1} = \mathbf{s^{-1}} \quad (or min^{-1}, time^{-1}) \]


Step 3: Evaluating for Zero Order Reaction (\(n = 0\)).

Substitute \(n = 0\) into the formula: \[ Unit = (mol L^{-1})^{1-0} s^{-1} = \mathbf{mol L^{-1} s^{-1}} \quad (or M s^{-1}) \]


Final Answer: (i) Unit for First Order: \(s^{-1}\). (ii) Unit for Zero Order: \(mol L^{-1}s^{-1}\). Quick Tip: Notice that the unit of \(k\) for a zero-order reaction is identical to the unit of the rate of reaction itself!

Concept:
Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion. It depends on:

Nuclear charge.
Electronic stability (half-filled and fully-filled configurations are exceptionally stable).



Step 1: Electronic configuration of Manganese species.


Manganese atom (\(Mn\), \(Z=25\)): \([Ar] \, 3d^5 \, 4s^2\)
\(Mn^+\) ion: \([Ar] \, 3d^5 \, 4s^1\)
\(Mn^{2+}\) ion: \([Ar] \, 3d^5\)



Step 2: Analyzing the third ionization step.

The third ionization enthalpy (\(\Delta_i H_3\)) corresponds to the removal of an electron from the \(Mn^{2+}\) ion to form \(Mn^{3+}\): \[ Mn^{2+}_{([Ar] 3d^5)} \xrightarrow{\Delta_i H_3} Mn^{3+}_{([Ar] 3d^4)} + e^- \]

\(Mn^{2+}\) has a \(3d^5\) electronic configuration, which is exactly half-filled.
Half-filled subshells possess high exchange energy and extra stability.
Removing an electron requires disrupting this extraordinarily stable subshell.



Conclusion:
A large amount of energy must be supplied to break the stable \(3d^5\) half-filled shell, resulting in an exceptionally high third ionization enthalpy.


Final Answer: The third ionization enthalpy is high because removing an electron requires disrupting the extra stable, half-filled \(3d^5\) configuration of the \(Mn^{2+}\) ion. Quick Tip: Always write the electronic configuration of the ion before the electron removal step to easily see if a stable \(d^5\) or \(d^{10}\) shell is being broken.

Concept:
According to the official IUPAC definition:

A transition element is defined as an element that has an incomplete d-subshell (partially filled d-orbitals) either in its ground state or in any of its common oxidation states.



Step 1: Examining electronic configurations.

Let's check the configurations of Group 12 elements:

Zinc (\(Zn\), \(Z=30\)):

Ground State: \([Ar] \, 3d^{10} \, 4s^2\)
\(Zn^{2+}\) Ion: \([Ar] \, 3d^{10}\)

Cadmium (\(Cd\), \(Z=48\)):

Ground State: \([Kr] \, 4d^{10} \, 5s^2\)
\(Cd^{2+}\) Ion: \([Kr] \, 4d^{10}\)

Mercury (\(Hg\), \(Z=80\)):

Ground State: \([Xe] \, 4f^{14} \, 5d^{10} \, 6s^2\)
\(Hg^{2+}\) Ion: \([Xe] \, 4f^{14} \, 5d^{10}\)




Step 2: Applying the definition.

In both their elemental ground states and their common oxidation states (+2), \(Zn\), \(Cd\), and \(Hg\) have completely filled d-subshells (\(d^{10}\)). They do not possess a partially filled d-orbital in any stable form.


Conclusion:
Because they lack incomplete d-subshells, they are classified as d-block elements, but not as transition elements.


Final Answer: They are not transition elements because they have completely filled d-orbitals (\(d^{10}\)) in their ground state as well as in their common oxidation states. Quick Tip: To remember: Transition elements must have partially filled d-orbitals (\(d^1\) to \(d^9\)). \(d^{10}\) means non-transition element!

Concept:
Ligands are atoms, ions, or molecules that donate electron pairs to a central metal atom/ion to form coordinate bonds.

Denticity refers to the number of donor groups bound to a single central metal atom.



Step 1: Definition.

A didentate (or bidentate) ligand is a species that possesses two donor atoms capable of simultaneously donating two pairs of electrons to a single central metal ion, forming two coordinate covalent bonds.


Step 2: Examples and structural illustration.

Common examples include:

Ethane–1,2–diamine (commonly abbreviated as 'en'):
\[ H_2\ddot{N}-CH_2-CH_2-\ddot{N}H_2 \]
Here, both Nitrogen atoms act as donor sites.

Oxalate ion (commonly abbreviated as 'ox'):
\[ (COO^-)_2 \quad or \quad C_2O_4^{2-} \]
Here, two negatively charged Oxygen atoms act as donor sites.



Final Answer: A didentate ligand has two donor atoms that bind to a central metal. Example: Ethane-1,2-diamine (\(en\)) or Oxalate ion (\(C_2O_4^{2-}\)). Quick Tip: Didentate ligands form cyclic ring structures with the metal called "chelate rings," which impart extra stability to the coordination complex.

Concept:
Complexes of Platinum(II) are typically 4-coordinate and adopt a **square planar** geometry.

Square planar complexes of the general formula \([MA_2BC]\) or \([MA_2B_2]\) exhibit Geometrical Isomerism.



Step 1: Analyzing the complex structure.

In \([Pt(NH_3)(H_2O)Cl_2]\):

Central metal = Platinum(II) (\(sp^2d\) or \(dsp^2\) square planar geometry).
Ligands = Two chlorido (\(Cl^-\)) ligands, one ammine (\(NH_3\)) ligand, and one aqua (\(H_2O\)) ligand.



Step 2: Identifying the geometric forms (cis and trans).

Because there are two identical ligands (\(Cl^-\)):

cis-Isomer: The two \(Cl^-\) ligands are located adjacent to each other (\(90^\circ\) angle).
trans-Isomer: The two \(Cl^-\) ligands are located opposite to each other (\(180^\circ\) angle).



Conclusion:
The complex exhibits Geometrical Isomerism (cis-trans isomerism).


Final Answer: The given complex exhibits Geometrical Isomerism cis-trans isomerism. Quick Tip: 4-coordinate tetrahedral complexes DO NOT show geometrical isomerism because all positions are adjacent to one another. Only square planar 4-coordinate complexes show geometrical isomerism!

Concept:
For a first-order reaction, the rate of reaction is directly proportional to the first power of the concentration of the reactant. The integrated rate equation that relates the rate constant (\(k\)), time (\(t\)), initial concentration (\([R]_0\)), and final concentration (\([R]\)) is given by: \[ k = \frac{2.303}{t} \log \left( \frac{[R]_0}{[R]} \right) \]
This formula allows us to calculate the rate constant if we know how much reactant is left after a certain period.


Step 1: Extracting the given values from the problem.


Initial concentration of reactant, \( [R]_0 = 0.6 mol L^{-1} \)
Final concentration of reactant, \( [R] = 0.2 mol L^{-1} \)
Time taken, \( t = 5 minutes \)



Step 2: Substituting the values into the integrated rate equation.
\[ k = \frac{2.303}{5 min} \log \left( \frac{0.6}{0.2} \right) \]


Step 3: Simplifying the logarithmic term and calculating the result.
\[ \frac{0.6}{0.2} = 3 \]
So the equation becomes: \[ k = \frac{2.303}{5} \log (3) \]
We are given \( \log 3 = 0.48 \). Substituting this value: \[ k = \frac{2.303 \times 0.48}{5} \] \[ k = \frac{1.10544}{5} \] \[ k = 0.221088 min^{-1} \]
Rounding off to three significant figures, we get \( 0.221 min^{-1} \).


Final Answer: The rate constant of the reaction is \( 0.221 min^{-1} \). Quick Tip: When units of time are given in minutes, the rate constant for a first-order reaction will have units of \(min^{-1}\). You don't need to convert to seconds unless explicitly asked!

Concept:
For a first-order reaction, the time required for a specific fraction of the reactant to decompose can be found using the integrated rate equation: \[ t = \frac{2.303}{k} \log \left( \frac{[R]_0}{[R]} \right) \]
When a reaction is "three-fourths complete," it means that \( 3/4 \) of the initial concentration has reacted, and therefore \( 1/4 \) of the initial concentration remains.


Step 1: Setting up the initial and final concentrations.

Let the initial concentration be \( [R]_0 = 100 \) (or \(a\)).
Since three-fourths (\(75%\)) of the reactant decomposes, the amount reacted is \( 75 \).
The remaining concentration \( [R] \) is: \[ [R] = 100 - 75 = 25 \quad \left(or [R] = a - \frac{3a}{4} = \frac{a}{4} \right) \]


Step 2: Substituting the values into the integrated rate equation.

Given rate constant \( k = 2.54 \times 10^{-3} s^{-1} \). \[ t = \frac{2.303}{2.54 \times 10^{-3}} \log \left( \frac{100}{25} \right) \]


Step 3: Simplifying the logarithmic term and calculating time.
\[ \frac{100}{25} = 4 \] \[ t = \frac{2.303}{2.54 \times 10^{-3}} \log (4) \]
We are given \( \log 4 = 0.60 \). Substitute this into the equation: \[ t = \frac{2.303 \times 0.60}{2.54 \times 10^{-3}} \] \[ t = \frac{1.3818}{2.54 \times 10^{-3}} \] \[ t = 544.015 s \]
Rounding to the nearest whole number gives \( 544 s \).


Final Answer: The time required for three-fourth decomposition is \( 544 s \). Quick Tip: For a first-order reaction, the time required for \( 75% \) completion (\(t_{3/4}\)) is exactly twice the half-life (\(t_{1/2}\)). You can also solve this by finding \( t_{1/2} \) first and multiplying by 2!

Concept:

When a non-volatile solute is added to a volatile solvent, the vapor pressure of the solvent decreases, which causes an elevation in the boiling point. This colligative property is expressed as:

Where:

• \(\Delta T_b\) = Elevation in boiling point \((T_{\text{solution}} - T_{\text{solvent}})\).
• \(K_b\) = Molal boiling point elevation constant.
• \(m\) = Molality of the solution \(=\dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}}\).

Step 1: Extract the given data and calculate the elevation in boiling point

• Mass of solute, \(W_B = 8.0\,\text{g}\)
• Mass of solvent, \(W_A = 100\,\text{g} = 0.1\,\text{kg}\)
• Boiling point of solution, \(T_b = 36.86^\circ\text{C}\)
• Boiling point of pure solvent, \(T_b^\circ = 35.60^\circ\text{C}\)
• Ebullioscopic constant, \(K_b = 2.02\,\text{K kg mol}^{-1}\)

Elevation in boiling point:

(The temperature difference is the same in Kelvin and degrees Celsius.)

Step 2: Set up the formula for molar mass

Molality is given by:

Substituting this into the boiling point elevation equation:

Rearranging to find the molar mass of the solute:

Step 3: Substitute the values

Final Answer: The molar mass of the non-volatile solute is \(128.25\,\text{g mol}^{-1}\).

Quick Tip: Always convert the mass of the solvent to kilograms when calculating molality. If the solvent mass is used in grams, remember to include the 1000 factor in the numerator.

Concept:
The electromotive force (EMF) of a cell under non-standard conditions is calculated using the Nernst Equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
Where:

\( E^\circ_{cell} \) is the standard cell potential.
\( n \) is the number of moles of electrons transferred.
\( Q \) is the reaction quotient.



Step 1: Writing the half-cell reactions and determining \(n\).


Anode (Oxidation, left side): \( Sn_{(s)} \longrightarrow Sn^{2+}_{(aq)} + 2e^- \)
Cathode (Reduction, right side): \( 2H^+_{(aq)} + 2e^- \longrightarrow H_{2(g)} \)

Overall cell reaction: \( Sn_{(s)} + 2H^+_{(aq)} \longrightarrow Sn^{2+}_{(aq)} + H_{2(g)} \)
The number of electrons exchanged is \( n = 2 \).


Step 2: Calculating the Standard Cell Potential (\( E^\circ_{cell} \)).
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ E^\circ_{cell} = E^\circ_{H^+/H_2} - E^\circ_{Sn^{2+}/Sn} \] \[ E^\circ_{cell} = 0.00 V - (-0.14 V) = +0.14 V \]


Step 3: Applying the Nernst Equation to find \(E_{cell}\).

The reaction quotient \( Q = \frac{[Sn^{2+}][P_{H_2}]}{[H^+]^2} \).
Since \( P_{H_2} = 1 bar \): \[ E_{cell} = 0.14 - \frac{0.0591}{2} \log \left( \frac{[Sn^{2+}]}{[H^+]^2} \right) \]
Substitute the given concentrations (\( [Sn^{2+}] = 0.001 M = 10^{-3} M \), \( [H^+] = 0.01 M = 10^{-2} M \)): \[ E_{cell} = 0.14 - 0.02955 \log \left( \frac{10^{-3}}{(10^{-2})^2} \right) \] \[ E_{cell} = 0.14 - 0.02955 \log \left( \frac{10^{-3}}{10^{-4}} \right) \] \[ E_{cell} = 0.14 - 0.02955 \log (10^1) \]


Step 4: Final Calculation.

Since \( \log 10 = 1 \): \[ E_{cell} = 0.14 - (0.02955 \times 1) \] \[ E_{cell} = 0.14 - 0.02955 = 0.11045 V \]
Rounding to appropriate significant figures, \( E_{cell} \approx 0.11 V \).


Final Answer: The emf of the given cell is \( 0.11 V \). Quick Tip: Don't forget the stoichiometric coefficient from the balanced equation becomes the power in the reaction quotient \( Q \). Here, \( 2H^+ \) means \( [H^+] \) must be squared!

Concept:
In the f-block elements, as atomic number increases across a period, the atomic and ionic radii progressively decrease. This phenomenon is termed "Lanthanoid contraction" for the 4f series and "Actinoid contraction" for the 5f series. This contraction is caused by the imperfect shielding effect of f-electrons.


Step 1: Understanding the shielding effect in f-orbitals.

As we move along the actinoid series, new electrons are added to the \(5f\) subshell. Similarly, in the lanthanoid series, electrons are added to the \(4f\) subshell. The shapes of f-orbitals are highly diffused, meaning they are spread out over a large volume of space. Because they are so diffused, f-electrons are very poor at shielding the outermost valence electrons from the attractive pull of the positive nucleus.


Step 2: Comparing 4f and 5f orbitals.

The \(5f\) orbitals in actinoids are even larger and more diffused (spread out) than the \(4f\) orbitals in lanthanoids.
Therefore, the shielding effect of a \(5f\) electron is significantly weaker than that of a \(4f\) electron.


Step 3: Concluding the effect on atomic size.

Due to the poorer shielding provided by the \(5f\) electrons, the effective nuclear charge (\(Z_{eff}\)) felt by the outermost electrons is much stronger in actinoids compared to lanthanoids. This stronger inward pull by the nucleus results in a more drastic decrease in size (a greater contraction) from one element to the next in the actinoid series.


Final Answer: Actinoid contraction is greater than lanthanoid contraction because \(5f\) electrons have a poorer shielding effect than \(4f\) electrons, leading to a stronger effective nuclear charge pulling the valence electrons inward. Quick Tip: Just remember the order of shielding effect: \(s > p > d > f\). Since \(5f\) is larger and more diffused than \(4f\), its shielding is even worse, leading to greater contraction.

Concept:

The magnetic nature of a coordination compound depends upon the number of unpaired electrons present in the central metal ion.


A compound containing one or more unpaired electrons is paramagnetic.
A compound having all electrons paired is diamagnetic.
Strong field ligands cause pairing of electrons, whereas weak field ligands generally do not.




Step 1: Analysis of \([\mathrm{Fe(H_2O)_6}]^{2+}\).



Oxidation state of Fe = \(+2\).
Electronic configuration of Fe:
\[ [Ar]\,3d^6\,4s^2 \]
Electronic configuration of \(Fe^{2+}\):
\[ [Ar]\,3d^6 \]
Water (\(\mathrm{H_2O}\)) is a weak field ligand.
It cannot force pairing of the \(3d\) electrons.
Therefore, the complex is a high-spin complex having four unpaired electrons.
Hence,
\[ [\mathrm{Fe(H_2O)_6}]^{2+} \]
is paramagnetic.




Step 2: Analysis of \([\mathrm{Fe(CO)_5}]\).



Carbon monoxide (\(\mathrm{CO}\)) is a neutral ligand.
Therefore, the oxidation state of Fe is \(0\).
Electronic configuration of Fe:
\[ [Ar]\,3d^6\,4s^2 \]
Carbon monoxide is a strong field ligand.
It causes pairing of the electrons in the \(3d\) orbitals.
The complex undergoes \(dsp^3\) hybridisation with all electrons paired.
Therefore,
\[ [\mathrm{Fe(CO)_5}] \]
contains no unpaired electrons and is diamagnetic.




Final Answer:
\[ [\mathrm{Fe(H_2O)_6}]^{2+} \]

contains four unpaired electrons because \(\mathrm{H_2O}\) is a weak field ligand. Hence, it is paramagnetic.
\[ [\mathrm{Fe(CO)_5}] \]

contains no unpaired electrons because \(\mathrm{CO}\) is a strong field ligand that causes electron pairing. Hence, it is diamagnetic. Quick Tip: Always determine the oxidation state of the metal ion first. Then identify whether the ligand is a weak-field ligand (\(\mathrm{H_2O}, \mathrm{F^-}, \mathrm{Cl^-}\)) or a strong-field ligand (\(\mathrm{CO}, \mathrm{CN^-}, \mathrm{NH_3}\)). Weak-field ligands generally produce high-spin (paramagnetic) complexes, whereas strong-field ligands produce low-spin (often diamagnetic) complexes.

Concept:
Crystal Field Theory (CFT) explains the splitting of degenerate d-orbitals into two sets of different energies when surrounded by ligands in an octahedral field:

\( t_{2g} \): The lower energy set consisting of 3 orbitals (\(d_{xy}, d_{yz}, d_{zx}\)).
\( e_g \): The higher energy set consisting of 2 orbitals (\(d_{x^2-y^2}, d_{z^2}\)).

The energy gap between them is the crystal field splitting energy (\( \Delta_0 \)). The term \( P \) stands for pairing energy (the energy required to force two electrons into the same orbital).


Step 1: Evaluating the given condition (\( \Delta_0 < P \)).

The condition \( \Delta_0 < P \) indicates that the crystal field splitting energy is less than the energy required to pair up electrons.

This situation arises in the presence of a weak field ligand.
Because the energy gap (\( \Delta_0 \)) is small, it requires less energy for an electron to jump to the higher \( e_g \) orbitals than to pair up with another electron in the lower \( t_{2g} \) orbitals.



Step 2: Writing the electronic configuration for \( d^4 \).

According to Hund's rule and the Aufbau principle under these specific conditions:

The first three electrons will singly occupy the three lower energy \( t_{2g} \) orbitals (Configuration: \( t_{2g}^3 \)).
For the fourth electron, since \( \Delta_0 < P \), it is energetically more favorable to jump to the higher \( e_g \) orbital rather than pairing up in the \( t_{2g} \) level.
Therefore, the fourth electron occupies an \( e_g \) orbital.



Final Answer: The electronic configuration for a \( d^4 \) ion under the condition \( \Delta_0 < P \) is \( t_{2g}^3 \, e_g^1 \). Quick Tip: If \( \Delta_0 < P \), it forms a High Spin complex (\( t_{2g}^3 \, e_g^1 \)). If \( \Delta_0 > P \), it forms a Low Spin complex, where electrons pair up instead of jumping (\( t_{2g}^4 \, e_g^0 \)).

Concept:
This reaction is a free radical substitution reaction. When an alkyl-substituted benzene (like ethylbenzene) is treated with a halogen in the presence of heat or UV light:

The substitution occurs at the benzylic position (the carbon atom directly attached to the benzene ring).
The benzylic free radical is highly stable due to resonance stabilization with the aromatic ring.
Substitution does not occur on the ring itself under these conditions.



Step 1: Identifying the reactive site.

In 1-ethyl-4-nitrobenzene, there are two carbons in the side chain: the benzylic carbon (\(-CH_2-\)) and the terminal carbon (\(-CH_3\)). The benzylic carbon is more reactive because the radical formed there is stabilized by the delocalization of electrons into the benzene ring.


Step 2: Determining the product.

One hydrogen atom from the benzylic position is replaced by a bromine atom. The presence of the nitro group (\(-NO_2\)) at the para position is a strong electron-withdrawing group, but the site of bromination is still governed by the stability of the benzylic radical.


Final Answer: The major product is 1-bromo-1-(4-nitrophenyl)ethane. Quick Tip: Halogenation in dark/Lewis acid (like \(FeCl_3\)) goes to the ring, but halogenation in heat/light goes to the side chain.

Concept:
Chirality is a geometric property that describes the symmetry (or lack thereof) of a molecule or object.


Step 1: The fundamental definition.

An object or molecule is said to be chiral if it is non-superimposable on its mirror image. This property is similar to our left and right hands; they are mirror images but cannot be perfectly laid on top of each other with all parts matching.


Step 2: Requirement for molecular chirality.

In organic chemistry, chirality usually arises when a carbon atom is bonded to four different atoms or groups. Such a carbon is called a chiral center or an asymmetric carbon.


Final Answer: Chirality is the property of non-superimposability of an object on its mirror image. Quick Tip: If a molecule has a plane of symmetry, it is always achiral (not chiral).

Concept:
The \(S_N1\) (Substitution Nucleophilic Unimolecular) mechanism involves two steps, and the intermediate formed dictates the stereochemistry of the product.


Step 1: Formation of the Carbocation.

In the first step, the leaving group departs, leaving behind a carbocation intermediate. This carbocation is \(sp^2\) hybridized and has a planar geometry.


Step 2: Nucleophilic attack from both sides.

Because the carbocation is flat (planar), the incoming nucleophile can attack from the front side (the side from which the leaving group left) or the back side with equal probability.


Step 3: Formation of the racemic mixture.

If the starting material was optically active, this dual-side attack leads to a 50:50 mixture of two enantiomers (retention and inversion). This resulting 1:1 mixture is optically inactive and is called a racemic mixture.


Final Answer: Racemisation occurs because the planar carbocation intermediate allows the nucleophile to attack with equal probability from both sides. Quick Tip: \(S_N2\) \(\rightarrow\) Complete Inversion (Walden Inversion).
\(S_N1\) \(\rightarrow\) Racemisation.

Concept:
Rosenmund's reduction is a specific method used to synthesize aldehydes from acyl chlorides.


Step 1: The Chemical Reaction.

In this reaction, an acyl chloride is hydrogenated over a catalyst of palladium supported on barium sulphate (\(Pd-BaSO_4\)). This catalyst is often "poisoned" with sulfur or quinoline to prevent the further reduction of the aldehyde into an alcohol.


Step 2: Equation.
\[ R-COCl + H_2 \xrightarrow{Pd-BaSO_4} R-CHO + HCl \]
Example: Benzoyl chloride to Benzaldehyde. \[ C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO + HCl \]


Final Answer: Rosenmund's reduction converts acyl chlorides to aldehydes using \(H_2/Pd-BaSO_4\). Quick Tip: \(BaSO_4\) acts as a catalyst poison. Without it, the reaction wouldn't stop at the aldehyde and would proceed to form a primary alcohol.

Concept:
This is a disproportionation (self-oxidation and reduction) reaction that occurs in aldehydes that lack \(\alpha\)-hydrogen atoms.


Step 1: Reaction Conditions.

Aldehydes like formaldehyde (\(HCHO\)) or benzaldehyde (\(C_6H_5CHO\)) are treated with concentrated alkali (\(NaOH\) or \(KOH\)).


Step 2: The Chemistry.

One molecule of the aldehyde is reduced to the corresponding alcohol, while another molecule is oxidized to the salt of the carboxylic acid.


Step 3: Equation.
\[ 2C_6H_5CHO + conc. NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \]
(Benzaldehyde \(\rightarrow\) Benzyl alcohol + Sodium benzoate)


Final Answer: Cannizzaro's reaction is the disproportionation of aldehydes without \(\alpha\)-H into an alcohol and a carboxylate salt. Quick Tip: With \(\alpha\)-H \(\rightarrow\) Aldol Condensation.
No \(\alpha\)-H \(\rightarrow\) Cannizzaro Reaction.

Concept:

The Hell--Volhard--Zelinsky (HVZ) reaction is used for the \(\alpha\)-halogenation of carboxylic acids containing at least one \(\alpha\)-hydrogen atom.
In this reaction, the \(\alpha\)-hydrogen atom of a carboxylic acid is replaced by a halogen atom (usually bromine or chlorine).
The reaction is carried out in the presence of red phosphorus (or \(PBr_3\)), which acts as a catalyst.




Step 1: Identify the reactants

A carboxylic acid having an \(\alpha\)-hydrogen reacts with bromine in the presence of red phosphorus.

Example:
\[ \mathrm{CH_3COOH \xrightarrow[Red P]{Br_2} CH_2BrCOOH} \]



Step 2: Write the general reaction

The general equation of the Hell--Volhard--Zelinsky reaction is:
\[ \boxed{ \mathrm{RCH_2COOH \xrightarrow[Red P]{Br_2} RCHBrCOOH} } \]

where \(R\) represents an alkyl group.



Step 3: State the significance

The reaction introduces a bromine atom at the \(\alpha\)-carbon of the carboxylic acid. The resulting \(\alpha\)-bromo acid is an important intermediate in the preparation of amino acids and many other organic compounds.



Final Reaction:
\[ \boxed{ \mathrm{CH_3COOH \xrightarrow[Red P]{Br_2} CH_2BrCOOH} } \] Quick Tip: HVZ reaction occurs only in carboxylic acids having at least one \(\alpha\)-hydrogen. Reagents used: Bromine (\(Br_2\)) and red phosphorus (or \(PBr_3\)). Product formed is an \(\alpha\)-bromocarboxylic acid.

Concept:

Proteins possess a specific three-dimensional structure which is responsible for their biological activity.
This structure is maintained by weak forces such as hydrogen bonds, ionic interactions and hydrophobic interactions.




Step 1: Define denaturation

Denaturation of proteins is the process in which the natural three-dimensional structure of a protein is destroyed due to the action of heat, acids, alkalis, alcohol, heavy metal salts or other chemicals.



Step 2: Effect of denaturation

During denaturation, the secondary and tertiary structures of the protein are disrupted, whereas the primary structure (sequence of amino acids) remains unchanged.

As a result, the protein loses its biological activity and its characteristic properties.



Examples:

Boiling of an egg causes egg albumin to coagulate due to denaturation.
Milk proteins are denatured during curd formation.



Quick Tip: Denaturation changes only the spatial arrangement of protein molecules. The primary structure (peptide bonds) remains intact. Denatured proteins usually lose their biological activity.

Concept:

Carbohydrates are classified as monosaccharides, oligosaccharides and polysaccharides on the basis of the number of monosaccharide units present.




Step 1: Definition

Oligosaccharides are carbohydrates that on hydrolysis produce two to ten monosaccharide units.

They are formed by the condensation of monosaccharides through glycosidic linkages.



Step 2: Examples

Common examples of oligosaccharides are:

Sucrose (Glucose + Fructose)
Maltose (Glucose + Glucose)
Lactose (Glucose + Galactose)


Among these, disaccharides are the most common type of oligosaccharides.



Quick Tip: Oligosaccharides yield 2--10 monosaccharide molecules on hydrolysis. Disaccharides are the simplest and most common oligosaccharides. They contain glycosidic linkages between monosaccharide units.

Concept:

Starch is the main storage polysaccharide in plants.
It consists of two components namely amylose and amylopectin.




Step 1: Amylose

Amylose is a linear, unbranched polysaccharide made up of \(\alpha\)-D-glucose units joined by \(\alpha(1\rightarrow4)\) glycosidic bonds.

It forms a helical structure and constitutes about \(20%-30%\) of starch.



Step 2: Amylopectin

Amylopectin is a highly branched polysaccharide.

Its glucose units are linked by:

\(\alpha(1\rightarrow4)\) glycosidic bonds in the main chain.
\(\alpha(1\rightarrow6)\) glycosidic bonds at the branching points.


It constitutes about \(70%-80%\) of starch.



Differences:


Amylose is linear, whereas amylopectin is highly branched.
Amylose contains only \(\alpha(1\rightarrow4)\) linkages, whereas amylopectin contains both \(\alpha(1\rightarrow4)\) and \(\alpha(1\rightarrow6)\) linkages.
Amylose forms a helical structure, whereas amylopectin has a tree-like branched structure.




Quick Tip: Amylose = Linear chain. Amylopectin = Branched chain. Both are polymers of \(\alpha\)-D-glucose.

Concept:

A nucleotide is the basic structural unit of DNA.
Each DNA nucleotide consists of three components:

A nitrogenous base,
A pentose sugar (2-deoxyribose),
A phosphate group.





Step 1: Identify the nucleotide

The given nucleotide contains the nitrogenous base thymine. Therefore, it is composed of:

Thymine
2-Deoxyribose sugar
Phosphoric acid




Step 2: Hydrolysis of the nucleotide

On complete hydrolysis, all the bonds connecting these components break, producing the individual constituents.

Hence, the products obtained are:


Thymine
2-Deoxyribose sugar
Phosphoric acid (\(H_3PO_4\))




Final Answer:

A nucleotide of DNA containing thymine on hydrolysis gives:
\[ \boxed{Thymine + 2-Deoxyribose Sugar + Phosphoric Acid} \]



Quick Tip: DNA nucleotides always contain deoxyribose sugar, whereas RNA nucleotides contain ribose sugar.

Concept:

Glucose contains one aldehyde group and five hydroxyl (\(-OH\)) groups.
The presence of hydroxyl groups is confirmed by its chemical reactions.




Step 1: Reaction with acetic anhydride

When glucose is treated with excess acetic anhydride in the presence of pyridine, it forms glucose pentaacetate.
\[ \mathrm{Glucose} \xrightarrow[Pyridine] {(CH_3CO)_2O} \mathrm{Glucose\;Pentaacetate} \]



Step 2: Interpretation

Formation of glucose pentaacetate shows that five acetyl groups are introduced into the molecule.

Since each acetyl group replaces one hydrogen atom of a hydroxyl group, glucose must contain five hydroxyl groups.

The fact that glucose pentaacetate does not react with hydroxylamine indicates that the aldehyde group is not free in this derivative.



Final Answer:

The formation of glucose pentaacetate on acetylation proves that glucose contains five hydroxyl (\(-OH\)) groups attached to five different carbon atoms.



Quick Tip: One molecule of glucose reacts with five molecules of acetic anhydride to form one molecule of glucose pentaacetate, confirming the presence of five alcoholic hydroxyl groups.

Concept:
The reduction of amides is a standard method for the preparation of amines.

Lithium aluminium hydride (\(LiAlH_4\)) is a very strong reducing agent.
When an amide reacts with \(LiAlH_4\), the carbonyl group (\(C=O\)) is completely reduced to a methylene group (\(CH_2\)).
The number of carbon atoms in the chain remains unchanged.



Step 1: Analyzing the starting material.

The compound is Ethanamide (Acetamide), represented as \(CH_3CONH_2\). It contains two carbon atoms and one amide functional group.


Step 2: Determining the product of reduction.

The \(LiAlH_4\) attacks the carbonyl carbon. After the workup with water (\(H_2O\)), the oxygen atom is removed and replaced by two hydrogen atoms. \[ CH_3-C(=O)-NH_2 \xrightarrow{reduction} CH_3-CH_2-NH_2 \]


Step 3: Final Identification.

The resulting product is Ethanamine (Ethylamine).


Final Answer: The product is \(CH_3CH_2NH_2\) (Ethylamine). Quick Tip: Remember: Reducing an amide with \(LiAlH_4\) gives an amine with the same number of carbons. Reducing a nitrile also gives an amine with the same number of carbons.

Concept:
The reduction of nitro compounds to primary amines is an important industrial and laboratory reaction.

Nitroarenes (\(Ar-NO_2\)) can be reduced using metals like Iron (Fe), Tin (Sn), or Zinc (Zn) in the presence of concentrated hydrochloric acid (HCl).
Reduction with Fe/HCl is preferred in industry because only a small amount of HCl is required to initiate the reaction, as the \(FeCl_2\) formed gets hydrolyzed to release HCl.



Step 1: Analyzing the reactive functional group.

The starting material is 4-Nitrotoluene. It has a methyl group (\(-CH_3\)) and a nitro group (\(-NO_2\)) on the benzene ring. Under these acidic reducing conditions, the methyl group remains inert.


Step 2: Identifying the transformation.

The nitro group (\(-NO_2\)) is reduced to an amino group (\(-NH_2\)). The oxygen atoms are removed and replaced by hydrogen atoms.


Step 3: Naming the product.

The product is 4-Methylaniline (also known as p-Toluidine). \[ O_2N-C_6H_4-CH_3 \xrightarrow{Fe/HCl} H_2N-C_6H_4-CH_3 \]


Final Answer: The product is 4-methylaniline (p-toluidine). Quick Tip: Fe/HCl is the best choice for this reduction because the byproduct \(FeCl_2\) produces \(HCl\) upon hydrolysis, making the process highly economical.

Concept:
The boiling point of a substance is determined by the strength of the intermolecular forces holding the molecules together in the liquid phase. For amines, the most significant force is intermolecular hydrogen bonding.


Step 1: Analyzing the structure of Primary Amines.

Primary amines (\(R-NH_2\)) have two hydrogen atoms directly attached to the highly electronegative nitrogen atom. This allows for extensive intermolecular hydrogen bonding between the lone pair of one nitrogen and the hydrogen of another molecule.


Step 2: Analyzing the structure of Tertiary Amines.

Tertiary amines (\(R_3N\)) have no hydrogen atoms attached to the nitrogen atom. While they have a lone pair, they lack the "active" hydrogen required to initiate a hydrogen bond with another tertiary amine molecule.


Step 3: Comparing the Energy requirement.

Because of the strong network of hydrogen bonds in primary amines, a large amount of thermal energy is required to break these bonds to convert the liquid into vapor. Tertiary amines only experience weaker dipole-dipole interactions and Van der Waals forces.


Final Answer: Primary amines have higher boiling points because they can form intermolecular hydrogen bonds, which are absent in tertiary amines. Quick Tip: Order of Boiling Points for isomeric amines:
Primary (\(1^\circ\)) \(>\) Secondary (\(2^\circ\)) \(>\) Tertiary (\(3^\circ\)).

Concept:

Amines are classified as primary (\(1^\circ\)), secondary (\(2^\circ\)), or tertiary (\(3^\circ\)) depending on the number of carbon-containing groups attached to the nitrogen atom.
If the nitrogen atom is attached to three carbon groups, the amine is called a tertiary amine.




Step 1: Identify the groups attached to nitrogen

In N,N-dimethylnaphthalen-2-amine, the nitrogen atom is bonded to:

One naphthyl group.
Two methyl (\(\mathrm{CH_3}\)) groups.


Thus, the nitrogen atom is attached to a total of three carbon-containing groups.



Step 2: Classify the amine

Since the nitrogen atom is bonded to three alkyl/aryl groups and has no hydrogen atom attached, it is classified as a tertiary (\(3^\circ\)) amine.



Final Answer:
\[ \boxed{N,N-dimethylnaphthalen-2-amine is a Tertiary (3^\circ) amine.} \]



Quick Tip: A tertiary amine has the general formula \(R_3N\), where all three hydrogen atoms of ammonia are replaced by alkyl and/or aryl groups.

Concept:

Amines are classified according to the number of carbon-containing groups attached to the nitrogen atom.
If the nitrogen atom is attached to only one alkyl or aryl group, it is called a primary amine.




Step 1: Identify the groups attached to nitrogen

In Naphthalen-2-amine, the amino group (\(-NH_2\)) is attached to the second carbon atom of the naphthalene ring.

The nitrogen atom is bonded to:

One naphthyl group.
Two hydrogen atoms.




Step 2: Classify the amine

Since only one hydrogen atom of ammonia is replaced by a naphthyl group, the compound is classified as a primary (\(1^\circ\)) amine.



Final Answer:
\[ \boxed{\textit{Naphthalen-2-amine is a Primary \((1^\circ)\) amine.}} \]



Quick Tip: A primary amine has the general formula \(RNH_2\), where only one hydrogen atom of ammonia is replaced by an alkyl or aryl group.

Concept:

The solubility of an organic compound in water depends mainly on two factors:


The ability of the molecule to form hydrogen bonds with water molecules.
The balance between the hydrophilic (polar) functional group and the hydrophobic (non-polar) hydrocarbon chain.


Both Butan-1-amine and Butan-1-ol contain a four-carbon hydrocarbon chain, so the major difference in their solubility arises from their functional groups, namely the amino group (\(-NH_2\)) and the hydroxyl group (\(-OH\)).



Step 1: Identify the functional groups present.

\[ Butan-1-amine : CH_3CH_2CH_2CH_2NH_2 \]
\[ Butan-1-ol : CH_3CH_2CH_2CH_2OH \]

Both compounds possess one polar functional group attached to the same four-carbon alkyl chain.

The hydrocarbon chain is non-polar and reduces water solubility, whereas the functional group increases water solubility.



Step 2: Compare their hydrogen bonding with water.


Both compounds are capable of forming hydrogen bonds with water molecules.


The amino group (\(-NH_2\)) can both donate and accept hydrogen bonds.
The hydroxyl group (\(-OH\)) can also donate and accept hydrogen bonds.


Therefore, both compounds interact well with water through hydrogen bonding.



Step 3: Apply the NCERT solubility trend.


According to the NCERT Chemistry textbook, lower aliphatic amines are more soluble in water than the corresponding alcohols having approximately the same molecular mass.

Since Butan-1-amine and Butan-1-ol have nearly the same molecular mass and identical carbon chain length, their solubility is compared directly on the basis of their functional groups.

Hence,
\[ \boxed{Butan-1-amine is more soluble than Butan-1-ol.} \]



Step 4: Reason for greater solubility of Butan-1-amine.


The amino group possesses a lone pair of electrons on the nitrogen atom, allowing effective interaction with surrounding water molecules through hydrogen bonding.

Although the hydrocarbon chain decreases the overall solubility, the amino group is sufficiently polar to make Butan-1-amine more soluble than the corresponding alcohol.

Thus, the interaction between Butan-1-amine and water molecules is stronger than that of Butan-1-ol for compounds of comparable molecular mass.



Step 5: Conclusion.


Since both compounds have the same carbon chain length but the amino group provides better overall interaction with water, Butan-1-amine exhibits higher water solubility.

Therefore, among the two compounds,
\[ \boxed{Butan-1-amine is more soluble in water.} \]



Final Answer:
\[ \boxed{Butan-1-amine is more soluble in water than Butan-1-ol.} \]

This is because lower aliphatic amines are more soluble in water than the corresponding alcohols of similar molecular mass, owing to their effective hydrogen bonding with water molecules. Quick Tip: Remember the NCERT trend: \[ \boxed{Primary Aliphatic Amines > Corresponding Alcohols (similar molecular mass)} \] Both can form hydrogen bonds, but lower aliphatic amines generally exhibit greater water solubility.

Concept:
The osmotic pressure (\(\pi\)) of a solution containing an electrolyte is given by the modified van't Hoff equation: \[ \pi = iCRT \quad or \quad \pi = i \left(\frac{w}{M \cdot V}\right) RT \]
Where \(w\) is the mass of solute, \(M\) is the molar mass, \(V\) is volume in Liters, and \(i\) is the van't Hoff factor.


Step 1: Identifying given values and converting units.


\(\pi = 0.70 \, atm\)
\(V = 2.46 \, L\)
\(i = 2.59\)
\(M = 111 \, g/mol\)
\(T = 27 + 273 = 300 \, K\)
\(R = 0.082 \, L atm K^{-1} mol^{-1}\)



Step 2: Setting up the calculation for mass (\(w\)).
\[ 0.70 = 2.59 \times \frac{w}{111 \times 2.46} \times 0.082 \times 300 \]
Rearranging for \(w\): \[ w = \frac{0.70 \times 111 \times 2.46}{2.59 \times 0.082 \times 300} \]


Step 3: Calculating the final value.
\[ w = \frac{191.142}{63.714} \approx 3.0 \, g \]


Final Answer: The amount of \(CaCl_2\) dissolved is 3.0 grams. Quick Tip: Always convert temperature to Kelvin and check if the van't Hoff factor (\(i\)) is required (it is required for all salts/electrolytes).

Concept:

Osmosis is the movement of solvent molecules (generally water) from a region of lower solute concentration (higher water potential) to a region of higher solute concentration (lower water potential) through a semipermeable membrane.

When water enters a cell or any biological material by osmosis, the process is called endosmosis. This causes the cell or object to swell.



Step 1: Identify the nature of the raisin.


A raisin is a dried grape that contains a large amount of dissolved sugars and other solutes inside its cells.

Its outer skin acts as a semipermeable membrane, allowing water molecules to pass through while restricting the movement of larger solute molecules.



Step 2: Compare the concentration inside and outside the raisin.


When the raisin is placed in pure water,


The concentration of solutes inside the raisin is much higher.
The surrounding water has a very low solute concentration.
Therefore, water moves from the surrounding solution into the raisin through the semipermeable membrane.


This movement occurs naturally because water always moves from a region of higher water potential to a region of lower water potential.



Step 3: Identify the phenomenon involved.


The movement of water molecules through the semipermeable membrane into the raisin is called osmosis.

Since water enters the raisin, this particular type of osmosis is known as endosmosis.



Step 4: Explain why the raisin swells.


As more and more water enters the cells of the raisin,


The cells become turgid.
The volume of the raisin increases.
Consequently, the raisin swells and becomes larger than its original size.


Thus, the swelling of raisins is a direct result of water entering the cells by endosmosis.



Step 5: Conclusion.


The swelling of raisins in water occurs because water enters the raisin through its semipermeable membrane by the process of osmosis. Since the movement of water is inward, the process is specifically called endosmosis.



Final Answer:

The phenomenon involved is
\[ \boxed{Osmosis (specifically Endosmosis).} \] Quick Tip: Remember the difference: \[ \boxed{Endosmosis \rightarrow Water enters the cell \rightarrow Cell swells} \] \[ \boxed{Exosmosis \rightarrow Water leaves the cell \rightarrow Cell shrinks} \] A raisin placed in pure water always undergoes endosmosis because its internal solute concentration is higher than that of the surrounding water.

Concept:
Osmotic pressure (\(\Pi\)) is the pressure required to stop the flow of solvent through a semi-permeable membrane. Its advantages in measuring molar mass include:

Use of Molarity (\(M\)) instead of Molality (\(m\)).
Measurement at a constant temperature (usually room temperature).
Large magnitude of values for high molar mass solutes.



Step 1: Suitability for Biomolecules.

Measurement is carried out at room temperature. This is a significant advantage for macromolecules like proteins, which are often unstable and can undergo denaturation at the higher temperatures required for measuring "Elevation of Boiling Point."


Step 2: Measurable Magnitude for Dilute Solutions.

Since polymers and proteins have very high molar masses, their solutions are usually very dilute. For such solutions, the changes in boiling point or freezing point are too small to be measured accurately. However, the osmotic pressure is large enough to be measured with high precision even at very low concentrations. Quick Tip: Remember that osmotic pressure is the preferred method for determining the molar mass of polymers because it provides a significant, readable value at \(298\) K.

Concept:
Desalination is the process of removing salts from saline water. It relies on:

Semi-permeable membrane (SPM): A membrane that allows only solvent molecules to pass.
External Pressure: Force applied to overcome natural osmotic flow.



Step 1: Defining the process.

If a pressure larger than the osmotic pressure is applied to the solution side (sea water), the pure solvent (water) flows out of the solution through the semi-permeable membrane. This is called Reverse Osmosis.


Step 2: Practical setup.

In desalination, sea water is placed on one side of a semi-permeable membrane (often made of cellulose acetate). When high pressure is applied, pure water is pushed through the membrane into the fresh water side, leaving the concentrated salt behind. Quick Tip: In Reverse Osmosis, the solvent moves from a region of high solute concentration to low solute concentration, which is opposite to natural osmosis.

Concept:

Alcohols react with active metals to release \(H_2\) gas.
Compounds with \(CH_3CH(OH)-\) group undergo the iodoform test.
Intermolecular dehydration of alcohols at \(413\) K yields ethers.
Ethers react with concentrated HI to form alkyl iodides.



Step 1: Identifying A, B, and C.

Molecular formula \(C_2H_6O\) and reaction with Na suggests 'A' is Ethanol (\(CH_3CH_2OH\)). \[ 2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa (B) + H_2 \uparrow \]
Reaction with \(I_2/NaOH\) (Iodoform test) yields a yellow precipitate of Iodoform (C): \[ CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3 (C) + HCOONa + 5NaI + 5H_2O \]


Step 2: Identifying D and E.

Heating Ethanol with \(H_2SO_4\) at \(413\) K leads to dehydration to form Diethyl ether (D): \[ 2CH_3CH_2OH \xrightarrow[413 K]{H_2SO_4} CH_3CH_2OCH_2CH_3 (D) + H_2O \]
Reaction of ether with excess HI yields Ethyl iodide (E): \[ C_2H_5OC_2H_5 + 2HI \rightarrow 2C_2H_5I (E) + H_2O \] Quick Tip: Remember the temperature dependence: Ethanol with \(H_2SO_4\) at \(413\) K gives an ether, but at \(443\) K it gives an alkene (ethene).

Concept:

The hydroxyl group (\(-OH\)) present in phenol strongly activates the benzene ring.
It directs incoming electrophiles to the ortho and para positions.
Therefore, phenol undergoes rapid bromination even without a catalyst.




Step 1: Identify the type of reaction

The conversion involves bromination of phenol through an electrophilic substitution reaction.



Step 2: Identify the reagent

Phenol reacts with bromine water to form 2,4,6-tribromophenol as a white precipitate.
\[ \mathrm{C_6H_5OH + 3Br_2(aq) \longrightarrow C_6H_2Br_3OH + 3HBr} \]



Final Answer:
\[ \boxed{Reagent: Bromine water (Br_2/H_2O)} \]



Quick Tip: Phenol decolourises bromine water immediately and forms a white precipitate of 2,4,6-tribromophenol.

Concept:

Propan-1-ol is obtained from propene by anti-Markovnikov hydration.
This reaction is carried out through hydroboration followed by oxidation.




Step 1: Identify the type of reaction

The conversion requires the addition of water across the double bond in an anti-Markovnikov manner.



Step 2: Identify the reagent

The required reagents are:
\[ \mathrm{BH_3\cdot THF} \]

followed by
\[ \mathrm{H_2O_2/OH^-} \]

This sequence converts propene into propan-1-ol.
\[ \mathrm{CH_3CH=CH_2 \xrightarrow[\mathrm{H_2O_2/OH^-}] {\mathrm{BH_3\cdot THF}} CH_3CH_2CH_2OH} \]



Final Answer:
\[ \boxed{\mathrm{BH_3\cdot THF \; followed\; by\; H_2O_2/OH^-}} \]



Quick Tip: Hydroboration--oxidation always gives the anti-Markovnikov alcohol.

Concept:

Ketones are converted into secondary alcohols by reduction.
Reducing agents supply hydrogen to the carbonyl group.




Step 1: Identify the type of reaction

The conversion involves the reduction of a ketone into a secondary alcohol.



Step 2: Identify the reagent

The commonly used reducing reagent is sodium borohydride (\(NaBH_4\)).

Alternatively, lithium aluminium hydride (\(LiAlH_4\)) may also be used.
\[ \mathrm{CH_3COCH_2CH_3 \xrightarrow{NaBH_4} CH_3CHOHCH_2CH_3} \]



Final Answer:
\[ \boxed{\mathrm{NaBH_4}} \]
\[ (Alternatively, \boxed{\mathrm{LiAlH_4}} may also be used.) \]



Quick Tip: Ketones are reduced to secondary alcohols using reducing agents such as \(NaBH_4\) or \(LiAlH_4\).

Concept:

Acid-catalysed hydration is an electrophilic addition reaction in which a molecule of water is added across the carbon-carbon double bond (\(C=C\)) of an alkene in the presence of a dilute mineral acid such as dilute sulphuric acid (\(H_2SO_4\)).

The reaction follows Markovnikov's rule, according to which the hydrogen atom attaches to the carbon atom already having more hydrogen atoms, while the hydroxyl group (\(-OH\)) attaches to the more substituted carbon atom.

The reaction proceeds through the formation of a carbocation intermediate.



Step 1: Protonation of the alkene (Formation of carbocation).


The \(\pi\)-electrons of the alkene act as a nucleophile and attack the proton (\(H^+\)) supplied by the acid.

For example, hydration of propene occurs as follows:
\[ CH_3-CH=CH_2 + H^+ \longrightarrow CH_3-\overset{+}{CH}-CH_3 \]

A secondary carbocation is formed because it is more stable than a primary carbocation.

According to Markovnikov's rule, the proton attaches to the carbon atom having more hydrogen atoms, producing the more stable carbocation.



Step 2: Attack of water molecule.


Water acts as a nucleophile due to the presence of lone pair electrons on the oxygen atom.

The water molecule attacks the positively charged carbocation to form a protonated alcohol (oxonium ion).
\[ CH_3-\overset{+}{CH}-CH_3 + H_2O \longrightarrow CH_3-CH(OH_2^+)-CH_3 \]

This step converts the unstable carbocation into a more stable intermediate.



Step 3: Deprotonation of the oxonium ion.


The protonated alcohol loses one proton (\(H^+\)) to another water molecule.
\[ CH_3-CH(OH_2^+)-CH_3 \longrightarrow CH_3-CH(OH)-CH_3 + H^+ \]

The proton released in this step regenerates the acid catalyst.

Thus, the catalyst is not consumed during the reaction.



Step 4: Overall reaction.


The overall acid-catalysed hydration of propene can be represented as
\[ CH_3CH=CH_2 + H_2O \overset{H^+}{\longrightarrow} CH_3CH(OH)CH_3 \]

The product obtained is propan-2-ol.



Step 5: Important observations.



The reaction is an electrophilic addition reaction.
It proceeds through the formation of a carbocation intermediate.
The addition follows Markovnikov's rule.
The acid acts only as a catalyst and is regenerated at the end of the reaction.
The major product is the alcohol formed through the more stable carbocation.




Final Answer:

Acid-catalysed hydration of an alkene occurs by electrophilic addition. First, the alkene is protonated to form the most stable carbocation. Next, a water molecule attacks the carbocation to produce a protonated alcohol, which finally loses a proton to give the alcohol. The reaction follows Markovnikov's rule, and the acid catalyst is regenerated. Quick Tip: Remember the three important steps of acid-catalysed hydration: \[ \boxed{ Alkene \;\xrightarrow{H^+}\; Carbocation \;\xrightarrow{H_2O}\; Oxonium ion \;\xrightarrow{-H^+}\; Alcohol } \] Always remember: Formation of the most stable carbocation. Addition follows Markovnikov's rule. Acid catalyst (\(H^+\)) is regenerated at the end of the reaction.

Concept:

Propanone is first reduced to the corresponding secondary alcohol.
The alcohol is then dehydrated to obtain the corresponding alkene.




Step 1: Reduction of propanone

Propanone is reduced to propan-2-ol using sodium borohydride (\(NaBH_4\)) or lithium aluminium hydride (\(LiAlH_4\)).
\[ \mathrm{CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CHOHCH_3} \]



Step 2: Dehydration of propan-2-ol

Propan-2-ol is heated with concentrated sulphuric acid at about \(443\,K\) to remove one molecule of water and form propene.
\[ \mathrm{CH_3CHOHCH_3 \xrightarrow[\;443\,K\;]{Conc.\ H_2SO_4} CH_3CH=CH_2 + H_2O} \]



Final Conversion:
\[ \boxed{ \mathrm{CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CHOHCH_3 \xrightarrow[\;443\,K\;]{Conc.\ H_2SO_4} CH_3CH=CH_2} } \]



Quick Tip: Ketones are first reduced to secondary alcohols and the alcohols are then dehydrated to obtain alkenes.

Concept:

Carboxylic acids cannot be reduced directly to aldehydes.
They are first converted into acid chlorides, followed by Rosenmund reduction.




Step 1: Formation of benzoyl chloride

Benzoic acid is treated with thionyl chloride (\(SOCl_2\)) to form benzoyl chloride.
\[ \mathrm{C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl} \]



Step 2: Rosenmund reduction

Benzoyl chloride is reduced using hydrogen gas in the presence of poisoned palladium catalyst (\(Pd/BaSO_4\)).
\[ \mathrm{C_6H_5COCl \xrightarrow[\;Pd/BaSO_4\;]{H_2} C_6H_5CHO} \]



Final Conversion:
\[ \boxed{ \mathrm{C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow[\;Pd/BaSO_4\;]{H_2} C_6H_5CHO} } \]



Quick Tip: Rosenmund reduction converts acid chlorides into aldehydes using hydrogen gas and poisoned palladium catalyst.

Concept:

The acetyl group (\(-COCH_3\)) is a deactivating and meta-directing group.
Therefore, acetylation is carried out first, followed by nitration to obtain the meta product.




Step 1: Friedel--Crafts Acylation

Benzene is treated with acetyl chloride in the presence of anhydrous aluminium chloride to form acetophenone.
\[ \mathrm{C_6H_6 \xrightarrow[\;AlCl_3\;]{CH_3COCl} C_6H_5COCH_3} \]



Step 2: Nitration

Acetophenone is treated with concentrated nitric acid and concentrated sulphuric acid.

Since the acetyl group is meta-directing, the nitro group enters the meta position.
\[ \mathrm{C_6H_5COCH_3 \xrightarrow[\;Conc.\ H_2SO_4\;]{Conc.\ HNO_3} m-NO_2C_6H_4COCH_3} \]



Final Conversion:
\[ \boxed{ \mathrm{C_6H_6 \xrightarrow[\;AlCl_3\;]{CH_3COCl} C_6H_5COCH_3 \xrightarrow[\;Conc.\ H_2SO_4\;]{Conc.\ HNO_3} m-NO_2C_6H_4COCH_3} } \]



Quick Tip: Always introduce the acetyl group before nitration because the \(-COCH_3\) group is a meta-directing group. If nitration is done first, Friedel--Crafts acylation becomes difficult due to the strongly deactivating nitro group.

Concept:

(i) Reactivity towards HCN:

The addition of hydrogen cyanide (HCN) to aldehydes and ketones is a nucleophilic addition reaction. The rate of the reaction depends upon two important factors:


Electronic effect: Alkyl groups donate electrons by the +I effect, thereby decreasing the positive charge on the carbonyl carbon and reducing its susceptibility to nucleophilic attack.
Steric effect: Bulky alkyl groups hinder the approach of the nucleophile towards the carbonyl carbon, thereby decreasing the reaction rate.


Hence,


Aldehydes are generally more reactive than ketones.
Less substituted carbonyl compounds are more reactive than highly substituted ones.




Step 1: Compare the structures of the given compounds.



Acetaldehyde (\(CH_3CHO\)) is an aldehyde containing only one alkyl group.
Propanone (\(CH_3COCH_3\)) is a ketone containing two methyl groups.
Di-tert-butyl ketone contains two bulky tert-butyl groups attached to the carbonyl carbon.


The number and size of alkyl groups increase in the order
\[ CH_3CHO < CH_3COCH_3 < (CH_3)_3CCO C(CH_3)_3 \]

which means steric hindrance and electron donation also increase in the same order.



Step 2: Determine the order of reactivity towards HCN.


Since aldehydes have less steric hindrance and a more electrophilic carbonyl carbon, they react faster with HCN than ketones.

Among the ketones, di-tert-butyl ketone is the least reactive because the two bulky tert-butyl groups greatly hinder the approach of the cyanide ion.

Therefore, the increasing order of reactivity is
\[ \boxed{Di-tert-butyl ketone < Propanone < Acetaldehyde} \]



(ii) Aldol Condensation

Concept:

Aldol condensation is shown by aldehydes and ketones that contain at least one \(\alpha\)-hydrogen atom. The \(\alpha\)-hydrogen is the hydrogen atom attached to the carbon adjacent to the carbonyl group.

Compounds lacking \(\alpha\)-hydrogen cannot undergo aldol condensation.



Step 3: Examine each compound for the presence of \(\alpha\)-hydrogen.



\(HCHO\) (Formaldehyde): No carbon adjacent to the carbonyl group, therefore no \(\alpha\)-hydrogen. It does not undergo aldol condensation.
\(C_6H_5CHO\) (Benzaldehyde): The carbon adjacent to the carbonyl group is part of the benzene ring and has no hydrogen attached. Hence, it does not undergo aldol condensation.
\(CH_3CHO\) (Acetaldehyde): Contains three \(\alpha\)-hydrogen atoms on the methyl group. Therefore, it undergoes aldol condensation.
Cyclohexanone: Contains several \(\alpha\)-hydrogen atoms on the carbon atoms adjacent to the carbonyl group. Hence, it also undergoes aldol condensation.




Step 4: Identify the correct compounds.


Therefore, the compounds that undergo aldol condensation are
\[ \boxed{CH_3CHO \quad and \quad Cyclohexanone} \]



Final Answer:

(i)

Increasing order of reactivity towards HCN:
\[ \boxed{Di-tert-butyl ketone < Propanone < Acetaldehyde} \]

(ii)

The compounds undergoing aldol condensation are
\[ \boxed{CH_3CHO and Cyclohexanone.} \] Quick Tip: For nucleophilic addition reactions: \[ \boxed{Aldehydes>Ketones} \] because aldehydes have less steric hindrance and a more positively polarized carbonyl carbon. For Aldol Condensation, always remember: \[ \boxed{Presence of at least one \alpha-hydrogen is essential.} \] No \(\alpha\)-hydrogen \(\Rightarrow\) No aldol condensation. Examples: \[ \boxed{HCHO,\; C_6H_5CHO \;\; do not undergo Aldol condensation} \] \[ \boxed{CH_3CHO,\; Cyclohexanone \;\; undergo Aldol condensation} \]

Concept:

Aldol condensation is a characteristic reaction of aldehydes and ketones that possess at least one \(\alpha\)-hydrogen atom. An \(\alpha\)-hydrogen is a hydrogen atom attached to the carbon adjacent to the carbonyl (\(C=O\)) group.

In the presence of a dilute base (such as NaOH or KOH), the compound first forms an enolate ion. The enolate ion then attacks the carbonyl carbon of another molecule to produce a \(\beta\)-hydroxy aldehyde or \(\beta\)-hydroxy ketone (called an aldol). On heating, this product loses water to form an \(\alpha,\beta\)-unsaturated carbonyl compound.

Therefore,
\[ \boxed{Presence of at least one \alpha-hydrogen is essential for Aldol condensation.} \]



Step 1: Examine Formaldehyde (\(HCHO\)).


Formaldehyde has only one carbon atom.

There is no carbon atom adjacent to the carbonyl group.

Hence, it has no \(\alpha\)-carbon and consequently no \(\alpha\)-hydrogen.

Therefore,
\[ \boxed{HCHO does not undergo Aldol condensation.} \]



Step 2: Examine Benzaldehyde (\(C_6H_5CHO\)).


In benzaldehyde, the carbonyl group is directly attached to a benzene ring.

The carbon adjacent to the carbonyl carbon is part of the aromatic ring and does not possess any hydrogen atom.

Thus, benzaldehyde has no \(\alpha\)-hydrogen.

Hence,
\[ \boxed{C_6H_5CHO does not undergo Aldol condensation.} \]



Step 3: Examine Acetaldehyde (\(CH_3CHO\)).


Acetaldehyde contains a methyl group attached to the carbonyl carbon.

The methyl carbon is the \(\alpha\)-carbon and contains three \(\alpha\)-hydrogen atoms.

Therefore, it can readily form an enolate ion in the presence of a base.

Hence,
\[ \boxed{CH_3CHO undergoes Aldol condensation.} \]



Step 4: Examine Cyclohexanone.


Cyclohexanone is a cyclic ketone.

The carbon atoms adjacent to the carbonyl group contain \(\alpha\)-hydrogen atoms.

Therefore, cyclohexanone can also form an enolate ion and undergo Aldol condensation.

Hence,
\[ \boxed{Cyclohexanone undergoes Aldol condensation.} \]



Step 5: Identify the required compounds.


Among the given compounds,
\[ \begin{aligned} HCHO &\rightarrow No
C_6H_5CHO &\rightarrow No
CH_3CHO &\rightarrow Yes
Cyclohexanone &\rightarrow Yes \end{aligned} \]

Thus, only acetaldehyde and cyclohexanone possess \(\alpha\)-hydrogen atoms and undergo Aldol condensation.



Final Answer:

The compounds which undergo Aldol condensation are
\[ \boxed{CH_3CHO and Cyclohexanone.} \] Quick Tip: The easiest way to identify compounds undergoing Aldol condensation is to check for the presence of an \(\alpha\)-hydrogen. \[ \boxed{\alpha-Hydrogen Present \;\Longrightarrow\; Aldol Condensation} \] \[ \boxed{\alpha-Hydrogen Absent \;\Longrightarrow\; No Aldol Condensation} \] Examples: \[ \boxed{HCHO,\; C_6H_5CHO \;\; do not undergo Aldol condensation} \] \[ \boxed{CH_3CHO,\; Cyclohexanone \;\; undergo Aldol condensation} \]

Concept:

Compounds containing the \(\mathrm{-COCH_3}\) group (methyl ketones) give a positive Iodoform test.
Acetophenone is a methyl ketone, whereas benzophenone is not.




Step 1: Perform the Iodoform Test

Treat both compounds with iodine (\(I_2\)) and sodium hydroxide (\(NaOH\)) solution.



Step 2: Observe the result


Acetophenone contains the \(\mathrm{-COCH_3}\) group and gives a yellow precipitate of iodoform (\(\mathrm{CHI_3}\)).
Benzophenone does not contain the methyl ketone group and therefore does not give the iodoform test.




Quick Tip: Only methyl ketones containing the \(\mathrm{-COCH_3}\) group give the iodoform test.

Concept:

Aldehydes are easily oxidised whereas ketones are resistant to mild oxidising agents.
Tollens' reagent is commonly used to distinguish aldehydes from ketones.




Step 1: Perform Tollens' Test

Add freshly prepared Tollens' reagent to both compounds and warm gently.



Step 2: Observe the result


Propanal is an aldehyde and reduces Tollens' reagent to metallic silver, producing a silver mirror.
Propanone is a ketone and does not react with Tollens' reagent.




Reaction:
\[ \mathrm{RCHO + 2[Ag(NH_3)_2]^+ +3OH^- \rightarrow RCOO^- +2Ag +4NH_3 +2H_2O} \]



Quick Tip: Tollens' reagent is known as the Silver Mirror Reagent because aldehydes produce a shining silver mirror on the inner wall of the test tube.

Concept:

Pentan-2-one is a methyl ketone because it contains the \(\mathrm{-COCH_3}\) group.
Pentan-3-one is not a methyl ketone.
Methyl ketones give a positive Iodoform test.




Step 1: Perform the Iodoform Test

Treat both compounds with iodine (\(I_2\)) and sodium hydroxide (\(NaOH\)) solution.



Step 2: Observe the result

Pentan-2-one:

Pentan-2-one contains the methyl ketone group \(\mathrm{(-COCH_3)}\). Therefore, it gives a positive Iodoform test and forms a yellow precipitate of iodoform (\(\mathrm{CHI_3}\)).



Pentan-3-one:

Pentan-3-one does not contain the methyl ketone group. Hence, it does not give the Iodoform test and no yellow precipitate is formed.



Final Answer:

The compounds can be distinguished by the Iodoform Test using iodine and sodium hydroxide.


Pentan-2-one gives a yellow precipitate of iodoform (\(\mathrm{CHI_3}\)).
Pentan-3-one does not give any yellow precipitate.




Quick Tip: Among ketones, only methyl ketones containing the \(\mathrm{-COCH_3}\) group give the Iodoform test.

Concept:

The strength of a carboxylic acid depends upon the ease with which it loses a proton (\(H^+\)).

A carboxylic acid is stronger if the conjugate base (carboxylate ion) formed after loss of proton is more stable.

Electron-withdrawing groups increase acid strength, whereas electron-releasing groups decrease acid strength.



Step 1: Compare the substituent groups.

\[ CH_2FCOOH \]

contains a fluorine atom.
\[ CH_3COOH \]

contains a methyl group.



Step 2: Study the electronic effect.


Fluorine is highly electronegative.

It exerts a strong
\[ \boxed{-I (electron-withdrawing) effect} \]

which pulls electron density away from the carboxyl group.

The methyl group exerts
\[ \boxed{+I (electron-donating) effect} \]

which pushes electrons toward the carboxyl group.



Step 3: Effect on conjugate base stability.


After losing a proton,
\[ CH_2FCOOH \]

forms a carboxylate ion whose negative charge is stabilized by the fluorine atom.

Greater stabilization means easier proton loss and hence stronger acidity.

In acetic acid, the methyl group destabilizes the carboxylate ion by donating electron density.



Final Answer
\[ \boxed{CH_2FCOOH>CH_3COOH} \]

because fluorine shows a strong electron-withdrawing (\(-I\)) effect, stabilizes the carboxylate ion, and increases the acidic strength. Quick Tip: Greater the electron-withdrawing (\(-I\)) effect, \[ \boxed{Greater the acidity} \] Greater the electron-donating (\(+I\)) effect, \[ \boxed{Lower the acidity} \]

Concept:

The boiling point of an organic compound depends mainly upon the strength of intermolecular forces.

The stronger the intermolecular forces, the higher is the boiling point.

The order of intermolecular forces is
\[ \boxed{Hydrogen bonding>Dipole-Dipole>London Dispersion} \]



Step 1: Identify the dominant intermolecular force in each compound.

 

Concept:

Molar conductivity: \(\Lambda_m = \frac{\kappa \times 1000}{C}\)
Degree of dissociation: \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\)
Current flows in the direction opposite to electron flow.



Step 1: Calculating Conductivity and Dissociation.

Molar Conductivity (\(\Lambda_m\)): \[ \Lambda_m = \frac{1.06 \times 10^{-2} \times 1000}{0.1} = 106 S cm^2 mol^{-1} \]
Limiting Molar Conductivity (\(\Lambda_m^\circ\)): \[ \Lambda_m^\circ = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} = 50.1 + 76.5 = 126.6 S cm^2 mol^{-1} \]
Degree of dissociation (\(\alpha\)): \[ \alpha = \frac{106}{126.6} \approx 0.837 \]


Step 2: Cell Logic and Batteries.

(b)(i) In the reaction, \(Zn\) is oxidized (anode) and \(Ag^+\) is reduced (cathode). Electrons flow from \(Zn \rightarrow Ag\). Thus, **current flows from Silver (Ag) to Zinc (Zn)**.
(b)(ii) **Primary batteries** cannot be recharged (reaction occurs once), e.g., Dry cell. **Secondary batteries** can be recharged by passing current, e.g., Lead storage battery. Quick Tip: Always check the units of conductivity (\(\kappa\)). If it's in S \(cm^{-1}\), use the factor of \(1000\) in the numerator to get \(\Lambda_m\) in S \(cm^2\) \(mol^{-1}\).

Concept:

A galvanic (voltaic) cell converts chemical energy into electrical energy through a spontaneous redox reaction.

In a galvanic cell:


Oxidation takes place at the anode.
Reduction takes place at the cathode.
Electrons always flow from the anode to the cathode through the external circuit.
Conventional current flows in the opposite direction to the flow of electrons.




Step 1: Identify oxidation and reduction half-reactions.


Given reaction:
\[ 2Ag^+ + Zn \longrightarrow 2Ag + Zn^{2+} \]

The oxidation half-reaction is
\[ Zn \longrightarrow Zn^{2+}+2e^- \]

Hence,
\[ \boxed{Zinc acts as the anode.} \]

The reduction half-reaction is
\[ 2Ag^+ +2e^- \longrightarrow 2Ag \]

Therefore,
\[ \boxed{Silver acts as the cathode.} \]



Step 2: Determine the direction of electron flow.


Since oxidation occurs at the zinc electrode, electrons are released there.

These electrons travel through the external wire towards the silver electrode where reduction takes place.

Hence,
\[ \boxed{Electrons flow from Zn electrode to Ag electrode.} \]



Step 3: Determine the direction of conventional current.


Conventional current always flows opposite to the direction of electron flow.

Therefore,
\[ \boxed{Current flows from Ag electrode to Zn electrode through the external circuit.} \]



Step 4: Role of the positive cell potential.


The given cell potential is
\[ E^\circ_{cell}=+1.56~V \]

A positive value of \(E^\circ_{cell}\) indicates that the reaction is spontaneous.

Hence, the direction of current predicted above is correct.



Final Answer:
\[ \boxed{Electrons flow from Zn (Anode) to Ag (Cathode).} \]
\[ \boxed{Current flows from Ag (Cathode) to Zn (Anode).} \] Quick Tip:

Always remember:

\[ \boxed{\text{Electrons: Anode} \rightarrow \text{Cathode}} \]

\[ \boxed{\text{Current: Cathode} \rightarrow \text{Anode}} \]

In every galvanic cell,

\[ \boxed{\text{Oxidation at Anode, Reduction at Cathode}} \]

Concept:

Batteries are electrochemical cells that convert chemical energy into electrical energy.

Based on the reversibility of the chemical reaction, batteries are classified into two types:


Primary batteries
Secondary batteries




Step 1: Primary Battery

A primary battery contains chemical reactions that are irreversible. Once the reactants are completely consumed, the battery becomes exhausted and cannot be recharged. These batteries are meant for single use only.

Examples include the Dry Cell and the Mercury Cell.



Step 2: Secondary Battery

A secondary battery contains chemical reactions that are reversible. After discharge, the original chemicals can be regenerated by passing an external electric current through the battery. Therefore, these batteries can be recharged and used repeatedly.

Examples include the Lead-acid Battery and the Lithium-ion Battery.



Step 3: Differentiate between the two batteries


A primary battery undergoes an irreversible chemical reaction, whereas a secondary battery undergoes a reversible chemical reaction.

A primary battery cannot be recharged after use, whereas a secondary battery can be recharged many times.

A primary battery is generally used only once, whereas a secondary battery is suitable for repeated use.

The initial cost of a primary battery is comparatively lower, whereas a secondary battery has a higher initial cost but is economical in the long run because it can be reused.

Examples of primary batteries are Dry Cell and Mercury Cell, whereas examples of secondary batteries are Lead-acid Battery and Lithium-ion Battery.




Final Answer:

Primary batteries undergo irreversible chemical reactions and cannot be recharged after use, whereas secondary batteries undergo reversible chemical reactions and can be recharged and reused multiple times. Quick Tip: Easy Trick to Remember: \[ \boxed{Primary Battery=Use Once} \] \[ \boxed{Secondary Battery=Recharge and Reuse} \] Examples: Primary: Dry Cell, Mercury Cell Secondary: Lead-acid Battery, Lithium-ion Battery

Concept:

Cell constant (\(G^*\)) is independent of the electrolyte. \(G^* = \kappa \times R\).
Mercury cell potential is constant because its net reaction involves no ions in solution.



Step 1: Conductivity Calculations.

Find Cell Constant (\(G^*\)) using \(0.1\) M solution: \[ G^* = \kappa \times R = 1.29 \times 10^{-2} S cm^{-1} \times 100 \Omega = 1.29 cm^{-1} \]
Calculate \(\kappa\) for \(0.01\) M solution: \[ \kappa = \frac{G^*}{R} = \frac{1.29}{300} = 4.3 \times 10^{-3} S cm^{-1} \]
Calculate \(\Lambda_m\) for \(0.01\) M solution: \[ \Lambda_m = \frac{4.3 \times 10^{-3} \times 1000}{0.01} = 430 S cm^2 mol^{-1} \]


Step 2: Fuel Cells and Mercury Cells.

(b)(i) **Advantages:** They have high efficiency (\(\approx 70%\)) and are eco-friendly (water is the only byproduct).
(b)(ii) The **mercury cell** potential remains constant throughout its life because the overall cell reaction does not involve any ions in the solution whose concentration can change during operation. Quick Tip: The cell constant (\(l/A\)) is a geometric property of the cell. Once calculated for one concentration, it can be used for any other concentration in the same cell.

Concept:

A hydrogen-oxygen fuel cell is an electrochemical cell that converts the chemical energy of hydrogen and oxygen directly into electrical energy. Unlike conventional batteries, the reactants are supplied continuously from outside, allowing the cell to produce electricity as long as fuel is available.

The overall reaction is
\[ 2H_2(g)+O_2(g)\longrightarrow 2H_2O(l) \]



Step 1: High efficiency.


Hydrogen-oxygen fuel cells convert chemical energy directly into electrical energy without involving combustion.

As a result, energy losses are very low, making them much more efficient than conventional thermal power plants.



Step 2: Environment friendly.


The only product formed during the operation of the fuel cell is pure water.

Since no harmful gases such as carbon dioxide, sulphur dioxide or nitrogen oxides are produced, the fuel cell is considered a clean and pollution-free source of energy.



Final Answer:

Any two advantages of the \(H_2-O_2\) fuel cell are:


It has high energy conversion efficiency.
It is eco-friendly because water is the only by-product and no harmful pollutants are produced. Quick Tip: Remember the advantages of the hydrogen-oxygen fuel cell: \[ \boxed{High Efficiency} \] \[ \boxed{Non-polluting (Only H_2O is formed)} \] Hence, hydrogen fuel cells are widely used in spacecraft and clean energy technologies.

Concept:

The cell potential of an electrochemical cell depends upon the concentration of the reacting species, as given by the Nernst equation.

If the concentrations of the reactants and products remain unchanged during the operation of the cell, the cell potential remains constant.



Step 1: Reaction occurring in the mercury cell.


In a mercury cell, the overall cell reaction is
\[ Zn(Hg)+HgO(s)\longrightarrow ZnO(s)+Hg(l) \]

The electrolyte used is a paste of potassium hydroxide (\(KOH\)).



Step 2: Reason for constant cell potential.


During the discharge of the mercury cell,


Zinc is converted into zinc oxide.
Mercury oxide is converted into mercury.
The concentration of the electrolyte (\(KOH\)) remains practically unchanged.


Since the concentrations of the reacting ions do not change appreciably, the cell potential remains almost constant throughout the life of the cell.



Step 3: Importance of constant voltage.


A constant cell potential ensures a steady supply of electrical energy.

Therefore, mercury cells were widely used in electronic devices such as hearing aids, watches and calculators where a stable voltage is required.



Final Answer:

The cell potential of a mercury cell remains constant throughout its life because the concentration of the electrolyte does not change during discharge. As a result, according to the Nernst equation, the cell potential remains nearly constant. Quick Tip: Mercury Cell Trick: \[ \boxed{No change in electrolyte concentration} \] \[ \Longrightarrow \] \[ \boxed{Constant Cell Potential} \] This is the main reason why mercury cells provide a nearly uniform voltage throughout their working life.