CBSE Class 12 Chemistry Question Paper 2024 (Set 3 - 56/5/3) with Answer Key

When \( K_2SO_4 \) dissociates completely in water, it dissociates as follows: \[ K_2SO_4 \rightarrow 2 K^+ + SO_4^{2-}. \]
Since it dissociates into 3 ions (2 potassium ions and 1 sulfate ion), the Van't Hoff factor (\( i \)) is 3. Hence, the correct answer is (B) 3. Quick Tip: The Van't Hoff factor (\( i \)) represents the number of particles into which a solute dissociates. For ionic compounds like \( K_2SO_4 \), \( i \) equals the number of ions produced upon complete dissociation.

When a solution is diluted, the conductivity (which is the ability of the solution to conduct electricity) typically decreases because the ions are spread out in a larger volume, leading to fewer collisions between the ions and less efficient movement of charge.

However, molar conductivity (which is the conductivity per mole of ions) increases with dilution because as the ions are spread out, they experience less ion pairing, allowing for greater mobility and, therefore, a higher molar conductivity.

This behavior is described by the relation: \[ \Lambda_m = \frac{\kappa}{c} \]
where \( \Lambda_m \) is the molar conductivity, \( \kappa \) is the conductivity, and \( c \) is the concentration of the solution.

Thus, conductivity decreases, while molar conductivity increases as the solution is diluted. Quick Tip: Dilution results in the decrease of conductivity due to fewer ions per unit volume. However, molar conductivity increases because ions move more freely when diluted.

The basicity of a species is determined by its ability to donate electrons (i.e., its nucleophilicity and its tendency to accept a proton). The stronger the conjugate base, the stronger the base. Here’s the analysis of the species:

1. \(OH-\): Hydroxide is a strong base, but it is less basic than alkoxides because it is the conjugate base of water, which is a relatively weak acid.

2. \( C_6H_5O^- \): Phenoxide ion is a weaker base compared to alkoxides because the negative charge on the oxygen is delocalized into the aromatic ring, stabilizing the anion but decreasing the availability of the electron pair for protonation.

3. \(RO-\): Alkoxide ions (such as methoxide, \( CH_3O^- \)) are stronger bases than hydroxide ions because the negative charge is localized on oxygen, and alkyl groups are electron-donating, increasing the basicity.

4. \( O_2N^-\): Nitro group has a strong electron-withdrawing effect, making this species a very weak base.

Thus, \(RO-\) (alkoxide) is the strongest base among these species. Quick Tip: Alkoxide ions are stronger bases than hydroxide and phenoxide due to the electron-donating effect of alkyl groups, which increases the negative charge on oxygen, making it more likely to accept a proton.

A chiral center is a carbon atom that is attached to four different substituents. Let’s analyze the molecules:

1. Molecule I: The carbon marked with an asterisk is attached to \(H\), \(CH_3\), \(Br\), and \(Cl\), all of which are different. Thus, it is chiral.

2. Molecule II: The carbon marked with an asterisk is attached to \(Cl\), \(Br\), and two \(Cl\) groups. Since two substituents are not identical (\(Cl\)), this carbon is chiral.

3. Molecule III: The carbon marked with an asterisk is attached to \(HO\), \(CH_3\), and \(CH_2CH_3\), all of which are different. Thus, it is chiral.

4. Molecule IV: The carbon marked with an asterisk is attached to \(D\), \(CH_3\), and two \(CH_3\) groups. Since two substituents are identical (\(CH_3\)), this carbon is not chiral.

Hence, the chiral molecules are I, II and III. Quick Tip: A chiral carbon must be attached to four different groups. If two or more groups are identical, the carbon is not chiral.

The rate law equation is given by: \[ Rate = k[A]^n. \]
When the concentration of the reactant increases 4 times, the rate increases 16 times. Using the relation: \[ \frac{Rate_2}{Rate_1} = \left( \frac{[A_2]}{[A_1]} \right)^n, \]
we substitute the known values: \[ 16 = 4^n. \]
Taking the logarithm of both sides: \[ \log 16 = n \log 4. \]
Since \( \log 16 = 1.2041 \) and \( \log 4 = 0.6021 \), solving for \( n \): \[ n = \frac{1.2041}{0.6021} = 2.0. \]

Therefore, the order of the reaction is \( 2.0 \). Quick Tip: The order of a reaction can be determined experimentally by measuring how the rate of reaction changes with the concentration of reactants.

The correct answer is Mercury cell. Mercury cells are used in hearing aids because they are compact and provide a steady and reliable voltage output for long periods of time. These cells are small, which makes them ideal for use in hearing aids. Quick Tip: Mercury cells are often used in small electronic devices, such as hearing aids, due to their high energy density and stable voltage output.

Isotonic solutions are solutions that have the same osmotic pressure. Osmotic pressure is the pressure exerted by a solvent when it moves through a semipermeable membrane to balance the concentrations of solute on both sides of the membrane. Isotonic solutions are crucial for maintaining the equilibrium in biological systems, especially for red blood cells.

For two solutions to be isotonic, the concentration of solute particles (ions, molecules) in the two solutions must be the same. This ensures that there is no net movement of water across the membrane, preventing cell shrinkage (crenation) or bursting (lysis).

For example, an isotonic saline solution used in medical treatments has the same osmotic pressure as the human blood plasma, ensuring that the cells do not undergo osmotic stress. Quick Tip: Osmotic pressure is the key property that defines isotonic solutions. This ensures that cells retain their shape by maintaining equal concentrations of solutes inside and outside the cell.

Interstitial compounds are formed when small atoms, such as carbon, hydrogen, or nitrogen, occupy the interstitial spaces between the larger metal atoms in the crystal lattice. This creates a more rigid structure as the smaller atoms obstruct the movement of the metal atoms.

For transition metals, which have relatively large atomic radii, the presence of smaller atoms in the interstitial sites restricts the mobility of the metal atoms, thus increasing the hardness of the metal. For example, the formation of steel (iron-carbon alloy) results in a much harder material than pure iron due to the carbon atoms occupying the interstitial spaces in the iron lattice.

The formation of interstitial compounds typically leads to increased hardness, but the material may become less ductile and more brittle, which makes it harder, not softer. Quick Tip: Interstitial compounds restrict the movement of metal atoms, increasing hardness but often decreasing ductility. This is why steel is harder than pure iron due to the carbon atoms filling interstitial spaces.

The auto-oxidation of chloroform \((CHCl_3)\) in the presence of air and light leads to the formation of phosgene \((COCl_2)\), a highly toxic gas. Phosgene is produced when chloroform is exposed to oxygen and ultraviolet light, which breaks down the chloroform and produces phosgene along with other by-products such as hydrogen chloride (HCl).

- Phosgene is used as a chemical warfare agent due to its toxicity, and it can cause severe damage to the lungs when inhaled. Quick Tip: Phosgene is a toxic gas and must be handled carefully, especially in the presence of chloroform and light. The reaction is a classic example of the dangers of auto-oxidation.

In acid-catalyzed hydration of alkenes, the reaction follows Markovnikov’s rule, which states that the proton \((H^+)\) adds to the carbon atom of the double bond that has the greater number of hydrogen atoms, while the hydroxyl group (OH) adds to the carbon with fewer hydrogen atoms.

For the given options:

- (D) \( (CH_3)_2 C = CH_2 \) is tert-butene, and when it undergoes acid-catalyzed hydration, the addition of the proton to the carbon in the double bond produces a tertiary carbocation (\( (CH_3)_3 C^+ \)), which is stable. The hydroxyl group then adds to this carbon, forming tertiary butyl alcohol.

- The other alkenes (A, B, C) will not form a tertiary carbocation, leading to a different set of products. Quick Tip: In acid-catalyzed hydration, Markovnikov’s rule governs the addition of \(H^+\) and \(OH^-\). Tertiary carbocations are the most stable and thus favored during the reaction.

The given reaction is a Sandmeyer’s reaction, which is used to prepare aryl halides and other aryl derivatives from diazonium salts. In this specific case, the aryl diazonium salt (\(Ar-N_2^+\)) reacts with copper (Cu) in the presence of potassium cyanide (KCN) to form an aryl cyanide (Ar-CN) along with the evolution of nitrogen gas \((N_2)\). This reaction is widely used to introduce various substituents onto aromatic rings, such as halides, cyanides, and hydroxyl groups.

- Reaction Mechanism: In this reaction, the diazonium group is first formed by the reaction of an arylamine with nitrous acid. The copper(I) chloride or copper(I) cyanide facilitates the substitution of the diazonium group by the cyanide ion (\(CN^-\)) to form the aryl cyanide. Quick Tip: Sandmeyer’s reaction is an important method in synthetic organic chemistry for replacing the diazonium group with various substituents, such as \(Cl\), \(CN\), and \(OH\), using different reagents like CuCl, CuCN, etc.

The primary structure of a protein refers to the linear sequence of amino acids in the polypeptide chain. This sequence is encoded by the genetic material (DNA) and is unique for each protein. The amino acids are connected by peptide bonds, which are covalent bonds formed through dehydration reactions between the amino group of one amino acid and the carboxyl group of another.

The sequence of amino acids in the primary structure determines the overall shape of the protein and, consequently, its function. Even a small change (mutation) in the amino acid sequence can drastically alter the function of a protein, leading to diseases such as sickle-cell anemia. Quick Tip: The primary structure is the foundation for all higher-level protein structures (secondary, tertiary, and quaternary). Any change in the amino acid sequence can potentially change the protein's function.

- Assertion (A) is true: In a zero-order reaction, the rate of reaction is independent of the concentration of reactants. The rate constant and rate of the reaction share the same unit of \( mol L^{-1} s^{-1} \).

- Reason (R) is true: In a zero-order reaction, the rate does not depend on the concentration of the reactant, meaning the rate is constant, and the rate law is written as:
\[ Rate = k[A]^0 = k \]
Thus, the rate of reaction is independent of the concentration, confirming the reason. Quick Tip: In a zero-order reaction, the rate remains constant regardless of the concentration of reactants, and the units of the rate constant and the rate of reaction are the same.

- Assertion (A) is true: Zirconium (Zr) and Hafnium (Hf) are in the same group of the periodic table, and their atomic radii are almost the same.

- Reason (R) is also true: The reason for this similarity is the Lanthanoid contraction, a phenomenon where the atomic radii of the lanthanoids decrease due to poor shielding of the nuclear charge by the f-electrons. As a result, elements following the lanthanoids (like Zr and Hf) experience a similar effective nuclear charge, leading to nearly the same atomic radii. Quick Tip: Lanthanoid contraction refers to the gradual decrease in size of the elements in the lanthanide series, affecting the elements in the same period after them, such as Zr and Hf.

- Assertion (A) is false: p-Nitrophenol is actually a stronger acid than p-methoxyphenol. The nitro group is an electron-withdrawing group, which stabilizes the negative charge on the conjugate base (phenoxide ion) more effectively, making it a stronger acid.

- Reason (R) is true: The methoxy group exhibits a +I effect (electron-donating), which increases the electron density on the aromatic ring and makes the phenol less acidic, whereas the nitro group exhibits a -I effect (electron-withdrawing), which makes the phenol more acidic. Quick Tip: Electron-withdrawing groups like the nitro group stabilize the conjugate base, making the compound more acidic, whereas electron-donating groups like methoxy make the compound less acidic.

Inversion of configuration occurs during an S\(_{N}\)2 reaction due to the backside attack of the nucleophile, which leads to the inversion of the leaving group. This is true for 1-bromobutane when it undergoes hydrolysis.

However, the reason given in the statement (R) is incorrect. The S\(_{N}\)2 mechanism involves a backside attack and results in the inversion of configuration. While a transition state is indeed formed, this alone doesn't explain the inversion of configuration; rather, it's the specific nature of the S\(_{N}\)2 mechanism (backside attack) that causes the inversion.

Therefore, Assertion (A) is true, but Reason (R) is false. Quick Tip: In an S\(_{N}\)2 reaction, the nucleophile attacks from the opposite side of the leaving group, leading to the inversion of configuration. A transition state is formed, but this doesn't directly explain the inversion.

(i) Nitrobenzene to Aniline:
To convert nitrobenzene to aniline, a reduction reaction is required. This can be achieved by reducing the nitro group (-NO\(_2\)) using reducing agents such as iron (Fe) and hydrochloric acid (HCl) or catalytic hydrogenation: \[ C_6H_5NO_2 + 3H_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \]
This process reduces the nitro group \((-NO_2)\) to an amino group \((-NH_2)\), resulting in the formation of aniline.


(ii) Aniline to Phenol:
To convert aniline \((C_6H_5NH_2)\) to phenol \((C_6H_5OH)\), the amino group must be oxidized to a hydroxyl group. This can be done using an oxidizing agent such as bromine in the presence of water: \[ C_6H_5NH_2 + Br_2 \xrightarrow{H_2O} C_6H_5OH + 2HBr \]
The reaction replaces the amino group \((-NH_2)\) with a hydroxyl group (-OH), forming phenol. Quick Tip: Reduction of nitrobenzene to aniline requires an electron-donating reduction, while the conversion of aniline to phenol involves oxidation using bromine in water.

To distinguish Dimethylamine \((CH_3NH_2)\) from Ethanamine \((C_2H_5NH_2)\), we can use the Nitrous acid test.

- Test for Dimethylamine:
When Dimethylamine is treated with nitrous acid \((HNO_2)\), it forms a blue solution due to the formation of a methyl diazonium salt. The reaction is as follows:
\[ CH_3NH_2 + HNO_2 \to CH_3N_2^+ + H_2O \]
The blue color indicates the presence of Dimethylamine.

- Test for Ethanamine:
When Ethanamine is treated with nitrous acid, it forms ethanol and nitrogen gas, and there is no color formation:
\[ C_2H_5NH_2 + HNO_2 \to C_2H_5OH + N_2 + H_2O \]
Hence, Ethanamine does not produce a color change.

Thus, Dimethylamine gives a blue color with nitrous acid, while Ethanamine does not. Quick Tip: Dimethylamine forms a blue-colored solution with nitrous acid, distinguishing it from Ethanamine, which produces no such color.

When benzene diazonium chloride \((C_6H_5N_2^+Cl^-)\) is treated with potassium iodide (KI), the diazonium group undergoes a nucleophilic substitution reaction with the iodide ion \((I^-)\). The product formed is iodobenzene \((C_6H_5I)\). The reaction is as follows:
\[ C_6H_5N_2^+Cl^- + KI \to C_6H_5I + N_2 + KCl \]


This reaction is an example of a diazonium substitution reaction, where the diazonium ion is replaced by an iodine atom, forming iodobenzene. Quick Tip: Benzene diazonium chloride reacts with potassium iodide to form iodobenzene, which is a useful halogenation reaction in organic synthesis.

It is defined as the conductivity of a solution containing one mole of electrolyte dissolved in a large volume of solvent. It is represented by the symbol \(\Lambda^\circ\) and can be determined experimentally by measuring the conductivity of the solution at infinite dilution. It is calculated using the formula: \[ \Lambda^\circ = \frac{\kappa}{c} \]
where \(\kappa\) is the conductivity of the solution and \(c\) is the concentration of the solution.



(b) Fuel cell
A fuel cell is an electrochemical cell that generates electrical energy by using a fuel (such as hydrogen) and an oxidant (such as oxygen) through a redox reaction. The most common type of fuel cell is the hydrogen fuel cell, where hydrogen gas reacts with oxygen to form water and produce electricity. The general reaction is: \[ 2H_2 + O_2 \rightarrow 2H_2O + electricity \]
Fuel cells are used in a variety of applications, including portable power supplies and electric vehicles. Quick Tip: Limiting molar conductivity refers to the maximum conductivity of an electrolyte at infinite dilution, whereas a fuel cell is a clean energy source that converts chemical energy directly into electrical energy.

- Monosaccharides:

- Fructose and Glucose are monosaccharides, meaning they consist of a single sugar unit and cannot be hydrolyzed into simpler sugars. Both are important sources of energy in the body and are involved in the synthesis of larger carbohydrates.

- Disaccharides:

- Lactose and Surcose are disaccharides, consisting of two monosaccharide units linked by a glycosidic bond. Lactose is composed of one molecule of glucose and one of galactose, while maltose is made up of two glucose molecules. Quick Tip: Monosaccharides are simple sugars like glucose and fructose, while disaccharides are made of two monosaccharide units, such as surcose (glucose-glucose) and lactose (glucose-fructose).

Step 1: Calculate the Cell Constant
The cell constant (G*) is given by: \[ Cell constant = Conductivity \times Resistance \]
Substituting values for 0.2 mol L\(^{-1}\) KCl solution: \[ G^* = 0.0248 \times 200 = 4.96 \, cm^{-1} \]

Step 2: Calculate Conductivity of 0.05 mol L\(^{-1}\) KCl Solution
Using the relation: \[ Conductivity (\kappa) = \frac{Cell Constant (G*)}{Resistance (R)} \] \[ \kappa = \frac{4.96}{620} = 0.008 \, S cm^{-1} \]

Step 3: Calculate Molar Conductivity
Molar conductivity \( \Lambda_m \) is given by: \[ \Lambda_m = \kappa \times \frac{1000}{C} \]
where \( C = 0.05 \) mol L\(^{-1}\): \[ \Lambda_m = 0.008 \times \frac{1000}{0.05} = 160 \, S cm^2 mol^{-1} \] Quick Tip: The cell constant remains the same for different solutions in the same conductivity cell. Conductivity decreases with dilution, but molar conductivity increases as the ions move more freely.

For a first order reaction, the integrated rate law is:
\[ t = \frac{2.303}{k} \log \left(\frac{[A]_0}{[A]}\right) \]

Where:

- \( t \) = time taken for the reaction (in this case, 40 min)

- \( [A]_0 \) = initial concentration

- \( [A] \) = concentration after time \( t \)

- \( k \) = rate constant


Since 75% decomposition occurs, 25% of the reactant remains, i.e., \( [A] = 25% \) and \( [A]_0 = 100% \).

Substituting the values:
\[ 40 \, min = \frac{2.303}{k} \log \left(\frac{100}{25}\right) \]

Now calculate \( \log (100/25) \):
\[ \log \left(\frac{100}{25}\right) = \log (4) = 0.60 \]

Now substitute into the equation:
\[ 40 \, min = \frac{2.303}{k} \times 0.60 \]

Solve for \( k \):
\[ k = \frac{2.303 \times 0.60}{40} = 0.0345 \, min^{-1} \] Quick Tip: For first order reactions, the rate constant can be calculated using the integrated rate law. Always use the correct logarithmic values and units for accurate calculations.

The given molecular formula \( C_3H_9N \) corresponds to an amine. The reaction with benzene sulfonyl chloride (\( C_6H_5SO_2Cl \)) leads to the formation of a sulfonamide derivative, which is insoluble in alkali.
The reaction follows:
\[ C_2H_5-NH-CH_3 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2N(C_2H_5)(CH_3) + HCl \]

- Here, \( X \) is identified as N-Ethylmethylamine.




- The IUPAC name of the product is N-Ethyl-N-methylbenzenesulfonamide.





- The sulfonamide formed is insoluble in alkali due to the absence of an acidic hydrogen. Quick Tip: Sulfonamides are generally \textbf{insoluble in alkali} due to the strong electron-withdrawing nature of the sulfonyl group, making the nitrogen non-basic.

Haloalkanes undergo a nucleophilic substitution reaction with silver cyanide (AgCN) to form isocyanides (also known as isonitriles). The cyanide ion (CN\(^-\)) acts as a nucleophile and attacks the carbon atom of the halide, displacing the halide ion. This reaction follows the \(S_N2\) mechanism, where the nucleophile directly attacks the electrophilic carbon from the opposite side of the leaving group.
\[ R-X + AgCN \rightarrow R-NC + AgX \]

Here, X is a halide ion such as Cl\(^-\), Br\(^-\), or I\(^-\), and the product formed is an isocyanide. Quick Tip: In nucleophilic substitution reactions, the cyanide ion (CN\(^-\)) can act as a strong nucleophile and react with alkyl halides to form isocyanides. The \(S_N2\) mechanism involves a backside attack by the nucleophile, displacing the leaving group.

Allyl chloride undergoes \(S_N1\) reactions readily because the resulting allyl carbocation is highly stabilized by resonance. When the chloride ion leaves, the positive charge formed on the carbon is delocalized onto the adjacent carbon atoms, stabilizing the carbocation intermediate. This resonance stabilization makes the formation of the carbocation energetically favorable, thus promoting the \(S_N1\) reaction mechanism. Quick Tip: In \(S_N1\) reactions, the formation of a stable carbocation intermediate is essential. Allyl chloride is highly reactive in \(S_N1\) reactions due to the resonance stabilization of the allyl carbocation formed after the chloride ion leaves.

Haloarenes, where a halogen (e.g., chlorine, bromine, or iodine) is bonded to a carbon atom of an aromatic ring, show very low reactivity in nucleophilic substitution reactions. This is because the halogen's lone pair of electrons participates in resonance with the aromatic ring, delocalizing the electron density and making the carbon-halo bond more stable. This resonance effect makes the carbon-halogen bond less susceptible to nucleophilic attack, thus hindering the substitution reaction. The halogen is less likely to leave due to the strong resonance stabilization of the ring. Quick Tip: In haloarenes, the lone pair on the halogen participates in resonance with the aromatic ring, stabilizing the bond and making the carbon-halo bond less reactive toward nucleophilic substitution. This resonance effect lowers the reactivity of haloarenes in nucleophilic substitution reactions.

(a) The compound \( [Co(NH_3)_4Cl(NO_2)]Cl \) is named as \textit{Tetraamminchloridonitrito-N-cobalt (III) chloride. In this compound, the central metal ion is cobalt (III), which is coordinated with four ammonia (NH\(_3\)) molecules, one chloride ion (Cl\(^-\)), and one nitrito (NO\(_2\)) ligand. The complex is balanced with the chloride ion as the counter ion. The ligand names are written with their respective prefixes (tetra- for four and chloro- for chloride). The nitrito ligand is described as nitrito-N because it is bound through the nitrogen atom.



(b) The compound \( [Ni(NH_3)_6]Cl_2 \) is named as \textit{Hexaamminenickel(II) chloride. In this case, nickel is in the +2 oxidation state, and it is coordinated with six ammonia molecules. The ligand ammonia is denoted as "ammine" (not ammonium), and the coordination number of nickel is six. Two chloride ions act as counterions to balance the charge of the complex cation.



(c) The compound \( K_3[Cr(C_2O_4)_3] \) is named as \textit{Potassium trioxalatochromate (III). The chromium in this complex is in the +3 oxidation state and is coordinated with three oxalate ions (C\(_2\)O\(_4\)) as ligands. The oxalate ligand is a bidentate ligand, meaning it binds through both oxygen atoms. Potassium ions are present to balance the negative charge of the complex anion.



(d) The compound \( [Co(en)_2Br_2]^+ \) is named as \textit{Dibromidobis(ethane-1,2-diamine)cobalt (III) ion. The central metal ion is cobalt in the +3 oxidation state, and it is coordinated with two ethylenediamine (en) ligands, each of which is a bidentate ligand. Two bromide ions are also coordinated to cobalt. The name follows IUPAC conventions for bidentate ligands and uses the appropriate prefixes to denote the number of each type of ligand. Quick Tip: - When naming coordination compounds, remember to list the ligands in alphabetical order.
- Use the appropriate prefixes for the number of each type of ligand (e.g., di-, tri-, tetra-, etc.).
- Always specify the oxidation state of the metal in Roman numerals inside parentheses.
- For bidentate ligands (such as ethylenediamine), use the correct nomenclature (e.g., "ethane-1,2-diamine").

(a) Reaction with \([Ag(NH_3)_2]^+ OH^-\) (Tollens' Test)

- The given compound Cyclohexanone undergoes oxidation by Tollens’ reagent.

- The aldehyde group in Cyclohexanone is oxidized to form a carboxylate anion.





- The reaction mechanism follows: \[ Cyclohexanone + [Ag(NH_3)_2]^+ + OH^- \rightarrow Cyclohexanone carboxylate + Ag + NH_3 + H_2O \]

Major Product: Cyclohexanone carboxylate ion.


(b) Iodoform Reaction with NaOI

- The given compound 4-Methylpent-3-en-2-one contains a methyl ketone \((-COCH_3)\) group.

- In the presence of NaOI, the iodoform test occurs, where the methyl ketone group is cleaved.

- This leads to the formation of Sodium propionate \((CH_3CH=CHCOONa)\) and Iodoform \((CHI_3)\).





Major Products: \[ Sodium propionate (CH_3CH=CHCOONa) + Iodoform (CHI_3) \]


(c) Cyanohydrin Formation with NaCN/HCl

- The given compound p-Carboxybenzaldehyde reacts with NaCN/HCl.

- The cyanohydrin reaction occurs, where the cyanide ion \((CN^)\) attacks the carbonyl group.

- The aldehyde (-CHO) is converted into a cyanohydrin (-CH(OH)CN).




Major Product: \[ 4-Carboxybenzaldehyde cyanohydrin (p-Carboxybenzylcyanohydrin) \] Quick Tip: 1. Tollens' test oxidizes aldehydes to carboxylate ions.
2. Iodoform test cleaves methyl ketones, forming carboxylates and iodoform \((CHI_3)\).
3. Cyanohydrin formation adds a cyanide (-CN) to an aldehyde, producing a cyanohydrin (-CH(OH)CN).

Using the Nernst equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^2} \]
Substituting the given values: \[ E_{cell} = 1.05 - \frac{0.059}{2} \log \frac{(0.1)}{(0.01)^2} \] \[ E_{cell} = 1.05 - \frac{0.059}{2} \times \log (0.1/0.0001) \] \[ E_{cell} = 1.05 - \frac{0.059}{2} \times \log (1000) \] \[ E_{cell} = 1.05 - \frac{0.059}{2} \times 3 \] \[ E_{cell} = 1.05 - (0.059 \times 1.5) \] \[ E_{cell} = 1.05 - 0.0885 \] \[ E_{cell} = 0.9615 V \] Quick Tip: 1. The Nernst equation relates cell potential to ion concentration.
2. The logarithmic term uses log base 10, which simplifies calculations.
3. For correct units, express the final answer in volts (V).

(a) Reaction of propanal with methyl magnesium bromide followed by hydrolysis:
The reaction of propanal (C\(_3\)H\(_6\)O) with methyl magnesium bromide (CH\(_3\)MgBr) is a Grignard reaction. The Grignard reagent acts as a nucleophile and attacks the carbonyl carbon in the propanal molecule, leading to the formation of an intermediate alkoxide. Upon hydrolysis, the alkoxide is converted into a secondary alcohol, specifically 2-butanol (CH\(_3\)CH(OH)CH\(_2\)CH\(_3\)).





(b) Reaction of phenol with Br\(_2\) in CS\(_2\):
The reaction of phenol (C\(_6\)H\(_5\)OH) with bromine (Br\(_2\)) in carbon disulfide (CS\(_2\)) results in a halogenation reaction. The bromine reacts with the hydroxyl group of phenol, leading to the substitution of a hydrogen atom on the benzene ring by a bromine atom. The major product is 2-bromophenol (C\(_6\)H\(_4\)BrOH), as bromination typically occurs at the ortho or para position relative to the hydroxyl group.



(c) Reaction of propene with diborane followed by oxidation:
The reaction of propene (C\(_3\)H\(_6\)) with diborane (B\(_2\)H\(_6\)) results in the formation of an organoborane intermediate. Upon oxidation, the organoborane undergoes hydrolysis to produce propan-2-ol (CH\(_3\)CH(OH)CH\(_3\)) as the major product. This is a typical hydroboration-oxidation reaction. Quick Tip: In Grignard reactions, a carbonyl compound reacts with a Grignard reagent to form an alcohol upon hydrolysis. In the case of phenol, bromine substitution occurs predominantly at the ortho and para positions due to the activating nature of the hydroxyl group.