The two compounds given are propanal (CH\(_3\)CH\(_2\)CHO), an aldehyde, and propanoic acid (CH\(_3\)CH\(_2\)COOH), a carboxylic acid.


The sodium bicarbonate (NaHCO\(_3\)) test is specifically used to detect the presence of a carboxylic acid group.


Carboxylic acids are acidic enough to react with a weak base like sodium bicarbonate, liberating carbon dioxide gas, which is observed as brisk effervescence.


The reaction is: CH\(_3\)CH\(_2\)COOH + NaHCO\(_3 \rightarrow\) CH\(_3\)CH\(_2\)COONa + H\(_2\)O + CO\(_2(\uparrow)\).


Aldehydes, like propanal, are not acidic and therefore do not react with sodium bicarbonate.


Thus, this test provides a clear distinction between the two compounds.
Quick Tip: The sodium bicarbonate test is a characteristic test for the carboxylic acid functional group (-COOH). The evolution of CO\(_2\) gas (effervescence) is a positive indicator. This test helps distinguish carboxylic acids from phenols, which are generally not acidic enough to react.

In an aqueous solution, the yellow chromate ion (CrO\(_4^{2-}\)) exists in equilibrium with the orange dichromate ion (Cr\(_2\)O\(_7^{2-}\)).


This equilibrium is pH-dependent and can be represented by the equation:

2CrO\(_4^{2-}\) (yellow) + 2H\(^+ \rightleftharpoons\) Cr\(_2\)O\(_7^{2-}\) (orange) + H\(_2\)O.


According to Le Chatelier's principle, adding an acid increases the concentration of H\(^+\) ions.


This shifts the equilibrium to the right, favoring the formation of the orange dichromate ion.


Since the yellow solution turned orange, the test tube must have contained an acidic substance.


Among the given options, HCl solution is a strong acid. Sodium hydrogen carbonate and sodium hydroxide are basic.


Therefore, the presence of HCl solution caused the conversion of yellow chromate to orange dichromate.
Quick Tip: Remember the pH-dependent equilibrium of chromate and dichromate ions. Chromate (CrO\(_4^{2-}\)) is stable in basic solutions (yellow), while dichromate (Cr\(_2\)O\(_7^{2-}\)) is stable in acidic solutions (orange).

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process.


It functions by providing an alternative reaction mechanism or pathway.


This new pathway has a lower activation energy (\(E_a\)) compared to the uncatalyzed reaction.


By lowering the activation energy barrier, a larger fraction of reactant molecules has sufficient energy to react, thus increasing the reaction rate.


A catalyst does not affect the thermodynamic state functions of the reaction, such as the enthalpy of reaction (\(\Delta H\)), Gibbs free energy of reaction (\(\Delta G\)), or the equilibrium constant (\(K_{eq}\)).


It only affects the kinetics, i.e., how fast the equilibrium is reached.
Quick Tip: A catalyst only changes the path of a reaction (kinetics), not the starting or ending points (thermodynamics). On a reaction coordinate diagram, a catalyst lowers the peak of the energy barrier but does not change the energy levels of reactants or products.

A molecule is chiral if it contains a chiral center and does not have a plane of symmetry.


A chiral center is typically a carbon atom bonded to four different atoms or groups.


Let's examine the structure of each option:


(A) 1-chloropropane: CH\(_3\)CH\(_2\)CH\(_2\)Cl. No carbon has four different substituents.


(B) 2-chloropropane: CH\(_3\)CHClCH\(_3\). The central carbon is attached to two identical methyl groups (-CH\(_3\)), so it's not chiral.


(C) 1-chlorobutane: CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)Cl. No carbon has four different substituents.


(D) 2-chlorobutane: CH\(_3\)CHClCH\(_2\)CH\(_3\). The second carbon atom (C-2) is bonded to:

1. a hydrogen atom (-H)

2. a chlorine atom (-Cl)

3. a methyl group (-CH\(_3\))

4. an ethyl group (-CH\(_2\)CH\(_3\))


Since all four groups attached to C-2 are different, it is a chiral center, making 2-chlorobutane a chiral molecule.
Quick Tip: To quickly identify a chiral molecule, look for a carbon atom (a stereocenter) bonded to four distinct groups. This is the most common source of chirality in organic molecules.

The conversion of ethanol (CH\(_3\)CH\(_2\)OH), a primary alcohol, to ethanal (CH\(_3\)CHO), an aldehyde, is an oxidation reaction.


This conversion requires a mild oxidizing agent to prevent further oxidation of the aldehyde to a carboxylic acid.


Let's analyze the reagents:


(A) Catalytic hydrogenation is a reduction process.


(B) LiAlH\(_4\) is a strong reducing agent.


(C) PCC (Pyridinium chlorochromate) is a mild and selective oxidizing agent that effectively oxidizes primary alcohols to aldehydes.


(D) KMnO\(_4\) is a strong oxidizing agent that would oxidize a primary alcohol all the way to a carboxylic acid (CH\(_3\)COOH).


Therefore, PCC is the appropriate reagent for this specific transformation.
Quick Tip: The choice of oxidizing agent for alcohols is crucial. Use mild agents like PCC or PDC (Pyridinium dichromate) to convert primary alcohols to aldehydes. Use strong agents like acidified KMnO\(_4\) or K\(_2\)Cr\(_2\)O\(_7\) to convert them to carboxylic acids.

The given compound is a tertiary amine. For IUPAC nomenclature of amines, we identify the longest carbon chain attached to the nitrogen atom as the parent alkane.


The alkyl groups attached to the nitrogen atom are: ethyl (-CH\(_2\)CH\(_3\)), methyl (-CH\(_3\)), and propyl (-CH\(_2\)CH\(_2\)CH\(_3\)).


The longest chain is the propyl group (3 carbons), so the parent name is propanamine.


The propyl group is attached to the nitrogen at its first carbon, so it is propan-1-amine.


The other two alkyl groups (ethyl and methyl) are treated as substituents on the nitrogen atom.


We use the prefix 'N-' to indicate their position.


Listing the substituents alphabetically, we get 'N-ethyl' and 'N-methyl'.


Combining everything gives the full IUPAC name: N-ethyl-N-methylpropan-1-amine.
Quick Tip: For naming secondary and tertiary amines, identify the largest alkyl group attached to the nitrogen as the parent alkane. The other alkyl groups are named as N-substituents and are listed alphabetically.

The graph shows a linear relationship between the concentration of the reactant [R] and time 't'.


The line has a negative slope and a positive y-intercept.


Let's consider the integrated rate laws for different orders of reaction:


Zero order: \([R] = -kt + [R]_0\). This is in the form \(y = mx + c\), where a plot of [R] vs. t is a straight line with slope = -k.


First order: \(\ln[R] = -kt + \ln[R]_0\). A plot of \(\ln[R]\) vs. t is linear.


Second order: \(1/[R] = kt + 1/[R]_0\). A plot of \(1/[R]\) vs. t is linear.


The given graph plots [R] versus t and is a straight line with a negative slope.


This corresponds exactly to the integrated rate law for a zero-order reaction.
Quick Tip: Memorize the linear plots associated with each reaction order: Zero Order: [R] vs. t is linear, slope = -k. First Order: ln[R] vs. t is linear, slope = -k. Second Order: 1/[R] vs. t is linear, slope = k.

The nitrite ion (NO\(_2^-\)) is an ambident nucleophile, meaning it can attack through two different atoms: nitrogen or oxygen.


When reacting an alkyl halide like ethyl bromide (CH\(_3\)CH\(_2\)Br) with a nitrite salt, the nature of the product depends on the cation (Ag\(^+\) vs. Na\(^+\) or K\(^+\)).


Silver nitrite (AgNO\(_2\)) has a significant covalent character in the Ag-O bond.


Therefore, the lone pair on the nitrogen atom is more available for nucleophilic attack.


The attack occurs through nitrogen, forming a C-N bond and resulting in a nitroalkane.


The reaction is: CH\(_3\)CH\(_2\)Br + AgNO\(_2 \rightarrow\) CH\(_3\)CH\(_2\)NO\(_2\) (nitroethane) + AgBr.


In contrast, ionic nitrites like NaNO\(_2\) or KNO\(_2\) would predominantly attack through the oxygen atom to form ethyl nitrite (CH\(_3\)CH\(_2\)ONO).
Quick Tip: Remember the rule for ambident nucleophiles: Reaction of alkyl halides with AgNO\(_2\) gives Nitroalkanes (R-NO\(_2\)), while reaction with NaNO\(_2\) or KNO\(_2\) gives Alkyl nitrites (R-ONO). A similar rule applies to KCN (gives nitriles, R-CN) and AgCN (gives isonitriles, R-NC).

The freezing point of a solution is lower than that of the pure solvent. This phenomenon is called depression in freezing point (\(\Delta T_f\)).


The formula for depression in freezing point is \(\Delta T_f = i \times K_f \times m\), where 'i' is the van't Hoff factor.


The highest freezing point will correspond to the lowest depression in freezing point (\(\Delta T_f\)).


Since the molality (approximated by molarity here) and \(K_f\) are constant, the lowest \(\Delta T_f\) will be for the substance with the lowest van't Hoff factor, 'i'.


Let's determine 'i' for each solute, assuming complete dissociation:


(A) KCl \(\rightarrow\) K\(^+\) + Cl\(^-\). Thus, i = 2.


(B) Na\(_2\)SO\(_4 \rightarrow\) 2Na\(^+\) + SO\(_4^{2-}\). Thus, i = 3.


(C) Glucose is a non-electrolyte and does not dissociate in water. Thus, i = 1.


(D) AlCl\(_3 \rightarrow\) Al\(^{3+}\) + 3Cl\(^-\). Thus, i = 4.


Glucose has the smallest van't Hoff factor (i=1), which will cause the least depression in the freezing point.


Therefore, the 1.0 M glucose solution will have the highest freezing point.
Quick Tip: For colligative properties, remember that the effect is proportional to the number of solute particles. A higher van't Hoff factor ('i') leads to a greater boiling point elevation and freezing point depression. To find the highest freezing point, look for the solute with the lowest 'i' value.

The Cannizzaro reaction is a disproportionation reaction undergone by aldehydes that do not have any hydrogen atoms on the \(\alpha\)-carbon.


The \(\alpha\)-carbon is the carbon atom adjacent to the carbonyl group (-CHO).


Let's analyze the \(\alpha\)-carbon for each option:


(A) CH\(_3\)–CH(CH\(_3\))–CHO: The \(\alpha\)-carbon is the CH group, which has one \(\alpha\)-hydrogen. It will not undergo the Cannizzaro reaction.


(B) (CH\(_3\))\(_3\)C CHO (2,2-Dimethylpropanal): The \(\alpha\)-carbon is the quaternary carbon atom bonded to three methyl groups. This carbon has zero \(\alpha\)-hydrogens. Therefore, it will undergo the Cannizzaro reaction.


(C) CH\(_3\)–CH\(_2\)–CHO (Propanal): The \(\alpha\)-carbon is the CH\(_2\) group, which has two \(\alpha\)-hydrogens. It will not undergo the Cannizzaro reaction.


(D) CH\(_3\)–CH(CH\(_3\))–CH(CH\(_3\))–CHO: The \(\alpha\)-carbon is the CH group, which has one \(\alpha\)-hydrogen. It will not undergo the Cannizzaro reaction.


Thus, only (CH\(_3\))\(_3\)C CHO lacks an \(\alpha\)-hydrogen and will undergo the Cannizzaro reaction.
Quick Tip: To quickly determine if an aldehyde will undergo the Cannizzaro reaction, locate the carbon atom next to the -CHO group (the \(\alpha\)-carbon). If this carbon has NO hydrogen atoms attached to it, the aldehyde is a candidate for the Cannizzaro reaction.

The color of transition metal ions is typically due to the presence of unpaired electrons in their d-orbitals, which allows for d-d electronic transitions.


Let's determine the electronic configuration and number of unpaired d-electrons for each ion:


I. Cu\(^+\) (Z=29): Electronic configuration is [Ar]3d\(^{10}\). All d-orbitals are fully filled, so there are zero unpaired electrons. Hence, Cu\(^+\) is colorless.


II. Ti\(^{4+}\) (Z=22): Electronic configuration is [Ar]3d\(^0\). There are no electrons in the d-orbital. Hence, Ti\(^{4+}\) is colorless.


III. Co\(^{2+}\) (Z=27): Electronic configuration is [Ar]3d\(^7\). It has 3 unpaired electrons in its d-orbitals. Hence, Co\(^{2+}\) is coloured (pink).


IV. Fe\(^{2+}\) (Z=26): Electronic configuration is [Ar]3d\(^6\). It has 4 unpaired electrons in its d-orbitals. Hence, Fe\(^{2+}\) is coloured (pale green).


Therefore, the group containing two coloured ions is III (Co\(^{2+}\)) and IV (Fe\(^{2+}\)).
Quick Tip: For transition metal ions, colour is associated with a partially filled d-orbital (d\(^1\) to d\(^9\)). Ions with empty (d\(^0\)) or completely filled (d\(^{10}\)) d-orbitals are generally colorless.

Let's match each cell with its common application:


i. Lead storage cell: This is a rechargeable secondary battery commonly used in automobiles and power inverters. So, (i) matches with (d).


ii. Mercury cell: It has a constant potential and is used in low-current devices like wristwatches, hearing aids, and cameras. So, (ii) matches with (c).


iii. Dry cell (Leclanché cell): This is a common primary battery used in portable electronics like torches, transistors, and wall clocks. So, (iii) matches with (a).


iv. Fuel cell: H\(_2\)–O\(_2\) fuel cells were used as a primary source of electrical energy in the Apollo space programme. So, (iv) matches with (b).


The correct set of matches is i-d, ii-c, iii-a, iv-b.
Quick Tip: Associate key applications with different types of electrochemical cells. Fuel cells are famous for their use in space missions, lead storage cells in cars/inverters, and mercury/dry cells in small portable devices.

Analyzing the Assertion (A): Lower aldehydes and ketones are soluble in water because they can form hydrogen bonds with water molecules via the oxygen atom of the carbonyl group. As the size of the alkyl group (a hydrophobic part) increases, the hydrophobic nature of the molecule increases, which hinders the formation of hydrogen bonds with water. This decreases solubility. So, Assertion (A) is true.


Analyzing the Reason (R): The carbonyl group (C=O) in aldehydes and ketones is polar. This polarity leads to dipole-dipole interactions between the molecules themselves. So, Reason (R) is true.


Analyzing the connection: The solubility in water (Assertion A) is primarily explained by the ability to form hydrogen bonds with water. The Reason (R), which talks about dipole-dipole interactions between aldehyde/ketone molecules, does not explain the solubility trend in water. The correct explanation for the assertion is the increasing size of the hydrophobic alkyl chain.


Therefore, both statements are true, but the reason is not the correct explanation for the assertion.
Quick Tip: Solubility of organic compounds in water is governed by the "like dissolves like" principle, which in this context means the ability to form hydrogen bonds with water. The solubility decreases as the non-polar (hydrophobic) part of the molecule becomes larger relative to the polar (hydrophilic) part.

Analyzing the Assertion (A): For a given alkyl group (R), the boiling point of alkyl halides increases with the increasing size and mass of the halogen atom. The order of mass and size is I > Br > Cl > F. Therefore, the boiling points follow the order RI > RBr > RCl > RF. Assertion (A) is true.


Analyzing the Reason (R): The boiling point depends on the strength of intermolecular forces. For alkyl halides, the dominant intermolecular forces are van der Waals forces. The magnitude of these forces increases with the size, mass, and number of electrons in the molecule. As we go from F to I, the size and mass of the halogen increase, leading to stronger van der Waals forces. Thus, the forces of attraction decrease in the order RI > RBr > RCl > RF. Reason (R) is true.


Analyzing the connection: The trend in boiling points (Assertion) is a direct consequence of the trend in the strength of intermolecular van der Waals forces (Reason). Stronger forces require more energy to overcome, resulting in a higher boiling point.


Therefore, both A and R are true, and R is the correct explanation of A.
Quick Tip: For molecules of similar polarity, boiling point is primarily determined by the strength of van der Waals forces. These forces increase with increasing molecular mass and surface area.

Analyzing the Assertion (A): It is a standard experimental technique to use an alternating current (AC) source, typically in a Wheatstone bridge setup, to measure the resistance of an electrolytic solution. So, Assertion (A) is true.


Analyzing the Reason (R): If a direct current (DC) source is used, it causes electrolysis. Ions migrate to the electrodes and undergo chemical reactions, which changes the composition and concentration of the solution near the electrodes. This phenomenon, known as polarization, alters the measured resistance. So, Reason (R) is true.


Analyzing the connection: The reason why an AC source is used is precisely to avoid the problems caused by a DC source. The rapid oscillation of the current in an AC source prevents the build-up of products at the electrodes and minimizes the effects of electrolysis and polarization. This ensures an accurate measurement of the solution's resistance.


Therefore, both A and R are true, and R is the correct explanation of A.
Quick Tip: When measuring the conductivity of ionic solutions, always use an AC source to prevent electrolysis and polarization effects, which would lead to inaccurate resistance readings if a DC source were used.

Analyzing the Reason (R): The dissolution of most gases in liquids is an exothermic process. According to Le Chatelier's principle, if the temperature of an exothermic process is increased, the equilibrium shifts in the endothermic (reverse) direction. This means less gas will dissolve. Thus, the solubility of gases in liquids generally decreases with an increase in temperature. Reason (R) is true.


Analyzing the Assertion (A): Henry's law is given by \(p = K_H \times x\), where \(p\) is the partial pressure, \(x\) is the mole fraction (a measure of solubility), and \(K_H\) is Henry's law constant.


From the equation, solubility \(x = p/K_H\). This shows that solubility (\(x\)) is inversely proportional to Henry's law constant (\(K_H\)).


Since we established that solubility (\(x\)) decreases with increasing temperature, for the inverse relationship to hold, the value of \(K_H\) must increase with increasing temperature.


Therefore, the statement that \(K_H\) decreases with an increase in temperature is false. Assertion (A) is false.
Quick Tip: Remember the inverse relationship between Henry's Law constant (\(K_H\)) and the solubility of a gas. Higher \(K_H\) means lower solubility. Since higher temperature leads to lower gas solubility, it must also lead to a higher \(K_H\) value.

The formula for elevation in boiling point is \(\Delta T_b = i \cdot K_b \cdot m\).


First, determine the van't Hoff factor (\(i\)).

Since CaCl\(_2\) undergoes complete dissociation, the reaction is:

CaCl\(_2 \rightarrow\) Ca\(^{2+}\) + 2Cl\(^-\).

One formula unit of CaCl\(_2\) gives 3 ions, so \(i = 3\).


Next, calculate the molality (\(m\)) of the solution.

Moles of CaCl\(_2\) = \(\frac{Given mass}{Molar mass} = \frac{3 g}{111 g/mol} = 0.0270 mol\).

Mass of solvent (water) = 260 g = 0.260 kg.

Molality (\(m\)) = \(\frac{Moles of solute}{Mass of solvent in kg} = \frac{0.0270 mol}{0.260 kg} = 0.1038 mol/kg\).


Now, calculate the elevation in boiling point (\(\Delta T_b\)).
\(\Delta T_b = i \cdot K_b \cdot m = 3 \times 0.52 K kg mol^{-1} \times 0.1038 mol kg^{-1}\).
\(\Delta T_b = 1.56 \times 0.1038 K\).
\(\Delta T_b = 0.162 K\).
Quick Tip: For colligative property calculations involving electrolytes, never forget to include the van't Hoff factor (\(i\)). The value of \(i\) is the number of ions the solute dissociates into, assuming complete dissociation.

According to Raoult's Law for an ideal solution, the total vapour pressure (\(P_{total}\)) is the sum of the partial pressures of the components.
\(P_{total} = P_X + P_Y = P_X^\circ \cdot x_X + P_Y^\circ \cdot x_Y\).


Given that the solution contains equal moles of 'X' and 'Y'.

Let the number of moles of X be \(n_X\) and the number of moles of Y be \(n_Y\).

Given \(n_X = n_Y\).


Calculate the mole fraction of X (\(x_X\)) and Y (\(x_Y\)).
\(x_X = \frac{n_X}{n_X + n_Y} = \frac{n_X}{n_X + n_X} = \frac{n_X}{2n_X} = 0.5\).
\(x_Y = \frac{n_Y}{n_X + n_Y} = \frac{n_Y}{n_Y + n_Y} = \frac{n_Y}{2n_Y} = 0.5\).


Now, substitute the values into Raoult's Law equation.
\(P_{total} = (120 mm Hg \times 0.5) + (160 mm Hg \times 0.5)\).
\(P_{total} = 60 mm Hg + 80 mm Hg\).
\(P_{total} = 140 mm Hg\).
Quick Tip: For an ideal solution with two components in equimolar amounts, the total vapour pressure is simply the average of the vapour pressures of the pure components: \(P_{total} = (P_X^\circ + P_Y^\circ) / 2\).

Variation of Conductivity (\(\kappa\)):

The data shows that conductivity decreases as the concentration decreases (i.e., on dilution).

Conductivity is defined as the conductance of ions present in a unit volume of the solution.

On dilution, the total number of ions increases, but the number of ions per unit volume decreases.

Since conductivity depends on the number of current-carrying ions per unit volume, it decreases with a decrease in concentration.


Variation of Molar Conductivity (\(\Lambda_m\)):

The data shows that molar conductivity increases as the concentration decreases (i.e., on dilution).

Molar conductivity is the conducting power of all the ions produced by one mole of an electrolyte.

\(\Lambda_m = \frac{\kappa \times 1000}{C}\).

On dilution, two factors affect molar conductivity:

1. The total volume V containing one mole of electrolyte increases.

2. The inter-ionic forces of attraction between ions decrease, leading to an increase in their mobility and speed.

This combined effect leads to an increase in molar conductivity with dilution, as the increase in ionic mobility more than compensates for the decrease in conductivity.
Quick Tip: Remember the key difference: Conductivity (\(\kappa\)) is about ions in a fixed volume (like 1 cm\(^3\)), so it decreases on dilution. Molar conductivity (\(\Lambda_m\)) is about the ions from a fixed amount of solute (1 mole), so it increases on dilution as ions move more freely.

Two key differences between the order and molecularity of a reaction are:


1. Definition and Determination:

Order: It is the sum of the powers of the concentration terms of the reactants in the experimentally determined rate law expression. It is an experimental quantity.

Molecularity: It is the number of reacting species (atoms, ions or molecules) that must collide simultaneously in an elementary reaction. It is a theoretical concept.


2. Possible Values:

Order: The order of a reaction can be a whole number, zero, or even a fraction.

Molecularity: The molecularity of a reaction must be a whole number (usually 1, 2, or rarely 3). It cannot be zero or fractional.
Quick Tip: A simple way to remember the difference is that 'order' comes from the experiment (rate law), while 'molecularity' comes from the proposed mechanism of an elementary step. Molecularity applies only to single-step reactions, while order applies to the overall reaction.

Given that the reaction is first order with respect to both X and Y.

The initial rate law can be written as:

Rate\(_1 = k[X]^1[Y]^1 = k[X][Y]\).


Now, the concentrations are changed:

New concentration of X, \([X]' = 2[X]\).

New concentration of Y, \([Y]' = 3[Y]\).


The new rate of reaction (Rate\(_2\)) will be:

Rate\(_2 = k[X]'[Y]' = k(2[X])(3[Y])\).


Simplifying the expression for the new rate:

Rate\(_2 = 6 \times (k[X][Y])\).


Since Rate\(_1 = k[X][Y]\), we can substitute this into the expression for Rate\(_2\):

Rate\(_2 = 6 \times Rate_1\).


Therefore, the rate of the reaction increases by 6 times.
Quick Tip: For a rate law Rate = k[A]\(^m\)[B]\(^n\), if [A] is changed by a factor of 'a' and [B] by a factor of 'b', the new rate will be (a\(^m\) \(\times\) b\(^n\)) times the original rate. Here, a=2, m=1, b=3, n=1, so the factor is 2\(^1 \times\) 3\(^1 = 6\).

The dehydration of ethanol to ethene at 443 K using concentrated H\(_2\)SO\(_4\) follows a three-step mechanism:


Step 1: Protonation of alcohol

The ethanol molecule acts as a Lewis base and accepts a proton from the strong acid (H\(_2\)SO\(_4\)) to form a protonated alcohol (ethyloxonium ion).

CH\(_3\)CH\(_2\)ÖH + H\(^+\) \(\rightleftharpoons\) CH\(_3\)CH\(_2\)Ö\(^+\)H\(_2\).


Step 2: Formation of carbocation

The C-O bond in the protonated alcohol is weak. It breaks heterolytically, releasing a stable water molecule and forming an ethyl carbocation. This is the slow, rate-determining step of the reaction.

CH\(_3\)CH\(_2\)Ö\(^+\)H\(_2\) \(\xrightarrow{Slow}\) CH\(_3\)C\(^+\)H\(_2\) + H\(_2\)O.


Step 3: Elimination of a proton (Deprotonation)

A proton is removed from the \(\beta\)-carbon (the carbon adjacent to the positively charged carbon) by a weak base (like HSO\(_4^-\) or H\(_2\)O) to form the alkene, ethene. The acid catalyst (H\(^+\)) is regenerated.

H-CH\(_2\)-C\(^+\)H\(_2\) \(\rightarrow\) CH\(_2\)=CH\(_2\) + H\(^+\).
Quick Tip: The mechanism of acid-catalyzed dehydration of alcohols involves three key steps: Protonation of the -OH group, Loss of water to form a carbocation, and Elimination of a proton to form an alkene. Remember that the temperature is crucial; at a lower temperature (413 K), ether formation dominates.

The acidity of a compound depends on the stability of its conjugate base formed after donating a proton (H\(^+\)).


For Carboxylic Acid (RCOOH):

It loses a proton to form a carboxylate ion (RCOO\(^-\)).

The carboxylate ion is highly stabilized by resonance. The negative charge is delocalized over two electronegative oxygen atoms.

R-C(=O)-O\(^-\) \(\leftrightarrow\) R-C(-O\(^-\))=O.

The two contributing resonance structures are equivalent, which leads to very high stability.


For Phenol (C\(_6\)H\(_5\)OH):

It loses a proton to form a phenoxide ion (C\(_6\)H\(_5\)O\(^-\)).

The phenoxide ion is also stabilized by resonance, but the negative charge is delocalized over one oxygen atom and the less electronegative carbon atoms of the benzene ring.

The resonance structures involve charge separation and are not equivalent. The negative charge resides on carbon in some structures, which is less stable.


Conclusion:

Since the carboxylate ion is significantly more resonance-stabilized than the phenoxide ion, the carboxylic acid has a greater tendency to donate a proton. Therefore, carboxylic acid is more acidic than phenol.
Quick Tip: When comparing acidity, always compare the stability of the conjugate bases. Greater delocalization of the negative charge, especially onto more electronegative atoms and in equivalent resonance structures, leads to a more stable conjugate base and a stronger acid.

The iodoform test can be used to distinguish between benzaldehyde and acetophenone.


Principle: This test is given by compounds containing a methyl ketone group (CH\(_3\)-C=O-) or an alcohol group (CH\(_3\)-CH(OH)-) which can be oxidized to a methyl ketone.


Procedure and Observations:

Acetophenone (C\(_6\)H\(_5\)COCH\(_3\)):

Acetophenone has a methyl ketone group. When warmed with an aqueous solution of iodine and sodium hydroxide (I\(_2\)/NaOH), it gives a positive iodoform test.


A yellow precipitate of iodoform (CHI\(_3\)), which has a characteristic antiseptic smell, is formed.

Reaction: C\(_6\)H\(_5\)COCH\(_3\) + 3I\(_2\) + 4NaOH \(\rightarrow\) CHI\(_3 \downarrow\) (yellow ppt) + C\(_6\)H\(_5\)COONa + 3NaI + 3H\(_2\)O.


Benzaldehyde (C\(_6\)H\(_5\)CHO):

Benzaldehyde does not contain a methyl ketone group.

When treated with I\(_2\)/NaOH, it does not give a positive iodoform test. No yellow precipitate is formed.


Conclusion:

The formation of a yellow precipitate with I\(_2\)/NaOH confirms the presence of acetophenone and distinguishes it from benzaldehyde.
Quick Tip: The iodoform test is a very reliable method for detecting the CH\(_3\)C=O group (or CH\(_3\)CH(OH)- group). Other tests like Tollens' test could also be used, as benzaldehyde (an aldehyde) would give a silver mirror while acetophenone (a ketone) would not.

The expected volume was 10 mL + 10 mL = 20 mL. The final volume is 20.2 mL, which is greater than expected.


This indicates that the solution shows a positive deviation from Raoult's law.


In such solutions, the intermolecular forces of attraction between the solute-solvent molecules (A-B) are weaker than the forces between the solute-solute (A-A) and solvent-solvent (B-B) molecules.


Due to these weaker forces, the molecules are held less tightly and move farther apart, leading to an increase in the total volume of the solution upon mixing (\(\Delta V_{mix} > 0\)).
Quick Tip: Remember the signs for non-ideal solutions: Positive Deviation: Weaker A-B interactions, \(\Delta V_{mix} > 0\), \(\Delta H_{mix} > 0\) (endothermic). Negative Deviation: Stronger A-B interactions, \(\Delta V_{mix} < 0\), \(\Delta H_{mix} < 0\) (exothermic).

The mixing process involves breaking stronger intermolecular forces (A-A and B-B) and forming weaker intermolecular forces (A-B).


More energy is absorbed to break the stronger bonds than is released when the weaker bonds are formed.


Therefore, the overall process is endothermic, and the enthalpy of mixing is positive (\(\Delta H_{mix} > 0\)).


An endothermic process absorbs heat from the surroundings, which will cause a decrease in the temperature of the solution.
Quick Tip: For positive deviations from Raoult's law, the mixing process is always endothermic (\(\Delta H_{mix} > 0\)), resulting in cooling. For negative deviations, it's exothermic (\(\Delta H_{mix} < 0\)), resulting in a temperature increase.

An example of a solution that shows positive deviation from Raoult's law is a mixture of ethanol and acetone.


In pure ethanol, molecules are held by strong hydrogen bonds.


When acetone is added, its molecules get in between the ethanol molecules, breaking some of the hydrogen bonds and weakening the overall intermolecular forces.


This leads to an increase in volume and an endothermic mixing process.


Another common example is a mixture of carbon disulfide (CS\(_2\)) and acetone.
Quick Tip: A good way to predict positive deviation is to mix a polar solvent with hydrogen bonds (like an alcohol) with a less polar solvent or one that disrupts those H-bonds.

This phenomenon is an application of the colligative property known as the depression of freezing point.


The freezing point of pure water is 0°C (273.15 K).


When salt (a non-volatile solute like NaCl or CaCl\(_2\)) is sprinkled on snow, it dissolves in the thin layer of water present on the surface of the ice.


This forms a salt solution, which has a lower freezing point than pure water.


For example, a NaCl solution can have a freezing point as low as -21°C.


As a result, the snow melts because the ambient temperature is now above the new, lower freezing point of the salt-water mixture.
Quick Tip: Colligative properties depend on the number of solute particles. Using salts like CaCl\(_2\) (\(i=3\)) is more effective than NaCl (\(i=2\)) for de-icing roads because it produces more ions per formula unit, causing a greater depression in the freezing point.

The fluid inside red blood cells (RBCs) has a salt concentration equivalent to a 0.9% (mass/vol) NaCl solution. This is called an isotonic solution.


The external solution is 0.5% NaCl, which has a lower solute concentration than the fluid inside the RBCs. Therefore, the 0.5% NaCl solution is hypotonic with respect to the RBCs.


Due to osmosis, the solvent (water) always moves from a region of lower solute concentration (hypotonic) to a region of higher solute concentration (hypertonic) across a semipermeable membrane.


In this case, water will move from the 0.5% NaCl solution into the red blood cells.


This influx of water will cause the red blood cells to swell and eventually burst, a process called hemolysis.
Quick Tip: Remember the effects of tonicity on cells: \textbf{Isotonic:} No net water movement, cell is stable. \textbf{Hypotonic:} Water enters cell, cell swells and may burst (hemolysis). \textbf{Hypertonic:} Water leaves cell, cell shrinks (crenation).

A major application of reverse osmosis (RO) is the desalination of seawater to obtain fresh, potable water.


In this process, a pressure greater than the osmotic pressure is applied to the seawater side of a suitable semipermeable membrane.


This high pressure forces the solvent molecules (water) to move from the region of higher solute concentration (seawater) to the region of lower solute concentration (fresh water), against the normal direction of osmosis.


The dissolved salts are left behind, producing pure drinking water.
Quick Tip: Reverse osmosis is essentially "pushing" a solvent through a semipermeable membrane, against its natural osmotic flow, by applying external pressure. It's a key technology for water purification worldwide.

Let the rate law for the reaction be: Rate = k[A]\(^x\)[B]\(^y\), where x and y are the orders with respect to A and B.


To find the order with respect to A (x):

Compare experiments 1 and 2, where [B] is constant (0.1 mol L\(^{-1}\)).
\(\frac{Rate_2}{Rate_1} = \frac{k(0.2)^x(0.1)^y}{k(0.1)^x(0.1)^y}\)
\(\frac{0.10}{0.05} = (\frac{0.2}{0.1})^x\)
\(2 = (2)^x\)

Therefore, \(x = 1\). The reaction is first order with respect to A.


To find the order with respect to B (y):

Compare experiments 1 and 3, where [A] is constant (0.1 mol L\(^{-1}\)).
\(\frac{Rate_3}{Rate_1} = \frac{k(0.1)^x(0.2)^y}{k(0.1)^x(0.1)^y}\)
\(\frac{0.05}{0.05} = (\frac{0.2}{0.1})^y\)
\(1 = (2)^y\)

Therefore, \(y = 0\). The reaction is zero order with respect to B.


To find the overall order of the reaction:

Overall order = \(x + y = 1 + 0 = 1\).

The overall order of the reaction is 1.
Quick Tip: When determining reaction orders from experimental data, always look for pairs of experiments where the concentration of only one reactant changes while the others are held constant. This isolates the effect of that single reactant on the rate.

Name: Ethylenediaminetetraacetate ion (EDTA\(^{4-}\)).


Structure:

It is a polydentate ligand with six donor sites: two nitrogen atoms and four oxygen atoms from the carboxylate groups.



The structure can be written as:

(\(^-\)OOC-CH\(_2\))\(_2\)N-CH\(_2\)-CH\(_2\)-N(CH\(_2\)-COO\(^-\))\(_2\).
Quick Tip: EDTA is a very important chelating agent used in coordination chemistry and analytical chemistry, particularly in complexometric titrations. It is also used medically to treat heavy metal poisoning.

For [Ni(CN)\(_4\)]\(^{2-}\):

The oxidation state of Ni is +2. (Let Ni be x, then x + 4(-1) = -2 \(\Rightarrow\) x = +2).

Ni\(^{2+}\) has the electronic configuration [Ar] 3d\(^8\).

CN\(^-\) is a strong field ligand, so it causes the pairing of the two unpaired electrons in the 3d orbitals.

This leaves one 3d orbital empty.

For coordination number 4, Ni\(^{2+}\) undergoes dsp\(^2\) hybridization (using one 3d, one 4s, and two 4p orbitals).

dsp\(^2\) hybridization results in a square planar geometry.


For [Ni(CO)\(_4\)]:

The oxidation state of Ni is 0.

Ni has the electronic configuration [Ar] 3d\(^8\) 4s\(^2\).

CO is a strong field ligand. In its presence, the 4s electrons are pushed into the 3d orbitals to pair up with the 3d electrons.

This results in a completely filled 3d orbital configuration, [Ar] 3d\(^{10}\).

For coordination number 4, Ni then uses its vacant 4s and three 4p orbitals for bonding.

This leads to sp\(^3\) hybridization.

sp\(^3\) hybridization results in a tetrahedral geometry.
Quick Tip: The geometry of a coordination complex depends on the hybridization of the central metal ion, which in turn is influenced by the oxidation state, coordination number, and the nature of the ligand (strong field vs. weak field).

Haloarenes are less reactive towards nucleophilic substitution reactions due to several reasons:

1. Resonance Effect: The lone pair of electrons on the halogen atom is in conjugation with the \(\pi\)-electrons of the benzene ring. This delocalization gives the C-X bond a partial double bond character, making it stronger and more difficult to break compared to the single C-X bond in haloalkanes.

2. Hybridization of Carbon Atom: The carbon atom attached to the halogen in haloarenes is sp\(^2\) hybridized. An sp\(^2\) hybridized carbon is more electronegative than an sp\(^3\) hybridized carbon (in haloalkanes). It holds the electron pair of the C-X bond more tightly, shortening the bond length and making it stronger.

3. Instability of Phenyl Cation: In a self-ionization pathway, the reaction would form a highly unstable phenyl cation, which makes the S\(_N\)1 mechanism unlikely.
Quick Tip: The key reason for the low reactivity of haloarenes in nucleophilic substitution is the partial double bond character of the C-X bond due to resonance. This makes the bond significantly stronger than in haloalkanes.

The melting point of a crystalline solid depends on the strength of the intermolecular forces and how well the molecules fit into the crystal lattice.


p-dichlorobenzene has a highly symmetrical, linear structure compared to the bent structures of the ortho and meta isomers.


This high degree of symmetry allows the para isomer to pack very closely and efficiently in the crystal lattice.


The closer packing results in stronger intermolecular forces of attraction (van der Waals forces) in the solid state.


More energy is required to break this well-packed, stable crystal lattice, which is why p-dichlorobenzene has a significantly higher melting point than its o- and m-isomers.
Quick Tip: For disubstituted benzene isomers, the para isomer is almost always the most symmetrical. This symmetry leads to better crystal packing and, consequently, a higher melting point and lower solubility compared to the ortho and meta isomers.

The S\(_N\)2 (bimolecular nucleophilic substitution) reaction proceeds via a mechanism involving a single transition state.


In this mechanism, the nucleophile attacks the carbon atom bearing the halogen from the side opposite to the leaving group (backside attack).


In a tertiary alkyl halide (R\(_3\)CX), the central carbon atom is bonded to three bulky alkyl groups.


These alkyl groups create significant steric hindrance around the carbon atom.


This steric crowding physically blocks the incoming nucleophile from approaching the carbon atom for the backside attack.


Due to this severe steric hindrance, tertiary alkyl halides are the least reactive towards the S\(_N\)2 mechanism. The order of reactivity for S\(_N\)2 is Primary > Secondary > Tertiary.
Quick Tip: Remember the role of steric hindrance in substitution reactions: \textbf{S\(_N\)2:} Favored by less steric hindrance (Methyl > 1° > 2° >> 3°). \textbf{S\(_N\)1:} Favored by more stable carbocations (3° > 2° > 1° > Methyl).

(i) Preparation of phenol from cumene (Cumene Process):

This process involves two main steps. First, cumene (isopropylbenzene) is oxidized in the presence of air to form cumene hydroperoxide. This is then treated with dilute acid to yield phenol and acetone.

C\(_6\)H\(_5\)CH(CH\(_3\))\(_2\) + O\(_2\) \(\xrightarrow{Oxidation}\) C\(_6\)H\(_5\)C(CH\(_3\))\(_2\)OOH \(\xrightarrow{H^+, H_2O}}\) C\(_6\)H\(_5\)OH + CH\(_3\)COCH\(_3\)

(Cumene) \(\qquad \qquad \qquad \quad\) (Cumene hydroperoxide) \(\qquad \qquad \quad\) (Phenol) \(\quad\) (Acetone)


(ii) Nitration of anisole:

Anisole reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid. The methoxy (-OCH\(_3\)) group is an ortho, para-directing group. This results in a mixture of o-nitroanisole and p-nitroanisole, with the para isomer being the major product due to less steric hindrance.

C\(_6\)H\(_5\)OCH\(_3\) + conc. HNO\(_3\) + conc. H\(_2\)SO\(_4\) \(\rightarrow\) o-(NO\(_2\))C\(_6\)H\(_4\)OCH\(_3\) + p-(NO\(_2\))C\(_6\)H\(_4\)OCH\(_3\) + H\(_2\)O

(Anisole) \(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad\) (o-nitroanisole, minor) (p-nitroanisole, major)
Quick Tip: The Cumene process is the primary industrial method for manufacturing phenol because it also produces acetone as a valuable by-product. When dealing with electrophilic substitution on substituted benzenes, always identify the directing effect of the existing group (-OCH\(_3\) is o,p-directing and activating).

This reaction involves the cleavage of an ether (tert-butyl methyl ether) by hydrogen iodide (HI).

The mechanism depends on the nature of the alkyl groups. Since one group is tertiary (tert-butyl), it can form a stable tertiary carbocation. Therefore, the reaction proceeds via an S\(_N\)1 mechanism.


Step 1: Protonation

The oxygen atom of the ether gets protonated by HI.

(CH\(_3\))\(_3\)C–O–CH\(_3\) + H\(^+\) \(\rightarrow\) (CH\(_3\))\(_3\)C–O\(^+\)H–CH\(_3\)


Step 2: Cleavage to form carbocation

The protonated ether cleaves to form the more stable tertiary carbocation and methanol.

(CH\(_3\))\(_3\)C–O\(^+\)H–CH\(_3\) \(\rightarrow\) (CH\(_3\))\(_3\)C\(^+\) + CH\(_3\)OH


Step 3: Nucleophilic attack

The iodide ion (I\(^-\)) attacks the tertiary carbocation.

(CH\(_3\))\(_3\)C\(^+\) + I\(^-\) \(\rightarrow\) (CH\(_3\))\(_3\)C–I


The final products are tert-butyl iodide and methanol.

(CH\(_3\))\(_3\)C–O–CH\(_3\) + HI \(\rightarrow\) (CH\(_3\))\(_3\)C–I + CH\(_3\)OH
Quick Tip: When cleaving ethers with HX, if one of the alkyl groups is tertiary, the reaction follows an S\(_N\)1 mechanism, and the halogen attaches to the tertiary carbon. If both groups are primary or secondary, it follows an S\(_N\)2 mechanism, and the halogen attaches to the smaller alkyl group.

(a) Reaction with Semicarbazide:

Acetaldehyde (CH\(_3\)CHO) reacts with semicarbazide (NH\(_2\)CONHNH\(_2\)) in a condensation reaction. The oxygen atom from the aldehyde and two hydrogen atoms from the terminal -NH\(_2\) group of semicarbazide are eliminated as a water molecule. The product is acetaldehyde semicarbazone.

CH\(_3\)CHO + NH\(_2\)CONHNH\(_2\) \(\rightarrow\) CH\(_3\)CH=NNHCONH\(_2\) + H\(_2\)O

(Acetaldehyde semicarbazone)


(b) Aldol Condensation:

Acetaldehyde (CH\(_3\)CHO) has \(\alpha\)-hydrogens. In the presence of a dilute base like NaOH, two molecules of acetaldehyde undergo aldol condensation. One molecule forms an enolate which then attacks the carbonyl carbon of the second molecule. The product is an aldol (a \(\beta\)-hydroxy aldehyde).

2 CH\(_3\)CHO \(\xrightarrow{dil. NaOH}\) CH\(_3\)CH(OH)CH\(_2\)CHO

(3-Hydroxybutanal)


(c) Hell-Volhard-Zelinsky (HVZ) Reaction:

Acetic acid (CH\(_3\)COOH) has an \(\alpha\)-hydrogen. It undergoes halogenation at the \(\alpha\)-position in the presence of red phosphorus and a halogen (Cl\(_2\)), followed by hydrolysis. The product is an \(\alpha\)-halo carboxylic acid.

CH\(_3\)COOH + Cl\(_2\) \(\xrightarrow{Red P}\) ClCH\(_2\)COOH + HCl

(\(\alpha\)-Chloroacetic acid)
Quick Tip: Recognize these named reactions: (a) is a standard condensation with an ammonia derivative. (b) is the Aldol reaction, characteristic of aldehydes/ketones with \(\alpha\)-hydrogens in dilute base. (c) is the HVZ reaction, specific for halogenating the \(\alpha\)-carbon of carboxylic acids.

The vitamin responsible for the coagulation of blood (blood clotting) is Vitamin K.

It is essential for the synthesis of prothrombin and other blood clotting factors in the liver.
Quick Tip: A simple way to remember is 'K' for 'Koagulation' (the German spelling, from where it was first identified).

Denaturation of protein: It is a process in which the native, biologically active conformation of a protein is disrupted, leading to the loss of its biological activity.


This involves the unfolding of the protein's secondary, tertiary, and quaternary structures, which are held together by weak forces like hydrogen bonds, disulfide bridges, and van der Waals forces.


The primary structure (the sequence of amino acids linked by peptide bonds) remains intact during denaturation.


Denaturation can be caused by physical changes (like heat) or chemical changes (like extreme pH, organic solvents, or salts of heavy metals).


Example: A common example is the coagulation of egg white upon boiling. The soluble globular protein, albumin, in the egg white gets denatured by heat and changes into an insoluble, rubbery fibrous form.
Quick Tip: Denaturation is like scrambling the complex 3D shape of a protein without breaking the main chain. Since a protein's function depends critically on its shape, denaturation almost always results in the loss of function.

Two chemical reactions can be used to demonstrate these features:


1. Presence of an aldehydic group:

Glucose, on reaction with a mild oxidizing agent like bromine water, gets oxidized to gluconic acid. This reaction is specific for the aldehyde group and does not affect the alcohol groups, thus confirming the presence of a -CHO group.

CHO-(CHOH)\(_4\)-CH\(_2\)OH + Br\(_2\)(aq) \(\rightarrow\) COOH-(CHOH)\(_4\)-CH\(_2\)OH

(Glucose) \(\qquad \qquad \qquad \qquad \qquad\) (Gluconic acid)


2. Presence of a straight chain of six carbon atoms:

When glucose is heated with concentrated hydroiodic acid (HI) and red phosphorus for a prolonged time, it undergoes complete reduction to form n-hexane. The formation of n-hexane indicates that all six carbon atoms in glucose are linked in a straight, unbranched chain.

CHO-(CHOH)\(_4\)-CH\(_2\)OH \(\xrightarrow{HI, Red P, \Delta}\) CH\(_3\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-CH\(_3\)

(Glucose) \(\qquad \qquad \qquad \qquad \qquad \qquad \qquad\) (n-Hexane)
Quick Tip: Remember key confirmatory tests for glucose structure: \textbf{-CHO group:} Tollens' test, Fehling's test, or Bromine water oxidation. \textbf{Straight chain:} Reduction with HI/Red P to give n-hexane. \textbf{Five -OH groups:} Acetylation with acetic anhydride to give a penta-acetate derivative.

Anomers are a pair of cyclic stereoisomers (specifically, diastereomers called epimers) of a monosaccharide that differ in their configuration only at the hemiacetal or hemiketal carbon.


This specific carbon atom (C-1 in aldoses, C-2 in ketoses) is called the anomeric carbon.


For example, \(\alpha\)-D-glucopyranose and \(\beta\)-D-glucopyranose are anomers of glucose. They differ only in the orientation (axial or equatorial) of the hydroxyl group attached to the C-1 anomeric carbon.
Quick Tip: Anomers are formed during the cyclization of a sugar. Remember, 'alpha' (\(\alpha\)) and 'beta' (\(\beta\)) designations are used to distinguish between anomers.

The structure of \(\beta\)-D-Glucopyranose in its Haworth projection is as follows:


It is a six-membered ring containing five carbon atoms and one oxygen atom.


The '\(\beta\)' designation means the hydroxyl (-OH) group on the anomeric carbon (C-1) is pointing up, on the same side as the -CH\(_2\)OH group at C-5.





The substituents are arranged as:

C-1: -OH group is up (\(\beta\) position).

C-2: -OH group is down.

C-3: -OH group is up.

C-4: -OH group is down.

C-5: -CH\(_2\)OH group is up (for D-sugar).
Quick Tip: For D-glucose pyranose ring (Haworth): The C-5 -CH\(_2\)OH group is always up. For the anomers, remember \(\alpha\) = -OH on C-1 is down (axial) and \(\beta\) = -OH on C-1 is up (equatorial).

Sucrose is a disaccharide that is dextrorotatory, meaning it rotates the plane of polarized light to the right. Its specific rotation is +66.5°.


Upon hydrolysis with dilute acid or the enzyme invertase, sucrose breaks down into an equimolar mixture of D-glucose and D-fructose.

Sucrose + H\(_2\)O \(\xrightarrow{H^+ or Invertase}\) D-Glucose + D-Fructose


D-glucose is dextrorotatory with a specific rotation of +52.5°.


D-fructose is strongly laevorotatory with a specific rotation of -92.4°.


The resulting mixture is laevorotatory overall because the negative rotation of fructose is greater in magnitude than the positive rotation of glucose.


Since the optical rotation of the solution changes from positive (dextro) to negative (laevo) during the hydrolysis, the process is called inversion of sugar, and the product mixture is called invert sugar.
Quick Tip: The term "inversion" refers specifically to the change in the sign of optical rotation during the hydrolysis of sucrose. This is a key property of sucrose and is utilized by bees to make honey.

(i) Structural Formula of the Complex:

The reaction with excess silver nitrate (AgNO\(_3\)) precipitates only the chloride ions that are present outside the coordination sphere as counter-ions.


Since one mole of the complex precipitates one mole of AgCl, there is only one chloride ion acting as a counter-ion.


The central metal ion, Cr, typically exhibits a coordination number (secondary valency) of 6.


The available ligands are four H\(_2\)O molecules and the remaining two Cl\(^-\) ions. These will be inside the coordination sphere to satisfy the coordination number of 6.


Therefore, the structural formula of the complex is [Cr(H\(_2\)O)\(_4\)Cl\(_2\)]Cl.


(ii) Secondary Valency of Cr:

The secondary valency corresponds to the coordination number of the central metal ion, which is the total number of ligands directly bonded to it.


In the complex [Cr(H\(_2\)O)\(_4\)Cl\(_2\)]Cl, the chromium ion is bonded to four water molecules and two chloride ions.


So, the secondary valency of Cr is 4 + 2 = 6.
Quick Tip: In problems like this, the amount of AgX precipitated tells you the number of halide ions outside the coordination sphere. The secondary valency is usually the fixed coordination number of the metal (often 6 for Cr, Co, Pt).

The key differences between a double salt and a complex are:


1. Identity in Solution:

Double Salt: A double salt, such as Mohr's salt [FeSO\(_4 \cdot\)(NH\(_4\))\(_2\)SO\(_4 \cdot\)6H\(_2\)O], exists only in the solid state. When dissolved in water, it completely dissociates into its constituent simple ions (e.g., Fe\(^{2+}\), SO\(_4^{2-}\), and NH\(_4^+\)).

Complex: A complex, such as potassium ferrocyanide [K\(_4\)[Fe(CN)\(_6\)]], retains its identity in solution. It dissociates into a complex ion and counter ions (e.g., [Fe(CN)\(_6\)]\(^{4-}\) and K\(^+\)). The complex ion itself does not break down further into its components.


2. Chemical Properties:

Double Salt: The chemical properties and tests of a double salt solution are the same as those of its constituent simple ions.

Complex: The chemical properties of a complex solution are different from its constituent ions because the complex ion is a distinct chemical entity with its own unique properties.
Quick Tip: The main test to distinguish them is dissociation in water. If all individual ions can be tested for, it's a double salt. If a stable complex ion is formed that doesn't give tests for its constituent metal ion, it's a coordination compound.

The molar conductivity of an electrolytic solution depends on the number of ions produced per formula unit of the compound upon dissociation in the solution. A greater number of ions leads to higher conductivity.


Let's analyze the dissociation of each complex:


1. [Cr(NH\(_3\))\(_3\)Cl\(_3\)]: This is a neutral complex with no counter-ions. It does not ionize in solution. Number of ions = 0.


2. [Cr(NH\(_3\))\(_5\)Cl]Cl\(_2\): This complex dissociates to give one complex cation and two chloride anions.

[Cr(NH\(_3\))\(_5\)Cl]Cl\(_2\) \(\rightarrow\) [Cr(NH\(_3\))\(_5\)Cl]\(^{2+}\) + 2Cl\(^-\). Total number of ions = 1 + 2 = 3.


3. [Cr(NH\(_3\))\(_6\)]Cl\(_3\): This complex dissociates to give one complex cation and three chloride anions.

[Cr(NH\(_3\))\(_6\)]Cl\(_3\) \(\rightarrow\) [Cr(NH\(_3\))\(_6\)]\(^{3+}\) + 3Cl\(^-\). Total number of ions = 1 + 3 = 4.


The number of ions produced is in the order: [Cr(NH\(_3\))\(_3\)Cl\(_3\)] < [Cr(NH\(_3\))\(_5\)Cl]Cl\(_2\) < [Cr(NH\(_3\))\(_6\)]Cl\(_3\).


Therefore, the increasing order of conductivity of their solutions is:

[Cr(NH\(_3\))\(_3\)Cl\(_3\)] < [Cr(NH\(_3\))\(_5\)Cl]Cl\(_2\) < [Cr(NH\(_3\))\(_6\)]Cl\(_3\).
Quick Tip: To compare the conductivity of complex solutions of the same concentration, simply count the total number of ions each formula unit produces upon dissociation. More ions = higher conductivity.

According to Werner's theory, the two types of valencies in coordination compounds differ as follows:


1. Ionization and Satisfaction:

Primary Valency: This corresponds to the oxidation state of the central metal ion. It is ionisable and is satisfied only by negative ions.

Secondary Valency: This corresponds to the coordination number of the central metal ion. It is non-ionisable and is satisfied by ligands, which can be negative ions, positive ions, or neutral molecules.


2. Directional Nature and Geometry:

Primary Valency: This is non-directional in nature. It does not play any role in determining the geometry of the complex.

Secondary Valency: This is directional. The ligands satisfying the secondary valency are directed towards fixed positions in space, which determines the overall geometry of the coordination complex (e.g., octahedral, tetrahedral, square planar).
Quick Tip: Think of primary valency as the charge-balancing role (oxidation state) and secondary valency as the structure-defining role (coordination number and geometry).

Pyrolusite ore is manganese dioxide (MnO\(_2\)).


(I) Formulae of 'A' and 'B':

When MnO\(_2\) is fused with KOH in the presence of air (an oxidizing agent like O\(_2\)), it forms potassium manganate, which is dark green. So, 'A' is K\(_2\)MnO\(_4\).

2MnO\(_2\) + 4KOH + O\(_2\) \(\rightarrow\) 2K\(_2\)MnO\(_4\) + 2H\(_2\)O

(A, dark green)


Potassium manganate (K\(_2\)MnO\(_4\)) is stable in alkaline solutions but undergoes disproportionation in a neutral or acidic medium to form potassium permanganate (purple) and manganese dioxide. So, 'B' is KMnO\(_4\).

3MnO\(_4^{2-}\) + 4H\(^+\) \(\rightarrow\) 2MnO\(_4^-\) + MnO\(_2\) + 2H\(_2\)O

(Manganate ion from A) \(\qquad\) (Permanganate ion from B, purple)

So, A = K\(_2\)MnO\(_4\) and B = KMnO\(_4\).


(II) Ionic equation for reaction of 'B' with Fe\(^{2+}\):

Potassium permanganate (compound B) is a strong oxidizing agent in acidic medium. It oxidizes Fe\(^{2+}\) ions to Fe\(^{3+}\) ions, while MnO\(_4^-\) is reduced to Mn\(^{2+}\).

The balanced ionic equation for the reaction is:

MnO\(_4^-\) + 5Fe\(^{2+}\) + 8H\(^+\) \(\rightarrow\) Mn\(^{2+}\) + 5Fe\(^{3+}\) + 4H\(_2\)O
Quick Tip: Remember the characteristic colours of manganese compounds in different oxidation states: MnO\(_4^-\) (permanganate, O.S. +7) is purple, and MnO\(_4^{2-}\) (manganate, O.S. +6) is green.

(I) Ce\(^{4+}\) as a good oxidising agent:

The common oxidation state for lanthanoids is +3. Cerium can show a +4 oxidation state due to its tendency to attain a noble gas configuration ([Xe]). However, the Ce\(^{3+}\) state is much more stable in aqueous solution. Therefore, Ce\(^{4+}\) has a strong tendency to gain an electron to revert to the more stable Ce\(^{3+}\) state. This property of easily accepting an electron makes it a strong oxidizing agent.


(II) Greater actinoid contraction:

Both lanthanoid and actinoid contractions are due to the poor shielding effect of f-electrons. The 5f orbitals in actinoids are more extended in space and have a poorer shielding effect than the 4f orbitals in lanthanoids. This results in a stronger effective nuclear charge experienced by the valence electrons in actinoids, leading to a greater and more pronounced contraction in atomic/ionic radii across the series.


(III) E\(^\circ\) values of Zn and Cu:

The standard electrode potential (E\(^\circ\)) depends on the net effect of enthalpy of atomization, ionization enthalpy, and hydration enthalpy.

For Zinc, its electronic configuration is 3d\(^{10}\)4s\(^2\). It has high atomization and ionization enthalpies, but its hydration enthalpy is also very high (due to small size), which compensates to a large extent. However, the overall E\(^\circ\) value remains negative (-0.76 V), indicating its reactive nature.

For Copper, the sum of its first and second ionization enthalpies is very high, and its hydration enthalpy is not high enough to compensate for this. This makes the overall E\(^\circ\) value positive (+0.34 V), indicating its relative inertness and resistance to oxidation.
Quick Tip: Electrode potentials are a delicate balance of three energy terms. For Cu, high IE is the deciding factor for its positive E\(^\circ\). For actinoids, remember 5f electrons are poorer shielders than 4f electrons, hence the contraction effect is stronger.

In a galvanic (voltaic) cell, the half-reaction with the higher (more positive or less negative) standard reduction potential (E\(^\circ\)) occurs as reduction at the cathode. The half-reaction with the lower E\(^\circ\) value is reversed and occurs as oxidation at the anode.


Comparing the given values:

E\(^\circ\)(Fe\(^{2+}\)/Fe) = -0.44 V

E\(^\circ\)(Cr\(^{3+}\)/Cr) = -0.74 V


Since -0.44 V > -0.74 V, the standard reduction potential of the Fe\(^{2+}\)/Fe half-cell is higher.


Therefore:

Reduction Reaction (at Cathode): The Fe\(^{2+}\)/Fe half-reaction will remain as the reduction reaction.

Fe\(^{2+}\) + 2e\(^-\) \(\rightarrow\) Fe(s)


Oxidation Reaction (at Anode): The Cr\(^{3+}\)/Cr half-reaction will be reversed to become the oxidation reaction.

Cr(s) \(\rightarrow\) Cr\(^{3+}\) + 3e\(^-\)
Quick Tip: Remember the rule for constructing a galvanic cell: MORE positive E\(^\circ_{red}\) = CATHODE (Reduction). LESS positive E\(^\circ_{red}\) = ANODE (Oxidation, reaction is reversed).

Cell Representation:

In the given reaction, Mg is oxidized (Mg \(\rightarrow\) Mg\(^{2+}\)) and Ag\(^+\) is reduced (Ag\(^+\) \(\rightarrow\) Ag).

Oxidation occurs at the anode, and reduction occurs at the cathode.

The cell is represented as: Anode | Anode ion || Cathode ion | Cathode

Mg(s) | Mg\(^{2+}\)(0.100 M) || Ag\(^{+}\)(0.001 M) | Ag(s)


Calculation of E\(_{cell}\):

We use the Nernst equation: E\(_{cell}\) = E\(^\circ_{cell}\) - \(\frac{0.0591}{n} \log Q\) at 298 K.

Here, n = 2 (two moles of electrons are transferred).

The reaction quotient, Q = \(\frac{[Mg^{2+}]}{[Ag^+]^2}\).

Q = \(\frac{0.100}{(0.001)^2} = \frac{10^{-1}}{(10^{-3})^2} = \frac{10^{-1}}{10^{-6}} = 10^5\).


Now, substitute the values into the Nernst equation:

E\(_{cell}\) = 3.17 V - \(\frac{0.0591}{2} \log(10^5)\).

E\(_{cell}\) = 3.17 - 0.02955 \times (5 \log 10)\(.
Since log 10 = 1:
E\)_{cell\( = 3.17 - 0.02955 \times 5.
E\)_{cell\( = 3.17 - 0.14775.
E\)_{cell\( = 3.02225 V.
Quick Tip: Be very careful when writing the expression for the reaction quotient, Q. The concentrations of products are in the numerator and reactants in the denominator, each raised to the power of its stoichiometric coefficient. Solids have an activity of 1 and are omitted.

Kohlrausch's Law of Independent Migration of Ions:

The law states that the limiting molar conductivity of an electrolyte (\(\Lambda_m^\circ\)) can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.

Mathematically, for an electrolyte A\(_x\)B\(_y\): \(\Lambda_m^\circ(A_xB_y) = x \cdot \lambda^\circ_+ + y \cdot \lambda^\circ_-\), where \(\lambda^\circ_+\) and \(\lambda^\circ_-\) are the limiting molar ionic conductivities of the cation and anion, respectively.


Two Applications:

1. Calculation of Limiting Molar Conductivity for Weak Electrolytes: The limiting molar conductivity of a weak electrolyte cannot be determined experimentally by extrapolation. It can be calculated using Kohlrausch's law from the \(\Lambda_m^\circ\) values of strong electrolytes. For example, \(\Lambda_m^\circ(CH_3COOH) = \Lambda_m^\circ(CH_3COONa) + \Lambda_m^\circ(HCl) - \Lambda_m^\circ(NaCl)\).


2. Calculation of Degree of Dissociation (\(\alpha\)): The degree of dissociation of a weak electrolyte at a given concentration 'c' can be calculated using the formula:
\(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\), where \(\Lambda_m\) is the molar conductivity at concentration 'c' and \(\Lambda_m^\circ\) is the limiting molar conductivity.
Quick Tip: Kohlrausch's law is a powerful tool for dealing with weak electrolytes. It allows us to find their properties at infinite dilution (where they are fully dissociated) by cleverly combining data from strong electrolytes.

The degree of dissociation (\(\alpha\)) is given by the formula: \(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\).


We are given the molar conductivity at the specific concentration:
\(\Lambda_m\) (for 10\(^{-2}\) M NH\(_4\)OH) = 9.33 S cm\(^2\) mol\(^{-1}\).


First, we need to calculate the limiting molar conductivity (\(\Lambda_m^\circ\)) of the weak electrolyte NH\(_4\)OH using Kohlrausch's law and the given data for strong electrolytes.
\(\Lambda_m^\circ(NH_4OH) = \Lambda_m^\circ(NH_4Cl) + \Lambda_m^\circ(NaOH) - \Lambda_m^\circ(NaCl)\).
\(\Lambda_m^\circ(NH_4OH) = (129.8 + 217.4 - 108.9)\) S cm\(^2\) mol\(^{-1}\).
\(\Lambda_m^\circ(NH_4OH) = (347.2 - 108.9)\) S cm\(^2\) mol\(^{-1}\).
\(\Lambda_m^\circ(NH_4OH) = 238.3\) S cm\(^2\) mol\(^{-1}\).


Now, calculate the degree of dissociation (\(\alpha\)).
\(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{9.33 S cm^2 mol^{-1}}{238.3 S cm^2 mol^{-1}}\).
\(\alpha = 0.03916\).

As a percentage, \(\alpha = 3.916%\).
Quick Tip: This is a classic problem combining two key concepts: using Kohlrausch's law to find \(\Lambda_m^\circ\) for a weak electrolyte, and then using that value to find the degree of dissociation at a specific concentration.

The molecular formula C\(_2\)H\(_7\)N corresponds to two possible isomers:

1. Ethylamine (CH\(_3\)CH\(_2\)NH\(_2\)), which is a primary (1°) amine.

2. Dimethylamine ((CH\(_3\))\(_2\)NH), which is a secondary (2°) amine.


The reaction is with C\(_6\)H\(_5\)SO\(_2\)Cl (benzenesulfonyl chloride), which is the Hinsberg's reagent test used to distinguish between primary, secondary, and tertiary amines.


A primary amine reacts with Hinsberg's reagent to form N-alkylbenzenesulfonamide. This product has an acidic hydrogen atom attached to the nitrogen, which makes it soluble in alkali (like NaOH).

RNH\(_2\) + C\(_6\)H\(_5\)SO\(_2\)Cl \(\rightarrow\) C\(_6\)H\(_5\)SO\(_2\)NHR \(\xrightarrow{NaOH}\) [C\(_6\)H\(_5\)SO\(_2\)NR]\(^-\)Na\(^+\) (Soluble)


A secondary amine reacts to form N,N-dialkylbenzenesulfonamide, which has no acidic hydrogen on the nitrogen and is insoluble in alkali.


Since the product formed from amine 'X' is soluble in NaOH solution, 'X' must be a primary amine.

Therefore, compound 'X' is Ethylamine (CH\(_3\)CH\(_2\)NH\(_2\)).
Quick Tip: Hinsberg's Test Summary: \textbf{1° amine:} Reacts, product is soluble in alkali. \textbf{2° amine:} Reacts, product is insoluble in alkali. \textbf{3° amine:} Does not react.

The pK\(_b\) value is inversely related to the basic strength of an amine. A lower pK\(_b\) value indicates a stronger base. Therefore, arranging in increasing order of pK\(_b\) is the same as arranging in decreasing order of basic strength.


The basicity of amines in the aqueous phase is determined by a combination of the +I (inductive) effect of alkyl groups, steric hindrance to solvation, and the extent of H-bonding with water.


Comparing the amines:

- C\(_6\)H\(_5\)NH\(_2\) (Aniline) is the weakest base because the lone pair on nitrogen is delocalized into the benzene ring, making it less available for protonation.

- Alkylamines are stronger bases than ammonia (NH\(_3\)) due to the electron-donating (+I) effect of alkyl groups.

- Among alkylamines in aqueous solution, the order of basicity is generally: Secondary > Primary > Tertiary. This is because (CH\(_3\))\(_2\)NH has the best balance of +I effect and stabilization of the conjugate acid by solvation (H-bonding). (CH\(_3\))\(_3\)N is less basic than expected due to steric hindrance to solvation.


So, the decreasing order of basic strength is:

(CH\(_3\))\(_2\)NH > CH\(_3\)NH\(_2\) > (CH\(_3\))\(_3\)N > NH\(_3\) > C\(_6\)H\(_5\)NH\(_2\)


Therefore, the increasing order of pK\(_b\) value is:

(CH\(_3\))\(_2\)NH < CH\(_3\)NH\(_2\) < (CH\(_3\))\(_3\)N < NH\(_3\) < C\(_6\)H\(_5\)NH\(_2\)
Quick Tip: Remember that lower pK\(_b\) = stronger base. The basicity order of methylamines in aqueous solution is a special case: 2° > 1° > 3°. Aniline is always a very weak base due to resonance.

Nitration is carried out in a strongly acidic medium (a mixture of conc. HNO\(_3\) and conc. H\(_2\)SO\(_4\)).


Aniline is a strong base. In this highly acidic medium, a significant portion of the aniline molecules get protonated to form the anilinium ion (C\(_6\)H\(_5\)NH\(_3^+\)).

C\(_6\)H\(_5\)NH\(_2\) + H\(^+\) \(\rightleftharpoons\) C\(_6\)H\(_5\)NH\(_3^+\)


The amino group (-NH\(_2\)) in aniline is an activating and ortho, para-directing group.


However, the anilinium group (-NH\(_3^+\)) is strongly deactivating and meta-directing due to its positive charge, which exerts a strong -I effect.


Therefore, the electrophilic substitution (nitration) occurs on both the activated aniline molecule (giving ortho and para products) and the deactivated anilinium ion (giving the meta product).


This is why a substantial amount of m-nitroaniline (around 47%) is formed in addition to the expected o- and p-nitroanilines.
Quick Tip: Whenever aniline undergoes a reaction in a strong acid (like nitration or sulfonation), always consider the formation of the anilinium ion, which changes the directing effect from ortho, para to meta.

(I) Aniline to p-bromoaniline:

The -NH\(_2\) group is highly activating, and direct bromination of aniline yields 2,4,6-tribromoaniline. To obtain the monobromo derivative, the amino group must first be protected by acetylation to reduce its activating effect.


Step 1: Protection (Acetylation): Aniline is treated with acetic anhydride in the presence of pyridine to form acetanilide.

C\(_6\)H\(_5\)NH\(_2\) + (CH\(_3\)CO)\(_2\)O \(\xrightarrow{Pyridine}\) C\(_6\)H\(_5\)NHCOCH\(_3\) + CH\(_3\)COOH


Step 2: Bromination: Acetanilide is treated with bromine in acetic acid. The para product is major due to steric hindrance.

C\(_6\)H\(_5\)NHCOCH\(_3\) + Br\(_2\) \(\xrightarrow{CH_3COOH}\) p-Br-C\(_6\)H\(_4\)-NHCOCH\(_3\)


Step 3: Deprotection (Hydrolysis): The resulting p-bromoacetanilide is hydrolyzed with acid or base to give p-bromoaniline.


p-Br-C\(_6\)H\(_4\)-NHCOCH\(_3\) \(\xrightarrow{H^+ or OH^-, \Delta}\) p-Br-C\(_6\)H\(_4\)-NH\(_2\)


(II) Aniline to phenol:

This conversion is done via the formation of a diazonium salt.

Step 1: Diazotization: Aniline is treated with nitrous acid (NaNO\(_2\) + dil. HCl) at low temperature (0-5°C or 273-278 K) to form benzenediazonium chloride.


C\(_6\)H\(_5\)NH\(_2\) + NaNO\(_2\) + 2HCl \(\xrightarrow{273-278 K}\) C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) + NaCl + 2H\(_2\)O


Step 2: Hydrolysis: The aqueous solution of the diazonium salt is warmed. The diazonium group is replaced by a hydroxyl group, releasing nitrogen gas and forming phenol.

C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) + H\(_2\)O \(\xrightarrow{Warm}\) C\(_6\)H\(_5\)OH + N\(_2\) + HCl
Quick Tip: Benzenediazonium salts are extremely versatile intermediates in organic synthesis. They are the gateway to converting an amino group on a benzene ring to many other functional groups like -OH, -X, -CN, and -H.

This reaction is the Carbylamine reaction, also known as the isocyanide test, which is a characteristic test for primary amines.


Ethylamine (a primary amine) reacts with chloroform (CHCl\(_3\)) and ethanolic potassium hydroxide (KOH) upon heating to form ethyl isocyanide.


Ethyl isocyanide is the foul-smelling gas observed.


The balanced chemical equation for the reaction is:

CH\(_3\)CH\(_2\)NH\(_2\) + CHCl\(_3\) + 3KOH (ethanolic) \(\xrightarrow{\Delta}\) CH\(_3\)CH\(_2\)NC + 3KCl + 3H\(_2\)O

(Ethylamine) \(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad\) (Ethyl isocyanide)
Quick Tip: The carbylamine test is a definitive way to distinguish primary amines (both aliphatic and aromatic) from secondary and tertiary amines, as only primary amines produce the characteristic foul odor of an isocyanide.

Identification of A:

The reagent H\(_2\)/Pd is used for catalytic hydrogenation.

When aniline (C\(_6\)H\(_5\)NH\(_2\)) is treated with H\(_2\) in the presence of a palladium catalyst, the aromatic benzene ring is reduced to a saturated cyclohexane ring.

The amino group (-NH\(_2\)) is not affected.

Therefore, A is Cyclohexylamine (C\(_6\)H\(_{11}\)NH\(_2\)).


Identification of B:

The reagent Br\(_2\)/NaOH is used for the Hofmann bromamide degradation reaction.

This reaction converts a primary amide into a primary amine containing one less carbon atom than the original amide.

To obtain aniline (C\(_6\)H\(_5\)NH\(_2\)) as the product, the starting material 'B' must be an amide with one more carbon atom, specifically benzamide.

C\(_6\)H\(_5\)CONH\(_2\) + Br\(_2\) + 4NaOH \(\rightarrow\) C\(_6\)H\(_5\)NH\(_2\) + Na\(_2\)CO\(_3\) + 2NaBr + 2H\(_2\)O

Therefore, B is Benzamide (C\(_6\)H\(_5\)CONH\(_2\)).
Quick Tip: Recognize these two fundamental reactions: Catalytic hydrogenation (H\(_2\)/catalyst) reduces rings and double/triple bonds. Hofmann bromamide degradation (Br\(_2\)/NaOH) is a step-down reaction that shortens a carbon chain by converting an amide to an amine.

(I) Aniline to Benzene:

This conversion involves diazotization followed by reduction (deamination).


Step 1: Diazotization. Aniline is treated with nitrous acid (NaNO\(_2\) + dil. HCl) at a low temperature (273-278 K) to form benzenediazonium chloride.


C\(_6\)H\(_5\)NH\(_2\) + NaNO\(_2\) + 2HCl \(\xrightarrow{273-278 K}\) C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) + NaCl + 2H\(_2\)O


Step 2: Reduction. The diazonium salt is treated with a mild reducing agent like hypophosphorous acid (H\(_3\)PO\(_2\)) or ethanol to replace the diazonium group with a hydrogen atom.


C\(_6\)H\(_5\)N\(_2^+\)Cl\(^-\) + H\(_3\)PO\(_2\) + H\(_2\)O \(\rightarrow\) C\(_6\)H\(_6\) + N\(_2\) + H\(_3\)PO\(_3\) + HCl

(Benzene)


(II) Aniline to Sulphanilic acid:

This is the sulfonation of aniline.

Aniline is reacted with concentrated sulfuric acid. Initially, an acid-base reaction occurs to form anilinium hydrogensulfate.


C\(_6\)H\(_5\)NH\(_2\) + H\(_2\)SO\(_4\) (conc.) \(\rightarrow\) C\(_6\)H\(_5\)NH\(_3^+\)HSO\(_4^-\)


(Anilinium hydrogensulfate)

Upon heating this salt at a high temperature (453-473 K), rearrangement and electrophilic substitution occur at the para position to yield p-aminobenzenesulfonic acid, commonly known as sulphanilic acid.

C\(_6\)H\(_5\)NH\(_3^+\)HSO\(_4^-\) \(\xrightarrow{453-473 K}\) p-NH\(_2\)-C\(_6\)H\(_4\)-SO\(_3\)H + H\(_2\)O

(Sulphanilic acid)

Sulphanilic acid exists predominantly as a zwitterion (p-\(^+\)NH\(_3\)-C\(_6\)H\(_4\)-SO\(_3^-\)).
Quick Tip: Deamination (removing -NH\(_2\)) is a key synthetic tool that proceeds via diazotization. For sulfonation of aniline, remember that direct reaction leads to the anilinium salt, which must be heated to induce the ring substitution, strongly favoring the para product.