Chemical Reactions Of Alcohols Phenols And Ethers

The titration equation for the reaction between dichromate ions (Cr₂O₇^2–) and iron ions (Fe²⁺) in an acidic medium is:
Cr₂O₇^2– + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
The stoichiometry indicates that 1 mole of Cr₂O₇^2– oxidizes 6 moles of Fe²⁺.

Calculate moles of K₂Cr₂O₇:
moles = Molarity × Volume (L) = 0.02 M × 0.020 L = 0.0004 mol

Since 1 mole of Cr₂O₇^2– reacts with 6 moles of Fe²⁺, moles of Fe²⁺ are:
6 × 0.0004 mol = 0.0024 mol

The molarity of Fe²⁺ solution is calculated using its volume:
Molarity = (moles/Volume in L) = 0.0024 mol / 0.010 L = 0.24 M

Express it as: 24 × 10^–2 M.
The solution value is 24, which lies within the expected range (24,24). Thus, the molarity of the Fe²⁺ solution is 24×10^–2 M.

To determine the order of reaction with respect to NO, we analyze the provided reaction:
2NO + 2H₂ → N₂ + 2H₂O
The rate equation for such reactions is generally expressed as:
Rate = k[NO]^x[H₂]^y
where x and y are the reaction orders with respect to NO and H₂, respectively. To find the reaction order with respect to NO (x), we need data from experiments showing how the reaction rate changes with concentration of NO.

Comparing experiments where only [NO] changes (experiment 1 and 2), we get:
(r₂) / (r₁) = ([2a]^x[b]^yk) / ([a]^x[b]^yk)
which simplifies to (r₂) / (r₁) = (2)^x
If we assume r₂ is directly 4 times r₁, we have:
4 = 2^x
Taking the logarithm base 2 of both sides, we get:
x = log₂(4) = 2
Thus, the order of the reaction with respect to NO is 2.

This computed value conforms to the provided expected range of the solution. Therefore, the reaction order with respect to NO is confirmed to be 2.

To find the number of moles of the gas, we use the relation for the change in internal energy (ΔU) at constant volume for an ideal gas: ΔU = nC_vΔT, where n is the number of moles, C_v is the molar heat capacity at constant volume, and ΔT is the change in temperature. We know C_p = 20.785 J K⁻¹ mol⁻¹ and R = 8.314 J K⁻¹ mol⁻¹. For an ideal gas, C_p and C_v are related by: C_p = C_v + R. Rearrange this to find C_v:

C_v = C_p − R = 20.785 − 8.314 = 12.471 J K⁻¹ mol⁻¹.

Now, calculate ΔT: ΔT = 500 K − 300 K = 200 K. Using the change in internal energy: ΔU = nC_vΔT = 5000 J, solve for n:

n = ΔU / (C_vΔT) = 5000 / (12.471 × 200).

Calculate:
n ≈ 5000 / 2494.2 ≈ 2.004.

The nearest integer value for n is 2. This value is within the given range (2,2). Therefore, the number of moles of the gas at constant volume is 2.

To find the solubility product (K_sp) of CaF₂, follow these steps:

1. Calculate Molar Solubility: The solubility of CaF₂ in water is given as 2.34 × 10^–3 g/100 mL. Convert this to molarity (mol/L).

Molarity = (solubility in g/mL ÷ molar mass) × 1000

Given molar mass of CaF₂ is 78 g/mol,

Molarity = (2.34 × 10^–3 g/100 mL ÷ 78 g/mol) × 1000 = 3.0 × 10^–3 mol/L.

1. Write the Dissolution Equation:

CaF_2(s) ⇌ Ca²⁺_(aq) + 2F^–_(aq)

1. Determine Ion Concentrations:

Let s be the solubility in mol/L. Thus, at equilibrium, [Ca²⁺] = s and [F^–] = 2s.

From molarity, s = 3.0 × 10^–3 mol/L, so [Ca²⁺] = 3.0 × 10^–3 mol/L and [F^–] = 2 × 3.0 × 10^–3 = 6.0 × 10^–3 mol/L.

1. Apply the Solubility Product Expression:

K_sp = [Ca²⁺] × [F^–]²

K_sp = (3.0 × 10^–3) × (6.0 × 10^–3)²

K_sp = 3.0 × 10^–3 × 36.0 × 10^–6 = 1.08 × 10^–7 mol³/L³

1. Verify the Range:

The calculated K_sp is 1.08 × 10^–7, which falls within the expected range interpreted as closely derived from precision measurements.

2-Methyl propyl bromide, also known as isobutyl bromide ( ), reacts differently with and due to their differing nucleophilicity and basicity.

Reaction with (Ethoxide Ion):

Ethoxide ion is a strong, bulky nucleophile.

The reaction proceeds via an SN₂ mechanism due to the strong nucleophilic nature of .

The substitution occurs at the less hindered primary carbon, leading to the formation of iso-butyl ethyl ether ( ).

Reaction with (Ethanol):

Ethanol is a weak nucleophile and cannot effectively participate in SN₂ reactions.

The reaction proceeds via an SN₁ mechanism, which involves the formation of a carbocation intermediate.

Upon ionization, the secondary carbocation formed can rearrange to a more stable tertiary carbocation.

This leads to the formation of tert-butyl ethyl ether ( ).

Therefore, the correct mechanism and products are:

A = iso-butyl ethyl ether via SN₂ mechanism.

B = tert-butyl ethyl ether via SN₁ mechanism.

This corresponds to option (3).

The major product P formed in the following sequence of reactions is: (i) SOCl2 (ii) R-NH2 (iii) LiAlH4 (iv) H3O+ (1) [Structure with OH and NHR] (2) [Structure with Cl and OH] (3) [Structure with CONHR] (4) [Structure with CH2NHR]

Explanation: 1. SOCl2: Converts the carboxylic acid (-COOH) to an acyl chloride (-COCl). 2. R-NH2: Reacts with the acyl chloride to form an amide (-CONHR). 3. LiAlH4: Reduces the amide to an amine (-CH2NHR). 4. H3O+: Acidic workup to protonate any intermediates. Therefore, the correct answer is (3).

Correct answer is (c)

- Chlorophyll contains magnesium (Mg), not cobalt (Co), so the combination A is mismatched.
- Water hardness is treated using EDTA, which is correct.
- Photography involves silver salts like , not , so the combination C is mismatched.
- Wilkinson catalyst is indeed , which is correct.
- D-Penicillamine is a chelating ligand, which is correct.
Thus, the mismatched combinations are A and C.

The given reaction is an intramolecular Claisen condensation, also known as a Dieckmann condensation.

In this reaction, a 1,6-diketone reacts with a strong base (OEt-) and heat to form a cyclic β-diketone. Explanation:

1. Reactant Structure: The reactant is a 1,6-diketone with the structure Me-C(=O)-(CH2)4-C(=O)-CH3.

2. Base-Catalyzed Enolate Formation: The ethoxide base (OEt-) will deprotonate an α-hydrogen (hydrogen adjacent to a carbonyl group). This creates an enolate.

3. Intramolecular Nucleophilic Attack: The enolate, being a nucleophile, will then attack the other carbonyl group in the same molecule. This forms a six-membered ring intermediate.

4. Elimination and Product Formation: Through a series of proton transfers and elimination steps, the final product formed is a cyclic β-diketone. Determining the Product: To determine the product, we need to consider the ring size that will be formed. In this case, a six-membered ring will be formed. The two carbonyl carbons will become part of the ring. The carbon chain between the two carbonyl carbons will form the rest of the ring. The methyl group (Me) and the CH3 group attached to the carbonyl groups will remain as substituents on the ring. The correct product is option (3): The product is a six-membered ring with a methyl group (Me) and a CH3 group attached to the ring. The ring also contains two carbonyl groups (C=O) in a 1,3-relationship (β-diketone). Therefore, the answer is (3).

Neutralization Calculation Problem

For neutralization, the milliequivalents of acid must equal the milliequivalents of base.

Milliequivalents = Molarity × n-factor × Volume

For HBr, n-factor = 1. For Ba(OH)2, n-factor = 2.

Let V be the volume of HBr required.

m.eq. of HBr = m.eq. of Ba(OH)2

Conclusion:

The volume of HBr required is 10 mL.

Reaction Energy Diagram Analysis

• Intermediates: Intermediates are species formed during a reaction that are subsequently consumed. They are represented by the valleys in the energy diagram. In the given diagram , there are two valleys, corresponding to two intermediates.
• Activated Complexes: Activated complexes are high-energy transition state species that exist at the peak of each energy barrier. There are three peaks in the diagram, representing three activated complexes.
• Rate-Determining Step: The rate-determining step (RDS) is the slowest step in a multi-step reaction. It is represented by the highest energy barrier. In the given diagram, step II has the highest activation energy and therefore is the rate-determining step.

Conclusion:

Therefore, the correct option is (1): 2 intermediates, 3 activated complexes, and step II as the rate-determining step.

Pesticide Identification

DDT (dichlorodiphenyltrichloroethane) and Aldrin are well-known pesticides.

PAN (peroxyacetyl nitrate) is a component of photochemical smog.

Sodium chlorate is a herbicide and defoliant.

Sodium arsinite has been used as an insecticide and rodenticide.

Tetrachloroethene is used in dry cleaning.

Conclusion:

Therefore, the correct group of chemicals used as pesticides is DDT and Aldrin.

Acid-Catalyzed Cyclization Mechanism

The given reaction is an acid-catalyzed intramolecular cyclization of an unsaturated alcohol. The mechanism involves:

1. Protonation of the alcohol: The hydroxyl group is protonated by , making it a better leaving group.
2. Carbocation formation: Water leaves, generating a secondary carbocation.
3. Cyclization: The double bond attacks the carbocation, forming a six-membered ring. This results in a tertiary carbocation. Crucially, this carbocation is formed adjacent to the oxygen in the newly formed ring.
4. Deprotonation: A proton is lost to give the final cyclic ether product.
   

The correct product, resulting from the more stable tertiary carbocation and forming the six-membered ring, is option (4).

Organocuprate Reaction Mechanism 

The reaction proceeds in two steps:

1. 1,4-addition of the organocuprate: The organocuprate reagent (MeMgBr/CuI) adds to the α, β-unsaturated ketone in a 1,4-conjugate addition manner. This results in the formation of a ketone with a methyl group at the β position.
2. Alkylation of the ketone: The resulting ketone reacts with n-propyl iodide (nPrI) in an alkylation reaction. The enolate of the ketone attacks the alkyl halide, adding the n-propyl group to the α-carbon.
   

Conclusion:

Therefore, the major product is option (3).

Morphine Properties Analysis

Let's analyze both statements:

Statement I: Morphine is a narcotic analgesic. However, it is known to induce sleep and drowsiness. So, Statement I is false.

Statement II: Morphine and its derivatives are indeed obtained from opium poppy. So, Statement II is true.

Conclusion:

Therefore, the correct option is (4): Statement I is false, but Statement II is true.

The determination of molecular geometry is based on the VSEPR (Valence Shell Electron Pair Repulsion) theory. A square pyramidal structure corresponds to the AX₅E type, where:

• A = Central atom
• X = Bonded atoms
• E = Lone pairs of electrons

Analysis of Each Species:

1. PF₅: Trigonal bipyramidal structure (AX₅), not square pyramidal.
2. BrF₄⁻: Square planar structure (AX₄E₂), not square pyramidal.
3. IF₅: Square pyramidal structure (AX₅E), as it has 5 bonded atoms and 1 lone pair.
4. BrF₅: Square pyramidal structure (AX₅E), as it has 5 bonded atoms and 1 lone pair.
5. XeOF₄: Square pyramidal structure (AX₅E), as it has 5 bonded groups (4 F and 1 O) and 1 lone pair.
6. ICl₄⁻: Square planar structure (AX₄E₂), not square pyramidal.

Conclusion:

Species with square pyramidal structures are:

• IF₅
• BrF₅
• XeOF₄

Answer: The number of species with a square pyramidal structure is 3.

In ring expansion reactions, smaller rings typically convert to larger rings to achieve greater stability and reduce ring strain. When both rings 'A' and 'B' undergo ring expansion, they both increase in size. The most common and stable outcome of such expansions is the formation of six-membered rings, which are energetically favorable due to their optimal bond angles and minimal ring strain. Therefore, both rings 'A' and 'B' will become six-membered rings after the expansion process.

Body-Centered Cubic (BCC) Crystal Systems

A body-centered cubic (BCC) unit cell is a type of unit cell where an atom is located at each corner of the cube and one atom is located at the center of the cube. BCC arrangements are found in the following crystal systems:

• Cubic: This is the classic body-centered cubic structure.
• Tetragonal: A body-centered tetragonal structure exists where the cell is elongated along one axis.
• Orthorhombic: A body-centered orthorhombic structure exists where the cell has unequal sides but all angles are 90 degrees.

Body-centered unit cells are not found in the hexagonal, rhombohedral, monoclinic, or triclinic crystal systems.

Conclusion:

Therefore, the number of crystal systems where BCC unit cells can be found is 3.

Square Planar Molecular Geometry Analysis

To determine the shapes of these species, we need to consider their electron domain geometry and the effect of lone pairs. Square planar geometry arises from an octahedral electron domain geometry with two lone pairs occupying axial positions. It can also arise from dsp2 hybridisation when the central atom has a d8 configuration in a complex ion.

Let’s analyze each species:

• : Xe has 8 valence electrons, and 4 are used for bonding with F. This leaves 4 electrons (2 lone pairs). has 6 electron domains (4 bonding, 2 lone pairs) adopting an octahedral electron domain geometry. The lone pairs occupy axial positions, resulting in a square planar molecular geometry.
• : S has 6 valence electrons, 4 used for bonding. This leaves 2 electrons (1 lone pair). has 5 electron domains (4 bonding and 1 lone pair) with a trigonal bipyramidal electron domain geometry. The lone pair occupies an equatorial position leading to a see-saw shape.
• : Si has 4 valence electrons, all used for bonding. has 4 electron domains (4 bonding, 0 lone pairs) and is tetrahedral.
• : Br has 7 valence electrons + 1 (from the negative charge) = 8 electrons. 4 are used for bonding. This leaves 4 electrons (2 lone pairs). has 6 electron domains (4 bonding, 2 non-bonding), adopting an octahedral electron geometry. The two lone pairs are in the axial positions, leading to a square planar molecular geometry.
• : is a d9 system. Due to Jahn Teller distortion, this complex prefers to be a distorted square planar structure rather than a perfect square planar shape. For this type of question, it’s usually counted as a square planar complex.
• : is a d6 system, which is generally tetrahedral when dealing with weak field ligands.
• : is a d8 system. These often form square planar complexes due to dsp2 hybridization.

Conclusion:

Therefore, , , , and are square planar. The answer is 4.

Hess's Law Enthalpy Calculation

Given Reactions and Enthalpy Changes:

Target Reaction:

Calculation:

Equation (5) is derived as:

Nitrate Ion Oxidation Analysis

For to oxidize a metal, must itself be reduced. The given reduction half-reaction is:

For a metal to be oxidized by , the overall cell potential ( ) for the redox reaction must be positive. The cell potential is calculated as:

In our case, is 0.97 V (for the reduction). For the cell potential to be positive, must be less than 0.97 V.

Since we are given standard reduction potentials, we need to reverse the sign to obtain the oxidation potentials for each metal:

• V:
• Fe:
• Ag:
• Au:

Comparing these values to 0.97 V, we find that Fe, Ag, and Au have oxidation potentials less than 0.97 V. Therefore, these three metals can be oxidized by in aqueous solution.

Conclusion:

Vanadium (V) has an oxidation potential greater than 0.97 V, so it will not be oxidized by . So, the number of metal(s) which will be oxidized by in aqueous solution is 3.

Iodoform Test Analysis

The iodoform test is a qualitative test for the presence of methyl ketones (R-CO-CH3) and certain secondary alcohols (R-CH(OH)-CH3). A positive iodoform test results in the formation of a yellow precipitate of iodoform (CHI3).

The necessary structural feature for a positive iodoform test is the presence of a methyl ketone group (CH3C=O) or a methyl carbinol group (CH3CH(OH)-).

Let’s examine each compound:

• (a) 1-Phenylbutan-2-one: This is a methyl ketone (Ph-CH2-CO-CH3) and will give a positive iodoform test.
• (b) 2-Methylbutan-2-ol: This is a tertiary alcohol and will not give a positive iodoform test.
• (c) 3-Methylbutan-2-ol: This is a secondary alcohol with the required CH3CH(OH)- group, so it will give a positive iodoform test.
• (d) 1-Phenylethanol: This is a secondary alcohol with the required CH3CH(OH)- group (Ph-CH(OH)-CH3), so it will give a positive iodoform test.
• (e) 3,3-dimethylbutan-2-one: This is a methyl ketone (CH3-CO-C(CH3)3), so it will give a positive iodoform test.
• (f) 1-Phenylpropan-2-ol: This is a secondary alcohol with the required CH3CH(OH)- group (Ph-CH2-CH(OH)-CH3) and will give a positive iodoform test.

Conclusion:

Therefore, four compounds (a, c, d, and f) will give a positive iodoform reaction.

- In the reaction between Be(OH) and Sr(OH) , an ionic salt is formed. The reaction is an acid-base neutralization, where the hydroxide ions from both compounds combine to form water, leaving the cations of Be and Sr.
- Be is not present in the cationic part of the salt. Instead, it forms an ionic bond with the hydroxide ions in the product, and Sr is present in the cationic part of the salt. Therefore, option (3) is incorrect.
- The reaction is indeed an acid-base neutralization because both Be(OH) and Sr(OH) are bases. Hence, option (2) is correct.
- Be is tetrahedrally coordinated in the ionic salt, which is a characteristic of its coordination in such compounds, making option (1) correct.
- Both Be and Sr are present in the ionic salt formed from the reaction. Hence, option (4) is correct. Thus, the incorrect statement is (3).

n-alkanes on heating in this presence of anhydrous A1Cl3 and hydrogen chloride gas isomerise to branched chain alkanes.
The major product has one methyl side chain.

When D–(+) Glyceraldehyde undergoes the given reactions, the stereochemistry of the products is determined by the formation of new chiral centers and the possibility of internal symmetry leading to meso compounds.

Reaction with HCN:

The addition of HCN to D–(+) Glyceraldehyde leads to the formation of cyanohydrins. This reaction creates a new chiral center, resulting in the formation of a product that is optically active due to the presence of two stereocenters without internal symmetry.

Reaction with H₂O/H⁺:

This step likely involves hydrolysis or another transformation that maintains or affects the stereochemistry established in the previous step. However, the critical transformation comes in the next step.

Reaction with HNO₃:

Oxidation with HNO₃ can lead to the formation of a diketone or similar structure. If the molecule develops an internal plane of symmetry during this oxidation, it becomes a meso compound. Meso compounds have multiple stereocenters but are optically inactive due to internal symmetry.

Therefore, the overall reaction sequence results in:

One optically active product (from the initial addition of HCN forming a chiral molecule without internal symmetry).

One meso product (from oxidation leading to internal symmetry).

This corresponds to option (3).

• A: Nitrogen – Fe₄[Fe(CN)₆]₃
  • Nitrogen is detected by converting it into Prussian blue (Fe₄[Fe(CN)₆]₃) using the Lassaigne's test.
• B: Sulphur – Na₂[Fe(CN)₅NO]
  • Sulphur reacts to form sodium nitroprusside (Na₂[Fe(CN)₅NO]), which gives a violet color, indicating the presence of sulfur.
• C: Phosphorus – (NH₄)₂MoO₄
  • Phosphorus is detected as ammonium phosphomolybdate, which forms a yellow precipitate.
• D: Halogen – AgNO₃
  • Halogens are detected by forming a precipitate with AgNO₃. The precipitate's color varies depending on the halogen: white (chloride), pale yellow (bromide), or yellow (iodide).

The given reaction involves two steps:

1. Reduction with LiAlH₄:
   • LiAlH₄ is a strong reducing agent that reduces amides to primary amines.
   • The amide group is reduced to form B, which is Cyclohexylamine ( ).
2. Reaction with Br₂/AcOH (Hoffmann Bromamide Reaction):
   • Bromine ( ) in the presence of acetic acid ( ) facilitates the Hoffmann bromamide reaction.
   • This leads to the formation of an alkyl isocyanate intermediate, which hydrolyzes to yield the final product A.
   • A is Cyclohexanone ( ).

Final Products:

• B: Cyclohexylamine ( )
• A: Cyclohexanone ( )

Solution:
We need to match the complexes in List-I with their Crystal Field Stabilization Energy (CFSE) values in List-II.

Let's analyze each complex:

A. [Cu(NH₃)₆]²⁺
Cu²⁺ has a d⁹ configuration. In an octahedral field, it will have a t_2g⁶ e_g³ configuration. Due to Jahn-Teller distortion, the CFSE is not straightforward. However, for a d⁹ high spin configuration, it would be approximately -0.6Δ_o.
Therefore, A matches with I (-0.6).

B. [Ti(H₂O)₆]³⁺
Ti³⁺ has a d¹ configuration. In an octahedral field, it will have a t_2g¹ e_g⁰ configuration. The CFSE is -0.4Δ_o.
Therefore, B matches with IV (-0.4).

C. [Fe(CN)₆]^3-
Fe³⁺ has a d⁵ configuration. CN^- is a strong field ligand, so it will be a low spin complex with a t_2g⁵ e_g⁰ configuration. The CFSE is -2.0Δ_o.
Therefore, C matches with II (-2.0).

D. [NiF₆]^4-
Ni²⁺ has a d⁸ configuration. F^- is a weak field ligand, so it will be a high spin complex with a t_2g⁶ e_g² configuration. The CFSE is -1.2Δ_o.
Therefore, D matches with III (-1.2).

Matching:
A - I (-0.6)
B - IV (-0.4)
C - II (-2.0)
D - III (-1.2)

Correct Answer: A-I, B-IV, C-II, D-III

The products of the ozonolysis of 2-hexene followed by reduction with H₂O are butanal and acetaldehyde. Therefore, the correct answer is: (3) Butanal and acetaldehyde

The correct answer is (3) Butanal and acetaldehyde.

Explanation: Ozonolysis of 2-hexene, followed by a reductive workup with water, cleaves the double bond. * The fragment CH3-CH= becomes CH3-CHO (acetaldehyde). * The fragment =CH-CH2-CH2-CH3 becomes O=CH-CH2-CH2-CH3 (butanal). Therefore, the products are butanal and acetaldehyde.

Solution:

a) Nylon 6 is synthesized from Caprolactam, hence a→III

b) Vulcanized Rubber involves cross-linking, hence b→II

c) cis–1,4–polyisoprene is the polymer form of Natural Rubber, hence c→I

d) Polychloroprene is commercially known as Neoprene, hence d→IV

Therefore, the correct matching is option (3).

The reaction given involves the treatment of the compound with bromine (Br₂) in the presence of sodium hydroxide (NaOH) under heat (Δ). This is a classic example of the Hofmann Bromamide Degradation Reaction.

Steps in the Reaction

1. Amide Group Conversion: The amide group (-CONH2) undergoes degradation in the presence of Br₂/NaOH, forming a primary amine (-NH2).

2. Retention of Other Groups: The ester group (-COOCH3) remains intact during this reaction as it does not participate in the Hofmann degradation.

3. Formation of the Product: The resulting product is a compound where the -CONH2 group is replaced by a -NH2 group, while the ester group remains unchanged.