Chemistry Test 248

CONCEPT:

Identifying Acidic Species in a Reaction

• In an acid-base reaction, acidic species are those that can donate protons (H⁺) or have a positive charge.
• In the given reactions, we examine the right side of the equilibrium to identify which species can act as acids.
• Species such as H₂F⁺ and HPO₄^2- can act as acidic species in certain conditions.

CALCULATION:

• Reaction 1: HF + HClO₄ → H₂F⁺ + ClO₄^-
  • On the right side, we have H₂F⁺ and ClO₄^-.
  • H₂F⁺ is acidic due to its positive charge and ability to donate a proton.
  • ClO₄^- is a stable anion and is not considered acidic.
• Reaction 2: H₂O + H₂PO₄^- → H₃O⁺ + HPO₄^2-
  • On the right side, we have H₃O⁺ and HPO₄^2-.
  • Although H₃O⁺ is acidic, HPO₄^2- can also act as a weak acid under certain conditions, as it can donate a proton.
• Conclusion: The acidic species on the right side are H₂F⁺ and HPO₄^2-.

CONCLUSION:

• The correct option is: Option 1 (H₂F⁺, HPO₄^2-).

CONCEPT:

Radioactive Decay and Half-Life

• Radioactive decay is a first-order process where the rate of decay is proportional to the number of radioactive atoms present.
• The decay constant (k) is related to the half-life (t_1/2) by the equation: 
• The rate of disintegration (decay rate) is given by: Rate = k ⋅ N, where N is the number of radioactive atoms present.
• For a given mass of a radioactive isotope, the number of atoms can be calculated using Avogadro's number and the molar mass of the isotope.

Conclusion: 

The calculations show that the decay constant for P₃₂ is indeed 5.6 × 10^-7 sec^-1, making option 1 the correct answer.

CONCEPT:

Calculating Oxidation State of Iodine in Complex Ions

• To determine the oxidation state of an element in a complex ion, we set up an equation using the known oxidation states of other atoms in the molecule.
• In H₄IO₆^-, we have:
  • Four hydrogen (H) atoms, each with an oxidation state of +1.
  • Six oxygen (O) atoms, each with an oxidation state of -2.
  • The overall charge of the ion is -1.

CALCULATION:

• Let the oxidation state of iodine (I) be x.
• Setting up the equation for the oxidation states:
  •  4(+1) + x + 6(-2) = -1 
  • Simplifying:  4 + x - 12 = -1 
  •  x - 8 = -1 
  •  x = +7 

CONCLUSION:

• The correct option is: Option 1 (+7)

CONCEPT:

Bond Order and Bond Strength

• Bond Order: Bond order is an indicator of bond strength, stability, and length. It is calculated based on the difference between bonding and antibonding electrons in molecular orbitals.
• The formula for bond order is: 
• Bond Strength: The bond strength is directly proportional to the bond order. A higher bond order indicates a stronger bond, while a lower bond order indicates a weaker bond.

• Since bond strength is directly related to bond order, the correct answer is the species with the highest bond order first.

CONCEPT:

Determining Oxidation Numbers in Compounds with Multiple Nitrogen Atoms

• In compounds like NH₄NO₂, there are two nitrogen atoms in different groups with distinct oxidation states.
• NH₄NO₂ consists of the ammonium ion (NH₄⁺) and the nitrite ion (NO₂^-).
• To determine the oxidation state of nitrogen in each part of the molecule, we analyze NH₄⁺ and NO₂^- separately.

CONCEPT:

Effect of Substituents on pKa Values in Phenols

• The pKa value of a compound is a measure of its acidity, with a lower pKa indicating a stronger acid. In phenols, the presence and type of substituent on the aromatic ring can significantly impact the pKa value by affecting the stability of the phenoxide ion (the conjugate base).
• Electron-Withdrawing Groups (EWG): Substituents like NO₂ (nitro group) are electron-withdrawing. They stabilize the negative charge on the phenoxide ion by delocalizing electrons through the aromatic ring. This increased stabilization makes the compound more acidic, resulting in a lower pKa value.
• Electron-Donating Groups (EDG): Substituents like OMe (methoxy group) are electron-donating. They destabilize the negative charge on the phenoxide ion by pushing electron density into the ring, making the compound less acidic and raising the pKa value.

EXPLANATION:

• To determine the order of pKa values, we analyze the effect of each substituent on the phenol ring:
  • Compound B: Contains a para-NO₂ group, a strong electron-withdrawing group. The NO₂ group in the para position is highly effective at stabilizing the phenoxide ion through resonance and inductive effects. This makes Compound B the most acidic among the options, giving it the lowest pKa value.
  • Compound C: Contains a meta-NO₂ group, which is still electron-withdrawing but less effective at stabilizing the phenoxide ion compared to the para position. Thus, Compound C is less acidic than B but more acidic than A and D, giving it a relatively low pKa value.
  • Compound A: This is a simple phenol with no additional substituents. It has a moderate pKa value as there is no stabilization or destabilization beyond the basic phenol structure.
  • Compound D: Contains a para-OMe group, which is an electron-donating group. The OMe group destabilizes the phenoxide ion by donating electron density into the ring, making the compound less acidic. As a result, Compound D has the highest pKa value among the options.

CONCLUSION:

The correct option is: Option 4 B < C < A < D

CONCEPT:

Arrangement of Fluorine Atoms in Molecules

• In molecular geometry, the arrangement of atoms affects the spatial orientation of atoms and bonds, which determines the shape of the molecule.
• The position of atoms, particularly those in the same plane, is influenced by the type of bonding and the geometry around the central atom.
• For certain compounds, like XeOF₄, the geometry allows for some atoms (like fluorine) to be in the same plane, while others (like oxygen) may be above or below this plane.

CONCLUSION:

• The correct answer is XeOF₄, where all fluorine atoms are in the same plane.

CONCEPT:

Reduction of Nitriles and Esters Using DIBAL-H

• DIBAL-H (Diisobutylaluminum hydride) is a selective reducing agent commonly used to reduce nitriles to aldehydes under controlled conditions, usually at low temperatures.
• However, DIBAL-H can sometimes reduce other functional groups, especially if conditions are not tightly controlled or if excess DIBAL-H is used.
• In certain reactions, DIBAL-H can partially reduce ester groups or lead to further reduction of other groups, resulting in a complex product.

CALCULATION:

• The compound provided has a nitrile (–CN) group and an ester group on an aromatic ring.
• Step 1: In the presence of DIBAL-H, the nitrile group (–CN) is initially reduced to an imine intermediate. Upon hydrolysis with H₃O⁺, this imine is converted to an aldehyde group (–CHO).
• Step 2: Due to the reactive nature of DIBAL-H, especially when used in excess, additional reductions can occur, particularly with the ester group, converting it into an alcohol group (–OH).

• Reaction:

CONCLUSION:

The correct option is: Option 3

CONCEPT:

Hofmann Elimination Reaction

• The given reaction is an example of the Hofmann elimination, where a quaternary ammonium hydroxide undergoes elimination upon heating (pyrolysis) to produce an alkene.
• In Hofmann elimination, the elimination generally follows the anti-Zaitsev rule, meaning the reaction favors the formation of the least substituted alkene as the major product.
• This is due to the bulky nature of the quaternary ammonium leaving group, which makes it difficult to form a more substituted double bond.

EXPLANATION:

• In the given structure, there is a quaternary ammonium hydroxide, which is ideal for undergoing Hofmann elimination.
• During pyrolysis, the hydroxide ion abstracts a proton from the β-carbon, leading to the formation of an alkene.

Reaction:

• Following the Hofmann rule, the elimination reaction results in the formation of 1-butene as the major product, as it is the less substituted alkene compared to 2-butene.

CONCLUSION:

The correct option is: Option 3

CONCEPT:

Oxidation of Toluene and Decarboxylation

• Toluene (P) undergoes oxidation when treated with alkaline potassium permanganate (KMnO₄/KOH).
• This oxidation converts the methyl group (–CH₃) of toluene into a carboxylic acid group (–COOH), resulting in the formation of benzoic acid (Q).
• Benzoic acid, when heated with soda lime (a mixture of NaOH and CaO), undergoes decarboxylation, where the –COOH group is removed as CO₂, yielding benzene (R).

EXPLANATION:

CONCLUSION:

The correct option is: Option 3