Chemistry Test 250

CONCEPT:

Types of Redox Reactions

• Intermolecular Redox Reaction: In this type, oxidation and reduction occur between different molecules.
• Intramolecular Redox Reaction: Both oxidation and reduction occur within the same molecule.
• Disproportionation Reaction: A specific type of redox reaction where an element in a single oxidation state is simultaneously oxidized and reduced, forming two different oxidation states.

EXPLANATION:

The reaction is: P₄ + 3NaOH + 3H₂O → 3NaH₂PO₂ + PH₃

• Phosphorus (P) in P₄ is initially in the zero oxidation state.
• In the products:
  • In NaH₂PO₂, phosphorus has an oxidation state of +1.
  • In PH₃, phosphorus has an oxidation state of -3.
• Thus, phosphorus in P₄ is both oxidized (to +1 in NaH₂PO₂) and reduced (to -3 in PH₃).

CONCLUSION:​

• The correct option is: Option 3 (Disproportionation Redox reaction)

CONCEPT:

Bond Dissociation Energy and pπ–pπ Interaction

• Bond dissociation energy is the energy required to break a bond between two atoms in a molecule.
• The bond dissociation energy can be affected by factors such as bond strength, atomic size, and additional bonding interactions like pπ–pπ interactions.
• In the case of BF₃ and CF₄:
  • BF₃ has a significant pπ–pπ interaction between the empty p-orbital of boron (B) and the lone pairs on fluorine (F).
  • This pπ–pπ interaction results in additional stabilization and strengthens the B–F bond in BF₃.
  • In CF₄, carbon does not have an empty p-orbital to participate in pπ–pπ interaction with fluorine, so such stabilization is absent.

EXPLANATION:

• Statement 1: Significant pπ–pπ interaction between B and F in BF₃ whereas there is no possibility of such interaction between C and F in CF₄.
  • This is correct because the additional pπ–pπ interaction between B and F in BF₃ contributes to the higher bond dissociation energy of the B–F bond.
• Statement 2: Lower degree of pπ–pπ interaction between B and F in BF₃ than that between C and F in CF₄.
  • This is incorrect because there is no pπ–pπ interaction between C and F in CF₄.
• Statement 3: Smaller size of B atom as compared to that of C atom.
  • While boron is indeed smaller than carbon, this does not explain the higher bond dissociation energy due to the specific role of pπ–pπ interaction in BF₃.
• Statement 4: Stronger bond between B and F in BF₃ as compared to that between C and F in CF₄.
  • This is too general and does not specify the role of pπ–pπ interaction, which is the key factor for the increased bond dissociation energy in BF₃.

CONCLUSION:

• The correct option is: Option 1 (Significant pπ–pπ interaction between B and F in BF₃ whereas there is no possibility of such interaction between C and F in CF₄)

CONCEPT:

Understanding Second Ionization Energy in Group 12 Elements (Zn, Cd, Hg)

• The second ionization energy is the energy required to remove an electron from a positively charged ion (after the first electron has been removed), making it a +2 ion.
• In Group 12 elements (Zn, Cd, Hg), the trend in ionization energy can be influenced by atomic size, electron shielding, and relativistic effects.
• General Trends in Group 12:
  • As we move down the group, atomic size increases due to the addition of electron shells, which typically results in lower ionization energy because the outer electrons are farther from the nucleus and more shielded.
  • However, mercury (Hg) shows relativistic effects due to its large atomic number. These effects cause the inner electrons to move closer to the nucleus, leading to a higher effective nuclear charge and slightly higher ionization energy than expected.

EXPLANATION:

• Zinc (Zn):
  • Located at the top of Group 12, Zn has a smaller atomic radius and lower shielding effect compared to Cd and Hg.
  • This results in Zn having the highest second ionization energy among Zn, Cd, and Hg.
• Cadmium (Cd):
  • Located below Zn, Cd has a larger atomic radius and increased electron shielding compared to Zn, which generally lowers its ionization energy.
  • Without relativistic effects (present in Hg), Cd’s second ionization energy is lower than Zn but expected to be lower than Hg as well.
• Mercury (Hg):
  • Hg is located below Cd in the group, so it would typically have the lowest ionization energy due to its larger atomic size and higher shielding.
  • However, due to relativistic effects, the inner electrons are pulled closer to the nucleus, increasing the effective nuclear charge felt by the outer electrons.
  • This results in a slightly higher second ionization energy for Hg than expected, often making it higher than Cd’s second ionization energy.

CONCLUSION:​

• The correct option is: Option 1 (Zn > Cd < Hg)

CONCEPT:

Reaction Sequence of Alcohols with Phosphorus Triiodide and Grignard Reagents

• The given sequence involves the conversion of ethanol (CH₃CH₂OH) to various intermediates through a series of reactions.
• In the first step, ethanol reacts with phosphorus triiodide (P + I₂) to form an alkyl iodide (ethyl iodide, A).
• The alkyl iodide then reacts with magnesium in ether to form a Grignard reagent (ethylmagnesium iodide, B).
• Grignard reagents react with formaldehyde (HCHO) to form primary alcohols with one additional carbon atom after hydrolysis.

CALCULATION:

• Step 1: Ethanol (CH₃CH₂OH) reacts with phosphorus triiodide (P + I₂) to form ethyl iodide (CH₃CH₂I), which is compound A.
• Step 2: Ethyl iodide (CH₃CH₂I) reacts with magnesium in ether to form ethylmagnesium iodide (CH₃CH₂MgI), which is compound B.
• Step 3: Ethylmagnesium iodide (CH₃CH₂MgI) reacts with formaldehyde (HCHO) to form a three-carbon compound after hydrolysis, resulting in n-propyl alcohol (CH₃CH₂CH₂OH), which is compound C.
• Step 4: Oxidation of n-propyl alcohol (C) would yield propanal (CH₃CH₂CHO), which is compound D.

CONCEPT:

Rate of Nucleophilic Addition Reactions (NAR) in Carbonyl Compounds

• The rate of nucleophilic addition reactions in carbonyl compounds depends on the electron-withdrawing or electron-donating effects of substituents attached to the carbonyl carbon.
• Electron-withdrawing groups increase the partial positive charge on the carbonyl carbon, making it more electrophilic and thus more susceptible to nucleophilic attack.
• Electron-donating groups decrease the electrophilicity of the carbonyl carbon, thereby reducing the rate of nucleophilic addition reactions.

CALCULATION:

• Compound I (Formaldehyde, HCHO): No alkyl or electron-donating groups are present, so the carbonyl carbon is highly electrophilic. This makes formaldehyde the most reactive toward nucleophilic addition.
• Compound II (Acetaldehyde, CH₃CHO): Contains one electron-donating methyl group, which slightly decreases the electrophilicity of the carbonyl carbon, making it less reactive than formaldehyde.
• Compound III (p-Nitrobenzaldehyde, C₆H₄(NO₂)CHO): Contains an electron-withdrawing nitro group (-NO₂) in the para position, which increases the electrophilicity of the carbonyl carbon. However, its effect is weaker compared to formaldehyde and acetaldehyde.
• Compound IV (p-Tolualdehyde, C₆H₄(CH₃)CHO): Contains an electron-donating methyl group (-CH₃) in the para position, which decreases the electrophilicity of the carbonyl carbon, making it the least reactive toward nucleophilic addition.

CONCLUSION:

The correct option is: Option 1

CONCEPT:

Reaction of Ethers with Hydrogen Iodide (HI)

• When an ether reacts with excess HI, the ether bonds (C–O bonds) are cleaved, resulting in the formation of alkyl iodides and phenols.
• In this reaction, each methoxy group (–OCH₃) will react with HI to produce methyl iodide (CH₃I) and a hydroxyl group on the benzene ring.
• The number of moles of HI required (n) will be equal to the number of methoxy groups present in the molecule since each methoxy group consumes one mole of HI.

EXPLANATION:

• The given compound contains three methoxy groups (–OCH₃) attached to the benzene ring.
• Each methoxy group will react with one mole of HI, resulting in the formation of three moles of CH₃I and replacing each –OCH₃ group with an –OH group.

CONCLUSION:

The correct option is: Option 1

CONCEPT:

Optical Isomerism in Compounds with Multiple Chiral Centers

• Optical isomerism occurs when a molecule has one or more chiral centers, leading to non-superimposable mirror images known as enantiomers.
• A chiral center is a carbon atom bonded to four different groups, creating two possible configurations (R and S) at each center.
• If there are multiple chiral centers, the maximum number of optical isomers is given by ( 2ⁿ), where ( n ) is the number of chiral centers.

EXPLANATION:

• In the given compound, CH₃–CH(OH)–CH(Cl)–CH₂–COOH, there are two chiral centers:
  • The carbon with –OH and –Cl substituents is a chiral center because it has four different groups attached: –CH₃, –OH, –Cl, and –CH₂–COOH.
  • The adjacent carbon with –CH₃ and –CH(OH)– groups is also a chiral center as it is bonded to four different groups: –CH₃, –CH(OH), –CH(Cl), and –CH₂–COOH.
• With two chiral centers, the maximum number of optical isomers is ( 2² = 4 ).

CONCLUSION:

The correct option is: Option 3