Chemistry Test 300

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Chemistry Test 300

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Weekly Quiz Competition

Question 1
4 / -1

Match List-I (Co(III) complexes) with List-II (crystal field splitting energy, Δ_o in cm^-1):

List-I: List-II:

A. [CoF₆]^3- I. 13,000

B. [Co(H₂O)₆]³⁺ II. 18,200

C. [Co(NH₃)₆]³⁺ III. 22,900

D. [Co(en)₃]³⁺ IV. 23,200

Choose the correct answer from the options given below:

A
A-I, B-II, C-III, D-IV

B
A-II, B-I, C-IV, D-III

C
A-I, B-III, C-II, D-IV

D
A-III, B-IV, C-I, D-II
Solution
- In the given question, the Δ_o (crystal field splitting energy) values for the Co(III) complexes can be matched based on the strength of the ligands in the spectrochemical series:
- List-I (Co(III) complexes) and their corresponding Δ_o values:
  
  [CoF₆]^3-: Fluoride (F^-) is the weakest ligand. It causes the smallest splitting energy. Therefore, Δ_o= 13,000cm⁻¹ (I).
  
  [Co(H₂O)₆]³⁺: Water (H₂O) is a stronger ligand than F^-. Hence, Δ_o= 18,200cm⁻¹ (II).
  
  [Co(NH₃)₆]³⁺: Ammonia (NH₃) is stronger than H₂O. Therefore, Δ_o= 22,900cm⁻¹ (III).
  
  [Co(en)₃]³⁺: Ethylenediamine (en) is the strongest ligand among the given options. Additionally, the chelate effect further increases the splitting energy. Thus, Δ_o= 23,200cm⁻¹ (IV).
- [CoF₆]^3- → I (13,000)
- [Co(H₂O)₆]³⁺ → II (18,200)
- [Co(NH₃)₆]³⁺ → III (22,900)
- [Co(en)₃]³⁺ → IV (23,200)
Therefore, the correct match is: A-I, B-II, C-III, D-IV.
Question 2
4 / -1

Given below are two statements:

Statement I: Aniline can be purified from a mixture containing high-boiling tar impurities by steam distillation.

Statement II: A mixture of ethanol and water forming a minimum-boiling azeotrope (95.6% ethanol) can be separated into pure components by simple fractional distillation.

In the light of the above statements, choose the correct answer from the options given below:

A
Both Statement I and Statement II are true

B
Both Statement I and Statement II are false

C
Statement I is true but Statement II is false

D
Statement I is false but Statement II is true
Solution
- Statement I:
  
  Aniline is slightly soluble in water and steam-volatile with a high boiling point (184°C).
  
  Steam distillation helps to purify aniline by allowing it to distil over at a temperature below 100°C, effectively separating it from non-volatile tar impurities.
  
  True: Statement I is correct.
- Statement II:
  
  A mixture of ethanol and water forms a minimum-boiling azeotrope at 95.6% ethanol composition.
  
  At this composition, the vapour produced during boiling has the same ratio of ethanol and water as the liquid mixture.
  
  Fractional distillation cannot separate the components further into pure ethanol and water.
  
  False: Statement II is incorrect.
Therefore, Statement I is true but Statement II is false.
Question 3
4 / -1

Given below are two statements:

Statement I: Neopentyl bromide reacts extremely slowly under S_N2 conditions despite being a primary alkyl halide.

Statement II: The bulky group adjacent to the reacting carbon in neopentyl bromide causes significant steric hindrance to backside attack by the nucleophile.

In the light of the above statements, choose the correct answer from the options given below:

A
Both Statement I and Statement II are true

B
Both Statement I and Statement II are false

C
Statement I is true but Statement II is false

D
Statement I is false but Statement II is true
Solution
- Statement I: Neopentyl bromide ((CH₃)₃C–CH₂Br) is a primary alkyl halide, which typically reacts faster in S_N2 reactions. However, in this case, the carbon atom adjacent to the alpha-carbon (the beta-carbon) is a quaternary carbon with three bulky methyl groups. These bulky groups create significant steric hindrance, making it extremely difficult for a nucleophile to access the alpha-carbon for a backside attack. As a result, neopentyl bromide reacts extremely slowly under S_N2 conditions. Therefore, Statement I is true.
- Statement II: The explanation provided in this statement is accurate. The bulky tert-butyl group on the beta-carbon of neopentyl bromide sterically shields the alpha-carbon, which prevents effective nucleophilic attack. This steric hindrance is the primary reason for the slow reaction rate in S_N2 conditions. Therefore, Statement II is true.
So, Both Statement I and Statement II are true, and Statement II correctly explains the slow reaction rate mentioned in Statement I.
Question 4
4 / -1

Choose the correct answer from the options given below :

A
A - I, B - III, C - IV, D - II

B
A - II, B - IV, C - III, D - I

C
A - II, B - III, C - I, D - IV

D
A - II, B - III, C - IV, D - I

Solution
Question 5
4 / -1

The correct IUPAC name of the compound CH₃–CH(OH)–CH₂–CHO is:

A
3-hydroxybutanal

B
1-hydroxybutan-3-al

C
2-hydroxybutanal

D
3-oxobutan-1-ol
Solution
- Number the longest chain to give the aldehyde carbon the lowest locant (C1).
- The structure is:
  
  C1(CHO)–C2(CH₂)–C3(CH(OH))–C4(CH₃)
- The OH group is located on C3, so it is named as '3-hydroxy'.
- The parent chain consists of 4 carbons with a terminal aldehyde group (CHO), making the base name 'butanal'.
Therefore, the correct IUPAC name is 3-hydroxybutanal.
Question 6
4 / -1

In acetanilide, the preferred position(s) for electrophilic aromatic substitution (e.g. nitration) is/are predominantly:

A
ortho only

B
meta only

C
ortho and para equally

D
predominantly para
Solution
- In N-phenylacetamide (acetanilide), the lone pair on nitrogen in the –NHCOCH₃ group is delocalized into the aromatic ring, activating both the ortho and para positions electronically.
- Due to steric bulk, the ortho positions adjacent to the substituent experience steric hindrance, which disfavors electrophilic attack at these positions.
- The para position is less sterically hindered while still being electronically activated, making it the preferred position for electrophilic aromatic substitution reactions.
Therefore, in N-phenylacetamide, electrophilic aromatic substitution predominantly occurs at the para position.
Question 7
4 / -1

The major product of the following reaction sequence is

A

B

C

D
Solution
- The given ketone is an aryl ketone having α-hydrogen, so it undergoes cross aldol reaction with excess HCHO in presence of NaOH.
- This leads to formation of a β-hydroxy ketone intermediate.
- Under strongly basic conditions with excess HCHO:
  
  Formaldehyde undergoes Cannizzaro reaction.
  
  The intermediate carbonyl compound gets reduced, resulting in formation of a diol (two −OH groups).
- In the second step (HCHO/H⁺, catalytic):
  
  The formed diol reacts with formaldehyde.
  
  This results in intramolecular cyclization forming a six-membered cyclic acetal (1,3-dioxane ring).
- The final product is therefore a cyclic acetal attached to the benzene ring, which is the most stable product under these conditions.
Thus, the major product formed is the cyclic acetal (1,3-dioxane derivative) obtained via aldol → Cannizzaro → acetal formation sequence.
Question 8
4 / -1

Arrange the following compounds in order of their increasing acid strength

A
I < II < III < IV

B
IV < III < II < I

C
I < IV < II < III

D
II < I < III < IV
Solution
- The given compounds are:
  
  I: HOOC(CH₂)₄COOH (Adipic acid)
  
  II: HOOCCH₂COOH (Malonic acid)
  
  III: HOOC–COOH (Oxalic acid)
  
  IV: HOOC(CH₂)₂COOH (Succinic acid)
- In compound I, the two –COOH groups are separated by four –CH₂– groups, so the −I effect is the weakest. Hence, it is the least acidic.
- In compound IV, the –COOH groups are separated by two –CH₂– groups, so its acidity is greater than that of compound I.
- In compound II, only one –CH₂– group separates the two –COOH groups. The stronger −I effect increases its acidity further.
- In compound III, the two –COOH groups are directly attached to each other, producing the maximum electron-withdrawing effect. Therefore, it is the most acidic.
- Hence, the order of increasing acid strength is: I < IV < II < III
Therefore, the increasing order of acid strength is I < IV < II < III.
Question 9
4 / -1

Match List-I (Vitamin) with List-II (Chemical name):

List-I: List-II: I

A. Vitamin A I. Retinol

B. Vitamin D II. Calciferol

C. Vitamin E III. Tocopherol,

D. Vitamin K IV. Phylloquinone

Choose the correct answer from the options given below:

A
A-I, B-III, C-II, D-IV

B
A-II, B-I, C-IV, D-III

C
A-III, B-IV, C-I, D-II

D
A-I, B-II, C-III, D-IV
Solution
- The given question asks for the correct matching between the vitamins and their chemical names:
  
  Vitamin A corresponds to Retinol (I).
  
  Vitamin D corresponds to Calciferol (II).
  
  Vitamin E corresponds to Tocopherol (III).
  
  Vitamin K corresponds to Phylloquinone (IV).
Therefore, the correct answer is: A-I, B-II, C-III, D-IV.
Question 10
4 / -1

What are X and Y respectively in the following reactions?

A
C₂H₅CI, CH₃COCI

B
C₂H₅COCI, CH₃CI

C
CH₃COCI, CH₃CI

D
CH₃C1, CH₃COCI
Solution
- Formation of Acetophenone:
  
  The reagent X required for this Friedel-Crafts acylation is Acetyl chloride (CH₃COCl) in the presence of anhydrous AlCl₃.
  
  To convert Benzene to Acetophenone, an acetyl group (-COCH₃) must be added.
  
  Acetophenone is a methyl phenyl ketone (C₆H₅COCH₃).
  
  Formation of Toluene:
  
  Toluene is methylbenzene (C₆H₅CH₃).
  
  To convert Benzene to Toluene, a methyl group (-CH₃) must be added.
  
  The reagent Y required for this Friedel-Crafts alkylation is Methyl chloride (CH₃Cl) in the presence of anhydrous AlCl₃.
Therefore, X is CH₃COCl and Y is CH₃Cl.