Acids And Bases

The correct answer is (A).
NH₄OH is a weak base and HCl is a strong acid. With the addition of HCl to NH₄OH, pH of solution will decrease gradually.
So, the correct graph should be

=

Step 1: Analyze Assertion (A)
α-halocarboxylic acids react with dilute ammonia (NH3) to give a good yield of α-amino carboxylic acids. This is because the carboxyl group (-COOH) increases the reactivity of the α-halo group towards nucleophilic substitution. In contrast, the yield of amines from simple alkyl halides reacting with ammonia is low due to overalkylation, where the initially formed amine can react further with the alkyl halide. Therefore, Assertion (A) is correct.
Step 2: Analyze Reason (R)
Amino acids exist as zwitterions in aqueous solutions and in the solid state. A zwitterion has both positive and negative charges within the same molecule, resulting in a net charge of zero. This is due to the acidic carboxyl group and the basic amino group present in
amino acids. Thus, Reason (R) is correct.
Step 3: Analyze the Relationship between (A) and (R)
While both statements are individually correct, Reason (R) doesn’t explain Assertion
(A). The higher yield of amino acids from α-halocarboxylic acids is due to the enhanced reactivity of the α-halo group, not the zwitterionic nature of amino acids.
The zwitterionic form is a characteristic of the product (amino acid) but doesn’t explain the higher yield compared to the reaction of alkyl halides with ammonia. The correct explanation is the neighboring group participation of the carboxylic group makes the reaction proceed via a two-step process leading to a higher yield.

To evaluate the given statements, we need to understand Seliwanoff's test and the behavior of ketoses and aldoses in it.

Assertion (A): Ketoses give Seliwanoff's test faster than Aldoses.
Reason (R): Ketoses undergo -elimination followed by formation of furfural.

Seliwanoff's test is a chemical test to distinguish ketohexoses (such as fructose) from aldohexoses (such as glucose). In this test, when heated with resorcinol and conc. hydrochloric acid, ketoses react faster to produce a deep red color compared to aldoses, due to the faster dehydration reaction of ketoses to form reactive intermediates, which then condense with resorcinol to give the colored complex.

The key points to assess:

• The assertion is indeed correct because ketoses generally react faster in Seliwanoff's test.
• However, the reason provided is not accurate. While furfural derivatives are formed, the mechanism described as '-elimination followed by formation of furfural' does not accurately capture the process at play in Seliwanoff's test as it involves the formation of hydroxymethylfurfural, not simply furfural.
  This process primarily involves dehydration, rather than a simple -elimination.
  Thus, (R) does not correctly explain (A).

Based on this analysis, the correct choice is:

(A) is true but (R) is false

The Correct answer is (A) +6 to +3

1. Cyclic Tripeptides:
- A cyclic tripeptide is a peptide consisting of three amino acids arranged in a cyclic structure without any free ends.
- Given two amino acids, A and B, the tripeptide can have different arrangements depending on the sequence of A and B.
2. Possible Arrangements:
- The cyclic nature of the tripeptide eliminates linear sequence redundancy.
- The distinct arrangements are:
1. A-A-A
2. A-A-B
3. A-B-A
4. A-B-B
Total: 4 unique cyclic tripeptides.

Solution:
We need to match the items in List-I with their corresponding descriptions in List-II.

Let's analyze each item:

A. Nitrogen oxides in air
Nitrogen oxides (NOx) contribute to acid rain. They react with water and oxygen in the atmosphere to form nitric acid (HNO₃), which lowers the pH of rainwater.
Therefore, A matches with IV (Acid rain).

B. Methane in air
Methane (CH₄) is a potent greenhouse gas that contributes to global warming by trapping heat in the atmosphere.
Therefore, B matches with III (Global warming).

C. Carbon dioxide
Carbon dioxide (CO₂) is a major greenhouse gas that contributes to global warming. It also causes the pH of rain water to be 5.6 because of the reaction with water.
Therefore, C matches with II (pH of rain water becomes 5.6).

D. Phosphate fertilisers in water
Phosphate fertilizers in water bodies can lead to excessive algal growth, causing eutrophication, which depletes oxygen and harms aquatic life.
Therefore, D matches with I (Eutrophication).

Matching:
A - IV (Acid rain)
B - III (Global warming)
C - II (pH of rain water becomes 5.6)
D - I (Eutrophication)

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

Acidity of Substituted Phenols Analysis

The strength of an acid is determined by the stability of its conjugate base.

Electron-withdrawing groups (-I and -M effects) stabilize the conjugate base by delocalizing the negative charge after the loss of . Electron-donating groups (+I and +M effects) destabilize the conjugate base.

In the given options, all are substituted phenols. The substituents are:

• (1) No substituent (phenol)
• (2) (chloro) - electron-withdrawing through -I effect
• (3) (methyl) - electron-donating through +I effect
• (4) (nitro) - strong electron-withdrawing through -I and -M effects

Since the nitro group ( ) has the strongest electron-withdrawing effect (-I and -M), it stabilizes the conjugate base of the phenol most effectively. Therefore, the compound with the substituent (option 4) is the strongest acid among the given options.

To solve this problem, we need to identify the most basic compound among the given options and then calculate the amount of HCl consumed by 1.0 g of that compound.

1. Identifying the Most Basic Compound:
We need to compare the basicity of 4-nitroaniline, benzylamine, N-phenylacetamide, and aniline. Basicity depends on the availability of the nitrogen lone pair for protonation.

2. Basicity Analysis:
4-Nitroaniline: Nitro group is electron-withdrawing, reducing basicity.
Benzylamine: Nitrogen attached to a benzyl group, more basic than aniline.
N-Phenylacetamide: Amide, very weakly basic due to resonance with the carbonyl group.
Aniline: Lone pair delocalized into the benzene ring, reducing basicity.
Therefore, benzylamine is the most basic compound.

3. Calculating Molar Mass of Benzylamine ( ):
Molar mass = (7 × 12) + (9 × 1) + (1 × 14) = 84 + 9 + 14 = 107 g/mol

4. Calculating Moles of Benzylamine:
Given 1.0 g of benzylamine, moles = mass / molar mass = 1.0 g / 107 g/mol = 1/107 mol

5. Reaction with HCl:
Benzylamine reacts with HCl in a 1:1 molar ratio:
So, 1/107 moles of benzylamine reacts with 1/107 moles of HCl.

6. Calculating Molar Mass of HCl:
Molar mass = 1 (for H) + 35.5 (for Cl) = 36.5 g/mol

7. Calculating Mass of HCl Consumed:
Mass of HCl = moles × molar mass = (1/107 mol) × (36.5 g/mol) = 36.5 / 107 g

8. Converting to Milligrams:
Mass of HCl (in mg) = (36.5 / 107) × 1000 mg ≈ 0.3411 × 1000 mg ≈ 341.1 mg

Final Answer:
1.0 g of benzylamine will consume approximately 341.1 mg of HCl.