Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Given below are two statements:
Statement (I): 1,2,3-Trihydroxypropane can be separated from water by simple distillation.
Statement (II): An azeotropic mixture cannot be separated by fractional distillation.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (1)
Key idea: Separation techniques: Simple distillation vs azeotropic mixtures
**Statement I Analysis:** 1,2,3-Trihydroxypropane is glycerol with b.p. = 290°C
Water has b.p. = 100°C. The large difference (190°C) allows separation by simple distillation
Glycerol and water don't form an azeotrope, so **Statement I is TRUE**
**Statement II Analysis:** Azeotropic mixtures have constant boiling composition where vapor and liquid have same composition
Since vapor and liquid compositions are identical in azeotropes, fractional distillation cannot separate them. **Statement II is TRUE**
Both statements are true, so answer is option **(1)**
Match List - I with List - II.
List - I (Vitamin) | List - II (Name)
A. Vitamin $B_1$ | I. Pyridoxine
B. Vitamin $B_2$ | II. Ascorbic acid
C. Vitamin $B_6$ | III. Thiamine
D. Vitamin C | IV. Riboflavin
Choose the correct answer from the options given below:
(1) A-II, B-I, C-III, D-IV
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-IV, C-I, D-II
(4) A-I, B-III, C-II, D-IV
Show answer and solution
Answer: Option (3)
Key idea: Vitamins and their chemical names - specifically water-soluble vitamins (B-complex and C)
Identify that Vitamin $B_1$ is **Thiamine**, so A → III
Identify that Vitamin $B_2$ is **Riboflavin**, so B → IV
Identify that Vitamin $B_6$ is **Pyridoxine**, so C → I
Identify that Vitamin C is **Ascorbic acid**, so D → II
Match the pattern: A-III, B-IV, C-I, D-II corresponds to option (3)
Identify the correct pair having amino acid (A) and the hormone (B) that is iodinated derivative of the amino acid (A).
(T and Y represent one letter code for amino acids)
Amino acid (A) Hormone (B)
(1) T Insulin
(2) T Thyroxine
(3) Y Thyroxine
(4) Y Insulin
Show answer and solution
Answer: Option (3)
Key idea: Amino acids and their derivatives; Thyroid hormones
Identify that the question asks for a hormone that is an iodinated derivative of an amino acid
Recall that thyroxine (T4) is a thyroid hormone containing iodine atoms
Remember that thyroxine is synthesized from the amino acid tyrosine through iodination
Identify the one-letter code: $Y$ represents tyrosine, $T$ represents threonine
Match the correct pair: Tyrosine ($Y$) is the amino acid and Thyroxine is its iodinated derivative
Given below are two statements:
**Statement I:** The structure of Maltose is given below:
[Figure: Chemical structure showing two glucose units connected by glycosidic linkage with OH groups and H atoms labeled]
Maltose is a non-reducing sugar.
**Statement II:** The structure of Lactose is given below:
[Figure: Chemical structure showing galactose and glucose units connected by glycosidic linkage with OH groups and H atoms labeled]
Lactose is a reducing sugar.
In the light of the above statements, choose the *correct* answer from the options given below
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (4)
Key idea: Reducing and Non-reducing Sugars - Disaccharides
Analyze Statement I: Maltose has two glucose units linked by $\alpha(1\to4)$ glycosidic bond
In maltose, one anomeric carbon is involved in the bond, but the other glucose has a FREE hemiacetal group
A free hemiacetal can open to form an aldehyde group, which can be oxidized. Thus, maltose IS a reducing sugar
Statement I is FALSE (claims maltose is non-reducing, but it's actually reducing)
Analyze Statement II: Lactose has galactose and glucose linked by $\beta(1\to4)$ glycosidic bond with a free hemiacetal on glucose
The free hemiacetal makes lactose a reducing sugar. Statement II is TRUE
Answer: Statement I is false but Statement II is true
Match the **LIST-I** with **LIST-II**
| | List-I | List-II |
|---|---|---|
| | Name of amino acid | One letter symbol/type |
| A | Arginine | I | D/Non-essential |
| B | Aspartic acid | II | R/Essential |
| C | Lysine | III | E/Non-essential |
| D | Glutamic acid | IV | K/Essential |
Choose the *correct* answer from the options given below:
(1) A-II, B-I, C-IV, D-III
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-IV, C-I, D-II
Show answer and solution
Answer: Option (1)
Key idea: Amino acids classification and one-letter symbols
Identify the one-letter symbol for each amino acid: Arginine = R, Aspartic acid = D, Lysine = K, Glutamic acid = E
Classify amino acids as essential or non-essential
Arginine (R) is essential → Match with II
Aspartic acid (D) is non-essential → Match with I
Lysine (K) is essential → Match with IV
Glutamic acid (E) is non-essential → Match with III
Identify the correct statements.
A. Glucose exists in two anomeric forms.
B. Anomers of glucose differ in configuration at C-1 in cyclic hemiacetal structure.
C. Melting point of $\alpha$ - anomer of glucose is greater than $\beta$ - anomer.
D. Specific rotation of $\alpha$ - anomer is $+19°$ while for $\beta$ - anomer is $+112°$
E. $\alpha$ and $\beta$ - anomers of glucose are prepared by crystallization of saturated glucose solution at 303 K and 371 K respectively.
Choose the correct answer from the options given below:
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Anomers of glucose - structure, properties, and preparation methods
Identify the incorrect statement about tertiary structure of proteins.
(1) They can be fibrous or globular in structure.
(2) The main forces that stabilize the structure are hydrogen bonding, disulphide links, van der Waals and electrostatic forces of attraction.
(3) The structure remains intact when exposed to pH changes.
(4) A linear polypeptide chain will convert to a secondary structure and then further folding of the secondary structure will convert to tertiary structure.
Show answer and solution
Answer: Option (3)
Key idea: Tertiary Structure of Proteins and Denaturation
Analyze statement (1): Proteins can indeed be fibrous or globular - this is **correct**
Analyze statement (2): Tertiary structure is stabilized by H-bonds, disulfide links, van der Waals forces, and electrostatic interactions - this is **correct**
Analyze statement (3): Tertiary structure is **highly sensitive** to pH changes, which can cause denaturation by disrupting ionic interactions and hydrogen bonds
Statement (3) claims structure remains intact with pH changes - this is **INCORRECT** (proteins denature with pH changes)
Analyze statement (4): The hierarchy Primary → Secondary → Tertiary structure is correct
Therefore, option **(3)** is the incorrect statement
Given below are two statements:
[Figure: Two cyclic structures of D-(+)-glucose shown with different OH orientations at one position]
**Statement I:** The two structures shown and are two anomers of D-(+)-glucose.
**Statement II:** The open chain forms of D-glucose and D-fructose contain three similar chiral carbons at $C_3$, $C_4$ and $C_5$.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Anomers of glucose and stereochemistry of monosaccharides
Match the LIST-I with LIST-II
| List-I | List-II |
| Deficiency Disease | Vitamin |
|---|---|
| A. Scurvy | I. Pyridoxine |
| B. Convulsions | II. Vitamin A |
| C. Cheilosis | III. Ascorbic Acid |
| D. Xerophthalmia | IV. Riboflavin |
Choose the correct answer from the options given below:
(1) A-I, B-III, C-II, D-IV
(2) A-I, B-III, C-IV, D-II
(3) A-III, B-I, C-IV, D-II
(4) A-III, B-I, C-II, D-IV
Show answer and solution
Answer: Option (3)
Key idea: Vitamin deficiency diseases and their corresponding vitamins
Identify that Scurvy is caused by deficiency of Vitamin C (Ascorbic Acid), so $A \rightarrow III$
Recognize that Convulsions are caused by deficiency of Vitamin B6 (Pyridoxine), so $B \rightarrow I$
Know that Cheilosis (cracks at mouth corners) is caused by deficiency of Vitamin B2 (Riboflavin), so $C \rightarrow IV$
Understand that Xerophthalmia (eye dryness, night blindness) is caused by deficiency of Vitamin A, so $D \rightarrow II$
Match the pattern: $A-III, B-I, C-IV, D-II$ corresponds to option (3)
Match the LIST-I with LIST-II
| List-I | List-II |
| Amino acid | Positive reaction/Test for functional group present in side chain of amino acid |
|---|---|
| A. Glutamine | I. Hinsberg's test |
| B. Lysine | II. Neutral $FeCl_3$ test |
| C. Tyrosine | III. Ceric ammonium nitrate test |
| D. Serine | IV. Hoffman bromamide degradation |
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-I, D-III
(2) A-IV, B-I, C-II, D-III
(3) A-III, B-II, C-I, D-IV
(4) A-IV, B-I, C-III, D-II
Show answer and solution
Answer: Option (2)
Key idea: Functional group tests for amino acid side chains
Identify functional groups in side chains: Glutamine has $-CONH_2$ (amide), Lysine has $-(CH_2)_4NH_2$ (1° amine), Tyrosine has phenolic $-OH$, Serine has alcoholic $-OH$
Glutamine (amide group) → Hoffman bromamide degradation converts $RCONH_2$ to $RNH_2$, so A-IV
Lysine (primary amine) → Hinsberg's test distinguishes amines using benzenesulfonyl chloride, so B-I
Tyrosine (phenolic OH) → Neutral $FeCl_3$ test gives violet color with phenols, so C-II
Serine (alcoholic OH) → Ceric ammonium nitrate test detects alcohols (red color), so D-III
The incorrect statement from the following with respect to carbohydrates is :
(1) All monosaccharides are reducing sugars.
(2) The monosaccharide units obtained from hydrolysis of oligosaccharides are always the same.
(3) Starch and cellulose are typical examples of polysaccharides, which are very high molecular weight compounds of more than ten monosaccharide units.
(4) Open chain and cyclic structures co-exist at equilibrium that are responsible for certain properties as in the case of $D-(+)-$ glucose.
Show answer and solution
Answer: Option (2)
Key idea: Carbohydrates - Oligosaccharides and their hydrolysis products
Analyze statement (1): All monosaccharides have free or potentially free carbonyl groups, making them reducing sugars. This is **correct**.
Analyze statement (2): Oligosaccharides can be homooligosaccharides (same units) or heterooligosaccharides (different units). Examples: sucrose → glucose + fructose, lactose → glucose + galactose. The word 'always' makes this **incorrect**.
Analyze statement (3): Polysaccharides are defined as carbohydrates with more than 10 monosaccharide units. Starch and cellulose fit this definition perfectly. This is **correct**.
Analyze statement (4): $D-(+)$-glucose exists in equilibrium between open chain and cyclic forms (mutarotation), which explains its reducing properties and optical rotation. This is **correct**.
Statement (2) is the incorrect statement because oligosaccharides do NOT always yield the same monosaccharide units upon hydrolysis.
