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.
An organic compound "$x$" where molar ratio of C, O and H are equal, on treatment with 50% KOH under reflux followed by acidification produced "$y$". The most likely structure of "$y$" is:
[Molar mass of '$x$' is 58 g mol$^{-1}$]
(1) $CH_2=CH-C-OH$ with double bonded O above C
(2) $CH_3-CH=CH-CH=O$
(3) $O=C-OH$ above $CH_2-OH$
(4) $CH_3-C-OH$ with double bonded O above C
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Answer: Option (3)
Key idea: Cannizzaro Reaction of Glyoxal
Given C:O:H = 1:1:1 and molar mass = 58 g/mol, find molecular formula: $(COH)_n$ where $29n = 58$, so $n=2$. Thus $x$ is $C_2H_2O_2$
Identify compound $x$: With formula $C_2H_2O_2$, the structure is glyoxal (OHC-CHO), which has two aldehyde groups and no α-hydrogens
Recognize the reaction: 50% KOH (concentrated base) with aldehydes having no α-H undergoes Cannizzaro reaction (disproportionation)
In Cannizzaro reaction, one aldehyde group is oxidized to $-COOH$ and another is reduced to $-CH_2OH$
Product $y$ after acidification: $HOCH_2COOH$ (glycolic acid), which matches option (3)
A molecule (X) with following structure under mild acidic condition is hydrolysed to produce (Y) and (Z). Identify the correct statements about (Y) and (Z).
[Figure: Structure showing a vinyl ether with formula containing C=C-O-C(CH3)=CH2, labeled as (X)]
A. Both (Y) and (Z) have same molar mass.
B. (Y) and (Z) can be distinguished from each other by NaHCO$_3$.
C. (Y) and (Z) react with HCN with same rates.
D. (Y) and (Z) undergo addition reaction with 2,4-DNP.
Choose the correct answer from the options given below:
(1) A, B and C Only
(2) B and C Only
(3) C and D Only
(4) A and D Only
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Answer: Option (4)
Key idea: Acid-catalyzed hydrolysis of vinyl ethers to carbonyl compounds and their chemical properties
Identify structure (X) as a vinyl ether: $CH_3CH=CH-O-C(CH_3)=CH_2$
Under mild acidic hydrolysis, vinyl ethers break to form carbonyl compounds: $Y = CH_3CH_2CHO$ (propanal) and $Z = CH_3COCH_3$ (acetone)
Check Statement A: Both have molecular formula $C_3H_6O$ with molar mass $58$ g/mol ✓
Check Statement B: Neither aldehyde nor ketone reacts with $NaHCO_3$ (only carboxylic acids do) ✗
Check Statement C: Aldehydes react faster with HCN than ketones due to less steric hindrance ✗
Check Statement D: Both aldehydes and ketones react with 2,4-DNP to form hydrazones ✓
Identify the colour of compound 'X' in the sequence of the reaction.
[Figure: A reaction sequence showing 2 molecules of phenol reacting with phthalic anhydride in the presence of conc. $H_2SO_4$ to form a triphenylmethane derivative with OH groups. This then reacts with NaOH to form a pink compound, which upon treatment with excess NaOH forms compound X with a quinonoid structure containing $COO^-$ groups]
(1) Violet
(2) Green
(3) Red
(4) Colourless
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Answer: Option (4)
Key idea: pH-dependent color changes of phenolphthalein indicator
Phenol + phthalic anhydride with conc. $H_2SO_4$ forms phenolphthalein (colorless lactone form)
With NaOH, the lactone ring opens forming a quinonoid structure with extended conjugation (pink/magenta color)
The pink color is due to the monoanionic quinonoid form of phenolphthalein
With excess NaOH, further deprotonation occurs forming the trianionic carbinol form
The quinonoid structure is disrupted, destroying the extended conjugation
Compound X (trianionic form in excess alkali) is colorless
Given below are two statements:
**Statement I:** The condensation reaction between $CH_3-CH=O$ and $H_2N-N-C-NH_2$ (with H below the first N and O below C) under optimum pH will produce $CH_3-CH=N-C-N-NH_2$ (with H below the second N and O below C)
**Statement II:** The molecule, $Ph-CH\begin{Bmatrix}O-H\\O-CH_3\end{Bmatrix}$ will generate $Ph-CH=O$ in the presence of dilute acid.
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
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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: Condensation reactions of carbonyl compounds with semicarbazide and hydrolysis of hemiacetals
'$x$' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333 K. '$y$' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from '$x$' and '$y$' in the presence of barium hydroxide followed by heating is:
(1) 2 - Methylpent-4-en-3-one
(2) 4 - Methylpent-3-en-2-one
(3) 4 - Methylpent-1-ene
(4) 2 - Methylpent-3-one
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Answer: Option (2)
Key idea: Aldol condensation of ketones; Hydration of alkynes; Grignard reaction with nitriles
Step 1: Hydration of prop-1-yne ($HC≡C-CH_3$) with $HgSO_4/H^+$ gives propanone: $HC≡C-CH_3 + H_2O → CH_3COCH_3$ (compound $x$)
Step 2: Reaction of ethane nitrile ($CH_3CN$) with $CH_3MgBr$ followed by hydrolysis gives propanone: $CH_3CN + CH_3MgBr → CH_3COCH_3$ (compound $y$)
Step 3: Both $x$ and $y$ are acetone ($CH_3COCH_3$). In presence of $Ba(OH)_2$, aldol addition occurs between two acetone molecules to form $CH_3COCH_2C(OH)(CH_3)_2$
Step 4: Upon heating, dehydration occurs to give the α,β-unsaturated ketone: $CH_3CO-CH=C(CH_3)_2$
Step 5: Numbering from carbonyl end: $CH_3-CO-CH=C(CH_3)-CH_3$ gives 4-Methylpent-3-en-2-one
Consider the following organic reaction sequence. Choose the final product (X) from the following (consider the major product in all intermediate reactions)
[Figure: Starting material showing CH₃CH₂-CO-OH structure]
(i) NH₃/Δ
(ii) NaOH/Br₂
(iii) HNO₂/H₂O, 0°C → (X) Major Product
(iv) C₆H₅N₂Cl, 0°C
(1) Benzene
(2) Phenol
(3) Propanol
(4) Chlorobenzene
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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: Hofmann Bromamide Degradation and Diazotization of Primary Aliphatic Amines
Consider the following reactions. Total number of electrons in the $\pi$ bonds and lone pair of electrons in the product (X) is:
$CHO$
$|$
$(CHOH)_4$ $\xrightarrow{\text{(i) HI, } \Delta \text{ (ii) } V_2O_5/10\text{-}20 \text{ atm}/773 \text{ K} \text{ (iii) Benzoyl chloride/Anhyd. AlCl}_3}$ (X) Major product
$|$
$CH_2OH$
(1) 12
(2) 16
(3) 14
(4) 18
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Answer: Option (4)
Key idea: Reduction of carbohydrates, catalytic aromatization, Friedel-Crafts acylation, and counting π electrons and lone pairs in benzophenone
Step 1: Treat glucose with $\text{HI}/\Delta$ to reduce all $\text{-OH}$ and $\text{-CHO}$ groups, forming n-hexane ($\text{C}_6\text{H}_{14}$)
Step 2: Catalytic dehydrogenation with $\text{V}_2\text{O}_5$ at high temperature converts n-hexane to benzene through cyclization and aromatization
Step 3: Friedel-Crafts acylation with benzoyl chloride and $\text{AlCl}_3$ gives benzophenone: $\text{C}_6\text{H}_5\text{-CO-C}_6\text{H}_5$
Count π electrons: First benzene ring has 3 π bonds = 6 electrons, second benzene ring has 3 π bonds = 6 electrons
Count carbonyl: $\text{C=O}$ has 1 π bond = 2 electrons
