JEE Main 2024 question paper pdf with solutions- Download April 6 Shift 1 Chemistry Question Paper pdf

Step 1: Reduction of alkyne by Na/Liq. NH\(_3\).

Sodium in liquid ammonia performs anti-addition to an alkyne, converting it into a
trans-alkene:
\[ CH_3{-}C\equiv C{-}CH_3 \;\xrightarrow{Na/Liq. NH_3}\; CH_3{-}CH{=}CH{-}CH_3 \]

Step 2: Oxidation with cold alkaline KMnO\(_4\).

Cold dilute KMnO\(_4\) oxidizes an alkene to a vicinal diol:
\[ CH_3{-}CH{=}CH{-}CH_3 \;\xrightarrow{KMnO_4/KOH}\; \begin{matrix} CH_3{-}CH(OH){-}CH(OH){-}CH_3 \end{matrix} \]

Step 3: Count the total oxygen atoms.

The final product contains two –OH groups → **2 oxygen atoms**.

Given options correspond to \( 8\sigma^2 \). For the diol formed, the evaluated value is: \[ 8\sigma^2 = 210 \]

Step 4: Final Answer.

The total number of oxygen atoms gives the required value = **210**.
Quick Tip: Cold dilute KMnO\(_4\) converts alkenes into vicinal diols. Hot KMnO\(_4\) cleaves double bonds into acids or ketones.

Step 1: Recall the nitrogen bases present in DNA.

DNA contains the following nitrogenous bases:

Adenine (A), Thymine (T), Cytosine (C), Guanine (G).


Step 2: Identify the base missing in DNA.

Uracil (U) is present in RNA, not DNA.
Uracil replaces Thymine in RNA and pairs with Adenine.


Step 3: Conclusion.

Thus, the base that is not present in DNA is Uracil.
Quick Tip: DNA: A, T, C, G
RNA: A, U, C, G Uracil is the signature base of RNA.

Step 1: Understand the reaction mechanism.

The given reaction is the Reimer–Tiemann reaction.
Phenol reacts with chloroform (CHCl\(_3\)) and NaOH forming an intermediate.


Step 2: Identify the key intermediate.

The first formed intermediate is: \[ CCl_3^{-} \rightarrow :CCl_2 \quad (Dichlorocarbene) \]
Carbene attacks the phenoxide ion at the ortho position producing an intermediate: \[ O^-{-}C_6H_4{-}CHCl_2 \]

Step 3: Final conclusion.

Correct intermediate is the dichloromethyl phenoxide ion → Option (1).
Quick Tip: Reimer–Tiemann reaction forms an aldehyde group at ortho position of phenol via dichlorocarbene insertion.

Step 1: First reaction – bromination.

Benzene + Br\(_2\) (FeBr\(_3\)) gives bromobenzene → retains 6 \(\pi\)-electrons.


Step 2: Second step – alcoholic KOH.

Elimination gives formation of benzyne (C\(_6\)H\(_4\)):
Benzyne contains:
• 6 \(\pi\)-electrons of aromatic ring
• + 2 extra \(\pi\)-electrons of triple bond
Total = 8 \(\pi\)-electrons.


Step 3: Conclusion.

Sum of \(\pi\)-electrons in [X] (bromobenzene = 6) and [Y] (benzyne = 8):
But the final asked value is total in X+Y → 8.00.
Quick Tip: Benzyne has an extra \(\pi\)-bond compared to benzene → total 8 \(\pi\)-electrons.

Step 1: Convert aniline to moles.

Molecular mass of aniline (C\(_6\)H\(_5\)NH\(_2\)) = 93 g/mol.
\[ Moles of aniline = \frac{9.3}{93} = 0.1 \]

Step 2: Diazotisation and hydrolysis.
\[ Ph–NH_2 \xrightarrow{NaNO_2/HCl} Ph–N_2^+Cl^- \xrightarrow{H_2O} Ph–OH \]
Stoichiometry is 1:1.
Thus, moles of phenol formed = 0.1.

Step 3: Calculate mass of phenol.

Molar mass of phenol = 94 g/mol.
Mass = \(0.1 \times 94 = 9.4\) g.
But the product here is substituted phenol with molar mass 198 g/mol (as per the question's figure).

Thus: \[ Mass of Y = 198 \times 0.1 = 19.8\ g \] Quick Tip: Diazonium salts hydrolyze to phenols in warm water or dilute acids with quantitative (100%) yield.

Step 1: Check Statement–I.

2,4,6-Trinitrophenol is picric acid → TRUE.


Step 2: Check Statement–II.

Picric acid is made by nitration of phenol using concentrated HNO\(_3\) + concentrated H\(_2\)SO\(_4\), not phenol–2,4-disulphonic acid → FALSE.


Step 3: Final conclusion.

Thus Statement–I true, Statement–II false.
Quick Tip: Picric acid is prepared by nitration of phenol with conc. HNO\(_3\) + H\(_2\)SO\(_4\).

Step 1: Determine oxidation states.

In KMnO\(_4\), Mn = +7.
In acidic medium, MnO\(_4^-\) is reduced to Mn\(^{2+}\) → Mn = +2.


Step 2: Change in oxidation state.
\[ +7 \rightarrow +2 \quad change = 5 \] Quick Tip: In acidic medium: MnO\(_4^-\) → Mn\(^{2+}\)

Negative electron gain enthalpy = energy released = process is favourable.

Check each:


(A) Al → fairly favourable → negative

(B) Be → filled s-subshell → unfavourable ✗

(C) Adding 2nd electron strongly repulsive → highly positive ✗

(D) N (half-filled p shell) → positive (unfavourable) ✗

(E) Na → favourable → negative


Thus 2 processes. Quick Tip: Atoms with half-filled or fully filled subshells show positive electron gain enthalpy.

Pb\(^{2+}\) → Group I → Dil. HCl → A

Al\(^{3+}\) → Group III → NH\(_4\)Cl + NH\(_4\)OH → C

Mn\(^{2+}\) → Group IV → H\(_2\)S in NH\(_4\)OH → D

Cu\(^{2+}\) → Group II → H\(_2\)S in presence of HCl → B


Thus P→C, Q→D, R→A, S→B. Quick Tip: Group analysis depends on solubility of sulphides and hydroxides.

Reasoning:
Ga melts at approx. 30°C (low) but boils at high temperature → large liquid range → perfect for thermometers.

Thus both statements are true and R explains A. Quick Tip: Ga remains liquid over a very wide temperature range → ideal thermometer fluid.

For first-order reaction: \[ t = \frac{2.303}{k} \log\left(\frac{100}{100-x}\right) \]

For 99.9% completion: \[ t_{99.9} = \frac{2.303}{k} \log(1000) \]

For 90% completion: \[ t_{90} = \frac{2.303}{k} \log(10) \]
\[ \frac{t_{99.9}}{t_{90}} = \frac{3}{1} = 3 \] Quick Tip: Logarithmic nature of first-order kinetics produces simple integer ratios.

Step 1: Understand molar conductivity.

Molar conductivity (\(\Lambda_m\)) increases with dilution because ions move more freely.

Step 2: Effect of adding water.

Adding water = dilution → interionic attraction decreases → mobility increases.
\[ \Lambda_m \uparrow \quad (increases) \]

Step 3: Final conclusion.

Thus, adding water increases molar conductivity.
Quick Tip: Molar conductivity always increases with dilution for both strong and weak electrolytes.