Consider the following reactions sequence
When the product (P) is subjected to Carius analysis using $\text{AgNO}_3$, $1.0\text{ g}$ of the product (P) will produce __________ g of the precipitate of AgBr. (Nearest Integer)
(Given : molar mass in $\text{g mol}^{-1}$ C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)
Step 1: Identify the starting compound
The given compound is p-nitrotoluene (–CH₃ and –NO₂ in para position).
Step 2: Apply the reaction sequence
(i) Sn/HCl, OH⁻ → Reduction
$-NO_2 \rightarrow -NH_2$
Product: p-toluidine
(ii) $(CH_3CO)_2O$ → Acetylation
$-NH_2 \rightarrow -NHCOCH_3$ (protection)
Product: p-methyl acetanilide
(iii) Br₂ / AlBr₃ → Bromination
Product: o-bromo p-methyl acetanilide
(iv) $H_3O^+$ → Hydrolysis
Removes acetyl group
→ Final product (P): o-bromo p-toluidine
Step 3: Number of Br atoms in product
Each molecule of P contains 1 Br atom
Step 4: Molar mass calculation
For $C_7H_8BrN$:
Total molar mass = 186 g/mol
Step 5: AgBr formed (Carius method)
1 mole Br → 1 mole AgBr
Molar mass of AgBr = $108 + 80 = 188$
Step 6: Final calculation
From 1 g of P:
$AgBr=188186≈1.01 g\text{AgBr} = \frac{188}{186} \approx 1.01 \, \text{g} AgBr=186188≈1.01g $
Final Answer (Nearest Integer):
1 g
Consider the following sequence of reactions to give the major product (X)
$\text{P}~\text{g}$ of the major product (X) formed is reacted with $\text{NaHCO}_3$ solution to liberate a gas which occupied $11.2~\text{dm}^3$ at STP.
$\text{P} = \_\_\_\_\_\_\_\_~\text{g}$.
(Given molar mass in $\text{g~mol}^{-1}$ $\text{H}: 1, \text{C}: 12, \text{O}: 16, \text{Cl}: 35.5$)
Among the following, how many hydrocarbon (s) with formula $C_{10}H_{16}$ which on reaction with acidic potassium permagenate yield(s) same product? ____________

On treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is :-

The total number of optically active compounds formed in the following reaction is :-

The major products U and V in the following reaction are
