JEE Advanced 2023 Paper 2 · Q13 · Named Reactions & Reagents

Question

The reaction of 4-methyloct-1-ene ($P$, $2.52$ g) with HBr in the presence of $(C_6H_5CO)_2O_2$ gives two isomeric bromides in a $9:1$ ratio, with a combined yield of $50\%$. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with aq. $K_2CO_3$ to give a non-ionic product $S$ in $100\%$ yield. The mass (in mg) of $S$ obtained is ___. [molar mass: H=1, C=12, N=14, Br=80]

SolveKar AI · Accepted Answer

$P = C_9H_{18}$, $M_P = 126$ g/mol; moles $P = 2.52/126 = 0.02$. With anti-Markovnikov HBr (peroxide effect): primary alkyl bromide formed in $90\%$, secondary in $10\%$, total yield $50\%$. So total bromides $= 0.02\cdot 0.5 = 0.01$ mol, primary $= 0.009$ mol. Primary bromide $C_9H_{19}Br + Et_2NH 	o C_9H_{19}NEt_2$ (a tertiary amine, non-ionic) after $K_2CO_3$ workup. Product $S = C_{13}H_{29}N$, $M_S = 13(12) + 29 + 14 = 199$ g/mol. Mass $= 0.009 \cdot 199 = 1.791$ g $= 1791$ mg.

JEE Advanced 2023 Paper 2 · Q13 · Named Reactions & Reagents

Solution