The reaction of 4-methyloct-1-ene (P, 2.52 ~g) with HBr in the presence of (C_6 H_5 CO)_2 O_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_2 CO_3 to give a non-ionic product S in 100 \% yield. The mass (in mg) of S obtained is \underline\hspace2cm . [Use molar mass (in g mol^-1 ) : H=1, C=12, ~N=14, Br=80 ]

Practical Organic Chemistry PYQ — JEE Advanced 2023 · Shift 2 · Q13 with Solution

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Correct answer: 1791

Given alkene: 4-methyloct-1-ene, $\mathbf{P}$

Its molecular formula is $\mathrm{C_9H_18}$ .

So, molar mass of $\mathbf{P}$ is $9\times 12 + 18\times 1 = 108 + 18 = 126\ \mathrm{g\,mol^{-1}}$

Given mass = $2.52\ \mathrm{g}$

Hence moles of $\mathbf{P}$ : $n_P = \frac{2.52}{126} = 0.02\ \mathrm{mol}$

In presence of peroxide, HBr adds by the radical (anti-Markovnikov) pathway, giving mainly the primary bromide and a minor isomeric bromide.

The two bromides are formed in a $9:1$ ratio, with combined yield = 50%.

Thus total moles of bromides formed: $0.02 \times 0.50 = 0.01\ \mathrm{mol}$

Now primary bromide is $\frac{9}{10}$ of this: $n(\text{primary bromide}) = 0.01 \times \frac{9}{10} = 0.009\ \mathrm{mol}$

A primary alkyl bromide reacts with diethylamine to give a tertiary amine:

N(C_2H_5)_2}$$ Then treatment with aqueous $\mathrm{K_2CO_3}$ neutralizes the ammonium salt and gives the free **non-ionic** amine $\mathbf{S}$. So product $\mathbf{S}$ is: $$\mathrm{R\!- N(C_2H_5)_2}$$ where $\mathrm{R}$ is the 2-bromo-4-methyloctyl chain minus Br, i.e. the alkyl group from the primary bromide. Primary bromide formed from anti-Markovnikov addition to 4-methyloct-1-ene is 1-bromo-4-methyloctane, so $$\mathbf{S} = \mathrm{N,N\!- diethyl\!- 4\!- methyloctan\!- 1\!- amine}$$ Its molecular formula can be obtained as: - alkyl part from bromide after replacing Br: $\mathrm{C_9H_{19}}$ - plus two ethyl groups from diethylamine: $\mathrm{C_4H_{10}}$ - plus one nitrogen Therefore, $$\mathbf{S} = \mathrm{C_{13}H_{29}N}$$ Check molar mass: $$13\times 12 + 29\times 1 + 14 = 156 + 29 + 14 = 199\ \mathrm{g\,mol^{-1}}$$ --- 4. **Moles of $\mathbf{S}$ formed** Given 100% yield from the primary bromide, moles of $\mathbf{S}$ formed = moles of primary bromide used: $$n_S = 0.009\ \mathrm{mol}$$ Mass of $\mathbf{S}$: $$m_S = 0.009 \times 199 = 1.791\ \mathrm{g}$$ Convert to mg: $$1.791\ \mathrm{g} = 1791\ \mathrm{mg}$$ --- 5. **Final answer** $$\boxed{1791}$$ --- 6. **Comparison with stored correct answer** Stored correct answer = $1791$ My derived answer matches it exactly.