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 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 eq. K$$_2$$CO$$_3$$ to give a non-ionic product S in 100% yield.

The mass (in mg) of S obtained is ____.

[Use molar mass (in g mol$$^{-1}$$): H = 1, C = 12, N = 14, Br = 80]

4-Methyloct-1-ene is $$C_9H_{18}$$ with molar mass
$$M_{\text{alkene}} = 9(12) + 18(1) = 126\text{ g mol}^{-1}$$.

Moles of the alkene taken:
$$n_{\text{alkene}} = \frac{2.52\text{ g}}{126\text{ g mol}^{-1}} = 0.020\text{ mol}$$.

In the presence of peroxide, $$HBr$$ adds anti-Markovnikov. Hence the major product is the primary bromide $$\text{1-bromo-4-methyloctane}$$; the minor is the corresponding secondary bromide. Given overall yield $$= 50\%$$, the moles of both bromides actually obtained are

$$n_{\text{bromides}} = 0.020 \times 0.50 = 0.010\text{ mol}$$.

The isomer ratio is $$9:1$$ (primary : secondary). Therefore moles of the primary bromide are

$$n_{\text{primary}} = 0.010 \times \frac{9}{10} = 0.009\text{ mol}$$.

This entire amount is allowed to react with diethylamine $$(C_2H_5)_2NH$$. An $$S_N2$$ substitution gives the tertiary amine

$$S = C_9H_{19}N(C_2H_5)_2 \;=\; C_{13}H_{29}N$$.

Its molar mass is
$$M_S = 13(12) + 29(1) + 14 = 156 + 29 + 14 = 199\text{ g mol}^{-1}$$.

Because the reaction yield of this step is 100 %:

$$m_S = n_{\text{primary}}\;M_S = 0.009\text{ mol}\;\times\;199\text{ g mol}^{-1} = 1.791\text{ g}$$.

Converting to milligrams:
$$1.791\text{ g} = 1791\text{ mg}$$.

Hence, the mass of the non-ionic product $$S$$ obtained is 1791 mg.