JEE Advanced 2023 — Paper 2 — CHEM-III solution

JEE Advanced2023Paper 2CHEM-III

The reaction of 4-methyloct-1-ene (P, $2.52$ g) with HBr in the presence of $(C_{6} H_{5} C O)_{2} O_{2}$ 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. $K_{2} C O_{3}$ to given 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 $= 80$ ]

Your answer

Solution

Moles of P. The molecular formula of 4-methyloct-1-ene is $C_{9} H_{18}$ , molar mass $= 9 (12) + 18 (1) = 126$ g mol $^{- 1}$ . Moles of P $= 2.52/126 = 0.02$ mol.

HBr/peroxide addition (anti-Markovnikov). The major (primary) bromide is $B r$ - $C H_{2}$ - $C H_{2}$ - $C H_{2}$ - $C H (C H_{3})$ - $C H_{2}$ - $C H_{2}$ - $C H_{2}$ - $C H_{3}$ , that is, 1-bromo-4-methyloctane. The combined yield is $50%$ and the primary isomer is $9/10$ of the product, so

$moles of primary bromide = 0.02 \times 0.5 \times 0.9 = 0.009$ mol.

Reaction with diethylamine then $K_{2} C O_{3}$ . $E t_{2} N H$ displaces $B r$ to form a quaternised ammonium salt; aqueous $K_{2} C O_{3}$ neutralises it to give the tertiary amine

$E t_{2} N - C H_{2} - C H_{2} - C H_{2} - C H (C H_{3}) - C H_{2} - C H_{2} - C H_{2} - C H_{3}$ ,

a non-ionic, neutral product S.

Molar mass of S. Formula $C_{13} H_{29} N$ : $13 (12) + 29 (1) + 14 = 156 + 29 + 14 = 199$ g mol $^{- 1}$ .

Mass of S. $0.009 \times 199 = 1.791$ g $= 1791$ mg.

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