JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Hydrocarbons

The reaction of 4-methyloct-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 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

]

Answer: 1791

Solution

4-Methyloct-1-ene (

C_{9} H_{18}

, molar mass

126

) amounts to

2.52/126 = 0.02

mol. With HBr and a peroxide the addition is anti-Markovnikov, so the primary bromide is the major (

9 : 1

) product; at

50%

combined yield its amount is

0.02 \times 0.5 \times 0.9 = 0.009

mol. This primary bromide reacts with diethylamine, and the subsequent

K_{2} C O_{3}

liberates the free base, giving the non-ionic tertiary amine

S

C_{13} H_{29} N

, of molar mass

199

. The mass of

S

0.009 \times 199 = 1.791

= 1791

mg.

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer