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JEE Advanced 2023Paper 2
Exam
JEE Advanced
Year
2023
Paper
Paper 2
Subject
Chemistry
Chapter
Haloalkanes and Haloarenes

JEE Advanced 2023 · Paper 2 · Chemistry · Haloalkanes and Haloarenes

The reaction of 4 -methyloct- 1 -ene ( P, 2 . 52g ) with HBr in the...

The reaction of 44-methyloct-11-ene (P,2.52g)( P, 2 . 52g ) with HBrHBr in the presence of (C6H5CO)2O2{\left({C}_{6}{H}_{5}CO\right)}_{2}{O}_{2} gives two isomeric bromides in a 9:19: 1 ratio, with a combined yield of 5050 %. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with aq. K2CO3{K}_{2}{CO}_{3} to give a non-ionic product SS in 100100 % Yield. The mass (in mgmg )ofSof S obtained is

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

The reaction of 4 -methyloct- 1 -ene ( P, 2 . 52g ) with HBr in the presence of ( C _ 6 H _ 5 CO) _ 2 O _ 2 gives twoThe reaction of 4 -methyloct- 1 -ene ( P, 2 . 52g ) with HBr in the presence of ( C _ 6 H _ 5 CO) _ 2 O _ 2 gives two

Correct answer

0

Step-by-step solution

The 1-alkene undergo addition reaction and major product formed according to anti-Markonikov's rule.

The reaction of 4 -methyloct- 1 -ene ( P, 2 . 52g ) with HBr in the presence of ( C _ 6 H _ 5 CO) _ 2 O _ 2 gives two

Moles of PP =MassofPMolarmassofP=2.52252=0.01mol= \dfrac{MassofP}{MolarmassofP} = \dfrac{2 . 52}{252} = 0 . 01mol

Given that isomeric bromides formed have the mole fractions 910\dfrac{9}{10} and 110\dfrac{1}{10}

Moles of AA(major product) =0.02×910×50100=0.009mol= 0 . 02 \times \dfrac{9}{10} \times \dfrac{50}{100} = 0 . 009mol

Now the alkyl bromide formed undergo nucleophilic substitution reaction with diethyl amine and potassium carbonate.

The reaction of 4 -methyloct- 1 -ene ( P, 2 . 52g ) with HBr in the presence of ( C _ 6 H _ 5 CO) _ 2 O _ 2 gives two

In the above reaction, the product SS formed has the molecular weight =199gmol−1= 199g{mol}^{- 1}

Mass of S=0.009×199S = 0 . 009 \times 199

=1.791g= 1 . 791g

=1791mg= 1791mg

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