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ChemistryInteger+4Question 47 of 51

JEE Advanced 2023 · Paper 2, Chemistry question 13

The reaction of 4-methyloct-1-ene (P, 2.52 g) with HBr in the presence of (CX6HX5CO)X2OX2\ce{(C6H5CO)2O2} 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. KX2COX3\ce{K2CO3} 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\mathrm{g\,mol^{-1}}): H = 1, C = 12, N = 14, Br = 80]

Answer and worked solution

Official answer: 1791

Starting moles. 4-Methyloct-1-ene is CX9HX18\ce{C9H18} (M=126M=126), so 2.52 g is 0.02 mol.

Bromination. With peroxide, HBr adds anti-Markovnikov.

  • The major (9 parts) product is the primary bromide 1-bromo-4-methyloctane.
  • The minor (1 part) product is the secondary bromide.

The combined yield is 50%, i.e. 0.01 mol, so there is 0.009 mol of the primary bromide.

Amination. Diethylamine alkylates it, and KX2COX3\ce{K2CO3} frees the base. S is the tertiary amine N,NN,N-diethyl-4-methyloctan-1-amine, CX13HX29N\ce{C13H29N}:
M=156+29+14=199.M=156+29+14=199.

Mass. 0.009×199=1.7910.009\times199=1.791 g =1791=1791 mg.

Topics:AlkenesAminesStoichiometry
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