Question

Spectroscopy. The following NMRwas obtained froma student's final isopentyl acetate product, revealing a small amount of impurity Using the spectrum provided, propose a structure of the impurity. Helpful Hint: think about common chemicals found in lab. a. S d S m m PPM 2H 0.48H 0.29H 3H 1H 2H 6H b. If you were to take an IR spectrum of the contaminated sample, what functional peaks would you see that would NOT be present in a pure sample of isopentyl acetate?

Answer 1

(a) yl acetate is -CH-CH-Cth-0 b) (c) (d) The NMR if is opentyl acetate is a 6H, doublet at 1.00ppm due to protons à IH, multiplet at N1.75ppm due to prolons b' 2H, quartet at ~1.50 ppm dure to protons é 2H, triplet at ~ 4.15ppm due to protons d 3H, singlet at ~ 2.00ppm due to protons é Extra (Impurity peaks) are 0.294, singlet at N 2.30ppm 0.4SH, muluplet at 7.00ppm Ralio of impurity protons. 0.48 - 5 0. 29 So, there is a 3H, singles at 2.30 ppm I SH, multiplet at no 7.00 ppm (aromatic) This NMR data is analogous to that of Toluene (13] - 34, singlet, ~ 2.30ppm 54, multiplet, ~ 7.00ppm. HTH

Impurity is Toluene.

b) In IR spectra, impurity will result in four absorption bands ~1650 cm^-1to ~1450 cm^-1. These will be exclusive to C=C stretching of aromatic carbons.

Other than this, =C-H peaks will be found at 3100 - 3000 cm^-1.