Question

Deduce the final structure consistent to the 1H-NMR data.

C4H8O2

1.2 ppm (triplet, 3H), 2.3 ppm (quartet, 2H), 3.6 ppm (singlet, 3H)

Answer 1

and 24 proton which comes. at 1.2 ppm and 2.3 ppm. respectively and are showing coupling to each other and after coupting one gune triplet and another give quartet by using (2nI+). acute i.e 2n I +1 - n - no. of neighbouring lydbeogen I I = spin of neighbouring Hybeogen. atom so for 34 Proton, nehich coupled with 24 preston.. so n. 2 I = 1/2 - 2nI tu triplet 2x2x2 + 1 - 3 for 34 proton which coupled o for with so 2 proton, 34 preston n-3 I= 1/2 2nI 1 4 quartet → 2x3 x + for 24 proton

and Now for 34 preoton (another enhich give means which not coupled with any neigh bowling atom. Now for oxygen attached peeston S- ppm value are coming at 3-4 ppr. while for ¿ attached to preston s-value comes at 2.5 – 3ppm. and CH3 proton peaks comes at 0.5 - 1.5 ppm. So the predictable stucture will be, CH3 – CH2 1.2ppm 2.3 ppm O-CH3 3.4 ppm

quien formula is. CAH & O2 feestly calculate the double bond equivalent (DBE) :- OBE - M - NH + N + 1 Me = No. of carbon. My – No. of Hydbergen. NN & No. of Nitrogen DBE - 4 - 8 + 0 +1 - 4 –4 + 1 DBE tells only one double sond is present in the predicting structure @ Now , we see the given Huma data for C4 H8 02 1.2 pm triplet , 3H) 2.3 ppm (quartet, 2H) 3.6 ppm (singlet, 3H) peem the above data which indicate that the 3H preston 0