Question

What is the structure for compound C8H16 with 1H NMR chemical shifts at 1.7(s); 1.9(t, J=0.89 Hz); 4.6(s, J=0.89 Hz); and 4.8(multiplet, J=1.4 Hz) with relative integral 1:1:2:3.

Answer 1

1) Determination of Unsaturation Index:

Given MF = C8H16

# of C (n) = 8, Actual # of H (m) = 16, then

U.I. = [(2n+2)-m]/2 = [(2x8+2)-16]/2 = 2/2 = 1

There is one unsaturation site which probably accounts for C=C as PMR signals are that much downfield.

2) Structure determination.

i) (1.7, s) possibly account for 3 -CH₃ groups with no proton on neighbourhood carbon. The only piece of structure is

-C(CH₃)₃.

ii) (1.9, t, J=0.89Hz) possibly -CH- with adjacent -CH2- piece. Moreover, this proton is mutually coupled with proton having signal 4.6 (s, J = 0.89Hz).

Very true that multiplicities are not appropriate but being a complicated PMR pattern one can neglect it.

(Error: Faulty analysis of PMR in terms of coupling pattern and multiplicity)

iii) 4.8(m, J = 1.4Hz) mutually coupled with above protons (ii).

Based on this the best possible structure proposed as,

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