[alrob8a]

This is my assignment question, which I've answered. Just hoping someone could weigh in and tell me if my reasoning is on track.

Q:

Compund X has a MW of 74. The IR spectrum of X contains a strong band between 1650 and 1800cm-1 but no bands at 1600cm-1 or 3500cm-1. X does not react with 2,4-DNP or with aq NaHCO3. The ¹³C NMR spectrum consists of three peaks - one above 80ppm and two below 80ppm. The ¹H NMR spectrum consists of 2 peaks of equal areas. Deduce the structure of X showing all your reasoning


My Answer:


MW of 74 equates to formulas of C₄H₁₀O, or C₃H₆O₂

IR absorption between 1650 and 1800cm-1 indicates a C=O bond.
No absorption at 1600cm-1 indicate absence of C=C bonds (occuring in the 1600-1650cm-1 spectrum). X is therefore not an alkene.
No absorption at 3500cm-1 indicates absence of O-H bonds (occuring in the 3300-3500cm-1 spectrum). X is therefore not an alcohol.

X has 1 degree of unsaturation in the form of a C=O, therefore follows the molecular formula of C_nH_2n

Since there is no reaction with 2,4-DNP or with aq NaHCO3, X cannot be an aldehyde or ketone, or an acid.

1¹³C NMR analysis tells us there are 3 carbon in the compound.
1 peak above 80ppm indicates an sp² C, confirming presence of C=O.
2 peaks below 80ppm indicates 2 sp³ C.

The ¹H NMR spectrum consisting of 2 peaks of equal areas tells us a 1:1 ratio of hydrogen in 2 different chemical environments.

Compound X must follow the formula C₃H₆O₂. Since X has no O-H bonds, is not an aldehyde or ketone, the 2 O must be part of an ester functional group.

(pardon crude structure here)

      H     O            H
       |    ||            |
H -- C -- C -- O -- C -- H
       |                   |
      H                   H


So how did I go?

Thanks heaps!
Alister

[fledarmus]

Looks good!