Question

The proton NMR spectrum of a compound with the formula C₇H₁₂O₂is

shown. The infrared spectrum displays a strong band at 1738 cm^-1 and

a weak band at 1689 cm^-1 . The normal carbon-13 and the DEPT experimental

results are tabulated. Draw the structure of this compound.

Short answer

DEPT- 90 spectrum indicates signal only for CH carbons. DEPT-135 spectrum indicates positive signals for CH and CH₃ carbons and negative signals indicative of CH₂ carbons.

CH₃carbons can be calculated by subtracting DEPT-90 from positive DEPT-135.

Step-by-step solution

Spectrum interpretation for DEPT-135 and DEPT-90

DEPT- 90 spectrum indicates signal only for CH carbons. DEPT-135 spectrum indicates positive signals for CH and CH₃ carbons and negative signals indicative of CH₂ carbons.

CH₃carbons can be calculated by subtracting DEPT-90 from positive DEPT-135.

Chart for IR absorption by specific functional group

The chart showsthe absorption of particular wavelength radiation by a specific functional group:

The functional group absorption band

From a chart and molecular formula, it can be concluded that the functional groups present are the ester group and the other group can be the alkene or aromatic group.

Proton spectrum reading

A single peak indicates no neighbor H atom is present for hydrogen. If the peak splits into two it indicates only one neighbor present and if into three then two neighbors present and if four peaks then three neighbor atoms and so on.

Table interpretation

Here one signal is observed in the given table at DEPT-90 indicating 1 CH carbon is present. 1 negative signals are present for DEPT-135 indicating 1CH₂carbon present.

4 positive signal is present for DEPT-135 indicating that threeCH₃carbon is present. 2 carbon with no signal indicate no H is present on it.

From below figure and given table it can be seen:

• Carbon at 18, 21, and 26 ppmis CH₃.
• Carbon at 61ppmindicatesit is CH₂.
• Carbon at 119 ppm indicates it is CH.
• Carbon at 139 ppm indicates it is not bonded with hydrogen and is in C double bond C form.
• As only one double bond is present so the structure can’t be aromatic instead is an alkene group.
• Carbon at 171ppm indicates it is a carbonyl group here carbonyl group is an ester.
• From this data two structures are possible one with agroup attached to carbonyl carbon and in other with CH₃attached to ester oxygen. This problem can be solved with peak analysis from the proton spectrum.
• 
  

Step 5:From proton NMR structure of given molecule

From the peak present at 2 ppm it can be concluded thatthegroup attached to the carbonyl carbonof ester RCOCH₃has a chemical shift between 2-2.3 ppm while ROCH₃

_{has 3.8 ppm.So C7H12O2given structure is the correct structure of the given molecule,}C₇H₁₂O₂

Structure ofC₇H₁₂O₂