Question

** please help with this **

8. CH1,0 IR: nothing interesting The spectrum displayed is a "decoupled" 13-C NMR spectrum. " But beside each coupled peak is a label that tells whether the carbon would be a singlet, doublet, triplet, or quartet if a "coupled" 13-C NMR was obtained. doublet triplet triplet quartet triplet 140 120 10080 60 40 20 PPM

Answer 1

Answer:

Compound formula is C₅H₁₀O. Let us calculate the double bond equivalent or degree of unsaturation.

Degree of unsaturation = No. of C - (No. of H/2) + (No. of N/2) + 1 = 5 - 10/2 + 1 = 1

Therefore, there is a possiblity of one ring or one double bond to be present.

The C13 peaks from higher ppm range to lower ppm range corresponds to doublet, triplet, triplet, quartet, triplet as per question.

Protons are spin 1/2 nuclei. Multiplicity of C13 peaks due to coupling is (n+1), where n is the no. of protons.

Hence from higher ppm range to lower ppm range the residues are -CH (above 130 ppm), -CH₂, -CH₂, -CH₃, -CH₂ (below 40 ppm).  

Above 110 ppm there are two peaks doublet and triplet. This is alkene region. So one residue will look like -CH=CH₂.

The peaks near 60 ppm and 80 ppm are for those carbons attached to Oxygen atom.

Hence there will be a residue like CH₃-O-CH₂-

The triplet just above 30 ppm is for aliphatic -CH₂- which is flanged between CH₃-O-CH₂- and -CH=CH₂ residues.

Hence the most probable structure is CH₃-O-CH₂-CH₂-CH=CH₂

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