Q54E - Organic Chemistry (Mcmurry)

Question

The NMR spectrum shown is that of 3-methyl-3-buten-1-ol. Assign

all the observed resonance peaks to specific protons, and account for

the splitting patterns.

Step-by-Step Solution

Answer

The peak absorbing at ( 3 H) is due to the d protons.

The peak absorbing at (1 H) is due to the $-\mathrm{OH}$ proton a.

The peak absorbing at (2 H) is due to the proton c.

The peak absorbing at ( 2 H) is due to the b protons. 

The peaks at and ( 2 H) are due to the protons of e and f.

Step 1: 1 H NMR spectrum

In ${}^{\mathbf{1}}\mathbf{H}$  NMR spectrum,  the alcoholic proton exhibit absorption in the range $\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\delta }$ to $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\delta }$ and the phenolic proton absorb in the range of 3.0 to 8.0 ppm. The multiplicity provides an idea about the protons present in the nearby carbon atom.

Step 2: Assigning the respective peaks to the protons

                                    Structure of the compound

The peak absorbing at $1.76\mathrm{\delta }$ ( 3 H) is due to the d protons. This peaks that is seen in the allylic region is unsplit.

The peak absorbing at $2.13\mathrm{\delta }$ (1 H) is due to the $-\mathrm{OH}$ proton a.

The peak absorbing at $2.30\mathrm{\delta }$ (2 H) is due to the proton c. The peak is a triplet due to the splitting by the adjacent b protons.

The peak absorbing at  $3.72\mathrm{\delta }$( 2 H) is due to the b protons. The adjacent oxygen causes the peak to be downfield. The adjacent $-{\mathrm{CH}}_2$ splits the peak into a triplet.

The peaks at $4.79\mathrm{\delta }$ and $4.85\mathrm{\delta }$ ( 2 H) are due to the protons of e and f.