Question

Hydroboration-oxidation of an alkene, C7H12, gives a single product that exhibits 13C spectral data below.

Answer 1

Answer:

The structure of the alkene is that shown on the first uploaded image

Explanation:

From the question the molecular formula of the compound is given as $C_{7}H_{12}$

In order to calculate the double bond we apply this formula

    DBE(double bond equation) = C + 1 - H/2

                                                    = 7 + 1 - 12/2

                                                    = 2

What these tells us is that the compound has either two double bond or one ring and a double bond

  We can describe Hydroboration - oxidation of an alkene  as that chemical reaction that results in the formation alcohol, this reaction follows an Anti-Markovnikov's rule where the nucleophile attacks at the less substituted carbon.

To interpret the $^{13} C-NMR$ spectrum of alcohol formed by the Hydroboration - oxidation reaction of an alkene we can say

1)  That the $^{13} C-NMR$ spectrum produces five peaks that the compound has  fine non-equivalent carbon relatively at 26.1 δ,26.9δ.29.9δ , 40.5δ and 68.2δ

2) DEPT - 90 produces only the peaks that corresponds to the -CH.it shows a peak at 40.5δ indicating that the compound has only one -CH present

3) DEPT - 135 produces positive peaks that corresponds to both -CH and $-CH_{3}$ and negative peaks that corresponds to -$-CH_{2}$.It shows positive peaks at 40.5δ that indicates the -CH group and the remaining negative peaks at  26.1 δ,26.9δ.29.9δ δ and 68.2δ.This indicates that the compound has  4 non-equivalent  $-CH_{2}$ and one -CH in the structure

Looking at the spectra data above the compound that is consistent with it is shown on the second uploaded image

The alcohol on the compound is being produced by the reaction shown on the third uploaded image

Hence the structure of the alkene is that shown on the first uploaded image.