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 [tex]C_{7}H_{12}[/tex]
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 [tex]^{13} C-NMR[/tex] spectrum of alcohol formed by the Hydroboration - oxidation reaction of an alkene we can say
1) That the [tex]^{13} C-NMR[/tex] 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 [tex]-CH_{3}[/tex] and negative peaks that corresponds to -[tex]-CH_{2}[/tex].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 [tex]-CH_{2}[/tex] 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.
