Chapter 20, Problem 20.71P

Treatment of compound E (molecular formula ${\text{C}}_4{\text{H}}_8{\text{O}}_2$) with excess

${\text{CH}}_3{\text{CH}}_2\text{MgBr}$ yields compound F (molecular formula ${\text{C}}_6{\text{H}}_{14}\text{O}$) after protonation with ${\text{H}}_2\text{O}$. E shows a strong absorption in its IR spectrum at ${\text{1743 cm}}^{-1}$. F shows a strong IR absorption at $\text{3600}-3200{\text{ cm}}^{-1}$. The ${}^1\text{H}$ NMR spectral data of E and F are given. What are the structures of E and F?

Compound E signals at $1.2\left(\text{triplet, 3H}\right)\text{, }2.0\left(\text{singlet, 3H}\right),$ and $4.1\left(\text{quartet, 2H}\right)\text{ppm}$.

Compound F signals at $0.9\left(\text{triplet, 6H}\right)\text{, 1}\text{.1}\left(\text{singlet, 3H}\right),\text{ 1}\text{.5}\left(\text{quartet, 4H}\right)$ and $1.55\left(\text{singlet, 1H}\right)\text{ppm}$.