Question

Treatment of compound E (molecular formula ${\mathbf{C}}_{\mathbf{4}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{2}}$ ) with excess${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{MgBr}$ yieldscompound F (molecular formula ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{14}}\mathbf{O}$) after protonation with${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ . E shows a strongabsorption in its IR spectrum at 1743${\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$. F shows a strong IR absorption at 3600–3200${\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$.The ${}^{\mathbf{1}}\mathbf{H}$NMR spectral data of E and F are given.

What are the structures of E and F?

Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppm

Compound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55(singlet, 1 H) ppm

Short answer

IR spectroscopy can be employed for various purposes. IR spectroscopy helps to recognize the pigments found in paints.

Step-by-step solution

IR spectroscopy

IR spectroscopy can be employed for various purposes. IR spectroscopy helps to recognize the pigments found in paints.

H1 NMR spectroscopy

The methyl protons possess lower chemical shift values in NMR spectroscopy and are regarded to be deshielded. The chemical shift proceeds downfield if the hydrogen atoms are linked to electronegative atoms.

Step 3:Identification of structures E and F

In molecule E, the IR peak at 1743 ${\mathrm{cm}}^{-1}$corresponds to the C=O group. In the ${}^1\mathrm{H}$NMR data, the triplet at 1.2 ppm corresponds to ${\mathrm{H}}_{\mathrm{c}}$, the singlet at 2 ppm corresponds to ${\mathrm{H}}_{\mathrm{a}}$, and the quartet at 4.1 ppm corresponds to ${\mathrm{H}}_{\mathrm{b}}$.

The structure of compound E can be given as:

Structure of compound E

In molecule F, the IR peak at 3600-3200${\mathrm{cm}}^{-1}$corresponds to the OH group. The triplet at 0.9 ppm corresponds to ${\mathrm{H}}_{\mathrm{a}}$, and the singlet at 1.1 ppm corresponds to the proton ${\mathrm{H}}_{\mathrm{c}}$.

The quartet at 1.5 ppm corresponds to theproton ${\mathrm{H}}_{\mathrm{b}}$,and the singlet at 1.55 ppm corresponds to proton ${\mathrm{H}}_{\mathrm{d}}$.The structure of compound F can be given as:

Structure of compound F

The compound E can react with ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{MgBr}$to yield compound F, and the reaction can be given as:

Reaction of compound E with${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{MgBr}$ to yield compound F