# Problem: Identify the following compound:C5H10ONMR: δ 9.8 (1 H, s), δ 1.1 (9 H, s)

###### FREE Expert Solution

Degree of unsaturation, IHD:

$\overline{){\mathbf{IHD}}{\mathbf{=}}\frac{\mathbf{\left(}\mathbf{2}\mathbf{C}\mathbf{-}\mathbf{N}\mathbf{+}\mathbf{H}\mathbf{\right)}\mathbf{+}\mathbf{2}}{\mathbf{2}}}$

$\mathbf{IHD}\mathbf{=}\frac{\mathbf{\left(}\mathbf{2}\left(5\right)\mathbf{-}\mathbf{0}\mathbf{+}\mathbf{10}\mathbf{\right)}\mathbf{+}\mathbf{2}}{\mathbf{2}}$

IHD = 1 → double bond or ring

• NMR:
• C-H  δ = 1 - 2
• δ = 1.1 9 H singlet 3 methyl groups (CH3)
• Adehyde -CHO  δ = 9 - 10
• δ = 9.8 1 H, singlet  → CH
• 1 O C = O (carbonyl)
84% (151 ratings) ###### Problem Details

Identify the following compound:
C5H10O

NMR: δ 9.8 (1 H, s), δ 1.1 (9 H, s)

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