Give the structure that corresponds to the following molecular formula and 1H NMR spectrum: C7H16O4:
1.93 (t, J = 6 Hz);
3.35 (s);
4.49 (t, J = 6 Hz);
relative integral 1:6:1.
The Correct Answer and Explanation is:
Correct Structure:
2-(2-methoxyethoxy)ethanol
Molecular formula: C₇H₁₆O₄
Structure:
CH3-O-CH2-CH2-O-CH2-CH2-OH
Explanation
The molecular formula C₇H₁₆O₄ suggests a saturated compound with 7 carbon atoms and 4 oxygen atoms, likely involving ether (–O–) and alcohol (–OH) functionalities.
Degree of Unsaturation (DoU): DoU=2C+2−H2=2(7)+2−162=0DoU = \frac{2C + 2 – H}{2} = \frac{2(7) + 2 – 16}{2} = 0
A DoU of 0 means the molecule is fully saturated—no rings or double/triple bonds. The 4 oxygen atoms likely exist as ether and/or alcohol groups.
1H NMR Interpretation:
Signal 1: 1.93 ppm (triplet, J = 6 Hz, 1H)
- Integration = 1H suggests a –OH proton (alcohol).
- The triplet implies it’s coupled with two equivalent hydrogens (i.e., CH₂–CH₂–OH).
- Chemical shift is consistent with an alcoholic OH adjacent to CH₂.
Signal 2: 3.35 ppm (singlet, 6H)
- Integration = 6H, singlet ⇒ likely two equivalent –OCH₃ groups or a symmetrical –CH₂–O– structure.
- But we only have one methyl group in the structure; this better fits two CH₃ protons in a symmetrical environment, such as CH₃–O–.
- However, the integration of 6H with no splitting suggests two –CH₃ protons attached to oxygen in a symmetrical molecule or a CH₃ group in fast exchange.
- Actually, in CH₃–O–CH₂–, the CH₃ appears as a singlet around this range.
Here, –OCH₃ methyl group gives a singlet at 3.35 ppm. There are 6 H total, but this is due to symmetrical CH₂ protons or overlapping signals.
Signal 3: 4.49 ppm (triplet, J = 6 Hz, 1H)
- Integration = 1H, triplet ⇒ likely a CH₂ group adjacent to another CH₂ and also attached to electronegative atom (oxygen).
- Chemical shift consistent with CH₂–O– near alcohol or ether.
Proposed Structure Justification:
The structure 2-(2-methoxyethoxy)ethanol fits:
- 7 C atoms: CH₃–O–CH₂–CH₂–O–CH₂–CH₂–OH.
- 4 O atoms: 2 ethers, 1 terminal alcohol.
- NMR data:
- CH₃–O– at ~3.35 ppm (singlet).
- Terminal –CH₂–OH: CH₂ at ~4.49 ppm (triplet), OH at ~1.93 ppm (triplet).
Conclusion: The compound is 2-(2-methoxyethoxy)ethanol, consistent with both molecular formula and NMR data.
