Two Nitrophenol Solids: Intramolecular vs Intermolecular Hydrogen Bonding
A student synthesizes nitrophenols by nitrating phenol and obtains two solids. Recrystallization gives Sample A, mp 45–46 °C, and Sample B, mp 113–115 °C. Both have the same molecular formula, C6H5NO3. A TLC plate developed in 3:1 hexanes/ethyl acetate shows A at Rf 0.62 and B at Rf 0.24. A solubility test shows A dissolves to 1.7 g/L in water, while B dissolves to 11.6 g/L. The student labeled both 'p-nitrophenol' because the textbook says para is the major product. The IR spectra both show O-H and NO2 stretches. The student has no NMR; the only standards are authentic 2- and 4-nitrophenol. A mixed melting point of A with authentic 4-nitrophenol is depressed; B with authentic 4-nitrophenol is unchanged.
1. Assign Sample A and Sample B to the ortho and para isomers and explain the melting-point and TLC differences using intermolecular forces. Include why 'A is less polar' is an incomplete explanation.
Read the worked answer
A is 2-nitrophenol (ortho); B is 4-nitrophenol (para). In the ortho isomer, the OH and NO2 groups are close enough to form an intramolecular hydrogen bond. That satisfies part of the OH hydrogen-bonding capacity internally, reducing intermolecular H-bonding, so the solid lattice is weaker (mp 45–46 °C) and the molecule interacts less strongly with silica's silanol groups, giving a higher Rf (0.62).
The para isomer cannot form an intramolecular H-bond; its OH and NO2 groups form intermolecular H-bonds, giving a stronger lattice (mp 113–115 °C), stronger adsorption to silica, and lower Rf (0.24). 'Less polar' is incomplete because both isomers have similar overall polarity; the key is the availability of the OH group for intermolecular H-bonding, not a simple dipole ranking.
2. Predict which sample is more soluble in water and which would be steam-volatile. Justify each prediction.
Read the worked answer
B (para) is more water-soluble (11.6 g/L vs 1.7 g/L) because its free OH and NO2 groups can hydrogen-bond with water; the ortho isomer's intramolecular H-bond reduces its ability to accept/donate H-bonds to water and makes it more internally satisfied, so it is less soluble.
A (ortho) should be steam-volatile: intramolecular H-bonding reduces intermolecular attraction and raises vapor pressure, so it can be carried with steam at temperatures below its boiling point. The para isomer's extensive intermolecular H-bonding lowers vapor pressure, so it is not efficiently steam-distilled. This is consistent with the lower melting point and higher TLC Rf of A.
3. The student has authentic samples of both isomers. Describe a mixed-melting-point test and one spectroscopic test that would confirm the assignment, and state the expected result.
Read the worked answer
Mixed melting point: mix A with authentic 4-nitrophenol; if A were para, the mixture would melt at the same temperature as pure B (unchanged). In the setup, A with authentic 4-nitrophenol gives a depressed, broadened melting range, while B with authentic 4-nitrophenol is unchanged; that confirms A = ortho and B = para.
Spectroscopic test: compare IR O-H stretching. The para isomer's intermolecular H-bond gives a broad O-H band whose position and shape change with concentration, while the ortho isomer's intramolecular H-bond gives a sharper, less concentration-dependent O-H band. Alternatively, 1H NMR in a nonpolar solvent: the ortho OH proton is strongly deshielded by the intramolecular H-bond and appears further downfield, while the para OH shift is concentration-dependent. Either result supports the assignment.