Consider the^1 H NMR spectrum of p-anisaldehyde (page 696 of the 6^th edition of
ID: 1064901 • Letter: C
Question
Consider the^1 H NMR spectrum of p-anisaldehyde (page 696 of the 6^th edition of the textbook, page 624 of the 5^th edition of the textbook). The aromatic hydrogen are separated into two peaks - one peak is right around 7.9 ppm, and the other group is right around 7.0 ppm. a) In a general sense, what does the chemical shift of a peak indicate about the amount of electron density around a nucleus? b) Considering the structure of p-anisaldehyde, why would the aromatic hydrogen be separated by such a large chemical shift. Support your answer with any relevant structures, if applicable. It may be useful for you to know that the chemical shift of the hydrogens In benzene is 7.26 ppm.Explanation / Answer
A large value of chemical shift indicates that the electron density around the nucleus is less, as a result the nucleus is less shielded ( de-shielded). Thus greater the chemical shift, lesser is the electron density around the nucleus.
As compared to benzene where all C-atoms have identical electron densities, p-anisaldehyde has two functional groups attached to the ring.
The CHO functional group is an electron withdrawing group while OCH3 is electron donating group. As a result the protons of C-atoms closer to the OCH3 group have higher electron density and hence lesser chemical shift of 7 ppm as compared to benzene, while in the protons of C-atoms near to CHO group the electron density is lesser so chemical shift is higher with value 7.9 ppm