# Draw the lewis structure of acetaminophen and all the intermolecular forces that occur in it?

##### 1 Answer
Dec 21, 2016

The name acetaminophen, which resembles "aceto", "amine", and "phenol", which suggests that it looks like an acetoamide and phenol, which it does:

1) $\text{London Dispersion}$

London dispersion is the weakest intermolecular force, and it is regarding an induced dipole.

Basically, the electron cloud of this molecule can be distorted by coming into close range with another acetaminophen molecule, which shifts around the partial charges and creates a weak temporary dipole.

2) $\text{Dipole-Dipole}$

Oxygen and nitrogen are both more electronegative than carbon. The more electronegative atom pulls "electron density" more towards it, making it more partially negative.

When we draw dipole arrows to approximate the direction to which the negative charge is pulled, we first get the following individual arrows, and then we add them together to form a net dipole:

The upper-left two dipole arrows add to give an approximately horizontal sum (slightly downwards) that is stronger than the individual ones, and that adds with the downward dipole arrow to give one that is less than ${45}^{\circ}$ from the horizontal.

Since the molecule is polarized to the lower-left, any dipole-dipole interactions would generally align the molecule (with respect to the others) so that the $O H$ of one molecule is near the carbonyl carbon of another, more or less.

3) "Hydrogen-Bonding" ("H-Bonding")

This molecule has several spots it can hydrogen-bond. With an $\text{OH}$ group, as well as an $\text{NH}$ and an $\text{O}$ with lone pairs, it has:

• $\text{OH}$ group: one H-bonding donor and (theoretically) two H-bonding acceptors
• $\text{NH}$ group: one H-bonding donor and (theoretically) one H-bonding acceptor
• $\text{C"="O}$ group: (theoretically) two H-bonding acceptors

Some reasonable possibilities on how this could work out are:

• The $\text{H}$ on the $\text{OH}$ interacts with the $\text{O}$ on the $\text{C"="O}$, donating an H-bonding interaction to the lone pair acceptors on the $\text{C"="O}$.
• The $\text{H}$ on the $\text{NH}$ donates an H-bonding interaction to the lone pair on the $\text{N}$ from another molecule.

However it works out, the molecule is going to try to maximize the number of H-bonding interactions that it can manage. Here is what I think it could look like:

In these orientations, acetaminophen can form two $\text{O":cdots"H"-"O}$ hydrogen-bonds in molecular interaction pairs, and these pairs can be connected by the remaining $\text{N"-"H"cdots:"N}$ interaction. Furthermore, the $\text{OH}$ is aligned near the carbonyl carbon, like mentioned when discussing dipole-dipole interactions.

This is a similar interaction scheme to the H-bonding in acetic acid, for example: