In this question, we're asked to identify a compound that has no dipole moment.

To answer this question, we need to understand what a dipole moment is. Some compounds have an uneven distribution of charge around their molecular structure. This uneven distribution is normally due to differences in electronegativity of the various atoms that make up that molecule. This uneven distribution of electrical charge is what is known as a molecule's dipole moment. However, while electronegativity differences are important, it is also vital to take into consideration the molecular structure of the compound in question because, many times, two dipole moments can cancel each other out and lead to a net dipole moment of zero.

Now, we'll need to look at each answer choice in order to see whether there will be an uneven distribution of electrical charge.

First, let's start with water. We know that water consists of an oxygen atom singly bound to two other hydrogen atoms. Furthermore, the oxygen has two lone pairs of electrons. This causes the water molecule to take on a bent shape. Moreover, the electronegativity of oxygen is much greater than that of hydrogen. Consequently, we can expect there to be a net dipole moment in water.

Next, let's take a look at dichloromethane, . In an ordinary molecule of methane, we know that the four hydrogen atoms are situated around the central carbon atom in a tetrahedral fashion. Likewise, dichloromethane also has a tetrahedral structure. The difference, however, is that two of the hydrogens are now chlorine atoms. As a result, two of the chlorine atoms will be pointed in the opposite direction to the two hydrogen atoms. Furthermore, chlorine and carbon have a massive difference in electronegativity. As a result, we would expect dichloromethane to have a net dipole moment.

Let's turn our attention to carbon monoxide. This compound consists of a carbon atom triple bonded to an oxygen atom, with a lone pair of electrons on both atoms. Since these are the only two atoms in the compound, the structure is relatively simple as it is just linear. Moreover, there is a substantial difference in the electronegativity of carbon and oxygen and, thus, we would expect carbon monoxide to have a net dipole moment.

Finally, let's look at carbon dioxide. In this compound, we have a central carbon atom which is double bonded to two other oxygen atoms. Just as with carbon monoxide, the structure of carbon dioxide is simply linear. And once again, due to the difference in electronegativity between carbon and oxygen, there is a dipole moment between the central carbon atom and each of the oxygen atoms. However, this doesn't mean that the molecule as a whole has a dipole moment!! Remember, we also need to take the structure of the molecule into account. Because the two oxygen atoms are arranged around the carbon in a linear fashion and are pointing directly away from each other, the two dipole moments between carbon and oxygen will exactly cancel each other out. Thus, carbon dioxide will not have a net dipole moment.

In this question, we're presented with the structure of pyrrole. We're told that the compound is aromatic and is stabilized by resonance. We're then given a number of resonance structures, and we're asked to decide which is a correct one.

In order to answer this question, we'll need to consider the structure of pyrrole, and see which resonance structures we can obtain. To do this, we'll need to do some electron pushing.

Also, it's critical to recognize that the nitrogen atom in pyrrole has a lone pair of electrons. This lone pair is situated in the nitrogen's p orbitals. This allows the p orbital electrons to participate in pi bonding with the other two carbon-carbon double bonds found in the molecule. Because there is a lone pair of electrons on the nitrogen atom, we can begin our "electron pushing" starting with these electrons first.

After pushing electrons as shown above, we end up with the following possible resonance structure.

Thus, this is the correct answer. All of the other resonance structures shown are not possible. You can do electron pushing to try it out for yourself.

 is the correct answer. In most cases, the deactivating substituents direct "meta." Halogens deactivate benzene rings, but are the exception, as they direct "ortho" or "para." A substituent is deactivating when it has a low electron concentration on the atom directly attached to the benzene ring. Carboxylic acid provides resonance, a delocalization of electrons. However, the amine group, for example, sees a lone pair on the nitrogen.

The functional group that is already on the phenyl is the group that dictates where any other substituent will be directed on the ring. In this case, we have a carboxylic acid as our director. Due to resonance, carboxylic acid deactivates (is an electron-withdrawing group) the benzene ring and directs the bromine to the meta position.

Here we have a classic EAS reaction.

If we examine the molecule, we see that the benzene ring already has two substituents on it: , an activating group, and , a deactivating group. Remember, when assigning the position of a new substituent to a benzene ring, activating groups always dictate the position of the new substituent.

We know that activating groups direct new substituents to the ortho or para positions. However, because of sterics, a new substituent will not be directed to the ortho position in between two substituents. Thus, the new substituent will be directed to the para position (or the other ortho position, but that is not one of our answer choices), and the correct answer is position 2.

With the exception of halogens, meta directors deactivate a benzene ring. In other words, they make the benzene ring less reactive, especially in an electrophilic aromatic substitution (EAS) reaction. Meta directors have little electron density at the point of contact with the benzene ring. For example, a carboxylic acid is a meta director because it experiences resonance, a delocalization of electrons. All of the answer choices in this problem have a lone pair of electrons on the point of contact with the benzene ring and they all are ortho/para directors.

Carbonyls (as in 4 and 5) are always electron withdrawing due to the Oxygen's electronegativity. Similarly, the oxygens on the nitrate (1) are electron withdrawing.

Phenol contains the hydroxide group, which is an electron donor, puts electron density into the benzene ring. Resonance structures are drawn as follows

 is the only electron-withdrawing substituent because it contains two electronegative oxygen atoms which pull electrons from the benzene ring towards itself. This effect is electron-withdrawing and makes the ring slightly positive in charge. All the other substituents are electron-donating groups, which activate the ring for electrophilic addition.

 is the only electron-donating group listed. Therefore, it will add to the ortho and para positions on phenol. The rest of the substituents are highly electron-withdrawing groups and will add to the meta positions on phenol.