Draw the major monoalkylation product(s) you would expect to obtain from reaction of p-chloroaniline with chloromethane and AlCl3.

a. You do not have to consider stereochemistry.

b. Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner.

c. Separate multiple products using the + sign from the dropdown menu.

d. If no reaction occurs, draw the starting material.

See explanation below

Explanation:

The reaction between the p-chloroaniline and the chlorometane in AlCl3, is a reaction of alkylation in an electrophilic aromatic sustitution reaction. This reaction is known as Friedel Craft Alkylation. This reaction is often used to put an alkyne group into an aromatic ring.

Now, with this said, let's explain what happen here. The amino group in this ring, is the NH2. This is a very strong activating group in the ring, therefore, when the reaction occurs, the methane will go to the orto position, but as NH" is very strong, this alylation would be in the whole ring, so it will occupy the other orto position in the ring. This is because the NH2 is a very strong activating group and promoves the excess of the alkylation in the whole molecule. For this reason, it will form only one product. The product is the 4-chloro-2,6-dimethyl-aniline. See picture for the product.

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots