Rank each of the following compounds in decreasing order of reactivity towards Friedel-Crafts alkylation. Most reactive = 1; if a compound will not react, rank it as "non" rather than assigning a numerical value. Compound A: iodobenzene Compound B: bromobenzene Compound C: fluorobenzene Compound D: benzoic acid

see explanation below

Explanation:

First, let's remember the principle of a friedel crafts alkylation. This kind of reaction are often used to alkylate an aromatic ring like bencene, or a mono substitued bencene. This reactions often involves reactions with alkyl halides or alkenes and aluminum or iron halides.

Now, in this case we have Iodobenzene, bromobenzene, fluorobenzene and benzoic acid.

First, the reactivity order always raise with the polarization of the C - X bond. Therefore, the one with the most polar bond, will react first. In this case, with the halides, the reactivity order would be RI > RBr > RCl >> RF

In the case of the benzoic acid, the COOH is a very strong deactivating group by resonance, therefore, the benzene it's really weak to do a reaction of alkylation, so this will most likely to not react.

In conclude, we have the following order:

Compound A: Iodobencene (1)

Compound B: Bromobencene (2)

Compound C: Fluorobencene (3)

Compound D: Benzoic acid (non)

true

explanation:

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.