# Identifying functional groups

I've spent a few days googling functional groups and reading my text assigned on organic functional groups. I'm understanding very little and I just need a little clarification. I understand that at the top right and bottom left, the hexagon structure signifies an aromatic, right?

Top Left: My first guess was carboxylic acid, but with further googling, I now think it's an amide, because it contains nitrogen and oxygen?

Top Right: Since the hexagon is aromatic, I know it can only be the last three choices, but I'm going with C because the $\ce{-NH2}$ is not connected to an oxygen.

Bottom Left: Because of the $\ce{Br}$, I know it's a bromide and, assuming that my thinking is correct with aromatic, the answer should be B.

But the bottom right, I am not sure what exactly to look at here to even guess, besides that it is not an aromatic ring.

To show that I've tried the last one, all I can find is that an alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$ and, the aromatic being the hexagon structure, I'm unable to grasp/understand which one the last group would fall into.

• Welcome to Chemistry.SE. Take the tour to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that ‎we aren't doing your homework for you. – user15489 Jul 1 '15 at 16:33
• @alyssa Is it multiple choice test? – Mithoron Jul 1 '15 at 17:48
• No, it's not a test or homework, these questions are coming from my book, so I can be more prepared for our final in a month. The functional groups just make no sense to me. – Alyssa Jul 1 '15 at 18:18
• Basically anything in this format is categorized as homework. It doesn't need to be homework itself. Self-study questions, puzzles etc. also count as homework. Regarding your edit, I'm really inclined to vote to reopen, but you've shown no evidence of your research. You shouldn't say I Googled but I couldn't find anything. that isn't helpful at all. You should say I Googled for blah blah, but it said bladiblah. so I can't understand the di in bladiblah. can you help me understand? Do as so and I'll vote to reopen. – M.A.R. Jul 1 '15 at 19:35
• en.wikipedia.org/wiki/Functional_group – Mithoron Jul 1 '15 at 20:01

Some pointers:

the hexagon structure signifies an aromatic

Not quite. The hexagon structure depicted with alternating carbon-carbon double and single bonds is aromatic. It is called benzene and is very commonly encountered in organic chemistry. It can be represented by two resonance structures as shown below and is sometimes also represented with a circle inside a ring of singly bonded carbons to show the delocalised $\pi$ system.

However, the are other molecules depicted as hexagons which are not aromatic such as cyclohexane.

Additionally, there are other molecules which are aromatic which are not benzene, and many which are not hexagonal but you probably won't come across these at an introductory level.

My first guess was carboxylic acid, but with further googling, I'm now thinking it's an amide, because it contains nitrogen and oxygen

You are correct that it is an amide. The amide functional group contains a nitrogen adjacent to a carbon-oxygen double bond. The carboxylic acid group is similar but the nitrogen is replaced by an $\ce{-OH}$ group. Amides can be primary, secondary or tertiary depending on the number of carbons that are bonded to the nitrogen.

alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$

These are the simplest example of alkanes, alkenes and alkynes. They are called methane, ethene and ethyne respectively. However, there are more complicated molecules involving these functional groups. In general, alkanes have only carbons-carbon single bonds and carbon-hydrogen single bonds. Alkenes are distinguished by having a carbon-carbon double bond denoted by two lines between the two carbons which look like an equals sign. Alkynes have a carbon-carbon triple bond which is denoted by three lines between the carbons. Therefore, your example is an alkene.