Many amines have the formula C₈H₁₁N and contain one benzene ring…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Many amines have the formula and contain one benzene ring. Chemists meet this formula when they analyse the products of a dye or a drug synthesis. A structure is a new isomer only when the connections between the atoms differ. The same compound drawn in another way is not counted twice. Count only the structural isomers in which the six ring carbons are kept and the other two carbon atoms and the nitrogen are placed in every possible way. Think of the compounds in groups. In some, the nitrogen is joined straight to the ring and the other carbons are an ethyl group or two methyl groups. In others, the nitrogen sits in a side chain, such as a CH2 group or a CH(CH3) group joined to the ring. For a ring with two or three groups, remember that the ortho, meta and para positions give different isomers. Find the number of primary amines, secondary amines and tertiary amines, and the total number of isomers.
Need a reminder? The key ideaShow
The class of an amine is the number of carbon groups on the nitrogen atom: one, two or three. An aromatic amine has the nitrogen joined straight to a benzene ring. A saturated amine with carbon atoms is . Count isomers by choosing alkyl groups: there are 1, 1, 2, 4, 8 and 17 different alkyl groups with 1 to 6 carbon atoms.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. List the ways the nitrogen can attach, then count the ring positions.
Step 1. Primary, NH2 on the ring. An ethyl group and an NH2 group: o, m and p, which is 3. Two methyl groups and an NH2 group: the six xylidines, which is 6. That is 9.
Step 2. Primary, NH2 in a side chain. 2-Phenylethanamine and 1-phenylethanamine give 2. A methyl group and a CH2NH2 group on the ring (o, m, p) give 3. That is 5.
Step 3. All primary. .
Step 4. Secondary. N-Ethylaniline (1), N-methylbenzylamine (1) and N-methyl-o-, m- and p-toluidine (3). That is 5.
Step 5. Tertiary. N,N-Dimethylaniline only. That is 1.
Answer: 14 primary, 5 secondary, 1 tertiary, 20 in all.
Check. The program turns the ring around and flips it, so patterns that match are counted once. It gives the same numbers.
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