Mod 7 nomenclature (1 Viewer)
- Thread starter NexusRich
- Start date
You can name it directly from that form if you are able to visualise it easily from that short hand form. That takes some practice and develops over time, so if you feel more comfortable drawing it out and then looking at it that way you should do that.
If you do need to expand it out, you know that CH3 must be the end of chain and CH2 is a linking group between (usually) 2 carbons, so I would start by looking at the parts that aren't either of these.
The first example, CH3CH2 CH2Br, is easily then seen to be
CH3- CH2- CH2-Br
and so is a 3 carbon chain with a bromine at the end, and so is 1-bromopropane.
The next example, (CH3)2CHCH2CHClCH3, has three CH3 groups and thus the chain must branch. The branch point will be short one hydrogen, and so will be a CH. The other CH has a Cl next to it, accounting for the missing H. So, we have
CH3
|
CH- CH2-CH(-Cl)- CH3
|
CH3
Our main chain has 5 carbons, with a methyl side chain and a chloro substituent. I would call it 2-chloro-4-methylpentane.
The third example, CH3(CH2)3CH2OH, has the CH2 linker group appearing three times in a row. So, we have:
CH3-CH2-CH2-CH2-CH2-OH
It is a primary alcohol on a 5 carbon chain, and hence is 1-pentanol.
The last example is strange: CH3(CH2)3CHNH2(CH2)2CH3. Starting from the ends, there can only be a methyl group followed by CH2 linkers, leaving us with
CH3-CH2-CH2-CH2-CHNH2-CH2-CH2-CH3
A nitrogen atom will typically have 3 bonds, and while it can have 4 and a positive charge, that interpretation with the NH2 in the chain would leave the adjacent C with only three bonds. The only plausible option is thus to have the NH2 as a branch, as:
CH3-CH2-CH2-CH2-CH(-NH2)-CH2-CH2-CH3
This leaves the C next to the N with four bonds and the N with 3 bonds. The main chain is then simply 8 carbons and we have 4-aminooctane.
The first example, CH3CH2 CH2Br, is easily then seen to be
CH3- CH2- CH2-Br
and so is a 3 carbon chain with a bromine at the end, and so is 1-bromopropane.
The next example, (CH3)2CHCH2CHClCH3, has three CH3 groups and thus the chain must branch. The branch point will be short one hydrogen, and so will be a CH. The other CH has a Cl next to it, accounting for the missing H. So, we have
CH3
|
CH- CH2-CH(-Cl)- CH3
|
CH3
Our main chain has 5 carbons, with a methyl side chain and a chloro substituent. I would call it 2-chloro-4-methylpentane.
The third example, CH3(CH2)3CH2OH, has the CH2 linker group appearing three times in a row. So, we have:
CH3-CH2-CH2-CH2-CH2-OH
It is a primary alcohol on a 5 carbon chain, and hence is 1-pentanol.
The last example is strange: CH3(CH2)3CHNH2(CH2)2CH3. Starting from the ends, there can only be a methyl group followed by CH2 linkers, leaving us with
CH3-CH2-CH2-CH2-CHNH2-CH2-CH2-CH3
A nitrogen atom will typically have 3 bonds, and while it can have 4 and a positive charge, that interpretation with the NH2 in the chain would leave the adjacent C with only three bonds. The only plausible option is thus to have the NH2 as a branch, as:
CH3-CH2-CH2-CH2-CH(-NH2)-CH2-CH2-CH3
This leaves the C next to the N with four bonds and the N with 3 bonds. The main chain is then simply 8 carbons and we have 4-aminooctane.