I need help on this.
How do you get the answer.
Thank you very much!
The first one is a straight chain alkane, so you just count the carbons (there’s 8), assign a prefix accordingly and add “-ane”. Then you get octane.
For the second you have a triple bond between the third and fourth carbon, so it’s an alkyne. So you count the carbons (there’s 7), assign the prefix (hept-) and add the number of the first atom with a triple bond (the third) and the suffix “yne”. Then you get hept-3-yne
Compound 1 is octane – 8 chain, single bonds – no problem
Compound 2: In the middle of the chain you hace 2 carbons together.
You write the compound like this:
H – C – C -C – C – C – C – C -H
You have aproblem between C3 and C4. This can be rectified by a triple bond here. Do this and all carbons have 4 bonds
7 carbon chain = hept
triple bond = yne
on C3, name is hept – 3 – yne
Properly notated: C8H18 = Octane
Look this up in Morrison and Boyd Chemistry Book…based on the proper molecular structure, you should be able to look it up.
Not certain on 2nd one but think it’s PropyleneSource(s): I know a Chemistry Brainiac but doesnt have any of his old college-books(sent to a Brainiac school)…he says you should be able to find all you need in that Morrison n Boyd book.
CH3CH2CCCH2CH2CH3 isalso called 3-heptyne