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Can you help with this please?​

Can you help with this please?​-example-1

1 Answer

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Answer:

A.) 2,4,5-trimethylhexan-2-ol

B.) 6-bromo-3-ethylhexan-2-one

C.) 3-isoproyl-4-propyloctan-1-al

Step-by-step explanation:

A.) Since there is an alcohol (-OH) group in the molecule, the name should have the ending -ol. Since the alcohol is on the 2nd carbon, there should be a 2 in front of the -ol. The longest carbon chain contains 6 single-bonded carbons, making this a hexane. There are 3 methyl (-CH₃) groups on the 2nd, 4th, and 5th carbons.

Therefore, the IUPAC name is: 2,4,5-trimethylhexan-2-ol

B.) Since there is a ketone (C=O) group in the molecule, the name should have the ending -one. Since the ketone takes priority, it is on the 2nd carbon. As such, there should be a 2 in front of the -one. The longest carbon chain contains 6 single-bonded carbons, making this a hexane. There is an ethyl (-CH₂CH₃) group on the 3rd carbon and a bromo (-Br) group on the 6th carbon.

Therefore, the IUPAC name is: 6-bromo-3-ethylhexan-2-one

C.) Since there is an aldehyde (HC=O) group in the molecule, the name should have the ending -al. Since the aldehyde is on the 1st carbon, you can add a 1 before the -al (technically not necessary). The longest carbon chain contains 8 single-bonded carbons, making this an octane. There is an isopropyl (-CHCH₃CH₃) group on the 3rd carbon and a propyl (-CH₂CH₂CH₃) on the 4th carbon.

Therefore, the IUPAC name is: 3-isoproyl-4-propyloctan-1-al (or 3-isoproyl-4-propyloctanal)

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