Question #36647

1 Answer
Truong-Son N.
Sep 19, 2015

It doesn't exist. I assume you mean:

#CH_3(CHBr)CH_3#

which is isopropyl bromide, or 2-bromopropane. Of course, you can count from 1 starting on either side because of symmetry. 2-bromopropane is the IUPAC naming, while isopropyl bromide is known as a common name.

(If you meant something else, then I wouldn't know what.)

You use isopropyl because the bromine is attached to the center carbon of a v-shaped substituent, and this substituent has 3 carbons. If there was one more carbon attached to carbon-2 between bromine and the v-shaped alkyl chain, then it would be isobutyl, and so on.

If the only explicit #H# were replaced with #CH_3#, it would be tert-butyl bromide, like "tertiary butyl". Or, it would be 2-bromo-2-methylpropane.

If carbon-1 had a #CH_3# coming off, it would be sec-butyl bromide, like "secondary butyl". Or, it would be 2-bromobutane. (In this case, it just so happens that there would be both R and S stereoisomers.)

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