How many 2º alkyl bromides,neglecting stereoisomers,exist with the formula C₆H₁₃Br?

Question

Quizplus · Accepted Answer

The formula of the alkyl bromide given is C(S1)(U1)B16S1U1B0HS1U1B113S1U1B0Br. Breaking this down, the alkyl bromide has:

- one bromine atom (Br)
- one carbon atom (C)
- one hydrogen atom (H)
- one tertiary butyl group (B113)
- one methyl group (B16)
- one methylene group (S1U1B0)
- one methine group (S1)

For a 2º alkyl bromide, we need two identical alkyl groups attached to the carbon atom bearing the bromine. The possible combinations of alkyl groups are:

- Two methyl groups (B16-B16)
- One methyl group and one tertiary butyl group (B16-B113 or B113-B16)
- Two tertiary butyl groups (B113-B113)

Therefore, there are 3 possible 2º alkyl bromides, neglecting stereoisomers:

- C(B16)2Br
- C(B16)(B113)Br
- C(B113)2Br

Thus, the answer is (C)6.

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How Many 2º Alkyl Bromides,neglecting Stereoisomers,exist with the Formula C₆H₁₃Br