How is $C_{6} H_{11} B r$ $C_6H_11Br$ made?

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How is $C_{6} H_{11} B r$ $C_6H_11Br$ made?

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Answer 1 · 2017-04-08T12:23:36

You might want to qualify this question. At the moment, I don't know what you ask.

Explanation:

Cyclohexane, $C_{6} H_{12}$ $C_6H_12$ , has one, so-called, $degree of unsaturation$ $"degree of unsaturation"$ . We refer to the saturated alkane, i.e. $C_{n} H_{2 n + 2}$ $C_nH_(2n+2)$ . The ring junction in cyclohexanol reduces the hydrogen count by 2.

As to how to make $cyclohexyl bromide$ $"cyclohexyl bromide"$ from $cyclohexane$ $"cyclohexane"$ ; the standard means is to halogenate the cycloalkane with $NBS$ $"NBS"$ under UV light:

$C_{6} H_{12} + B r_{2} h ν - - \to C_{6} H_{11} B r + H B r$ $C_6H_12 + Br_2 stackrel(hnu)rarr C_6H_11Br + HBr$

Hydrolysis of the cyclohexyl bromide would yield the alcohol.

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How is $C_{6} H_{11} B r$ $C_6H_11Br$ made?

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