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  2. 1-Bromohexane - Wikipedia

    en.wikipedia.org/wiki/1-Bromohexane

    Most 1-bromoalkanes are prepared by free-radical addition of hydrogen bromide to the 1-alkene. These conditions lead to anti-Markovnikov addition, giving the 1-bromo derivative. [2] 1-Bromohexane undergoes reactions expected of simple alkyl bromides. It can form Grignard reagents. [3]

  3. Bromobenzene - Wikipedia

    en.wikipedia.org/wiki/Bromobenzene

    Bromobenzene is prepared by the action of bromine on benzene in the presence of Lewis acid catalysts such as aluminium chloride or ferric bromide. [3]Bromobenzene is used to introduce a phenyl group into other compounds.

  4. Bromochlorobenzene - Wikipedia

    en.wikipedia.org/wiki/Bromochlorobenzene

    All three have been synthesized by various routes: 1-Bromo-2-chlorobenzene: from 2-chloroaniline, via diazotization followed by a Sandmeyer reaction [1]; 1-Bromo-3-chlorobenzene: by (3-chlorophenyl)trimethylgermanium by electrophilic substitution [2] [better source needed]

  5. 2-Phenylethyl bromide - Wikipedia

    en.wikipedia.org/wiki/2-Phenylethyl_bromide

    2-Phenylethyl bromide is an organobromide with the formula C 6 H 5 CH 2 CH 2 Br. It is a colorless liquid, although older samples appear yellow. Analogous to the preparation of most 1-bromoalkanes, it is prepared by free-radical addition of hydrogen bromide to styrene.

  6. Cyclohexylbenzene - Wikipedia

    en.wikipedia.org/wiki/Cyclohexylbenzene

    Cyclohexylbenzene is produced by the acid-catalyzed alkylation of benzene with cyclohexene. [1] [2] The process can proceed using benzene as the exclusive organic precursor.. Its partial hydrogenation gives cyclohexene, which alkylates the unhydrogenated benz

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  8. Why the Royal Kids Might Not Be Allowed to Keep Some of Their ...

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    The royal kids might not be allowed to keep some of their Christmas presents from the public this year thanks to the family's strict rules about gift giving.

  9. 1-Bromobutane - Wikipedia

    en.wikipedia.org/wiki/1-Bromobutane

    1-Bromobutane is the precursor to n-butyllithium: [4] 2 Li + C 4 H 9 X → C 4 H 9 Li + LiX where X = Cl, Br. The lithium for this reaction contains 1-3% sodium. When bromobutane is the precursor, the product is a homogeneous solution, consisting of a mixed cluster containing both LiBr and LiBu.