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Write Brief Answer · Q33

Q.How is propyne prepared from an alkylene dihalide?

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Step 1. Start from the alkylene (vicinal) dihalide of propane, 1,2-dibromopropane, CH3-CHBr-CH2Br (two bromines on adjacent carbons).

Step 2. Treat with alcoholic KOH: this eliminates the first molecule of HBr (dehydrohalogenation), giving the vinylic bromide 1-bromopropene, CH3-CH=CHBr (or the regiochemical alternative CH2=CBr-CH3, depending on which beta-hydrogen is removed).

Step 3. Treat this bromoalkene with a stronger base, NaNH2 (sodium amide), which is needed because eliminating HBr from a vinylic (sp2 C-Br) position requires a much stronger base than the first, ordinary alkyl elimination did; this removes the second molecule of HBr, installing the triple bond and giving propyne, CH3-C≡\equivCH.

Step 4. This is exactly the same two-step 'vicinal dihalide -> vinyl halide -> alkyne' logic used earlier in the chapter's general alkyne-preparation method, just applied starting directly from the dihalide rather than starting from the alkene that would first need halogenating to reach it.

✓Final answer

1,2-dibromopropane --alc. KOH (-HBr)--> 1-bromopropene --NaNH2 (-HBr)--> propyne (CH3-C≡\equivCH).

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