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Terminal alkyne is very weak acid, it forms salt with very strong base such as `NaNH_(2)` and sodium metal. `R-C-=C-Hunderset(NH_(3)(l))overset(NaNH_(2))toR-C-=overset(ө)(C)overset(o+)(N)a+(1)/(2)H_(2)` Sodium salt of alkyne is known as shown alkynide. sodium alkynide is hydrolysed with water because it is salt. Sodium salt behaves as nucleophile as well as strong base. for P-alkyl halides it behaves as a cucleophile. thus primary alkylhalides gives SN reaction halids it behaves as strong base hence they undergo elimination reaction. Q. `R-CH_(2)-C=ClIunderset(NH_(3)(l))overset(Na)toPoverset(CH_(3)-CH_(2)Br)toR`. R isA. `R-C-=C-CH_(2)-CH_(2)-CH_(3)`B. `R-CH_(2)-CH_(2)-C-=C-CH_(3)`C. `R-CH_(2)-C-=C-CH_(2)-CH_(3)`D. `R-CH_(2)-CH_(2)-CH_(2)-C-=CH` |
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Answer» Correct Answer - C |
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