Question

Give reasons for the following:
(a) Potassium cyanide reacts with R -X to give alkyl cyanide. While silver cyanide forms an isocyanide as a major product.
(b) Silver nitrite reacts R-X to give a mixture of nitroalkane and alkyl nitrite.
(c) R-OH does not react NaBr but on adding `H_(2)SO_(4)` . if forms R-Br.
(d) Alcohols do not undergo substitutionl in neutrol or alkaline solution.

Answer 1

(a) Potassium cyanide is an ionic compound and provide cyanide ions in solution, in which each of carbon and nitrogen carry a lone pair of elctrons [CN is an ambident nucleophile(ligand)].
As lone pair on carbon is more reactive, the carbocation, `R^(+)`, preferentially attacks the carbon atom and therby forms an alkyl cyanide predominantly.

On the other hand , AgCN (being insoluble) is a covalent compound and only nitrogen has the lone pair of electrons. ltbr. Thus, carbocation attacks through nitrogen and thereby forms an alkyl isocyanide as a major product.
`Ag -C-= N: +RX to RNC: + AgX`

sites i.e., N and either of the oxygen atoms . the carbocation can attack either N or O, thereby forming a mixture of nitroalkane `R-NO_(2)` and alkyl nitrite `R-ONO`.
(c) `Br^(-)` ion is a very weak Bronsted base and thus it can not displace the strong base `-OH^(-)` . When `H_(2)SO_(4)` acid is added ,`H^(+)` are available which form `overset(+)(ROH)_(2)`. Under this condition, `Br^(-)` displaces `H_(2)O` which is a very weak base.
`overset(+)(ROH_(2))) +B^(-) to R^(+) +H_(2)O + Br^(-) to RBr +H_(2)O`
(d) A leaving group must be a weak bse. The leaving `-OH` group of alcohols in neutral or alkaline solution would be stronger base, hence cannot be removed by weaker bases like `CI^(-), Br^(-) , I^(-)`.