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.

Explanation

(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^(-)`.