Q. 105.0( 2 Votes )

The best reagent

Answer :

• When 2–phenyl propanamide reacts with LiAlH4 in ether it undergoes reduction and produces correspondent amine i.e. 2-phenyl propanamine , without changing the carbon number and remaining the main carbon chain intact as the parent compound


• Amongst the other 3 options


(i)excess H2 (ii) Br2 in aqueous NaOH and (iii) iodine in the presence of red phosphorus none of them brings the change of amide to amine with the carbon numbers remaining the same as the reactant compound 2-phenyl propanamine.


• (i)excess H2 would not be able to bring the change because a catalyst would be required to activate the compound.


• (ii) Br2 in aqueous NaOH is actually the Hoffman bromamide degradation reaction, where migration of the alkyl or aryl group takes place from carbonyl carbon to the N atom, thus by producing amine 1 carbon less than the reactant amide compound.



• And at last (iii) iodine in the presence of red phosphorus do not bring the desired change of amides.

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