Answer:
(i)
Since liquid 'A' forms a white crystalline solid on treatment with NaHS03, it
may be an aldehyde or a methyl ketone.
Further since liquid 'A' forms a bright silver mirror on treatment with a
freshly prepared ammoiml solution of therefore,
liquid 'A' is an aldehyde.
(ii) Since liquid 'B' forms a white crystalline solid on
treatment with NaHS03, it may be an aldehyde or a methyl
ketone.
Further since liquid 'B' does not give test with
ammoniacal solution, therefore,
liquid 'B' must a methyl ketone.
Chemical equations for the reactions discussed are :
No silver mirror
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