A compound ′X′ having molecular formula C5H8, reacts with ammoniacal AgNO3 to give a white precipitate and reacts with excess of KMnO4 to give the acid, (CH3)2CH−COOH. Therefore, X is:
A
CH2=CH−CH=CH−CH3
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B
CH3−CH2−C≡C−CH3
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C
(CH3)2CH−C≡CH
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D
(CH3)2C=C=CH2
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Solution
The correct option is C(CH3)2CH−C≡CH Terminal alkynes contain acidic hydrogen. These terminal alkynes react with ammoniacal solutions of silver nitrate to give white precipitate of silver alkynide. Since compound 'X' on reacting with ammoniacal AgNO3 gives white precipitate. The given compound will have terminal alkyne part. From the given options we can see that compound of option (c) has terminal alkyne part. (CH3)2CH−C≡CH+[Ag(NH3)2]+NO−3→(CH3)2CH−C≡C−Ag+↓+NH+4NO−3+NH3
When this compound (c) is treated with excesss KMnO4, the cleavage will occur at alkyne part and they get oxidise to form carboxylic acid. Thus, it gives (CH3)2CH−COOH and CH3COOH (CH3)2−CH−C≡CHKMnO4−−−−−→(CH3)2−CH−COOH+HCOOH