13.
A compound ‘A’ having molecular formula C HyBr on reaction with alcoholic KOH gives a
compound 'B'. Bromination of 'B' gives compound Cº. Compound 'C' on treatment with soba
amide gives a gaseous compound 'D'. The gas 'D' when passed through ammonical silver nitrea
solution forms white precipitate.
Identify compounds A, B, C and D and write down the reactions involved,
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Br−CH
2
−CH
2
−CH
2
−CH
3
alc.
KOH
CH
2
=CH−CH
2
−CH
3
Br
2
CCl
4
Br−CH
2
−
∣
CH
−CH
2
CH
3
A B Br C
↓(Sodalime)(NaOH+CaO)
CH
3
CH
2
−C≡C
−
Ag
+
↓
[Ag(NH
3
)
2
]
+
(ammonicalsilvernitrate)
terminalalkyne
H−C≡C−CH
2
−CH
3
Alkyl bromide (A) on reaction with alcoholic KOH gives a alkene B which on bromination gives dibromobutane (C). C on treatment with soda lime gives a terminal alkyne D which gives white ppt with ammonical silver nitrate solution where alkynes should be terminal.
Br−CH
2
−CH
2
−CH
2
−CH
3
alc.
KOH
CH
2
=CH−CH
2
−CH
3
Br
2
CCl
4
Br−CH
2
−
∣
CH
−CH
2
CH
3
A B Br C
↓(Sodalime)(NaOH+CaO)
CH
3
CH
2
−C≡C
−
Ag
+
↓
[Ag(NH
3
)
2
]
+
(ammonicalsilvernitrate)
terminalalkyne
H−C≡C−CH
2
−CH
3
Alkyl bromide (A) on reaction with alcoholic KOH gives a alkene B which on bromination gives dibromobutane (C). C on treatment with soda lime gives a terminal alkyne D which gives white ppt with ammonical silver nitrate solution where alkynes should be terminal.
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