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Question
Hydrocarbon ′A′(C4H8) on reaction with HCl gives a compound ′B′(C4H9Cl), which on reaction with 1mol of NH3 gives compound ′C′(C4H11N). On reacting with NaNO2 and HCl followed by treatment with water. the compound C yields an optically active alcohol, ′D′. Ozonolysis of ′A′ gives 2 mol of acetaldehyde. Identify compound ′D′.
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Solution
AOzonolysis−−−−−−→2CH3CHOC4H9HCl−−→C4H9Cl(A)(B)Addition reaction, A is alkene.C4H9ClNH3−−−→C4H11N(C)−Cl is substituted by NH2[C]NaNO2/HCl−−−−−−−−→H2O[D]
⟹[C] is an aliphatic amine. Number of carbon atoms in amine is same as in compound A Since, A⟶CH3−CH=O+O=CH−CH3 on ozonolysis,A is CH3−CH=CH−CH3(A)↓HClCH3−CH2−CH|Cl−CH3(B)↓NH3CH3−CH2−CH|NH2−CH3(C)↓NaNO3/HCl/H2OCH3−CH−OH|CH2−CH3(D) optically active
C4H9HCl−−→C4H9Cl(A)(B)
Addition reaction, A is alkene.
C4H9ClNH3−−−→C4H11N(C)
−Cl is substituted by NH2
[C]NaNO2/HCl−−−−−−−−→H2O[D]
⟹[C] is an aliphatic amine. Number of carbon atoms in amine is same as in compound A
⟹[C] is an aliphatic amine. Number of carbon atoms in amine is same as in compound A
Since, A⟶CH3−CH=O+O=CH−CH3 on ozonolysis,
A is CH3−CH=CH−CH3(A)↓HClCH3−CH2−CH|Cl−CH3(B)↓NH3CH3−CH2−CH|NH2−CH3(C)↓NaNO3/HCl/H2OCH3−CH−OH|CH2−CH3(D) optically active
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