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A hydrocarbon 'A' (C(4)H(8)) on reaction...

A hydrocarbon 'A' `(C_(4)H_(8))` on reaction HCl gives a compound 'B', `(C_(4)H_(9)Cl)` which on reaction with 1 mol of `NH_(3)` gives compounds 'C' `(C_(4)H_(11)N)`. On reacting with `NaNO_(2)` and HCl followed by treatment with water compound 'C' yields an optically active alcohol, 'D'. Ozonolysis of 'A' given 2mols of acetyldehyde. Identify compound 'A' to 'D'. Explain the reaction involved.

Text Solution

Verified by Experts

(i) M.F. of compound `A(C_(4)H_(8))` corresponds to the general formula `C_(n)H_(2n)` where n=4, therefore, A is an alkene. Since ozonolysis of compound `A(C_(4)H_(8))` gives two moles of acetaldehyde, therefore, alkene (A) must be symmetrical, i.e., but-2-ene.
`underset("But-2-ene")(CH_(3)CH=CHCH_(3)) underset((ii)Zn//H_(2)O)overset((i)O_(3)//CH_(2)Cl_(2))to underset("Acetaldehyde (A)")(2CH_(3)CH=O)`
(ii) Since compounds (A), i.e., but-2-ene reacts with HCl to form compound (B), therefore, (B) must be 2-chlorobutane.
`underset("But-2-ene")(CH_(3)CH=CHCH_(3)) +HCl to underset("2-Chlorobutane (B)")(CH_(3)CH_(2)-CHCl-CH_(3))`
(iii) Since compound (B), i.e., 2-chlorobutane reacts with one mole of `NH_(3)` to give compound (C), therefore, compound (C) must be a `1^(@)` amine, i.e., butan-2-amine.
`underset("2-Chlorobutane (B) (Optically active)")(CH_(3)CH_(2)-underset(Cl)underset(|)overset(**)(C)H-CH_(3)) underset(-HCl)overset(NH_(3)" (1 mole)")to underset("Butan-2-amine (C)")(CH_(3)CH_(2)-underset(NH_(2))underset(|)overset(**)(C)H-CH_(3))`
(iv) Since compound (C) reacts with `NaNO_(2)//HCl`, to give an alcohol (D), therefore, compound (D) must be butan-2-ol. please note that 2-clorobutane (B), butan-2-amine (C) and butan-2-ol (D) contain a chiral carbon, therefore, all these compounds are optically active.
`underset("Butan-2-amine (C) (Optically active)")(CH_(3)CH_(2)-underset(NH_(2))underset(|)overset(**)(C)H-CH_(3)) overset (NaNO_(2)//HCl)to underset("Butan-2-ol (D) (Optically active)")(CH_(3)CH_(2)-underset(OH)underset(|)overset(**)(C)H-CH_(3))`
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Knowledge Check

  • A hydrocarbon A(C_4H_8) on reaction with HCl gives a compound B(C_4H_9Cl) which on reaction with 1 mole of NH_3 gives a compound C(C_4H_(11)N) . On reacting with NaNO_2 and HCl followed by treatment with water, compound C yields an optically active alcohol,D. Ozonolysisi of A gives 2 moles of acetaldehyde. The structure of compound C is

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  • C_(5)H_(13)N reacts with HNO_(2) to give an optically active alcohol The compound is

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    C
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