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
AI Generated Solution
To solve the problem step by step, we will identify compounds A, B, C, and D based on the given reactions and information.
### Step 1: Identify Compound A
The question states that hydrocarbon A (C₄H₈) undergoes ozonolysis to yield two moles of acetaldehyde (CH₃CHO). The ozonolysis of an alkene typically breaks the double bond and forms carbonyl compounds.
Given that ozonolysis produces two moles of acetaldehyde, we can deduce that compound A must be 2-butene (CH₃-CH=CH-CH₃). This is because the ozonolysis of 2-butene results in two moles of acetaldehyde.
**Compound A:** 2-butene (CH₃-CH=CH-CH₃)
...
Topper's Solved these Questions
AMINES
NCERT EXEMPLAR ENGLISH|Exercise SATQ|37 Videos
ALDEHYDE, KETONES AND CARBOXYLIC ACIDS
NCERT EXEMPLAR ENGLISH|Exercise Long Answer Type Questions|4 Videos
BIOMOLECULES
NCERT EXEMPLAR ENGLISH|Exercise Long Answer Type Questions|5 Videos
Similar Questions
Explore conceptually related problems
Compound (x)(m.f=C_(7)H_(8)N) , on reaction with NaNO_(2) and conc. HCl at O^(@)C followed by beta - naphthol gives orange coloured dye. Compound (x) is :
Optically active isomer (A) of (C_(5)H_(9)Cl) on treatment with one mole of H_(2) gives an optically inactive compound (B) compound (A) will be :
An alkene (A)C_(16)H_(16) on ozonolysis gives only one product (B)(C_(8)H_(8)O) . Compound (B) on reaction with NH_(2)OH followed by reaction with H_(2)SO_(4), Delta gives N - methyl benzamide the compound 'A' is -
An organic compound A (C_(4)H_(9)Cl) on the reaction with Na/diethly ether gives a hydrocarbon which on monochlorination gives only one chloro derivative then, A is
Treatment of benzene with "CO"//HCl in the presence of anhydrous AlCl_(3)//CuCl followed by reaction with Ac_(2)"O"//NaOAc gives a compound X. Reaction of X with H_(2)//Pd//c followed by treatment with H_(3)PO_(4) gives another compound Y. The compound Y is :
Knowledge Check
Compound A (C_(7)H_(14)) decolourises Br_(2) in C Cl_(4) and reacts with Hg(OAc)_(2) followed by reduction with NaBH_(4) to produce a resolvable compound B. A undergoes reductive ozonolysis to give as one of the compound. Identify A and B.
A
A is
B
A is
C
B is
D
B is
Optically active isomer (A) of (C_5H_9 Cl) on treatment with one mole of H2 gives an optically inactive compound (B) compound (A) will be :
