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1) CH3NH2 converts using CHCL3 and KOH gives (A) converts to (B) using Na n C2H5OH . 2) CH3CH2CN converts to (A) using Na/C2H5OH and convert to (B) using NaNO2 and HCL. plz rly the answer pf this 2 questions...

1) CH3NH2 converts using CHCL3 and KOH gives (A) converts to (B) using Na n C2H5OH .


2) CH3CH2CN converts to (A) using Na/C2H5OH and convert to (B) using NaNO2 and HCL.


plz rly the answer pf this 2 questions...

Grade:12

3 Answers

NISHARG GOLASH
31 Points
10 years ago

1) This is a Carbalamine reaction  and A will be CH3CN which when reacts to a metal will give CH3CH2NH2. You can read more about it here http://en.wikipedia.org/wiki/Carbylamine_reaction

 

2) CH3CH2CN + Na + CH3CH2OH ---> CH3CH2NH2  (A)

so, now

CH3CH3NH2 + NaNO2 + HCl ----> CH3CH2N2+}Cl- (B)

ankit singh
askIITians Faculty 614 Points
3 years ago
Silver nitrate test does not show existence of any HCl in chloroform (at least by ... step with sodium carbonate (instead of calcium carbonate) once using chloroform ... is converted slowly to phosgene (COCl2), releasing HCl in the process. 2 ... Further, if traces amounts of HCl are present the latter may react with Ca(CO3)2 to ...
Srinivasan S
27 Points
3 years ago
HI Upasana,
 
I am sorry but the queries seem wrongly formulated...
 
For 1) CH3NH2 is a carbylamine reaction giving an isocyanide CH3NC, which on reduction with Na/EtOH would give CH3NHCH3 (dimethylamine). I am not sure if the question is if CH3NC gets reduced or its an independent question.
 
For 2) CH3CH2CN on reduction would give propyalmine (CH3CH2CH2NH2) and which with NaNO2/HCl give CH3CH2CH2OH (n-propanol). HOwever, here too I am unsure if the 2nd part of question is a follow-up on the 1st  part
 
CH3CH3CN cannot give A (CH3NC) by reduction...!!

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