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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: orgabi chemistry question..plz help!!
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sinjan.j (574)

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CH3COCH3  LAH  ) ()  H+/Br2+CCl4___)()NaNH2liq NH3 _)() Na liq NH3___)X
X IS :
 
A) Trans CH3CHCHCH3
B) Cis CH3CHCHCH3
C) CH3CCH
D) CH3CCNa




    
rhd92781 (686)

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is the answer A

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sinjan.j (574)

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please explain ur answer..!!




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rhd92781 (686)

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in the first step CH3CHOHCH3 WILL BE FORMED
I'VE SOME CONFUSION IN THE NEXT STEP
IF WE GET CH3CCCH3 IN THE THIRD STEP, Na/NH3 WILL REDUCE IT PARTIALLY AND IN THE ANTI SENSE FORMING
TRANS CH3CHCHCH3

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pratikanand (568)

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first step is correct by rhd92781. but after than that, there will be substituition of Br which will be reduced (dehydrohalogenation) to alkene n then on......


let me see then i will post detailed soln....

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lokeshsardana (675)

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I think d will be the answer..........

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lokeshsardana (675)

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see.......in the first step, CH2CHCH3 ,propene will be formed........
 
second step results in trans addition of bromine...........
 
in third step propyne will be formed...........
 
and at last Na will replace terminal hydrogen............
 
and final product will be D..........
 
that's my solution...........
 
pls tell if i m wrong any where............
 
other wise rate...................
 
sinjan , if i am correct...........then post it on 123........

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tango_goat (143)

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i agree with lokeshsardana, but i dont think that LAH will convert acetone to propene,but instead i think 2-propanol will be formed
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lokeshsardana (675)

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first propan-2-ol will be formed then it will lose one water molecule to give propene............

hey guys atleast rate my efforts yaar!!!!!!

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Greatdreams (3083)

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propene should be the final product

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lokeshsardana (675)

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@ great dreams

I know yaar! as Na + NH3 is reducing agent .......and will reduce.....

propyne to propene................but it is not given in options.........

so, I formed the product of rxn with only Na...........


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khinart (198)

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I agree with lokeshsardana.In the first lithium aluminum hydride will reduce keytone top a secondary alcohol which is CH3CHOHCH3.Now the acid will convert it to prop-1-ene and then there will be a trans addition of bromine forming CH3CHBRCH2BR.Then Sodium amide will convert it into Prop-1-yne and Na/NH3 will replace the terminal hydrogen.so answer is d.

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