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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Dec 2007 12:46:48 IST
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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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