Tuesday, December 3, 2013

Chapter 9 Problem 36

This problem is a synthesis one where you have 1-butene as the starting material and are trying to make (cis) 3-hexene. First thing is to figure out how to get 2 more carbons onto the reactant. In order to add carbons to an alkene or an alkane you have to first create a terminal alkyne so that a strong base such as NaNH2 can deprotonate the alkyne and make it readily available to accept a carbon or carbons. Since we have an alkene as the starting material and we want an alkyne, we have to first go backwards and make an alkane with two bromide substituents on carbons one and two so that dehydrohalogenation can take place. By doing this, the reactant will be able to turn into an alkyne. Hence, starting with 1-butene, we add Br2 first yielding 1,2-dibromo-butane and then add excess NaNH2 so that the alkyl dihalide can undergo two consecutive dehydrohalogenations, ultimately giving us our alkyne.  Since there is excess NaNH2 our terminal alkyne is also deprotonated, so that it readily accepts the next species that's added - our two carbons - CH3CH2Br. The carbons on this chain have a partial positive charge (because the bromide inductively attracts the electrons) and are attracted to the nucleophilic character of the alkyne. We are now left with 3-hexyne and so the only thing left to do is turn the alkyne into an alkene by using H2 and Lindlar's because our desired product is cis.


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  2. Although a picture of the problem would be helpful, this is a great explanation. I actually went through the whole problem to follow all this and understood it more. Great review for me. Thanks for sharing!