The question i'm attempting is as follows:
Treatment of 1-iodod-3, 3-dimethylbutane with sodium cyanide (NaCN) in acetone gives as the major product:
a. 3,3-dimethylbutanol
b. 3,3-dimethylbutanenitrile
c. 2-iodopentanenitrile
d. 4,4-dimethylpentanenitrile
e. 3,3-dimethyl-1-butene
I'v gotten as far as knowing that Iodide is a very good leaving group...any further help greatly appreciated
Treatment of 1-iodod-3, 3-dimethylbutane with sodium cyanide (NaCN) in acetone gives as the major product:
a. 3,3-dimethylbutanol
b. 3,3-dimethylbutanenitrile
c. 2-iodopentanenitrile
d. 4,4-dimethylpentanenitrile
e. 3,3-dimethyl-1-butene
I'v gotten as far as knowing that Iodide is a very good leaving group...any further help greatly appreciated
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You got it! CN- is an excellent nucleophile and will substitute for the iodine atom. Here's the hint you may need: The C in the cyano group is now the first carbon atom in a new chain, Both the chain name and the numbering will be different. Draw it out and you will find the product on your list.