A compound with formula C_7H_14O undergoes dehydration to form only ONE compound

Question

A compound with formula C_7H_14O undergoes dehydration to form only ONE compound with the formula C_7H_12 (Hoffman product not observed). This product is less dense than water, reacts with Br_2 solution to turn it from orange to clear, a with KMnO_4 to turn the solution from purple to brown. Provide the IUPAC names and Lewis structures for the starting material and product that is formed. Show, using a step by step mechanism, how the dehydration occurs to form product. DRAW EACH STEP INDIVIDUALLY!

Explanation / Answer

By the given information , we can say it is a ketone (lighter than water) which is in enol form..

i.e. CH3-CO-CH2-CH2-CH2-CH2-CH3 <=====> CH3-CH(OH)=CH-CH2-CH2-CH2-CH3 -------->

CH3-CH(CH)-CH2-CH2-CH2-CH3

  The "Methyl pentyl ketone"(starting material) exists in enol form undergoes dehydration(darkened atoms will dehydrate) to form product 2-heptyne or hept-2-yne wehich satisfies the formula C7H12.

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