Draw the structure of two alkenes that would yield 1-methylcyclohexanol when treated with hg(oac)2 in water, then nabh4:

a) methylidenecyclohexane

b) 1-methylcyclohex-1-ene

Explanation:

The given reaction is oxymercuration-demercuration.

This reaction gives alcohol (according to Markovnikov's rule) from alkenes.

The reaction of alkene to give desired  1-methylcyclohexanol is shown in the figure.

The alkenes are:

a) methylidenecyclohexane

b) 1-methylcyclohex-1-ene

Both will give alcohol where hydroxyl group will be on the more substituted double bonded carbon atom.

The possible alkenes to produce 1-methylcyclohexanol are  and . (Refer to the attached image)

Further explanation:

Oxymercuration-demercuration reaction:

Oxymercuration-demercuration converts alkene into alcohol in two steps. In the initial step, the alkene is reacted with salt and nucleophile such as alcohol and water to produce an organomercury intermediate. In second step, the carbon-Hg bond in organomercury intermediate is converted into C-H bond.

In this method the product is obtained in accordance with Markovnikov’s rule, and rearrangement of carbocation does not take place.

The mechanism of Oxymercuration-demercuration is as follows(Refer to the attached image):

Step 1: break and form mercuroacetic cation and ions.

Step 2: The mercuroacetic cation reacts with a double bond and forms a cyclic cation.

Step 3:The cyclic intermediate is attacked by the nucleophile to the most substituted side according to Markovnikov’s rule and forms organomercury intermediate.

Step 4: the carbon-Hg bond in organomercury intermediate is converted into C-H bond.

The methylcyclohexanol has alcohol on tertiary carbon thus the double bond must be attached to the tertiary carbon in initial alkene. This means there are only two possible alkenes that can have double bond attached to the tertiary carbon. These alkenes are 1-methylcyclohexene and methylenecyclohexane.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

2. Determine the product formed when 2-propanol react with NaH:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter:Oxymercuration-demercuration reactions

Keywords: Oxymercuration-demercuration reaction, alkenes, structure of two alkenes, yield 1-methylcyclohexanol, Hg(OAc)2, NaBH4.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.