Question

Consider the following reactions.

a. When${\mathbf{\text{C}}}_{\mathbf{\text{5}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}$is reacted with${\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$in the presence of ultraviolet light, four different monochlorination products form. What is the structure of${\mathbf{\text{C}}}_{\mathbf{\text{5}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}$in this reaction?

b. When${\mathbf{\text{C}}}_{\mathbf{\text{4}}}{\mathbf{\text{H}}}_{\mathbf{\text{8}}}$is reacted with${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{O,}}$a tertiary alcohol is produced as the major product. What is the structure of${\mathbf{\text{C}}}_{\mathbf{\text{4}}}{\mathbf{\text{H}}}_{\mathbf{\text{8}}}$in this reaction?

c. Whenis reacted with1-chloro1-methylcyclohexane is produced as the major product. What are the two possible structures for${\mathbf{\text{C}}}_{\mathbf{\text{7}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}$in this reaction?

d. When a hydrocarbon is reacted with${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}$and the major product of this reaction is then oxidized, acetone (2-propanone) is produced. What is the structure of the hydrocarbon in this reaction?

e. When${\mathbf{\text{C}}}_{\mathbf{\text{5}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}\mathbf{\text{O}}$is oxidized, a carboxylic acid is produced. What are the possible structures for${\mathbf{\text{C}}}_{\mathbf{\text{5}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}\mathbf{\text{O}}$in this reaction?

Short answer

a.Isopentene

b. 2-methylprop-2-ol

c. 1-methylcyclohex-l-ene and methylenecyclohexane

d. Prop-l-ene

e. Pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol or 2,2 -dimethylpropan-1-ol.

Step-by-step solution

Step 1:Monochlorination reaction

(a)

Monochlorination is a reaction with a single chlorine atom.

To give four different products, the carbon atom chain needs to be long. So, isopentane will give four different products in monochlorination because of its long chain.

Hence,the structure of ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$along with the monochlorinated products is as shown below:

Alkene compound

(b)

The molecular formula${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$indicates either that we have an alkene or a ring structure.

Since alkenes are the ones that yield alcohol in reaction with water, we can assume that the starting compound is an alkene.

Isomers of${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}$that are alkenes are shown in the figure below.

Step 3:2-methylprop-l-ene compound

But-l-ene and but-2-ene will give secondary alcohols in reaction with water.

The 2-methylprop-l-ene will give tertiary alcohol as a product.

Hence,the structure is 2-methylprop-2-ol

Step 4:Methylenecyclohexane compound

(c)

Since 1-chloro-l-methylcyclohexane is formed by reaction with we can immediately conclude that the starting compound is an alkene.

There are two possible starting compounds, 1-methylcyclohex-1-ene and methylenecyclohexane.

The proton adds to the end of the double bond that is less substituted to give the more substituted carbocation (the more stable carbocation). In the second step, the halide ion attack carbocation and gives the addition product.

Hence,the structure is 1-methylcyclohex-1-ene and methylenecyclohexane

Acetone preparation reaction

(d)

The reduction of acetone yields propan-2-ol.

And propan-2-ol can be obtained by adding water to propan-l-ene.

The sequence of acetone preparation reactions is shown in the figure below.

Hence,the structure is Prop-l-ene

Carboxylic acid

(e)

We can assume from the molecular formula, that the starting compound is an alcohol which is then oxidized to a carboxylic acid.

Possible ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$structures are shown in the figure below.

Reactions in which the corresponding carboxylic acids are formed are also shown.

Hence,the structure of ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$is Pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol or 2,2 -dimethylpropan-1-ol.