Question

Give an example reaction that would yield the following products. Name the organic reactant and product in each reaction. a. alkane b. monohalogenated alkane c. di halogenated alkane d. tetra halogenated alkane e. monohalogenated benzene f. alkene

Short answer

(a) The reactant is "butane" and the product of the reaction is "2-chloro-butane

(b)The reactant is "propane" and the product of the reaction is "3-chloro-propane

(c) The reactant is "propane" and the product of the reaction is "1,3-chloro-propane”

(d) The obtained product is “3,4 tetrachloride propane”

(e) The obtained product is chloro-benzene

(f) The obtained product is “3,4-chloro-butene”

Step-by-step solution

To give an example reaction for   Alkane

To obtain the alkane one have to start from alkene, which will be transformed in the reaction with an acid.

The following reaction "butene" with hydrochloric acid is shown

-First, the double bond will break because hydrogen (atom from the hydrochloric acid) will take one electron from carbon atom according to the "Markovnikov rule",

-Then the hydrogen atom and the carbon atom will bond which will leave the chlorine atom with one extra electron,

-The chlorine atom will give its extra electrons to the other carbon atom and form a single bond there.

Hence The reactant is "butene" and the product of the reaction is "2-chloro-butane

To give an example reaction for Monohalogenated alkane:

To obtain the mono halogenated alkane one will start with propane, which will be transformed in the reaction with a gas.

The following reaction "propane" with chlorine.

-One single bond between carbon and hydrogen will break,

-Then one chlorine atom will break its bond with another chlorine atom and forms a bond with the carbon atom that has one free orbital.

-The hydrogen atom that was left out will react with the other chlorine atom to form hydrochloric acid.

Hence The product of the addition of hydrogen halide on but-1-ene is 2-bromobutane.

Hence the reactant is "propane" and the product of the reaction is "3-chloro-propane

Step 3:To give an example reaction for  Di halogenated alkane

To obtain the di halogenated alkane one will start with propane, which will be transformed in the reaction with a gas.

The following reaction "propane" with chlorine is represented as follows;

- One single bond between carbon and hydrogen will break,

- Then one chlorine atom will break its bond with another chlorine atom and form a bond with the carbon atom that has one free orbital.

- Since it is "(di)halogenated" the reaction will be repeated so that another chlorine atom gets attached to the structure.

-The hydrogen atom that was left out will react with the other chlorine atom to form hydrochloric acid.

Hence the reactant is "propane" and the product of the reaction is "1,3-chloro-propane

Step 4:To give an example reaction for Tetra Halogenated alkane

To obtain the tetra halogenated alkane one will start from alkyne, which will be transformed in the reaction with the gas.

The following reaction of "butyne" with chlorine gas;

-First, the triple bond will break because chlorine (atom from the chlorine gas) will take one electron from carbon atom according to the "Markovnikov rule",

-This will leave another chlorine atom with one extra electron,

-The chlorine atom will give its extra electrons to the other carbon atom and form a double bond there.