Question

Question: Devise a synthesis of each compound using CH₃CH₂CH=CH₂as the starting material. You may use any other compounds or inorganic reagents.

a.

b.

c.

d.

e.

(+ enantiomer)

Short answer

Answer

Step-by-step solution

Step-by-Step SolutionStep 1: Reactivity of alkynes

Alkynes are reactive due to the presence of an easily breakable bond. Terminal alkynes are reactive in nature due to the highly acidic carbon-hydrogen bonds.

Alkynes can undergo a variety of reactions. They can undergo addition reactions because of the presence of a weak bond.

The acetylide anions which are formed by the loss of a proton from terminal alkynes are strong nucleophiles and react with a variety of electrophiles.

Due to the strong nucleophilic nature of the acetylide anions, they are able to react with epoxide and undergo the opening of the epoxide rings.

Synthesis of compounds a and b using CH3CH2CH=CH2

The alkyne${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}$can be prepared by the halogenation reaction of the alkene${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CH}}_2$, i.e., but-1-ene. In the next step, the vicinal dihalide is treated with a strong base to produce the desired alkyne.

a. The geminal dihalide 2,2-dibromobutane can be prepared by the treatment of the alkyne ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}$produced in part a with hydrogen bromide, as shown in the following reaction.

b. The vicinal dihalide 1,2-dibromobutane is prepared by the chlorination of the alkyne ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}$produced as shown in the following reaction:

Synthesis of compounds c and d using  CH3CH2CH=CH2

a. The compound Hex-3-yne-1-ol is prepared by the reaction of alkyne from part a with a strong reducing agent NaH to form ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}$the acetylide, as shown below:

The acetylide ion is then treated with the epoxide to produce the desired compound, as shown below:

Formation of hex-3-yne-1-ol

b. The (+ enantiomer) of the given compound can be prepared by first treating the acetylide ion ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}$with epoxide, as shown in the following reaction:

Formation of + enantiomer