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Q52.

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Organic Chemistry
Found in: Page 452

Short Answer

Question: Devise a synthesis of each compound using CH3CH2CH=CH2 as the starting material. You may use any other compounds or inorganic reagents.

a.

b.

c.

d.

e.

(+ enantiomer)

Answer

See the step by step solution

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.

Step 2: Synthesis of compounds a and b using

The alkyne can be prepared by the halogenation reaction of the alkene , 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 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 produced as shown in the following reaction:

Step 3: Synthesis of compounds c and d using  

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 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 with epoxide, as shown in the following reaction:

Formation of + enantiomer

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