Chapter 7: Q28P (page 311)

How could the following compounds be synthesized from acetylene?

Short Answer

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Step by step solution

Some conclusions

1. NaNH₂ =With this we have to take any alkyl bromide then alkyl group will attach with triple bonded carbon atom by replacing hydrogen.

2. H₂ in presence of lindlar catalyst stops the reaction at alkene stage. It does reduction.

3. Hydroboration-oxidation= In this anti markoniv addition takes place.

Final answer

a) NaNH₂ =With this we have to take any alkyl bromide then alkyl group will attach with triple bonded carbon atom by replacing hydrogen.

b) NaNH₂ =With this we have to take any alkyl bromide then alkyl group will attach with triple bonded carbon atom by replacing hydrogen. After replacing both hydrogens with alkyl group then add H₂ in presence of lindlar catalyst which stops the reaction at alkene stage.

c) NaNH₂ =With this we have to take any alkyl bromide then alkyl group will attach with triple bonded carbon atom by replacing hydrogen. After replacing both hydrogens with alkyl group then add H₂ in presence of lindlar catalyst which stops the reaction at alkene stage.

d) NaNH₂ =With this we have to take any alkyl bromide then alkyl group will attach with triple bonded carbon atom by replacing hydrogen. After that excess HCl is used which do markoniv addition and replaces all hydrogens by chlorine.

e) NaNH₂ =With this we have to take any alkyl bromide then alkyl group will attach with triple bonded carbon atom by replacing hydrogen. After that Hydroboration-oxidation takes place in which anti markoniv addition will be there.

f)NaNH₂ =With this we have to take any alkyl bromide then alkyl group will attach with triple bonded carbon atom by replacing hydrogen. After replacing both hydrogens with alkyl group then add H₂ in presence of lindlar catalyst which stops the reaction at alkene stage. After that HBr is used which do markoniv addition.