Question

Draw the structures of organic compounds A and B. Indicate stereochemistry where applicable.Draw the structures orga C Compounds A ama P. Indicate StereoChemistry where applicao 1) NaNH2 H--C C- 2) CH3CH2CH2Br CH2 (1 equiv.) CH3 CH2 H2 Lindlar cat H20, OH H2O CH3 H3C

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Answer #1
Concepts and reason

The concept used is to draw the structures of A and B in the given reaction for the conversion of alkynes to the desired alcohol.

Fundamentals

NaNH2{\rm{NaN}}{{\rm{H}}_2} is a strong base that abstracts the acidic protons. Alkynes have two acidic protons.

Terminal alkynes on reaction with NaNH2{\rm{NaN}}{{\rm{H}}_2} produce the corresponding acetylide ion followed by the reaction with alkyl halide produces the corresponding internal alkynes.

Alkynes undergo cis-hydrogenation upon reaction with Lindlar’s catalyst. The two H atoms are added on the same side producing cis-alkenes.

Hydration of alkenes produce the corresponding alcohols. The reaction follows Markovnikov’s rule.

Acetylene reacts with NaNH2{\rm{NaN}}{{\rm{H}}_2} followed by alkyl halide produces the internal alkyne.

The reaction is as follows:

H-C=c-H - 1. NaNH,
H-C=C:
H-C=C—CH2CH2CH3
2. CH3CH2CH2Br
(1 equiv.)

Hydrogenation of alkyne A produces the corresponding cis-alkene B.

The reaction is as follows:

Hd
H.
-CH2CH2CH3
Lindlar cat.
CH2CH2CH3

Hydration of the alkene produces the desired alcohol.

The reaction is as follows:

OH
H20, H304,
HzC—CH
H
CH2CH2CH3
CH2CH2CH3

Ans:

Therefore, the structures of A and B are as follows:

H-C=C-H
1. NaNH,
-
2. CH3CH2CH2Br
(1 equiv.)
H-C=C-CH2CH2CH3
Α
Lindlar cat.
CH2CH2CH3
H20, H30+
OH
H3C—CH
CH2CH2CH3

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