70
Question
In order to complete the reaction
1−Pentyne a→ 4−Octane b→ cis−4−Octane a and b will be
(1) NaNH2; aCH3CH2Br:H2b(one mole)Pd or Ni
(2) NaNH2;CH3CH2CH2Br:H2 (two moles) Pd or Ni
(3) NaNH2;CH3CH2CH2Br:H2,(one−mole)Pd or Ni
(4) NaNH2;CH3CH2CH2Br:BH3,H2O2,OH−
1−Pentyne a→ 4−Octane b→ cis−4−Octane a and b will be
(1) NaNH2; aCH3CH2Br:H2b(one mole)Pd or Ni
(2) NaNH2;CH3CH2CH2Br:H2 (two moles) Pd or Ni
(3) NaNH2;CH3CH2CH2Br:H2,(one−mole)Pd or Ni
(4) NaNH2;CH3CH2CH2Br:BH3,H2O2,OH−
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Solution
The correct option is C 3
c)
CH3−CH2−CH2−C≡CHNaNH2−−−−−→
CH3−CH2−CH2−C≡C−NaCH3CH2CH2Br−−−−−−−−−−→
CH3−CH2−CH2≡C−CH2−CH2−CH34−OctyneH2−→Pd
c)
CH3−CH2−CH2−C≡CHNaNH2−−−−−→
CH3−CH2−CH2−C≡C−NaCH3CH2CH2Br−−−−−−−−−−→
CH3−CH2−CH2≡C−CH2−CH2−CH34−OctyneH2−→PdSuggest Corrections
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