Higher Alkynes from Alkynes
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Q. Acetylene is treated with excess sodium in liquid ammonia. The product is reacted with excess of methyl iodine. The final product is
- But-1-yne
- but-2-yne
- but-1-ene
- but-2-ene
Q. 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−
- 1
- 2
- 3
- 4
Q. In the reaction H−C≡CH(1) NaNH2/liq.NH3−−−−−−−−−−−−→(2) CH3CH2BrX(1) NaNH2/liq.NH3−−−−−−−−−−−−→(2) CH3CH2BrY, Xand Y are:
- X = 1 - Butyne; Y = 3 - Hexyne
- X = 2 - Butyne; Y = 3 - Hexyane
- X = 2 - Butyne; Y = 2 - Hexyane
- X = 1 - Butyne; Y = 2 - Hexyane
Q. What is the final product C, of the following reaction sequence?
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