Direction (Q. Nos. 1 - 8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. How many distinct alkynes exist for C6H10 which gives effervescence on heating with Na?
Which of the following improperly describes the physical property of an alkyne?
What two atomic or hybrid orbitals overlap to form the carbon-carbon σ (sigma) bond in ethyne?
The correct answer is option C
The chemical bonding in acetylene (ethyne) (C2H2) consists of sp-sp overlap between the two carbon atoms forming a sigma bond, as well as two additional pi bonds formed by p-p overlap. Each carbon also bonds to hydrogen in a sigma s-sp overlap at 180 degree angles.
Which of the following statements correctly describes the general reactivity of alkynes?
The correct answer is option D
The clouds of electrons surrounding the sigma bond makes an alkyne an electron-rich molecule. They are therefore nucleophiles that react with electrophiles. Thus alkynes, like alkenes, undergo electrophilic addition reactions because of their weak pi bonds. When a proton adds to an alkyne, a vinylic cation is formed.
How many different isomeric alkynes on catalytic hydrogenation gives the same 3-ethyl hexane?
An organic compound X (C6H13Br) is optically active. X on treatment with (CH3)3COK in (CH3)3COH gives Y (C6H12), a major product. Y on treatment with Br2— CCI4 in the presence of FeBr3 gives a dibromide which on further treatment with NaNH2 gives C6H10 which is still optically active. Hence, X and Y respectively are
When 2,3-dibromobutane is treated with KOH in ethanol, 2-bromo-2-butene is formed which does not undergo further dehydrobromination to form 2-butyne under similar condition because
Because vinyl bromide is less reactive and Alc. KoH is not that strong a base; we need a strong base like NaNH2 .
When vicinal dibromide is heated with KOH in ethanol (~ 200°C), double dehydrohalogenation takes place giving alkyne. Which of the following fails to give alkyne according to this procedure?
The correct answer is option C
C* has no H attached to it hence the next Bromine cannot perform dehydrohalogenation using this Carbon. Therefore, triple bond formation is not possible in this compound.
Direction (Q. Nos. 9 -12) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.
Q. In which of the following reaction(s), an alkyne product is formed?
The correct answer are options B & C
(A) An alkyne would be formed if both the bromines are on adjacent carbons.
(B) CaC₂ + H₂O → C₂H₂ + Ca(OH)₂
(C) Mg₂C₃ + H₂O → C₃H₄ + Mg(OH)₂
(D) Al₄C₃ + H₂O → CH₄ + Al(OH)₃
Therefore alkynes are produced only in (B) and (C).
Which of the following reagents gives an effervescence when reacted with 1-butyne?
Which of the following reagent(s) can be used to distinguish between 1-hexyne and 2-hexyne?
The correct answers are Options A and B.
AgNO3 is called Tollen's reagent.
It reacts with terminal alkynes (1-hexyne) but not with alkenes (1-hexene).
2CH≡CCH2CH2CH3 + 2AgNO3 + 2NH4OH——›
2CH≡CCH2CH2CH3Ag + 2NH4NO3 + 2H2O
Terminal alkynes (hex-1-yne) react with ammoniacal
Cu2Cl2 solution. Tollens reagent, etc.
Terminal alkyne being therm odynamically less stable than an internal alkyne, instead when 2,2-dibromo butane is treated with NaNH2,1-butyne is formed as major product, not 2-butyne because
Direction (Q. Nos. 13 - 16) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.
Statement I : When 1-butyne is treated with NaNH2, sodium salt is formed.
Statement II : NH3 is w eaker acid than .
Statement I : Treatment of either 1,2-dibromobutane or 2,2-dibromobutane with NaNH2 gives the same 1-butyne.
Statement II : NaNH2 forms salt with terminal alkyne which makes possible the above observation.
Statement I : Propyl lithium on reaction with 1-bromo-1-pentyne gives 4-octyne.
Statement II : Propyl lithium acts as a very strong nucleophile, brings about SN2 reaction of halides.
Statement l : C6H5MgBr with propyne gives 1-phenyl propyne.