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3 Methyl 2 Pentene Hbr

 Multiple Choice QuestionsMultiple Choice Questions

21.

Which of the following, upon treatment with tert-BuONa followed past add-on of bromine water, fails to decolourize the colour of bromine?


A.


22.

The major production obtained in the following reaction is

  • C6H5CH (OtBu) CH2C6Hv

  • Chalf-dozenHfiveCH=CHC6H5

  • (+)C6HfiveCH(OtBu)CH2H5

  • (+)Chalf dozenHfiveCH(OtBu)CH2H5


B.

C6H5CH=CHC6H5

Elimination reaction is highly favoured if
(a) Bulkier base is used
(b) College temperature is used
Hence in given reaction biomolecular ellimination reaction provides major product


23.

The increasing order of the reactivity of the following halides for the SNorthone reaction is

  • (3) < (II) < (I)

  • (2) < (I) < (III)

  • (I) < (3) < (II)

  • (I) < (Iii) < (Ii)


B.

(Two) < (I) < (III)

For any Due southNorth1 reaction, reactivity is decided by ease of dissociation of alkyl halide

R – 10 ⇌ R+ + X-

Higher the stability of R+ (carbocation) higher would exist reactivity of SN1 reaction.Since the stability of cation follows the gild.


24.

Which of the following compounds will the meaning amount of meta product during mono-nitration reaction?


C.

(a) Nitration reactions take place in presence of concentrated HNO3 + concentrated HiiAnd theniv.
(b) Aniline acts equally a base. In presence of H2SO4, its protonation takes identify and anilinium ion is formed
(c)Anilinium ion is a strongly deactivating group and meta directing in nature so it gives meta nitration product in a pregnant corporeality.


25.

The formation of which of the following polymers involves hydrolysis reaction?

  • Nylon 6

  • Bakelite

  • Nylon vi, half dozen

  • Nylon 6, 6


A.

Nylon six

Formation of Nylon-6 involves hydrolysis of its monomer (caprolactum) in the initial land.


26.

Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?


A.

Ester in presence of Aqueous KOH solution give SNAE reaction and then following reaction takes place


27.

The correct sequence of reagents for the following conversion will be

  • [Ag(NHiii)2]+ OH, H+/CHthreeOH, CHiiiMgBr

  • CH3MgBr, H+/CHiiiOH, [Ag(NHthree)ii]+ OH

  • CHthreeMgBr, [Ag(NH3)2]+ OH, H+/CH3OH

  • CH3MgBr, [Ag(NHiii)ii]+ OH, H+/CH3OH


A.

[Ag(NH3)two]+ OH, H+/CHiiiOH, CHthreeMgBr


28.

Given,

straight E subscript Cl subscript 2 divided by Cl to the power of minus end subscript superscript 0 space equals 1.36 space comma  straight E subscript Cr to the power of 3 plus end exponent divided by Cr end subscript superscript 0 space equals space minus 0.74 space straight V  straight E subscript Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript end subscript superscript 0 space equals space 1.33 space straight V comma  straight E subscript MnO subscript 4 superscript minus divided by Mn to the power of 2 plus end exponent end subscript superscript 0 space equals space 1.51 space straight V
Amongst the following, the strongest reducing agent is

  • Cr

  • Mnii+

  • Crthree+

  • Criii+


A.

Cr

Cr+3 is having least reducing potential, therefore Cr is the all-time Reducing agent.


29.

The major product obtained in the post-obit reaction is


B.

DIBAL – H is electrophilic reducing agent reduces cynide, esters, lactone, amide,carboxylic acid into corresponding Aldehyde (partial reduction)


30.

3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product.The number of possible stereoisomers for the product is

  • Half-dozen

  • 2

  • Nix

  • Zero


D.

Zero

iii-Methyl-pent-two-ene on reaction with HBr in presence of peroxide forms two Bromo-3-methyl pentane.As the molecule is nonsymmetric therefore according to Anti Markownikov rule, due to 2 chiral centre 4 stereo isomers are possible.


  • 2
  • three

3 Methyl 2 Pentene Hbr,

Source: https://www.zigya.com/previous-year-papers/JEE/12/Chemistry/2017/JEE2017001/CHENJE12152215/20

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