3 Methyl 2 Pentene Hbr

Multiple 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?

22.

The major production obtained in the following reaction is

C₆H₅CH (O^tBu) CH₂C₆H_v
C_half-dozenH_fiveCH=CHC₆H₅
(+)C₆H_fiveCH(O^tBu)CH₂H₅
(+)C_{half dozen}H_fiveCH(O^tBu)CH₂H₅

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 S_Northone reaction is

(3) < (II) < (I)
(2) < (I) < (III)
(I) < (3) < (II)
(I) < (Iii) < (Ii)

(Two) < (I) < (III)

For any Due south_North1 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 S_N1 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?

(a) Nitration reactions take place in presence of concentrated HNO₃ ⁺ concentrated H_iiAnd then_iv.
(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

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?

Ester in presence of Aqueous KOH solution give SN^AE reaction and then following reaction takes place

27.

The correct sequence of reagents for the following conversion will be

[Ag(NH_iii)₂]⁺ OH^–, H⁺/CH_threeOH, CH_iiiMgBr
CH₃MgBr, H⁺/CH_iiiOH, [Ag(NH_three)_ii]+ OH^–
CH_threeMgBr, [Ag(NH₃)₂]⁺ OH^–, H⁺/CH₃OH
CH₃MgBr, [Ag(NH_iii)_ii]⁺ OH^–, H⁺/CH₃OH

[Ag(NH₃)_two]⁺ OH^–, H⁺/CH_iiiOH, CH_threeMgBr

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⁺
Cr^three+
Crⁱⁱⁱ⁺

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

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

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.