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The major product of the following reaction is : 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Alcohols reaction with hydrogen halides - Forms alkyl halides. - wherein     Alcohols reaction with hydrogen halides - Forms alkyl halides. - wherein       CN is strong electron withdrawing group, therefore it will create  charge on adjacent carbon and thus (  ) will attack on this carbon by  mechanism. Thus, NOTE: answer provided by NTA is wrong. The answer provided by us is...

The major product of the following reaction is :

  • Option 1)

       

  • Option 2)

     

  • Option 3)

       

  • Option 4)

       

  Alcohols reaction with phosphorus trihalides - Form alkyl halides  - wherein X = Cl, Br, I     Option 1)   Option 2) Option 3)   Option 4)   

But-2-ene on reaction with alkaline KMnO_{4} at elevated temperature followed by acidification will give:

  • Option 1)

     2 molecules of CH_{3}COOH

  • Option 2)

    2 molecules of CH_{3}CHO

  • Option 3)

    one molecule of CH_{3}CHO and one molecule of CH_{3}COOH

  • Option 4)

Oxidation of alkene - Alkene on reaction with cold dilute aqueous solution of KMnO4 (Bayers Reagent) produce vicinal glycol. Decolourisation  of KMnO4 solution is used as a test of unsaturation. - wherein            ...

The electrons are more likely to be found: 

  • Option 1)

    in the region a and c

  • Option 2)

    in the region a and b 

  • Option 3)

    only in the region a 

  • Option 4)

    only in the region c

  The energy (E) of a quantum of radiation - Where h is plank’s constant and  is frequency -     Bohr's model - 1. Force of attraction between the nucleus and an electron is equal to centripetal force. 2.      n = principal quantum number. 3. Energy can be absorbed or emitted when electron transfer orbit   -   We knew that probability of finding an electron is proportional  Now on converting...

Which of the following is a thermosetting polymer? 

 

 

  • Option 1)

    PVC

  • Option 2)

    Nylon 6 

  • Option 3)

    Buna-N

  • Option 4)

     Bakelite

  Thermosetting Polymers - They are highly cross - linked, hard, infusible and insoluble polymers.They cannot be reused. - wherein Eg:-  Bakellite, Urea,  Formaldehyde, Resin, etc       Bakelite is an example of the thermosetting polymer.  Bakelite is another name of Phenal formaldehyde. It is a thermosetting  polymer because it has cross linked structure. It can't be revised once it is...

The correct statement among the following is : 

  • Option 1)

    \left ( SiH_{3} \right )_{3}N is pyramidal and less basic than \left (CH_{3} \right )_{3}N.
     

  • Option 2)

    \left ( SiH_{3} \right )_{3}Nnis planar and less basic than  \left (CH_{3} \right )_{3}N.

  • Option 3)

    \left ( SiH_{3} \right )_{3}N is planar and more basic than \left (CH_{3} \right )_{3}N.

     

  • Option 4)

    \left ( SiH_{3} \right )_{3}N is pyramidal and more basic than \left (CH_{3} \right )_{3}N.

  Electronegativity of Nitrogen family - Gradually decreases down the group. More electronegative than group 14 -       Positive oxidation state of Nitrogen family - Shows +3 and +5 oxidation states (except N2) Stability of +3O.S increases down the group due to inert pair effect - wherein +3 when only p electron are involved and +5 when all valence five (s and p) electrons are...

The major product of the following addition reaction is 

H_3-CH=CH_2\overset{Cl_2/H_2O}{\rightarrow}

 

 

  • Option 1)

    H_3-CH-CH_2

                    \mid               \mid 

                 Cl          OH

  • Option 2)

  • Option 3)

  • Option 4)

    HC_3-CH-CH_2

                     \mid                 \mid 

                  OH          Cl

  Addition of Hydrogen Halide on alkene - hydrogen halides add up to alkenes to form alkyl halides. The order of reactivity of the hydrogen halides is HI>HBr>HCl - wherein       Now,    Option 1)                                                         Option 2)Option 3)Option 4)                                                             

What is the molar solubility of Al\left ( OH \right )_{3} in 0.2M\: \: \: \: NaOH solution? Given that, solubility product of Al\left ( OH \right )_{3}=2.4\times 10^{-24}:

 

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

Solubility and K{H} values - Gases having more values will have less solubility. -         Option 1)Option 2)Option 3)Option 4)

The degenerate orbitals of  \left [ Cr\left ( H_{2}O \right )_{6} \right ]^{3+} are

  • Option 1)

    d_{xz}\; and\; d_{yz}             

  • Option 2)

     d_{yz}\; and\; d_{z^{2}}

  • Option 3)

    d_{z^{2}}\; and\; d_{xz}

  • Option 4)

    d_{x^{2}}-_{y^{2}} \; and \; d_{xy}

  Crystal field splitting in octahedral complex -  orbitals are more energetic as compared to  - wherein     CFSE in octahedral complex -   -      : degenerate orbitals =? Which after CFS bliching energy  divided into 2-set of degenerate orbitals-  & . From option given degenerate orbitals are -  & .  Option 1)             Option 2) Option 3)Option 4)

The major product(s) obtained in the following reaction is/ are:

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Ozonolysis - Ozonolysis of alkenes involved the addition of ozone molecules to alkene to form ozonide and then cleavage of the ozonide by Zn - H2O to smalter molecules.This reaction is highly useful in detecting the position of the double bond in alkene or other unsaturated compounds. - wherein   option (4) is correctOption 1)Option 2)Option 3)Option 4)

Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? ( relative orbital energies not on scale).

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Valence bond theory (VBT) - Ligands should have atleast one donor atom with a lone pair of elements. Transition metal atom provide required number of vacant orbitals. These vacant orbitals undergo hybridiation. These hybridised orbitals overlap with filled orbitals of ligands to form coordinaate compound. -     Hybridisation - sp3d2 - square bipyramidal or octahedral  d2sp3 -...

What will  be the major product when m-cresol is reacted with propargyl bromide (HC\equiv C-CH_{2}Br) in the presence of K_{2}CO_{3}  in acetone?

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Ethers preparation by alcohol dehydration - It is nucleophilic bimolecular reaction (SN2) having primary alkyl group.Follows SN1 pathway with secondary or tertiary alcohol. - wherein        +  major product = ? This process is also known as Williamson's synthesis of ether  Option 1)Option 2)Option 3)Option 4)

The strength of 11.2 volume solution of H_{2}O_{2} is [ Given that molar mass of H=1 g mol-1 and O=16 g mol-1]

  • Option 1)

    13.6%

  • Option 2)

    3.4%

  • Option 3)

    34%

  • Option 4)

    1.7%

  volume strength of H2O2 - Volume or percentage strength of hydrogen peroxide is a term to express concentration of H2O2 in terms of volumes of oxygen gas ,based on its decomposition to form water and oxygen. 2H2O2 = 2 H2O + O2 Suppose we have a solution of M moles of H2O2 in 1L solution From the reaction stoichiometry, 2 moles of H2O2 gives 1mol of oxygen or 22.4L at S.T.P 1mole of H2O2 =...

The increasing order of the pK_b of the following compound is : 

  • Option 1)

    (C)< (A)< (D)<(B)

  • Option 2)

    (B)<(D)<(C)<(A)

  • Option 3)

    (A)<(C)<(D)<(B)

  • Option 4)

    (B)<(D)<(A)<(C)

  Inductive effect - Displacement of  electron towards more electronegative atom is called inductive effect - wherein It is a permanent effect     -I - effect (Negative inductive effect) - Those atoms or group of atoms having more tendency to displace e - towards itself as compared to Hydrogen atom is called Negative Inductive effect - wherein       The more the value of , the weaker is the...

For a reaction,

N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g);\, \, \, \, \, \: \: \; \: \:

identify dihydrogen \left ( H_{2} \right ) as a limiting reagent in the following reaction mixtures.

  • Option 1)

     56\, g\: of\: N_{2}\: +\: 10\, g\: of\: H_{2}              

  • Option 2)

    35\: g\: of\: N_{2}\: +\: 8\: g\: of\: H_{2}

  • Option 3)

    28\: g\: of\: N_{2}\: +\: 6\: g\: of\: H_{2}

  • Option 4)

    14\: g\: of\: N_{2}\: +\: 4\: g\: of\: H_{2}

  Concept of limiting reagent and excess reagent - The reactant which gets consumed and thus limits the amount of product formed is called the limiting reagent. - wherein  CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) ; In this reaction, if 1 mole of methane and 1 mole of oxygen are taken then oxygen would be limiting reagent.       For :- Identifying  as a limiting reagent                     ...

The major product of the following reaction is :

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

Oxidation of primary alcohol or an aldehyde - Primary alcohol oxidizes in the presence of oxidising agent to give aldehyde which further oxidises to give the carboxylic acid.  - wherein   Option (3) is correctOption 1)Option 2)Option 3)Option 4)

The correct of the following polymer is :

 

  • Option 1)

    Polyisobutane 

  • Option 2)

    Polytert - butylene

  • Option 3)

    Polysioprene

  • Option 4)

    Polyisobutylene

  Addition Polymers - Formed by repeated addition of monomers having double or triple bonds. - wherein Eg:- Polythene (from ethene), Polypropylene (from propane)       Free radical mechanism of Chain growth Polymerisation - 1.  Chain initiation step   - Production of free radical   - Addition of this radical to a monomer to produce chain initiating species 2. Chain propagating step   -...

Consider the following reactions:

'A' is : 

  • Option 1)

    CH\equiv CH

     

     

     

  • Option 2)

    CH_{3}-C\equiv C-CH_{3}

  • Option 3)

    CH_{3}-C\equiv CH

  • Option 4)

    CH_{2}=CH_{2}

  Addition of water on alkyne - By the addition of water on alkyne in presence of HgSO4 and H2SO4 alkyne give aldehyde or ketone. - wherein                                                      (kucherov...

An 'Assertion' and a 'Reason' are given below. Choose the correct answer from the following options:

Assertion (A): Vinyl halides do not undergo nucleophilic substitution easily.

Reason (R): Even though the intermediate carbocation is stabilized by loosely held the cleavage is difficult because of strong bonding.

 

  • Option 1)

    Both (A) and (R) are wrong statements

  • Option 2)

    Both (A) and (R) are the correct statement and (R) is the correct explanation of (A).

     

  • Option 3)

    Both (A) and (R) are correct statements but (R) isnot the correct explanation of (A).

     

  • Option 4)

    (A) is a correct statement but (R) wrong statement.

     

 

Vinayl Chloride -

Halogen present on double bonded carbon atom.

- wherein

 

 

 

 

Reactivity Order of alkyl halides -

1^{\circ}Alkyl\: Halide> 2^{\circ}Alkyl\: Halide> 3^{\circ}Alkyl \: Halide

 

- wherein

1^{\circ}Alkyl\: Halide> 2^{\circ}Alkyl\: Halide> 3^{\circ}Alkyl \: Halide

Inhibition by steric hindrance SN2 reactions are particularly sensitive to steric factors, since they are greatly retarded by steric hindrance (crowding) at the site of reaction. In general, the order of reactivity of alkyl halides in SN2 reactions is: methyl  > 1> 2o> 3o

 

 

 

 

vinyl halides do not undergo a substitution reaction. It is a fact so, it is true statement 

(R) 


Option 1)

Both (A) and (R) are wrong statements

Option 2)

Both (A) and (R) are the correct statement and (R) is the correct explanation of (A).

 

Option 3)

Both (A) and (R) are correct statements but (R) isnot the correct explanation of (A).

 

Option 4)

(A) is a correct statement but (R) wrong statement.

 

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The increasing order of reactivity of the following compounds towards aromatic electrophilic substitution reaction is :

  • Option 1)

      D< A< C< B

  • Option 2)

     B< C< A< D

  • Option 3)

     A< B< C< D

  • Option 4)

     D< B< A< C

Order of strength  for electrophilic substitution                                                      So ,    Option 1)  Option 2) Option 3) Option 4) 
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