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Engineering
1189 Views   |  

The IUPAC name of the compound shown below is

  • Option 1)

    2-bromo-­6-chlorocyclohex­-1-­ene

  • Option 2)

    6-­bromo­-2-­chlorocyclohexene

  • Option 3)

    3­-bromo­-1-­chlorocyclohexene

  • Option 4)

    1-­bromo­-3-chlorocyclohexene.

 

As we learnt in

Nomenclature of organic compound -

1)    Some of locant value or position no. of substituent should be minimum.

 

2)    Senior functional group written in the form of suffix.

 

3)    Junior  functional group written in the form of prefix.

- wherein

1)

2)Butan-2-ol

3)3 -Hydroxy Hexanal

Image result for 3 hydroxy hexanal

 

 The IUPAC name of is  

3­-bromo­-1-­chlorocyclohexene


Option 1)

2-bromo-­6-chlorocyclohex­-1-­ene

This option is incorrect

Option 2)

6-­bromo­-2-­chlorocyclohexene

This option is incorrect

Option 3)

3­-bromo­-1-­chlorocyclohexene

This option is correct

Option 4)

1-­bromo­-3-chlorocyclohexene.

This option is incorrect

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Engineering
208 Views   |  

The increasing order of stability of the following free radicals is

  • Option 1)

    (CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H< (C_{6}H_{5})_{3}\dot{C}

  • Option 2)

    (C_{6}H_5)_3\dot{C}< (C_{6}H_5)_2\dot{C}H< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

  • Option 3)

    (C_{6}H_5)_2\dot{C}H< (C_{6}H_5)_3\dot{C}< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

  • Option 4)

    (CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H

 

As we learnt in

Stability of alkyl free radical -

Electron donating group stabilises alkyl free radicals and withdrawing group destabilises alkyl free radicals.

- wherein

 

 

Stability of alkyl free radical due to resonance -

More the no of resonating structure more is the stability.

- wherein

 

 Free radical is stabilised due to resonance, hyperconjugation and inductive effect. In given problem the order of stability of free radical is 

(CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H< (C_{6}H_{5})_{3}\dot{C}


Option 1)

(CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H< (C_{6}H_{5})_{3}\dot{C}

This option is correct

Option 2)

(C_{6}H_5)_3\dot{C}< (C_{6}H_5)_2\dot{C}H< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

This option is incorrect

Option 3)

(C_{6}H_5)_2\dot{C}H< (C_{6}H_5)_3\dot{C}< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

This option is incorrect

Option 4)

(CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H

This option is incorrect

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Engineering
444 Views   |  

Which one of the following sets of ions represents a collection of isoelectronic species?

  • Option 1)

    K^{+},Cl^{-},Ca^{2+},Sc^{3+}\;

  • Option 2)

    \; Ba^{2+},Sr^{2+},K^{+},S^{2-}\;

  • Option 3)

    \; N^{3-},O^{2-},F^{-},S^{2-}\; \;

  • Option 4)

    \; Li^{+},Na^{+},Mg^{2+},Ca^{2+}

 
As we leant in Isoelectronic species - A series of atom, ions and molecules in which each species contains the same number of electrons but different nuclear charge. - wherein    Let's state the number of electrons in each ion: K+ : 18          Ba2+ : 54           N3- : 10          Li+ : 2 Cl- : 18          Sr2+ : 36           O2 : 10           Na+ : 10 Ca2+ : 18       K+ : 18             ...
Engineering
680 Views   |  

The increasing order of the first ionisation enthalpies of the elements B, P, S and F (lowest first) is

  • Option 1)

    F<S<P<B\;

  • Option 2)

    \; P<S<B<F\;

  • Option 3)

    \; B<P<S<F\;

  • Option 4)

    \; B<S<P<F

 
As we learnt in Ionization enthalpy - It is defined as the minimum amount of energy required to remove the outer most shell electron from an isolated gaseous atom (X) to form the gaseous ion. - wherein       Along a period from left to right, I.E increases with increasing atomic number, and ionization enthalpy decreases down the group. But P has a half filled stable configuration as compared to...
Engineering
86 Views   |  

The decreasing values of bond angles from NH_{3} (106°) to SbH_{3} (101°) down group­-15 of the periodic table is due to

  • Option 1)

    increasing bond-­bond pair repulsion

  • Option 2)

    increasing p-orbital character in sp^{3} 

  • Option 3)

    decreasing lone pair­-bond pair repulsion

  • Option 4)

    decreasing electronegativity.

 
As we learnt in Electronegativity - A qualitative measure of the ability of an atom in a chemical compound to attract shared electron is electronegativity. - wherein It is not a measurable quantity.    Because of increased bond length of Sb-H bond as compared to N-H bond, the repulsion between lone pair - bond pair electron decreases, resulting in lesser bond angle. Option 1) increasing...
Engineering
197 Views   |  

Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture?

  • Option 1)

    The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group

  • Option 2)

    In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group

  • Option 3)

    Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens

  • Option 4)

    In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group

 
As we learnt in Electropositive or metallic character of alkaline earth metal - On account of their relatively low ionisation energies, the alkaline earth metal has a strong tendency to lose both the valence electrons to form dispositive cations.thus these elements show strong electropositive or metallic character. -    and Periodic trend and chemical reactivity - The maximum chemical...
Engineering
1746 Views   |  

Phenyl magnesium bromide reacts with methanol to give

  • Option 1)

    a mixture of anisole and Mg(OH)Br

  • Option 2)

    a mixture of benzene and Mg(OMe)Br

  • Option 3)

    a mixture of toluene and Mg(OH)Br

  • Option 4)

    a mixture of phenol and Mg(Me)Br

 
As learnt in Zerewitinoff Method - Reaction of alcohol with grignard reagent. - wherein       Reaction of Grignard reagent with Alcohol - Alkane is obtained.     - wherein     Option 1) a mixture of anisole and This option is incorrect. Option 2) a mixture of benzene and This option is correct. Option 3) a mixture of toluene and This option is incorrect. Option 4) a mixture of phenol and...
Engineering
173 Views   |  

The structure of the compound that gives a tribromoderivative on treatment with bromine water is

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
As we learnt in Halogenation when Heated with bromine water - While precipitate of  2, 4, 6 - tribromophenol is formed. - wherein    Bromine adds to ortho and para position of phenol derivatives.   SN2 Substitution Nucleophilic bimolecular - These reactions proceed in only one step. - wherein    m-cresal due to pheroxide ion in solvent gives tribromoderivativeat at  para positions.   Option...
Engineering
479 Views   |  

The electrophile involved in the above reaction is

  • Option 1)

    dichloromethyl cation 

  • Option 2)

    dichlorocarbene  (:\; CCl_2)

  • Option 3)

    trichloromethyl anion

  • Option 4)

    formyl cation

 
As we learnt in Reimer Tiemann's reaction - The electrophile used is CCl2  (Dichloromethylene) -CHO is introduced in the ortho position. - wherein    This is Riemer - Tiemann reaction and electrophile is dichlorocarbene. Option 1) dichloromethyl cation  This option is incorrect. Option 2) dichlorocarbene  This option is correct. Option 3) trichloromethyl anion This option is...
Engineering
119 Views   |  

Fluorobenzene (C_6H_5F) can be synthesised in the laboratory

  • Option 1)

    by heating phenol with HF and KF

  • Option 2)

    from aniline by diazotization followed by heating the diazonium salt with HBF_4

  • Option 3)

    by direct fluorination of benzene with F_2 gas

  • Option 4)

    by reacting bromobenzene with NaF solution.

 
As we learnt in  By Sandmeyer's reaction - In the reaction benzene diazonium chloride is treated with cuprous chloride cuprous bromide.     Option 1) by heating phenol with  and This option is incorrect. Option 2) from aniline by diazotization followed by heating the diazonium salt with This option is correct. Option 3) by direct fluorination of benzene with gas This option is...
Engineering
331 Views   |  

Reaction of trans-2-phenyl­-1­-bromocyclopentane on reaction with alcoholic KOH produces

  • Option 1)

    4-­phenylcyclopentene

  • Option 2)

    2-­phenylcyclopentene

  • Option 3)

    1-­phenylcyclopentene

  • Option 4)

    3-­phenylcyclopentene.

 
As we learnt in  Reaction of alkyl halide with KOH (alc) - - elimination reaction take place and produces alkenes.     Option 1) 4-­phenylcyclopentene This option is incorrect. Option 2) 2-­phenylcyclopentene This option is incorrect. Option 3) 1-­phenylcyclopentene This option is incorrect. Option 4) 3-­phenylcyclopentene. This option is correct.
Engineering
124 Views   |  

HBr reacts with CH_2=CH-OCH_3  under anhydrous conditions at room temperature to give

  • Option 1)

    CH_3CHO  and CH_3Br

  • Option 2)

    BrCH_2CHO  and CH_3OH

  • Option 3)

    BrCH_2-CH_2-OCH_3

  • Option 4)

    H_3C-CHBr-OCH_3

 
As we learnt in Chemical properties of aldehydes and ketones - Undergo nucleophilic addition reactions at carbon-oxygen double bond, hybridization of carbon changes from sp2 to  sp2.  Aldehydes are more reactive than ketones due to steric hindrance and  +I effect of an alkyl group. -     Reactivity order of hydrohalic acid - Towards electrophilic addition reaction reactivity order is -...
Engineering
126 Views   |  

The increasing order of the rate of HCN addition to compounds A-D is

A.\; \; HCHO

B.\; \; CH_3COCH_3

C.\; \; PhCOCH_3

D.\; \; PhCOPh

  • Option 1)

    A<B<C<D

  • Option 2)

    D<B<C<A

  • Option 3)

    D<C<B<A

  • Option 4)

    C<D<B<A

 
As we learnt in  Chemical properties of aldehydes and ketones - Undergo nucleophilic addition reactions at carbon-oxygen double bond, hybridization of carbon changes from sp2 to  sp2.  Aldehydes are more reactive than ketones due to steric hindrance and  +I effect of an alkyl group. -    Rate of addition of HCN is determined by steric hindrance. Thus, D<C<B<A Option 1) This option is...
Engineering
109 Views   |  

Among the following the one that gives positive iodoform test upon reaction with I_2 and NaOH is

  • Option 1)

    CH_3CH_2CH(OH)CH_2CH_3

  • Option 2)

    C_6H_5CH_2CH_2OH

  • Option 3)

  • Option 4)

    PhCHOHCH_3

 
As we learnt in  Test for alcohols by Victor - Meyer's test - Reagents used  - wherein    Only those alcohols which contain group undergo haloform reaction. Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is incorrect Option 4) This option is correct
Engineering
132 Views   |  

The correct order of increasing acid strength of the compounds

(A)\; \; CH_3CO_2H

(B)\; \; MeOCH_2CO_2H

(C)\; \; CF_3CO_2H

(D)\; \;  

is

  • Option 1)

    B<D<A<C

  • Option 2)

    D<A<C<B

  • Option 3)

    D<A<B<C

  • Option 4)

    A<D<C<B

 
As we learnt in  Effects of substituents on acidity of phenols - 1.  Electron withdrawing group like    2.  Electron donating group like     - wherein 1.  Increase the acidic strength of phenol. 2.  Decrease the acidic strength of phenol.    Electron withdrawing group increases the acid strength and electron donating group decreases the acid strength. D < A < B < C Option 1) This option is...
Engineering
164 Views   |  

The term anomers of glucose refers to

  • Option 1)

    isomers of glucose that differ in configurations at carbons one and four  (C­-1 and C-­4)

  • Option 2)

    a mixture of (D)­-glucose and (L)­-glucose

  • Option 3)

    enantiomers of glucose

  • Option 4)

    isomers of glucose that differ in configuration at carbon one (C-­1).

 
As we learnt in Anomers - The two forms of glucose,   glucose and   glucose , are called anomers . Both are optically active and differ at   position .  - wherein Specific rotation of   glucose is  and of  glucose is     Isomers of glucose differ in configuration at C1 position due to the presence of chiral carbon. Option 1) isomers of glucose that differ in configurations at carbons one...
Engineering
304 Views   |  

The pyrimidine bases present in DNA are

  • Option 1)

    cytosine and adenine

  • Option 2)

    cytosine and guanine

  • Option 3)

    cytosine and thymine

  • Option 4)

    cytosine and uracil.

 
As we learnt in Deoxyribonucleic acid (DNA) - Made up of nucleotides consisting of pentose sugar which is  deoxyribose, Phosphoric acid and bases (Adenine , Guanine , cytosine and thymine) -    Pyrimidine bases in DNA = Cytosine +Thymine Purine bases in DNA = Adenine+Guanine Option 1) cytosine and adenine This option is incorrect Option 2) cytosine and guanine This option is incorrect Option...
Engineering
120 Views   |  

Which of the following statements is true?

  • Option 1)

    H_{3}PO_{3}  is a stronger acid than H_{2}SO_{3}

  • Option 2)

    In aqueous medium HF is a stronger acid than HCl

  • Option 3)

    HClO_4  is a weaker acid than HClO_3

  • Option 4)

    HNO_3  is a stronger acid than HNO_2

 
As we learnt in  Acidity of nitric acid - Strong acid, in aqueous solution gives hydronium and nitrate ions - wherein    Let's look at conjugate bases of HNO3 and HNO2 The negative charge is more delocalized on NO3- making it a more stable conjugate base. Option 1)   is a stronger acid than This option is incorrect. Option 2) In aqueous medium HF is a stronger acid than This option is...
Engineering
357 Views   |  

What products are expected from the disproportionation reaction of hypochlorous acid?

  • Option 1)

    HClO_3\; \; and\; \; Cl_2O

  • Option 2)

    HClO_2\; \; and\; \; HClO_4

  • Option 3)

    HCl\; \; and\; \; Cl_2O

  • Option 4)

    HCl\; \; and\; \; HClO_3

 
As we learnt in Oxidising property of Chlorine - - wherein Oxidising property is due to nascent oxygen        Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is incorrect option Option 4) This is correct option
Engineering
321 Views   |  

Which of the following factors may be regarded as the main cause of lanthanide contraction?

  • Option 1)

    Poor shielding of one of  4f electron by another in the subshell

  • Option 2)

    Effective shielding of one of  4f electrons by another in the subshell

  • Option 3)

    Poorer shielding of 5d electrons by 4f electrons

  • Option 4)

    Greater shielding of 5d electrons by 4f electrons

 
As we learnt in Lanthanoid contraction - The filling of if orbits before 5d orbits result regular decrease in atomic radii called Lanthanoid contraction or can say the size of radii not much increase. - wherein      In lanthanoid contraction shielding of one 4f electron on another 4f electron is less than shielding of one 5d electron on another 5d electron.  Option 1) Poor shielding of one...
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