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84551.

All carbonates in alkali metals are

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1
7
4
2

Solution :ALKALI metals valence SHELL configuration `=NS^(1)`
84552.

All carbon atom present in KH_(3)(C_(2)O_(4) )_(2).2H_(2)O weighing 762 gm is converted to CO_(2).How many gm of CO_(2) were obtained.

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ANSWER :`[w_(Co_(2))=528gm]`
84553.

All but one of the following compounds reacts with aniline to give acetanilide. Which one does not?

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`CH_(3)OVERSET(O)overset(||)C-Cl`

`CH_(3)overset(O)overset(||)C-H`

ANSWER :D
84554.

All carbohydrates contain

Answer»

-CHO GROUP
gtC=O group
-COO-group
-CONH-group

Answer :B
84555.

All bonds in SF_(4) are not equivalent. Comment.

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SOLUTION :S in `SF_(4)` is `sp^(3)` d-hybridized. Therefore, `SF_(4)` has trigonal bipyramid structure or see-saw GEOMETRY. Two of the S-F bonds occypy equatorial positions and are 155 pm LONG while the third equatorial position is occupied by a lone pair of electrons. The remaining two S-F bonds occypy axial positions and are little longer (165 pm) due to repulsions between the axial bond pairs and the equatorial lone pair. Thus, all the four S-F bonds in `SF_(4)` are not equivalent.
84556.

All bonds in benzene are equal due to :

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Tautomerism
I-effect
Resonance
Isomerism

Answer :C
84557.

All bond angles are exactly equal to 109^(@) 28' in:

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methyl chloride
iodoform
chloroform
carbon tetrachloride

Solution :The bond ANGLE will be EXACTLY `109^(@) 28.` when the central atom is `sp^(3)` hybridised and all bonds have same atom. `C CI_(4)` has all identical bonds (C-Cl) and central carbon is `sp^(3)` hybridised. So, it has bond angle exactly `109^(@) 28.`
84558.

All axial distances are unequal as well as all axial angles are unequal in the system

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MONOCLINIC
TRIGONAL
triclinic
hexagonal

ANSWER :C
84559.

All ammonium salt liberate ammonia when

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Heated with HCI
Heated with caustic soda
Heated with `H_(2)SO_(4)`
Heated with `NaNO_(2)`

SOLUTION :Any ammoniumsalt onheatingwith are alkline LOSSES `NH_(3)`
84560.

All amines have general formula

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`C_(N)H_(2n)NH_(2)`
`C_(n)H_(2n)N`
`C_(n)H_(2n+3)N`
`C_(n)H_(2n+2)N`

ANSWER :C
84561.

All amines are basic in nature because

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they possess ONE pair of electron on nitrogen
they give `OH^(-)` ionsinaqueousmedium
they FORM saltwith ACID
all of these

ANSWER :D
84562.

All alumns contain

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ONE MONOVALENT and one TRIVALENT metal
Both movovalent metal
One DIVALENT and monovalent metal
Both divalent metals

Answer :A
84563.

All aminesaresoluble in

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WATER
HCl
`CS_(2)`
`C Cl_(4)`

Solution :Amineare basicin NATURE. Hence these aresolublein ACID formingsalt.
84564.

Describe reaction of alkaline earth metals with acids.

Answer»
84565.

All alkali metals dissolve in anhydrous liquid ammonia to give blue colour solution. It is the ammoniated electron whichis reponsible for the blue colour of the solution, and the electrical conductivity is due to the ammoniated cation, [M(NH_(3))_(x)]^(+) as well as the ammoniated electron, [e(NH_(3))_(y)]^(-), value of x and y depend on the extent of solvation by NH_(3). Dilute solutions are paramagnetic due to free ammoniated electrons. Q.Ammoniated solutions of alkali metals are reducing agents due to the presence of free ammoniated or solvated electrons that can reduce: (I)O_(2) "to" O_(2)^(2-)(II) K_(2)[Ni(CN)_(4)] " to "K_(4)[Ni(CN)_(4)] (III) Aromatic ring(IV) Non-terminal alkyne Choose the correct code:

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III and IV
II and III
I, II, III and IV
I, III and IV

SOLUTION :`M(s)+UNDERSET("excess")(NH_(2))(l) to M^(+)(NH_(3)_(2)+overline(e)(NH_(3))_(2)`
ammoniated `overline(e)` responsible for blue colour and reducing character.
If conc. Of solution is increased, then ASSOCIATED of solvated electrons gas started hence, paramagnetism DECREASE and solution changes to BRONZE colour.
84566.

How alkaline earth metal dissolve in liquid ammonia ?

Answer»
84567.

All alkali metals crystallize into BCC strucutres. (a) Find an equation relating the metallic radius to the density of a BCC solid in terms of its molar mass. (b) Determine the relative size of Cs atom compared to a Li atom, if their densities are 1.87 g/cc and 0.53 g/cc, respectively.

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Solution :(a) For a BCC solid of an element, there are two atoms per UNIT cell.
`implies` Density (rho)=(2xxM)/(6.023xx10^(23)a^(3))`
Also, in BCC of an atomic solid,
`rho=(2Mxx3sqrt(3))/(6.023xx10^(23)xx64r^(3))=2.696xx10^(-23)(M)/(r^(3))`
(B)`(rho_(Cs))/(rho_(LI))=(1.87)/(0.53)=(133)/(7)(r_(Li)/(r_(Cs)))^(3)`
`(r_(Cs))/(r_(Li))=1.753 implies r_(Cs)=1.753 r_(Li)`
84568.

All aliphatic amines are more basic than ammonia but due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituents. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength. the base strenghening effect of the electron-donating groups and base weakening effect of the electron-withdrawing groups is, however, more pronounced at p-than at m-position. However, due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron-withdrawing. Q. Among the following, the weakest base is

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`C_(6)H_(5)NH_(2)`
`p-CH_(3)O-C_(6)H_(4)NH_(2)`
`m-CH_(3)O-C_(6)H_(4)NH_(2)`
`o-CH_(3)O-C_(6)H_(4)NH_(2)`

Answer :C
84569.

All aliphatic amines are more basic than ammonia but due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituents. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength. the base strenghening effect of the electron-donating groups and base weakening effect of the electron-withdrawing groups is, however, more pronounced at p-than at m-position. However, due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron-withdrawing. Q. What is the order of basicity of I. p-methylaniline II. m-methylaniline III. aniline IV. o-methylaniline.

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IgtIIgtIIIgtIV
IgtIIgtIVgtIII
IVgtIgtIIgtIII
IIgtIgtIVgtIII

Answer :A
84570.

All aliphatic amines are more basic than ammonia but due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituents. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength. the base strenghening effect of the electron-donating groups and base weakening effect of the electron-withdrawing groups is, however, more pronounced at p-than at m-position. However, due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron-withdrawing. Q. Among the following, weakest base is

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`C_(6)H_(5)CH_(2)NH_(2)`
`C_(6)H_(5)CH_(2)NHCH_(3)`
`O_(2)N-CH_(2)NH_(2)`
`CH_(3)NHCHO`

Solution :AMIDES are weaker bases than AMINES.
84571.

All aliphatic amines are more basic than ammonia but due to delocalisation of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent whereas electron donating group tend to increase, electron-with drawing groups tend to decrease the basic strength. The base trengthening effect of the electron-donating group and the base weaken effect of the electron withdrawing group is, however, more pronouned at p-than at m-position. However due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron withdrawing. Select the weakest base among the following

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`C_(6)H_(5)CH_(2)NH_(2)`
`C_(6)H_(5)CH_(2)NHCH_(3)`
`O_(2)N-CH_(2)NH_(2)`
`CH_(3)NHCHO`.

Answer :C
84572.

All aliphatic amines are more basic than ammonia but due to delocalisation of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent whereas electron donating group tend to increase, electron-with drawing groups tend to decrease the basic strength. The base trengthening effect of the electron-donating group and the base weaken effect of the electron withdrawing group is, however, more pronouned at p-than at m-position. However due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron withdrawing. Choose among the following amines the correct decreasing order of their basic strength

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`I GT II gt III gt IV`
`I gt IV gt II gt III`
`II gt I gt IV gt III`
`I gt IV gt III gt II`.

ANSWER :D
84573.

All aliphatic amines are more basic than ammonia but due to delocalisation of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent whereas electron donating group tend to increase, electron-with drawing groups tend to decrease the basic strength. The base trengthening effect of the electron-donating group and the base weaken effect of the electron withdrawing group is, however, more pronouned at p-than at m-position. However due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron withdrawing. The correct order of increasing basic nature of the following bases is

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`2 gt 1 lt 3 lt 4`
`2 lt 1 gt lt 4`
`2 lt 1 gt 4 lt 3`
`2 lt 1 lt 3 lt 4`.

SOLUTION :Electron DONATING group increase while electron with drawing groups decrease the BASICITY, i.e., basicity increases in the order `2 lt 1 lt 3 lt 4`
84574.

All aldehdes can made to undergo the Cannizzaro reaction by treatment with aluminium ethoxide. Under these conditions, the acid and alcohol are combined to form an ester. The reaction is called

Answer»

Claisen REACTION
PERKIN reaction
ALDOL CONDENSATION
Tischenko reaction

ANSWER :D
84575.

All allums contain

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ONE MONOVALENT and one TRIVALENT metal
Both monovalent metals
One DIVALENT and one monovalent metal
Both divalent metals

Answer :A
84576.

How 1^@ amine reacts with nitrous acid ?

Answer»


ANSWER :T
84577.

All adsorptions are exothermic. Explain.

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84578.

Alkynes undergo acid catalysed addition of water across the triple bond in the presence of mercuric ion as catalyst. A mixture of HgSO_(4) and aqueous H_(2)SO_(4) is used and addition product follows Markownikoff's rule R-C-=CH+H_(2)O underset(H_(2)SO_(4)"Step 1")overset(HgSO_(4))toR-underset(OH)underset(||)(C)=CH_(2) underset("Step 2")to underset("Step 2")to R-underset("Ketone")underset(O)underset(||)(C CH_(3) Hydration of 2 pentyne by a mixture of HgSO_(4) and H_(2)SO_(4) forms

Answer»

2-pentanone
3-pentanone
both A and B
None of these

Answer :C
84579.

All 1° amines on diazotisation with NaNO_2(s) and HCl(aq) at low temperature give diazonium salts.

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ANSWER :F
84580.

Alkynes undergo acid catalysed addition of water across the triple bond in the presence of mercuric ion as catalyst. A mixture of HgSO_(4) and aqueous H_(2)SO_(4) is used and addition product follows Markownikoff's rule R-C-=CH+H_(2)O underset(H_(2)SO_(4)"Step 1")overset(HgSO_(4))toR-underset(OH)underset(||)(C)=CH_(2) underset("Step 2")to underset("Step 2")to R-underset("Ketone")underset(O)underset(||)(C CH_(3) Select the correct statement(s)

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Enol form is the hydration of alkyne is called vinyl alcohol
Conversion of enol to KETONE (step 2)is called tautomerisation and enol and ketone are called tautomer
Step 2 is acid catalysed
`HgSO_(4)` forms vinyl cation `R-overset(OPLUS)(C)-underset(Hg^(oplus))underset(|)(CH) "with" RC-=CH`

ANSWER :A,B,C,D
84581.

Alkynes can be reduced to alkenes hydrogenation in presence of:

Answer»

Rancy Ni
Anhy `AICI_3`
Pd
Lindlar's catalyst

Answer :D
84582.

Alkynes (A) and (B) hav ethe molecular formula C_(8)H_(14) giving following reactions: (A) or (B) overset(H_(2)//Pt)rarr "n-Octane". (A) overset(O_(3)//H_(2)O)rarr CH_(3)CH_(2)CH_(2)CO_(2)H (B) overset(Ag(NH_(3))_(2)OH)rarr silver containing white ppt. Write the structure of (A) and (B).

Answer»


Solution :Since `(A)` and `(B)` by reduction using `H_(2)//Pt` gives `n`-Octane, hence both have straight chain
`A overset(O_(3)//H_(2)O)rarr CH_(3)CH_(2)CH_(2)CO_(2)H` ("and no other compound")
Hence, there should be two mole of this ACID. THUS `A` is `CH_(3)CH_(2)CH_(2)C-=C CH_(2)CH_(2)CH_(3)`
`B` gives white PPT. containing `Ag` hence, `B` is terminal alkyne.
Structure of `B` is `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(2)C-=CH`
84583.

Alkyl substituted chlorosilane, R_n SiCl_((4-n)), on hydrolysis and then condensation polymerisation yield silicones What is the value of n is the alkyl substituted chlorosilane used for the preparation of cross linked silicone.

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SOLUTION :`RSiCl_3` on hydrolysis GIVES a CROSS LINKED SILICONE.
84584.

Alkyl nitrite on reduction withSn//HCl gives :

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Alcohol
Hydroxylamine
Both (a) and (B)
Hydrazine

Answer :C
84585.

Alkyl isocyanides (Roverset(o+)(N)-=overset(ɵ)(C )) are reduced to 2^(@) amines (R-NH-CH_(3)) with:

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a.`LAH`
b.`NaBH_(4)`
c.`HI+P`
d.`H_(2)+Pt`

ANSWER :A::C
84586.

Alkyl isocyanides can be converted into cyanides by .....

Answer»

SOLUTION : PROLONGED HEATING
84587.

Alkyl iodides are often prepared by the reaction of alkyl chlorides/bromides with Nal in dry acetone. This reaction is known as

Answer»

WURTZ REACTION .
FINKELSTEIN reaction .
Gngnard reaction .
Sandmeyer's reaction .

Answer :B
84588.

Alkyl iodides reacts with NaCN to form alkyl cyanides plus a little amount of alkyl isocyanides. The reason for the formation two types of products is-

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Ionic CHARACTER of NaCN
Nucleophilic character of cyanide ION
AMBIDENT character of cyanide ion
Electrophilic character of cyanide

ANSWER :C
84589.

Alkyl Iodide can be prepared by

Answer»

`R -CH_2COOAg + I_2 overset("CCl"_4)(to)`
`R - CH_2 - Cl + NaI overset("acetone")(to)`
`R - OH + HI to `
`RCOOH + I_2 overset(hv)(to)`

Solution :Alkyl IODIDE can be prepared by
`a) R - CH_2COOAg + I_2 overset("CCl"_4)(to) R- CH_2 - I + CO_2 + AGI`
`b) R - CH_2 - Cl +NaI to NaCl + R - CH_2 - I` these RXN is a Frankelstain RXn
`c) R - OH + HI to R - I + H_2O`
84590.

Alkyl isocyanide on reduction withZn-Hg//HClgives

Answer»

Primary amine
TERTIARY amine
N-Alkyl ALKANAMINE
N- Methyl alkanamine

Answer :D
84591.

Alkyl iodide react with NaCN to give alkyl cyanide and small amount of alkyl isocyanide. Formation of these two product is due to the

Answer»

Ionic character of NaCN
Nucleophilic character of `CN^(-)`
Ambidentate character of `CN^(-)`
Electrophilic character of `CN^(-)`

Solution ::`overset(ө)(C)overset(.)(N)`: has ambidentate character because: `overset(ө)(C)overset(.)(N):` has double attacking SITE
`R-Ioverset(NaCN)tounderset("MAJOR")(RCN)+underset("Minor")(RNC)`.
84592.

Alkyl halides undergoing substitution nucleophilic bimolecular reaction involves

Answer»

Formation of carbocation
Racemic mixture
INVERSION of configuration
Retention of configuration

Solution :`S_(N^(2))` reactions take PLACE VIA inversion of configuration
84593.

Alkyl halides undergo elimination when heated with strong base. A concerted reaction in which both the groups to be eliminated, leave silumtaneously is E2 reaction. In another case ionization of alkyl halide takes place losing X^(-) and forms carbocation in the first stage followed by the loss of H^(+) . this is substituted alkene) and Hofmann (less substituted alkene) are formed depending on the condition of the reaction. Q. Which of the following is not correct regarding E2 reaction?

Answer»

KINETIC isotopic effect is observed.
When the concentration of BASE is doubled, rate of elimination also doubles.
Rearranged PRODUCT is formed.
All of these

Solution :In `E_(2)` REACTION no intermediate is formed. It is a single step bimoleuclar reaction in which both the LEAVING groups are lost simultanesouly.
84594.

Alkyl halides undergo elimination when heated with strong base. A concerted reaction in which both the groups to be eliminated, leave silumtaneously is E2 reaction. In another case ionization of alkyl halide takes place losing X^(-) and forms carbocation in the first stage followed by the loss of H^(+) . this is substituted alkene) and Hofmann (less substituted alkene) are formed depending on the condition of the reaction. Q. Identify the reaction the reaction in which Hofmann product could be the major product?

Answer»

2-fluorohexane `underset(C_(2)H_(5)OH)OVERSET(C_(2)H_(5)O^(-))to`
2-chlorohexane `underset(C_(2)H_(5)OH)overset(C_(2)H_(5)O^(-))to`
2-bromohexane `underset(C_(2)H_(5)OH)overset(C_(2)H_(5)O^(-))to`
2-iodohexane`underset(C_(2)H_(5)OH)overset(C_(2)H_(5)O^(-))to`

Solution :Fluoro alkene gives mainly Hofmann product due to E1cB mechanism
84595.

alkyl halides, though polar, are immiscible with water?

Answer»

Solution :Since the new forces of attraction between WATER and alkyl halide molecules are WEAKER than the forces of attraction already existing between alkyl halide-alkyl halide molecules and water-water molecules. Therefore, alkyl HALIDES are immiscible (not SOLUBLE) with water.
84596.

Alkyl halides reacts with dialkyl copper reagents to give (A). The reaction is called B,(A) and (B) are :

Answer»

alkenes, corey HOUSE synthesis
alkanes, corey house synthesis
alkanes, Rosenmund's synthesis
aleknyl halides, elemination reaction

Solution :Corey house synthesis yields ALKANE on TREATING alkyl halides with lithium dialkyl cuprate.
`R_(2)CuLi+R'XtoR R'Rcu+LiX`
84597.

Alkyl halides react with sodium in dry ether to give hydrocarbons containing _________ the number of carbon atoms present in the halide.

Answer»


ANSWER :DOUBLE
84598.

Alkyl halides react with sodium in dry ether to give hydrocarbons.

Answer»


ANSWER :1
84599.

Alkyl halides react with lithium dialkyl copper reagents to give :

Answer»

alkenes
ALKYL copper halides
alkanes
alkenyl halides

Solution :`R_(2)CuLi+CH_(3)CH_(2)BrrarrRCH_(2)CH_(3)+RCu+LiBr`
84600.

Alkyl halides react with dialkyl copper reagents to give:

Answer»

Alkenes
Alkyl COPPER halides
ALKANES
Alkenyl halides

ANSWER :3