Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Dehydrochlorination of (CH_(3))_(2)CHCHCHlCH_(3) gives major product,

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`(CH_(3))_(2)C=CHCH_(3)`
`(CH_(3))_(2)CH-CH=CH_(2)`
bot 'a' and 'b'
`(CH_(3))_(2)CHCHOHCH_(3)`

Solution :Major product is formed ACCORDING to SAYTZEFF RULE. It state that `beta`- hydrogen atom is PREFERENTIALLY eliminatd from that carbon atom which have LESS number of hydrogen atom. Carbon `(**)` has only number of haydrogenatom.
2.

Dehydration of ethanol gives:

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ACETIC acid
Ethane
Ethylene
Acetylene

Answer :C
3.

Dehydration require an acid catalyst to protonate the hydroxy group of the alcohol and convert and convert it into good leaving group. Loss of water followed by a loss of a proton, given the alkene an equilibrium is established between reactants and products. Q. Which alcohol is most reactive towards dehydration of alcohol in acid catalysed reaction.

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SOLUTION :
4.

Dehydration require an acid catalyst to protonate the hydroxy group of the alcohol and convert and convert it into good leaving group. Loss of water followed by a loss of a proton, given the alkene an equilibrium is established between reactants and products. Q. total number of alpha-hydrogen in A+B is

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13
15
17
19

Solution :
5.

Dehydration require an acid catalyst to protonate the hydroxy group of the alcohol and convert and convert it into good leaving group. Loss of water followed by a loss of a proton, given the alkene an equilibrium is established between reactants and products. Q. To improve the yield of above reaction which of following is correct.

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High temperature
Distillation (removal of alkene)
ADDITION of `H_(2)O`
Both A and B

Solution :[E' REACTION favoured at high T & remval of alkene, Equilibrium reaction, FORCED in forward direction by removal of alkene, Le-chatlier's PRINCIPLE.
6.

Dehydration of tertiary alcohols with Cu at 573 K gives:

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Aldehydes
ketons
Alkenes
None of these

Answer :C
7.

Dehydration of primary amides with SOCl_2 gives :

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Nitriles
Isocyanides
Amines
NITRO compounds

Answer :A
8.

Dehydration of neophentyl alcohol gives

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2-methyl but-1-ene
3-methyl but-1-ene
2-methyl but-2-ene
no PRODUCT will be FORMED

ANSWER :C
9.

Dehydration of methanol with conc. H_2SO_4 at 410K gives

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DIMETHYL ether
Ethane
HCHO
All

Answer :A
10.

Dehydration of glycerol produces

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Propanone
Allyl alcohol
Acrolein
Benzene

Solution :DEHYDRATION of glycerol with `KHSO_(4)` PRODUCES acrolein.
`{:(CH_(2)OH""CH_(2)),(|""|),(CHOHunderset(Delta)overset(KHSO_(4))to CH+2H_(2)O),(|""|),(CH_(2)OH ""UNDERSET("Acrolein")(CHO)):}`
11.

Dehydration of formic acid with sulphuric acid gives

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`CO`
`C`
`CO and CO_(2)`
`C_(2)H_(4)O_(4)`

SOLUTION :`HCOOH overset(H_(2)SO_(4))rarr CO+H_(2)O`
12.

Dehydration of ethanol with excess amount of conc.sulphuric acid gives .

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SOLUTION :Deithl ETHER is obtaned.
13.

Dehydration of ethanol yields

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ACETIC ACID
Ethylene
Ethane
Acetylene

Answer :B
14.

Dehydration of ethanol with conc. Sulphuric acid gives acetylene. Is it true or false?

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Solution :DEHYDRATION of ETHANOL with conc.sulphuric ACID gives ETHYLENE.
15.

Dehydration of ethanol with conc. H_2 SO_4 at 413 K gives ethane

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ANSWER :F
16.

Dehydration of ethanolgives

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Acetic acid
Ethane
Ethylene
Acetylene

Solution :`CH_(3)-CH_(2)-OH underset(170^(@)C) overset("conc." H_(2)SO_(4))to CH_(2) = CH_(2)+ H_(2)O`
17.

Dehydration of an amide gives :

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CYANIDE
Amine
Isocyanide
FATTY acid

Answer :A
18.

Dehydration of alcohols to form alkenes is always carried out with conc. H_(2)SO_(4) and not witth conc. HCl or HNO_(3), explain why?

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Solution :DEHYDRATION of alcohols to form alkenes occurs through carbocation INTERMEDIATES. If HCl is used, then chloride `(CL^(-))` ion beinng a good nucleophile brings about substitution reaction to give an alkyl chloride.
`underset("Alcohol")(R-CH_(2)CH_(2)-OH) underset(-H_(2)O)overset(H^(+))to underset(R-CH_(2)CH_(2)-Cl" Alkyl chloride (substitution product)")underset(darr+Cl^(-))underset("Carbocation")(R-CH_(2)-overset(+)(C)H_(2))underset(-H^(+))to underset("Alkene (elimination product)")(R-CH=CH_(2))`
If `H_(2)SO_(4)` is used, then the bisulphate ion `(HSO_(4))^(-)` being non-nucleophilic cannot bring about the substitution reaction. instead the carbocation loses a proton to form an alkene. conc. `HNO_(3)` is not used because it being a strong oxidising AGENT, preferentially brings abou the oxidation of alcohols first to aldehydes or ketones and then to the carboxylic acids. thus, dehydration of alcohols to form alkenes is always carried out with such strong acids (e.g. `H_(2)SO_(4),H_(3)PO_(4),KHSO_(4)`) which have non-nucleophilic anions.
19.

Dehydration of alcohols take place more rapidly with POCl_(3) than with H_(2)SO_(4) select the correct statement(s) about the following dehydration reaction.

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It does not involve carbocation
It involves `R-OPOCl_(2)` with `-OPOCl_(2)` as a batter leaving group.
It involves `E2` mechanism as pyridine BASE abstrcts PROTON from the ADJACENT carbon as the same time at which `-OPOCl_(2)` is leaving.
It is E1 reaction WITHOUT formation of carbocation.

Answer :A::B::C
20.

Dehydration of alcohols to give alkenes is catalysed by :

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`"CONC. "H_(2)SO_(4)" at 413 K"`
`"Conc. "H_(2)SO_(4)" at 443 K"`
`"Conc. "H_(2)SO_(4)" at 383 K"`
NONE of these.

Answer :B
21.

Dehydration of alcohols on give ethers is catalyzed by :

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CONE.`H_2 S0_4` at 413 K
Cone.`H_2 S0_4` at 443 K
Cone.`H_2 S0_4` at 483 K
None of these

Answer :A
22.

Dehydration of alcohols is suitable for preparing ethers having primary alkyl groups only. Comment

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Solution :During DEHYDRATION of alcohol the ALKYL group should be unhindered and the temperature be kept low. Otherwise the reaction FAVOURS the formation of alkene. The reaction follows `S_N1` pathway when the alcohol is SECONDARY or tertiary.
However, the dehydration of secondary and tertiary alcohols to give corresponding ethers is unsuccessful as elimination competes over substitution and as a CONSEQUENCE, alkenes are easily formed.
23.

Dehydration of alcohol take place more rapidlywith POCL_(3)then with H_(2)SO_(4). Select the correct statements about the following dehydration reaction :

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It does not involve carbocation .
It involves R-`OPOCl_(2)" with " -OPOCl_(2)`as a BETTER leaving group.
It involves E2 mechanism as pyridine BASE abstracts PROTON from the adjacent carbon as the same TIME at which -`OPOCl_(2)` is leaving .
It is E1 reaction without FORMATION of carboction.

Solution :N//A
24.

Dehydration of alcohol produces alkene, the reaction intermediate is

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CARBONIUM ion
CARBANION
carbon FREE RADICAL
carbene

Answer :A
25.

Dehydration of alcohol by conc. H_(2)SO_(4) takes place according to following steps: CH_(3)-overset(CH_(3))overset("|")underset("H")underset("|")"C "-CH_(2)overset(* *)underset(* *)OHoverset(H^(+))underset("step - 1")rarrCH_(3)-overset(CH_(3))overset("|")underset("H")underset("|")"C "-CH_(2)overset(+)OH_(2)overset(-H_(2)O)underset("step - 2")rarr CH_(3)-overset(CH_(3))overset("|")underset("H")underset("|")"C "-overset(+)CH_(2) overset(1,2-H^(-)" shift")underset("step - 3")Rarr CH_(3)-overset(CH_(3))overset("|")underset("+")C-CH_(3)overset(-H^(+))underset("step - 4")rarr CH_(3)=overset(CH_(3))overset("|")"C "-CH_(3) The lowest and fastest steps in the above reaction are

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step 1 is lowest, by 3 is fastest
step 2 is lowest while 3 is fastest
step 2 is lowest, while 4 is fastest
all steps PROCEED at equal rate

Solution :Step 2 INVOLVES the formation of carbonium ion by the LOSS of weakly basic `H_(2)O` molecule. It is slowest step. Step 4 involves the conversion of an unstable (or intermediate) into a quite stable product, HENCE it is fastest step.
26.

Dehydration of alcohol involves :

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CARBONIUM ion
Carbanion
Free radical
Carbene

Answer :A
27.

Dehydration of 3-phenyl butan-2-ol gives

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2-phenyl but-2-ene
4-phenyl 2-methyl but-I-ene
1-phenyl 3-methyl but-1-ene
3-phenyl but-1-ene

ANSWER :A
28.

Dehydration of alcohol can be prepared from rectified spirit by____distillation.

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SOLUTION :SYMMETRICAL ETHERS.
29.

Dehydration of 3^(@) alcohol undergoes in presence of:

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20% of `H_(3)PO_(4)`
85 % `H_(3)PO_(4)`
50% pf `H_(3)PO_(4)`
conc. `H_(2)SO_(4)`

Solution :20% of `H_(3)PO_(4)`
30.

Dehydration of 1 -propanol by the use of H_(2)SO_(4) and subsequent treatment with HI gives :

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`CH_(3)CH_(2)CH_(2)I`
`CH_(3)CH(I)CH_(3)`
`CH_(2)=CHCH_(2)I`
`ICH=CHCH_(3)`

Solution :`CH_(3)CH_(2)CH_(2)OHoverset(H_(2)SO_(4))rarrCH_(3)CH=CH_(2)OVERSET(HI)rarrCH_(3)underset(I)underset("| ")(CH)-CH_(3)`
31.

Dehydration of 2-butanol yield

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1-butene
2-butene
2-butyne
Both (a) and (b)

SOLUTION :`UNDERSET("2-butanol")(CH_(3)-CH_(2)-OVERSET(OH)overset(|)CH-CH_(3))overset("dehydration")to underset("1-butene")(CH_(3)-CH_(2)-CH)=underset("2-butene Major product")(CH_(2)+CH_(3)-CH=CH-CH_(3))`
32.

Dehydration of 1-butanol gives 2-butene as a major product, by which of the following intermediate the compound 2-butene is obtained.

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SOLUTION :
HENCE, (C) is the CORRECT ANSWER.
33.

Dehydration method used for

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INDUCE sleep
hypnotic
food preservation
relive pain

Answer :C
34.

Dehydration is most easy for

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primary alcohols
TERTIARY alcohols
SECONDARY alcohols
ethanol

ANSWER :B
35.

Dehydration by conc. H_(2)SO_(4) is the most difficult in

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ANSWER :3
36.

Degree of unsturation in C_7H_8 is two. The value of y is….........

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ANSWER :5
37.

Dehydrated phosphorus trichloride in water gives

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`HPO_(3)`
`H_(3)PO_(4)`
`H_(3)PO_(2)`
`H_(3)PO_(3)`

Solution :`3H_(2)O +PCl_(3) to H_(3)PO_(3) +3HCl`
38.

Degree of unsaturation of this compound is

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10
9
8
7

Answer :C
39.

Degree of unsaturation in oils and fats is measured in terms of :

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Saponification value
Iodine value
ACID value
Acetyl value

ANSWER :B
40.

Degree of unsaturation of product form after complete hydrogenation of the following compound:

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`0`
`2`
`6`
`4`

SOLUTION :has four rings so Degree of unsaturation `= 4`
41.

Degree of unsaturation for

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12
13
14
15

Solution :
42.

Degree of unsaturation for

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12
13
14
15

Solution :
43.

Degree of ionisation of a soltuion produces

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Solution :The degree of ionization depend UPON the NATURE of the solute the SIZE of the solute molecules and the CONCENTRATION of the solution.
44.

Degree of hydrolysis of a salt of weak acid and a seak base:

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INCREASES with concentration
decreases with concentration
Independent of concentration
None of these

Answer :C
45.

Degree of hydrolysis for a salt of strong acid and weak base is

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Independent of dilution
Increases with dilution
Increases with decrease in `K_(b)` of the bases
Decreases with decrease in temperature

Solution :Let BA be this salt `BA to B^(+)+A^(-)`
`A^(-)` does not undergo hydrolysis because HA is strong acid. `B^(+)` undergoes hydrolysis
`B^(+) +H_(2)O=BOH+H^(+)`
`h=sqrt((K_(h)V)/(c))-(1)h=` degreee of hydrolysis where
`K_(h)= " Hydrolysis constant "= (K_(w))/(K_(b))`
`h propsqrt(K_(h))-(3)`greater the `K_(b)` lesser the h
`h propsqrtV-(4)V=` volume of salt solution hence h increases with dilution. `K_(h)=(K_(w))/(K_(b)). Both K_(w)and B_(b)` CHANGE with temperature, so `K_(h)` change with temperature. Hence, STATEMENT (b) is correct, h increases if `K_(b)` decreases, statement (c) correct.
It is found that as TEMPERATUE, increases, `k_(w) and K+_(b)` increase but increase in `K_(w)` is greater than increase in `K_(b).` Hence, h increase with increase in temp. or h decreases with decrease in temp., hence statement (d) is correct, (a) is not correct from explanation of (b).
46.

Degree of dissociation of pure water is 1.9xx10^(-9). Molar ionic conductances of H^(+) and OH^(-) ions at infinite diluton are 200 S cm^(2)mol^(-1) and 250" S "cm^(2)mol^(-1) respectively. Molar conductance of water is

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`3.8xx10^(-7)" S "cm^(2)MOL^(-1)`
`5.7xx10^(-7)" S "cm^(2)mol^(-1)`
`9.5xx10^(-7)" S "cm^(2)mol^(-1)`
`1.045xx10^(-6)" S "cm^(2)mol^(-1)`

Solution :Degree of dissociation `(alpha)=(wedge_(m)^(c))/(wedge_(m)^(0))`
`wedge_(m)^(c)=alphaxxwedge_(m)^(0)=(1.9xx10^(-9))xx(200+350)" S "cm^(2)mol^(-1)`
`=1.9xx550xx10^(-9)=1.045xx10^(-6)" S "cm^(2)mol^(-1)`
47.

Degree of dissociation of 0.1 N CH_(3)COOH is (Dissociation constant = 1 xx 10^(-5))

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`10^(-5)`
`10^(-4)`
`10^(-3)`
`10^(-2)`

SOLUTION :Degree of dissociation `alpha = ?`
NORMALITY of solution `= 0.1 N = (1)/(10) N`
Dissociation constant `K = 1 xx 10^(-5)`
`K = C alpha^(2), alpha = sqrt((K)/(C)) = sqrt((1 xx 10^(-5))/(0.1)) , alpha = 1 xx 10^(-2)`.
48.

Degree of dissociation of pure water is 2.8 times 10^(-9), Molar ionic conductance of H^(+) and OH^(-) ions at infinite dilution are 300 and 350 S cm^(2) respectively. The molar conductance of water is _________.

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`1.5 times 10^(-6)S" "CM^(2)MOL^(-1)`
`2.8 times 10^(-7)S" "cm^(2)mol^(-1)`
`1.82 times 10^(-6)S" "cm^(2)mol^(-1)`
`3.8 times 10^(-6)S" "cm^(2)mol^(-1)`

Solution :`alpha=WEDGE^(C)/wedge^(@)`
`wedge^(C)=alpha times wedge^(@)`
`""=2.8 times 10^(-19) times (300+350)`
`""=1.820 times 10^(-6)`
49.

Degree of dissociation alpha is

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`ALPHA=(K_a)/(C )`
`alpha=(C^2)/(K_a)`
`alpha=sqrt((K_a)/(C ))`
`alpha=sqrt((C )/(K_a))`

ANSWER :C
50.

Degree of dissociation (alpha) alpha are the number of moles which are dissociating from 1 mole of given reactants and gas density measurements can be used to determine the degree of dissociatin. Let us take a general case where one molecule of a substance A splits up into n molecules of A(g) on heating i.e., A_(n)(g)hArrnA(g) t=0a t =t_(eq)a-x nx ""alpha=x/aimpliesx=a alpha a-a alpha n a alpha Total number of Moles =a-a alpha +n a alpha =[1+(n-1)alpha]a Observed molecular weight of molar mass of the mixture M_("mixture")=(M_(A_(n)))/([1+(n-1)alpha]),M_(A_(n))= Molar mass of A_(n) If the t otal mass of the mixture in question (1) is 300 gm, then moles of C(g) present are

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`1/4`
`4/3`
`3/4`
`1/2`

ANSWER :C