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NEET Chemistry MCQ
Alcohols, Phenols And Ethers Mcq Neet Chemistry
Quiz 2
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Q.1
ion Q29) The major product of the following reaction is ..[ IIT 2008]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Nucleophilic substitution of an alkyl halide is easier as compared to that of an aryl halide PhS- is a strong nucleophile and dimethyl formamide is highly polar aprotic solvent. These reagent favour SN2 reactions at 2° benzylic carbon In SN2 reaction, the major product formed is inversion product. The option (a) is correct Answer: (a)
Q.2
Bottle containing C6H5I and C6H5CH2I lost their original labels. They were labelled A and B for testing. A and B were separately taken in test tube and boiled with NaOH solution. The end solution in each tube was made acidic with dilute HNO3 and some AgNO3 solution was added. Substance B gave yellow precipitate. Which one of the following statements is true for this experiment? [ AIEEE 2003]
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a) A and C6H5CH2I
0%
b) B and C6H5I
0%
c) Addition of HNO3 was unnecessary
0%
d) A was C6H5I
Explanation
I) C6H5I + NaOH → C6H5ONa + HNO3/H+ → C6H5OH + AgNO3→ NO yellow ppt II) C6H5CH2I + NaOH → C6H5 CH2ONa+ HNO3/H+ → C6H5OH + AgNO3 → yellow ppt. Since benzyl iodide give yellow ppt. hence this is compound B and A was phenyl iodide (C6H5I) Answer: (d)
Q.3
The compound formed on heating chlorobenzene with chloral in presence of concentrated sulphuric acid, is [ AIEEE 2004]
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a) freon
0%
b) DDT
0%
c) gammexene
0%
d) hexachloroethane
Explanation
DDT is prepared by heating chlorobenzene and chloral with concentrated sulphuric acid. Answer:(b)
Q.4
Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of [ AIEEE 2005]
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a) steric hindrance
0%
b) inductive effect
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c) instability
0%
d) insolubility
Explanation
Answer: (a)
Q.5
Phenyl magnesium bromide reacts with methanol to give [ AIEE 2006]
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a) a mixture of toluene and Mg(OH)Br
0%
b) a mixture of phenol and Mg(Me)Br
0%
c) a mixture of anisol and Mg(OH)Br
0%
d) a mixture of benzene and Mg(OMe)Br
Explanation
CH3OH + C6H5MgBr → CH3O .MgBr + C6H6 Answer:(d)
Q.6
benzene (C6H5F) can be synthesized in the laboratory..[ AIEEE 2006]
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a) by direct fluorination of benzene with F2 gas
0%
b) by reacting bromobenzene with NaF solution
0%
c) by heating phenol with HF and KF
0%
d) From aniline by diazotisation followed by heating the diazonium salt with HBF4
Explanation
Answer: (d)
Q.7
Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces..[ AIEEE 2006]
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a) 1-phenylcyclopentene
0%
b) 3-phenylcyclopentene
0%
c) 4-phenylcyclopentene
0%
d) 2-phenylcyclopentene
Explanation
The reaction is dehydrogenation Answer: (a)
Q.8
ion Q38) The structure of the major product formed in the following reaction ... [ AIEEE 2006]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Nuclear substitution will not take place Answer: (b)
Q.9
Which of the following is the correct order of decreasing SN2 reactivity [ AIEEE 2007]
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a) R2CHX > R3CX > RCH2X
0%
b) RCHX > R3CX > R2CHX
0%
c) RCH2X > R2CHX >R3CX
0%
d) R3CX >R2CHX >RCH2X
Explanation
In SN2 mechanism transition state is pentavalent. For bulky alkyl group it will have sterical hindrance and smaller alkyl groups favours the SN2 mechanism. So the decreasing order of reactivity of alkyl halides is RCH2X > R2CHX >R3CX Answer:(c)
Q.10
The organic chloro compound, which shows complete sterochemical inversion during a SN2 reaction is ..[ AIEEE 2008]
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a) (C2H5)2CHCl
0%
b) (CH3)3CCl
0%
c) (CH3)2CHCl
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d) CH3Cl
Explanation
SN2 reaction is shown to maximum extent by primary halides. The only primary halide given is CH3Cl Answer: (d)
Q.11
Ethyl alcohol is heated with conc. H2SO4 the product formed is .. [ IIT 1980]
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a)
0%
b) C2H6
0%
c) C2H4
0%
d) C2H2
Explanation
Answer: (c)
Q.12
Which of the following is basic [ IIT 1980]
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a) CH3CH2OH
0%
b) OHCH2CH2OH
0%
c) H-O-O-H
0%
d)
Explanation
Alcohol can act as both an acid and a base, similar to water. Alcohols are slightly weaker acids than water but still react with strong bases such as sodium hydride and metals such as sodium. However, the oxygen atom makes alcohol a weaker base in the presence of strong acid-like sulfuric acid Answer: (a)
Q.13
The compound which reacts fastest with Lucas reagent at room temperature is [ IIT 1981]
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a) butan-1-ol
0%
b) butan-2-ol
0%
c) 2-methylpropan-1-ol
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d) 2-methylpropan-2-ol
Explanation
The order of Lucas reagent is tert > sec. > pri. Answer:(d)
Q.14
A compound that gives a positive iodoform test is [ IIT 1982]
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a) 1-pentanol
0%
b) 2-pentanone
0%
c) 3-pentanone
0%
d) pentanal
Explanation
Compound having -CO-CH3 gives iodoform test formula for 2-penttenone is Answer: (b)
Q.15
Diethyl ether on heating with concentrated HI gives two moles of ..[ IIT 1983]
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a) ethanol
0%
b) iodoform
0%
c) ethyl iodide
0%
d) methyl iodide
Explanation
C2H5OC2H5 + 2HI → 2C2H5I + H2O Answer: (c)
Q.16
n industrial method of preparation of methanol is .. [ IIT 1984]
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a) catalytic reduction of carbon monoxide in presence of ZnO-Cr2O3
0%
b) by reacting methane with steam at 900°C with nickel catalyst.
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c) by reducing formaldehyde with lithium aluminium hydride
0%
d) by reacting formaldehyde with aqueous sodium hydroxide solution
Explanation
Answer: (a)
Q.17
When phenol is treated with excess bromine water, it gives [ IIT1984]
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a) m-bromophenol
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b) o- and p- bromophenol
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c) 2,4- dibromophenol
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d) 2,4,6 - tribromophenol
Explanation
Answer:(d)
Q.18
HBr reacts fastest with [ IIT 1986]
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a) 2-methylpropan-2-ol
0%
b) propan-1-ol
0%
c) propan-2-ol
0%
d) 2-methylpropan-1-ol
Explanation
Reactions involving cleavage of carbon-oxygen bond, (C-OH) follows the following order tertiary > secondary > primary Answer: (a)
Q.19
Which of the following compounds is oxidised to prepare methyl ethyl ketone? [ IIT 1987]
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a) 2-propanol
0%
b) 1-nutanol
0%
c) 2-butanol
0%
d) t-butyl alcohol
Explanation
Secondary alcohol oxidise to produce ketone Answer: (c)
Q.20
Phenol reacts with bromine in carbon disulphate at low temperature to give [ IIT 1988]
0%
a) m-bromphenol
0%
b) o- and p- bromophenol
0%
c) p- bromophenol
0%
d) 2,4,6 - tribromophenol
Explanation
Note in absence of CS2, polyhalogination in o- and p-positions takes place Answer: (b)
Q.21
Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives [ IIT 1990]
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a) o-cresol
0%
b) p-cresol
0%
c) 2,4-dihydrozytoluene
0%
d) benzoic acid
Explanation
Answer:(d)
Q.22
When phenol is reacted with CHCl3 and NaOH followed by acidification salicyladehyde is obtained. Which of the following species are involved in the above mentioned reaction as intermediate? [ IIT 1995]
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a)
0%
b)
0%
c)
0%
d)
Explanation
Riemer-Tieman reaction involves electrophilic substitution of the highly reactive phenoxide ring . HCCl3 + OH- → H2O + -:CCl3 -:CCl3 → Cl- + :CCl2 Note only C has a sextet of electrons Answer: (d)
Q.23
ion Q53) The order of reactivity of the following alcohol towards conc.HCl is [ IIT 1997]
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a) I > II > III > IV
0%
b) I > III > II > IV
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c) IV > III > II > I
0%
d) IV < III < II < I
Explanation
The order of reactivity depends on the stability of the carbocation formed We know that Phenoxy ion is more stable due to resonance and electron withdrawing group F is near to Carbon attached to OH group in structure (I) thus order of increasing stability will be I < II < III < IV Hence order of reactivity of the alcohols will be option "c" Answer: (c)
Q.24
The compound that will react most readily with NaOH to form methanol is [ IIT 2001]
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a) (CH3)4N+ I-
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b) CH3OCH3
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c) (CH3)3S+ I-
0%
d) (CH3)3CCl
Explanation
Compound (CH3)4N+ I- is most reactive due to i) better leaving group , I- and ii) due to the fact that methyl group, with positive N, is more electron deficient. Hence this group is more reactive towards nucleophile OH- Answer: (a)
Q.25
1-propanol and 2-propanol can be best distinguished by [ IIT 2001]
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a) oxidation with alkaline KMnO4 followed by reaction with Fehling solution
0%
b) oxidation with acidic dichromate followed by reaction with Fehling solution
0%
c) oxidation by heating with copper followed by reaction with Fehling solution
0%
d) oxidation with concentrated H2SO4 followed by reaction with Fehling solution
Explanation
Fehling solution is weak oxidising agent which can oxidise aldehyde but not ketone. Primary alcohols undergoes oxidation with alkaline KMnO4, acidic dichromate and Conc. H2SO4 to give acid, whereas with Cu they give aldehydes. Answer:(c)
Q.26
ion Q56) [IIT 2003]
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a) C6H5OC2H5
0%
b) C2H5OC2H5
0%
c) C6H5OC6H5
0%
d) C6H5I
Explanation
This reaction is an example of Williamson's synthesis . C2H5O- will abstract proton from phenol converting the latter into phenoxide ion. This would then make nucleophilic attack on the methylene carbon of alkyl iodide forming C6H5OC2H5. But if C2H5O- is in excess the C2H5OC2H5 will be form , as C2H5O- is better nucleophile than phenoxide ( C6H5O-) ion because of negative charge is localised on oxygen on C2H5O- while phenoxide is stabilized due to resonance so, it is C2H5O- ion that would make nucleophilic attack at ethyl iodide to give diethyl ether ( Williamson's synthesis) Answer: (b)
Q.27
The product of acid catalyzed hydration of 2-phenylpropene is [ IIT 2004]
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a) 3-phenyl-2-propanol
0%
b) 1-phenyl-2-propanol
0%
c) 2-phenyl-2-propanol
0%
d) 2-phenyl-1-propanol
Explanation
Addition of water to 3-phenylpropene follows Markownikov's rule Answer: (c)
Q.28
The best method to prepare cyclohexene from cyclohexanol is by using [ IIT 2005]
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a) conc. HCl + ZnCl2
0%
b) conc. H3PO4
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c) HBr
0%
d) Conc. HCl
Explanation
conc. HCl , Hbr and ZnCl2 are all nucleophiles, thus convert alcohol to alyl halides. However conc.H3PO4 is good dehydrating agent which converts an alcohol to alkene Answer: (b)
Q.29
The reaction : C2H5OH + SOCl2 → C2H5Cl + SO2 + HCl[ in presence of pyridine] is known as [ AIIMS 2002]
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a) kharasch effect
0%
b) Williamson's synthesis
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c) Darzen's procedure
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d) Hunsdiecker reaction
Explanation
Alkyl halides can be prepared by treating alcohol with SOCl2. This reaction is known as Darzen's procedure. Answer: (c)
Q.30
The increasing order of boiling points of the below mentioned alcohol is I) 1,2-dihydroxbenzene II) 1,3-dihydroxybenzene III) 1,4-dihydroxybenzene IV) Hydroxybenzene [IIT 2006]
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a) I < II < IV < III
0%
b) I < II < III < IV
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c) IV < II < I < III
0%
d) IV < I < II < III
Explanation
Among the given compounds, hydroxybenzene (IV) has least molar mass and therefore posses least boiling point. Among the three isomeric dihydroxybenzenes 1,2-dihydroxybenzene(I) forms intramolecular H-bonding with the result it will not form H-bonding leading to lowest boiling point. On the other hand 1,3-dihdroxybenzene (II) and 1,4-dihydroxybenzene(II) form intermolecular hydrogen bond leading to higher boiling point. Intermolecular bonding is stronger in the p-isomer than the m-isomer hence former has highest boiling point. Thus the correct option is "d" Answer:(d)
0 h : 0 m : 1 s
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