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CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 13 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Alcohols, Phenols And Ethers
Quiz 13
Which of the following is unsymmetrical ether?
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$$C_6H_5-O-C_6H_5$$
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$$CH_3-O-C_2H_5$$
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$$C_2H_5-C_2H_5$$
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$$CH_3-O-CH_3$$
Phenol on treatment with $$CO_{2}$$ in the presence of $$NaOH$$ followed by acidification produces compound $$X$$ as the major product.$$X$$ on treatment with $$(CH_{3}CO)_{2}O)$$ in the presence of catalytic amount of $$H_{2}SO_{4}$$ produces :
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Which of the following will react most readily with $$HI$$?
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$$CH_3OH$$
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$$(CH_3)_3COH$$
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$$CH_3CH_2OH$$
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$$(CH_3)_2CHOH$$
Identify the compound '$$X$$'.
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Toluidine
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Benzamide
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Para-Cresol
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Oleic acid
Explanation
Both Para-Cresol and Oleic acid form salt with $$10\% NaOH$$, but Para-Cresol salt is soluble whereas Oleic acid salt is insoluble due to very long unsaturated carbon chain.
Which of the following reactions are feasible ?
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$$I,II\,and\, IV$$
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$$I,II\,and\, III$$
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Only $$III$$
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Only $$I$$
Which of these dehydrates most readily when reacts with conc.$${H}_{2}{PO}_{4}$$
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An organic compound $$'X'$$ showing the following solubility profile is:
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m-Cresol
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Oleic acid
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o-Toluidine
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Benzamide
Explanation
Solution:- (A) m-Cresol
*Oleic acid is also soluble in $$NaHCO_3$$
*o-toluidine is not soluble in $$NaOH$$ as well as $$NaHCO_3$$
*Benzamide is also not soluble in $$NaOH$$ & $$NaHCO_3$$
IUPAC name of compound 'X' is
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Cyclohex -1- en -3-ol
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Cyclohex -2-en-1-ol
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Cyclohex-5-en-2-ol
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Cyclohex-5-en-1-ol
The IUPAC name of
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1-ethoxypropane-2-ol
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3-ethoxypropane-2-ol
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1-ethoxy-2-hydroxy propane
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None of these
What will be the major product when m-cresol is reacted with propargyl bromide. $$(HC\equiv C-CH_2Br)$$ in presence of $$K_2CO_3$$ in acetone?
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Explanation
Hence option D is correct.
When vapours of methyl alcohol are passed under pressure over heated $$Al_2O_3$$ at $$250^o C$$, the product obtained is:
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dimethyl ether
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ethyl methyl ether
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diethyl ether
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none of these
Explanation
When vapours of alcohol are passed under pressure over heated $$Al_2O_3$$ at $$250^0C$$,
the product obtained is
dimethyl ether
$$CH_3OH+HO-CH_3\overset{Al_2O_3}{\rightarrow}CH_3-O-CH_3+H_2O$$
Which is dehydrated to a maximum extent using conc.$$ H_2 SO_4 $$?
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Explanation
Extent of dehydration using conc. $$\mathrm{H_2SO_4}$$ depends upon the stability of intermediate carbonium ion formed in the reaction.
Among the given options, Intermediate carbonium ion formed in the above reaction has highest stability.
Hence, Option "D" is the correct answer.
State whether the following statement is True or False.
Primary alcohols are dehydrated easily than secondary and teritary alcohols.
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True
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False
Explanation
Secondary and tertiary alcohols are dehydrated easily than primary alcohols.
Because in dehydration intermediate carbocation forms, so more stable carbocation more is the reactivity.
Order of dehydration is: Tertiary alcohol > secondary alcohol >primary alcohol
Heating together of sodium ethoxide and ethyl iodide will give :
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ethyl alochol
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acetadehyde
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diethyl ether
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acetic acid
Dehydration of alcohol involves:
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free radical
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carbocation
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carbanion
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carbene
Explanation
A mechanism for dehydration of alcohol is illustrated as above.
From above mechanism, we can see that it involves carbocation in its mechanism.
Hence, Option "B" is the correct answer.
Which of the following pairs of reagents will not form ether?
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$$C_2H_5Br + CH_3COONa$$
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$$C_2H_5Br + C_2H_5ONa$$
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$$C_2H_5 Br + (CH_3 )_2 CHONa$$
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$$(CH_3)_2CHBr + CH_3ONa $$
Explanation
The reaction $$\mathrm{C_2H_5Br + CH_3COONa}$$ results in the formation of $$\mathrm{CH_3COOCH_2CH_3}$$
Reaction of $$\mathrm{RBr + R'ONa}$$ results in the formation of Ether $$\mathrm{(ROR')}$$
Hence, Option "A" is the correct answer
Main product(s) of the reaction is/are:
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$$ CH_3COOH+ H_2 $$
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$$ CH_3CH_2COOH $$
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$$ CH_3CH(OH)CH_3 $$
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$$ CH_3CH_2OH $$
State, whether the following statements are True or False:
2-Aminoethanol is formed when ethylene oxide combines with ammonia.
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True
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False
State, whether the following statements are True or False:
Phenol gives salicyclic acid on treatment with $$CHCl_{3}/ KOH$$.
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True
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False
Explanation
Phenol gives salicylaldehyde(o-Hydroxy benzaldehyde) on treatment with $$CHCl_3/KOH$$
State, whether the following statements are True or False:
Ethers are more polar than the isomeric alcohols.
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True
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False
Which of the following reactions will give good yield of ether?
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Dehydration of methyl alcohol gives:
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Methane
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Ethane
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Methyl hydrogen sulphate
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Dimethyl ether
Explanation
Dehydration of methyl alcohol gives Dimethyl ether in the presence of concentration $$ H_2SO_4 $$ and at $$413$$k temperature. The reaction follows the nucleophilic bimolecular reaction($$ SN_2 $$) mechanism. Mechanism of reaction is given below:
Step-1: $$ CH_3-OH + H^+ \rightarrow CH_3-O^+H_2 $$
Step-2: $$ CH_3-OH + CH_3-O^+H_2 \rightarrow CH_3-O^+H-CH_3 + H_2O$$
Step-3: $$ CH_3-O^+H-CH_3 \rightarrow CH_3-O-CH_3 + H^+ $$
Ortho-nitropenol is less soluble in water than p and m-nitrophenols becauses:
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O-nitrophenol is more volatile steam than those of m- and p-isomers
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O-nitrophenol shows intramolecular H-bonding
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O-nitrophenol shows intermolecular H-bonding
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Melting point of o-nitrophenol is lower than those m- and p-isomer
Explanation
In o-nitrophenol, there is intramolecular hydrogen bonding, whereas in m-nitrophenol and p-nitrophenol, there is no intramolecular hydrogen bonding. o-nitrophenol has less tendency for hydrogen bonding with water than m- and p-nitrophenol.
The organic products formed in the reaction, $$C_{6}H_{5}COOCH_{3}\xrightarrow {LiAlH_{4}, H^{+}}$$ are
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$$C_{6}H_{5}CH_{2}OH$$ and $$CH_{3}OH$$
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$$C_{6}H_{5}COOH$$ and $$CH_{4}$$
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$$C_{6}H_{5}CH_{3}$$ and $$CH_{3}OH$$
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$$C_{6}H_{5}CH_{3}$$ and $$CH_{4}$$
Which of the following can work as dehydrating agent for alcohols?
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$$H_2SO_4$$
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$$Al_2O_3$$
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$$H_3PO_4$$
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All of these
Explanation
Secondary and tertiary alcohols are best dehydrated by dilute sulfuric acid. By heating an alcohol with concentrated sulfuric acid. Other dehydrating agents like phosphoric acid and anhydrous zinc chloride, aluminium oxide may also be used.
Hence, the answer is option $$D$$.
An ether is more volatile than an alcohol having the same molecular formula, This is due to:
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dipolar character of ethers
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alcohols having resonance structures
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intermolecular hydrogen bonding in ethers
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intermolecular hydrogen bonding in alcohols
Explanation
An ether is more volatile than an alcohol having same molecular formula because intermolecular hydrogen bonding in alcohols.
The $$-OH$$
group of alcohol can form intermolecular hydrogen bonds whereas $$-O$$
group of ethers cannot form hydrogen bonds.
Hydrogen bond formation is possible when $$H$$ atom is attached to electronegative $$N$$, $$O$$ or $$F$$ atom. Intermolecular hydrogen bonds leads to molecular association.
Hence correct answer is option D.
Which of the following will be most easily dehydrated in acidic conditions?
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When reaction undergoes dehydration reaction in presence of concentrated $$H_2SO_4$$ then what will be the major product?
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Anisole is the product obtained from phenol by the reaction
known as
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Coupling
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Etherification
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Oxidation
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Esterification
Explanation
Phenol reacts with alkyl halides in alkali solution to form phenyl ethers (Williamson's synthesis). The phenoxide ion is a nucleophile and will replace halogen of alkyl halide.
Tert- butyl alcohol is:
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$$2$$-methyl propan-$$2$$-ol
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$$2$$-methyl propan-$$1$$-ol
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$$3$$-methyl butan-$$1$$-ol
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$$3$$-methyl butan-$$2$$-ol
Which of the following reactions will yield propan-$$2$$-ol?
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$$H_2C=CH-CH_3+HOH\overset{H^+}{\rightarrow}$$
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$$CH_3-CHO\overset{CH_3MgBr}{\underset{HOH}{\rightarrow}}$$
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$$CH_2O\overset{(i)C_2H_5MgI}{\underset{(II)HOH}{\rightarrow}}$$
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$$H_2C=CH-CH_3\overset{Neutral KMnO_4}{\rightarrow}$$
Alcohol can be prepared by which of the following methods?
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By hydration of alkene
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By reduction of carbonyl compounds
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By reaction of primary aliphatic amines with nitrous acid
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By hydrolysis of esters
Which of the following will react with NaOH to form methanol as a major product?
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$$(CH_3)N^+I^-$$
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$$(CH_3)_3S^+I_-$$
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$$(CH_3)_3CCl$$
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$$CH_3-O-CH_3$$
Explanation
$$3^0$$ R-X will afford alkene as major product with NaOH. Due to greater electronegativity of N over S, positive charge on N will make the methyl groups more electron-deficient therefore,$$(CH_3)_4N^+I^-$$ will undergo nucleophilic substitution more readily.
$$HO^-+CH_3-N^+(CH_3)_3\rightarrow CH_3OH+(CH_3)_3N$$
Which of the following compound/s have some solubility in water?
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Explanation
After polarisation of bond gives stable aromatic systems containing charges.
All compounds have polarity so solubility is water.
In which case first has higher solubility than second?
(I) Phenol, Benzene
(II) Nitrobenzene, Phenol
(III) o-Hydroxybenzaldehyde, p-Hydroxy benzaldehyde
(IV) Ethanal, Dimethylether
(V) o-Nitrophenol, p-Nitrophenol
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only $$I$$
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$$III, V$$
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$$I, IV$$
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$$I, IV$$
$$IUPAC$$ name of the compound is
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$$4-$$ethyl$$-2-$$pentanol
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$$5-$$methyl$$-2-$$hexanol
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$$2-$$ethyl$$-2-$$pentanol
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$$3-$$methyl$$-2-$$hexanol
The IUPAC name of $$CH_3-CHBr-CH_2OH$$ is
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$$3-$$hydroxy $$-2-$$bromopropane
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$$2-$$bromopropan$$-1-$$ol
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$$2-$$bromo-$$3$$-propanol
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$$3-$$hydroxy isopropyl bromide
The $$IUPAC$$ name of $$(C_2H_5)_2CHCH_2OH$$ is
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2-ethyl-1-butanol
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2-methyl-1-pentanol
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2-ethyl-1-pentanol
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3-ethyl-1-butanol
Which of the following are correct statements?
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Both the ethers obtained by the two routes have opposite but equal optical rotation.
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One of the ether is obtained as a racemic mixture
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Step II and III both are $$S^2_N$$ reaction and both have inversion.
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Step II has inversion but step III has retention
Explanation
Option A is correct.
The product is :
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None
To remove last traces of water from alcohol, the metal used is
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Sodium
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Potassium
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Calcium
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Aluminium
Explanation
The metal which is used to remove last traces of water from alcohol is Calcium.
Hence, Option "C" is the correct answer.
$$IUPAC$$ name of the given compound is
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$$1-$$chloro $$-4-$$methyl$$-2-$$hexanol
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$$1-$$chloro $$-4-$$ethyl$$-2-$$pentanol
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$$1-$$chloro $$-4-$$methyl$$-2-$$hexanol
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$$1-$$chloro $$-2-$$hydroxy$$-4-$$methyl hexanol
$$IUPAC$$ name of the compound
$${C}H_2 = {C}H-{C}H_2-{C}H_2OH$$ is
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$$1-$$buten$$-4-$$ol
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$$3-buten-1-ol$$
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$$4-$$hydroxy-$$1-$$butene
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$$1-$$butene$$-4-ol$$
The name of $$\begin{matrix} { H }_{ 3 }C-CH-CH-{ CH }_{ 3 } \\ |\qquad | \\ { CH }_{ 3 }\qquad OH \end{matrix}$$
The $$IUPAC$$ nomenclature is:
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Butanol
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$$2-$$methyl butanol-3
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$$3-$$methyl butan 2-ol
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Pentanol
The $$IUPAC$$ name of the compound $$\begin{matrix} { CH }_{ 3 }-C={ CH }_{ 2 }{ CH }_{ 2 }OH \\ | \\ { CH }_{ 3 } \end{matrix}$$ is
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$$2-$$methyl$$-2-$$butenol
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$$2-$$methyl$$-3-$$butenol
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$$3-$$methyl$$-2-$$butenol
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$$2-$$methyl byt-$$-2-$$enol
The $$IUPAC$$ name of the compound $$\begin{matrix} \quad \quad \quad \quad \quad{ CH }_{ 3 }-{ CH }-{ { CH }_{ 2 } }-{ CH }_{ 2 }-OH \\ | \\ { CH }_{ 3 } \end{matrix}$$ is
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1-pentanol
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Pentanol
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2-methyl-4-butanol
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3-methyl-1-butanol
Propene, $$ CH_3-CH=CH_2 $$ can be converted to l-propanol by oxidation. Which set of regents among the following is ideal to effect the conversion
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Alkaline $$ KMnO_4 $$
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$$ B_2H_6 $$ and alkaline $$ H_2O_2 $$
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$$O_3 $$ /Zn dust
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$$ OsO_4/CH_4,CL_2 $$
Explanation
Hydroboration oxidation (Industrial preparation of alcohol)
$$ 3CH_3CH=CH_2 + \frac {1}{2}B_2H_6 \xrightarrow [ether]{Dry}(CH_3CH_2CH_3)_3 B (CH_3CH_2CH_3)_3B \xrightarrow []{H_2O_2} 3CH_3CH_2CH_2-OH $$
Commercially methanol is prepared by
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Reduction of CO in presence of $$ ZnO.Cr_2O_3 $$
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Methane reacts with water vapours at 900 $$^oC$$ in presence of Ni catalyst
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Reduction of HCHO by $$ LiAlH_4 $$
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Reduction of HCHO by aqueous NaOH
Explanation
$$ CO+H_{ 2 }\xrightarrow [ 573K,200atm ]{ CuO-ZnO-Cr_{ 2 }O_{ 3 } } \begin{matrix} CH_ 3OH \\ Methanol \end{matrix} $$
Methanol is prepared commercially by the reduction of Carbon monoxide in presence of $$ZnO-Cr_{ 2 }O_{ 3 }$$
Option A is correct.
The reaction, water gas $$ (CO+H_2)+H_2 $$ 673K,300 atmosphere in presence of the catalyst $$ Cr_2O_3 /ZnO $$ is used for the manufacture of
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HCHO
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HCOOH
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$$ CH_3OH $$
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$$ CH_3COOH $$
Explanation
$$ \underbrace { CO+H_2 }_{ Water \ gas } +H_2 \xrightarrow [ 672 K, 200atm]{Cr_2O_2/ZnO} CH_3OH $$
Option C
In which case methyl-t-butyl ether is formed
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$$ (C_2H_5)_3CONa+CH_3Cl $$
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$$ (CH_3)_3CONa+CH_3Cl $$
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$$ (CH_3)_3CONa+C_2H_5Cl$$
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$$ (CH_3)_3CONa+CH_3Cl$$
Explanation
$$(CH_3)_3CONa+CH_3Cl$$ are required to form Methyl-t-butyl ether.
Option B is correct.
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