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CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 14 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Alcohols,Phenols And Ethers
Quiz 14
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: $$ CH_3-O^+H-CH_3 \rightarrow CH_3-O-CH_3 + H^+ $$
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 $$D$$.
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
When reaction undergoes dehydration reaction in presence of concentrated $$H_2SO_4$$ then what will be the major product?
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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.
$$HO^-+CH_3-N^+(CH_3)_3\rightarrow CH_3OH+(CH_3)_3N$$
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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$$I, IV$$
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.
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.
An organic compound $$(X)$$ with molecular formula $$C_{7}H_{8}O$$ is insoluble in aqueous $$NaHCO_{3}$$ but dissolves in $$NaOH$$. When treated with bromine water $$(X)$$ rapidly gives $$(Y)$$, $$C_{7}H_{5}OBr_{3}$$. The compound $$(X)$$ and $$(Y)$$ respectively are
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Benzyl alcohol and $$2,4,6-tribromo-3-methoxy$$ benzene
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Benzyl alcohol and $$2,4,6-tribromo-3-methoxy$$ phenol
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$$o-cresol$$ and $$3,4,5-tribromo-2-methyl$$ phenol
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Methoxy benzene and $$2,4,6-tribromo-3-methoxy$$ benzene
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$$m-cresol$$ and $$2,4,6-tribromo-3-methyl$$ phenol
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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$$H_2C=CH-CH_3\overset{Neutral KMnO_4}{\rightarrow}$$
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.
$$IUPAC$$ 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
$$IUPAC$$ 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 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 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 $$(C_2H_5)_2CHCH_2OH$$ 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
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
The compound which gives the most stable carbonium on dehydrogenation
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$$CH_3-\underset{CH_3}{\underset{|}{CH}}-CH_2OH$$
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$$CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-OH$$
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$$CH_3-CH_2-CH_2-CH_2OH$$
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$$CH_3-\underset{CH_3}{\underset{|}{CH}}-CH_2-CH_3$$
Explanation
Tert-butanol loses an $$−OH$$ group to give most stable carbonium ion. The positive charge on carbon atom is stabilized by $$+I$$ inductive effect of 3 methyl groups.
An organic compound X on treatment with acidified $$ K_2Cr_2O_7 $$ gives a compound Y which reacts with $$ I_2 $$ and sodium carbonate to form tri-iodomethane. The compound X is
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$$ CH_3OH $$
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$$ CH_3-CO-CH_3 $$
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$$ CH_3CHO $$
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$$ CH_3CH(OH)CH_3 $$
The reagent used for the dehydration of an alcohol is
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Phosphorus pentachloride
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Calcium chloride
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Aluminium oxide
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Sodium chloride
Sodium ethoxide is a specific reagent for
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Dehydration
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Dehydrogenation
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Dehydrohalogenation
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Dehalogenation
Explanation
$$CH_3 - CH_2 - CH_2 - Br \xrightarrow[\text{Dehydrohalogenation}]{C_2H_5ONa} CH_3 - CH = CH_2 + HBr$$
Dehydrogenation of gives
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Acetone
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Acetaldehyde
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Acetic acid
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Acetylene
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)
$$ 3CH_3CH=CH_2 + \frac {1}{2}B_2H_6 \xrightarrow [ether]{Dry}(CH_3CH_2CH_3)_3 B (CH_3CH_2CH_3)_3B \xrightarrow []{H_2O_2} 3CH_3CH_2CH_2-OH $$
Which of the following statements is correct regarding case of dehydration in alcohols?
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Primary > Secondary
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Secondary > Tertiary
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Tertiary > Primary
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None of these
Explanation
Dehydration of alcohol forms carbocation as intermediate. More stable is the carbocation, ease of reaction is more.
The order of stability of carbocation is:
$$tertiary>secondary>primary$$
So, the dehydration of tertiary alcohol is faster than the dehydration of the secondary alcohol and
the dehydration of the secondary alcohol is faster than the dehydration of primary alcohol.
Hence, the option $$C$$ is correct.
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
Phenol at $$ 25^0 c $$ is
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A white crystalline solid
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A transparent liquid
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A gad
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yellow solution
Explanation
Phenol is an aromatic organic compound which is a white crystalline solid that is volatile at $$25^0C$$.
So, the answer is option $$A$$.
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.
The best method to prepare cyclohexene from cyclohexanol is by using
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Conc. HCl + ZnCl
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Conc. $$H_3PO_4$$
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HBr
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Conc. HCl
Explanation
Because conc. $$ H_3PO_4 $$ acts as dehydrating agent.
Option C is correct.
Ether which is liquid at room temperature is
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$$ C_2H_5OCH_3 $$
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$$ CH_3OCH_3 $$
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$$ C_2H_5OC_2H_5 $$
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None of these
Explanation
$$ CH_3OCH_3 $$ and $$ C_2H_5OCH_3 $$ are gases at room temperature while $$ C_2H_5OC_2H_5 $$ (b.p 308K) is low boiling liquid.
Option B is correct.
The compound which gives the most stable carbonium ion on dehydration is
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Explanation
Tertiary carbonium ion is the most stable and it will be given by dehydration of tertiary alcohol.
Option B is correct.
In which of the following reaction, phenol or sodium phenoxide is not formed:
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$$ C_6H_5N_2^+Cl^- + alc. KOH \rightarrow $$
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$$ C_6H_5OCl + NaOH \rightarrow $$
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$$ C_6H_5OCl + aq. NaOH \rightarrow $$
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$$ C_6H_5N_2^{+}Cl^- \ \xrightarrow [ \Delta ]{ H_{ 2 }O } $$
The reaction shown is the example of:
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Sulphonation
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Dehydration
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Alkylation
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Decomposition
Explanation
The reaction shown is the example of dehydration as a water molecule is removed from the reactant molecule to give the product.
$$\underset{\text{2 Methyl-2-hydroxypropane}}{CH_3 - \overset{CH_3}{\overset{|}{\underset{OH}{\underset{|}{C}}}}} \!\!\!\!\!\!\!\!\!\! - CH_3 \xrightarrow[\text{Dehydration}]{H_2SO_4} \underset{\text{Ethane}}{CH_3 - \overset{CH_3}{\overset{|}{C}} = CH_2 + H_2O}$$
Isopropyl alcohol is obtained by reacting which of the following alkenes with conc. $$H_{2}SO_{4}$$ and $$H_{2}O$$
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Ethylene
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Propylene
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2-methyl propene
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Isoprene
Explanation
$$CH_{3}-CH=CH_{2}+H_{2}O\xrightarrow[Markownikoffs rule]{Conc.H_{2}SO_{4}}CH_{3}-\underset{\underset{lIsopropyl alcohol}{OH}}{\underset{|}{CH}}-CH_{3}$$
Maximum solubility of alcohol in water is due to
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Covalent bond
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Ionic bond
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H-bond with $$ H_2O $$
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None of the above
Explanation
Alcohol is soluble in water due to H-bonding.
Option C is correct.
Which will dehydrate easily?
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3-methyl-2-butanol
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Ethyl alcohol
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2-methyl propane-2-ol
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2-methyl butanol
Explanation
The more stable carbocation is generated, thus more easily it will be dehydrated.
The order of stability of carbocation is:
$$tertiary>secondary>primary$$
Primary alcohol produces primary carbocation, secondary alcohol produces secondary carbocation and tertiary alcohol gives tertiary carbocation.
So, tertiary alcohol will dehydrate easily.
Hence, option $$C$$ is correct.
Which of the above paths is/are feasible for the preparation of ether(E)?
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Path $$I$$ is feasible.
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Path $$II$$ is feasible.
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Both paths are feasible.
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None is feasible.
Explanation
These reactions are also known as Wiliiamson's ether synthesis and are a commercial method for the preparation of ethers.
The product of acid catalyzed hydration of 2-phenyl propene is
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3-phenyl-2-propanol
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1-phenyl-2-propanol
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2-phenyl-2-propanol
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2-phenyl-1-propanol
Explanation
The major product obtained on acid - catalysed hydration of 2-phenylpropene is 2-Phenylpropan-2-ol. A molecule of water is added to $$C=C$$ double bond.
Propylene on hydrolysis with sulphuric acid forms
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n-propyl alcohol
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Isopropyl alcohol
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Ethyl alcohol
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Butyl alcohol
Explanation
Propylene on hydrolysis with sulphuric acid forms isopropyl alcohol.
$$\underset{\text{Propylene}}{CH_2 = CH - CH_3 + H_2O} \xrightarrow[]{H_2SO_4} \underset{\text{Isopropyl alcohol}}{CH_3 - {\overset{OH}{\overset{|}{C}}H - CH_3}}$$
The only alcohol that can be prepared by the indirect hydration of alkene is
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Ethyl alcohol
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Propyl alcohol
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Isobutyl alcohol
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Methyl alcohol
Explanation
Except ethyl alcohols, no other primary alcohol can be prepared by this method as the addition of $$H_2SO_4$$ follows Markownikoff's rule. Generally secondary and tertiary alcohols are obtained.
Ethyl hydrogen sulphate is obtained by the reaction of $$H_{2}SO_{4}$$ on
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Ethylene
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Ethane
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Ethyl chloride
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Ethanol
Explanation
$$\underset{Ethanol}{CH_{3}-CH_{2}OH}+\underset{conc.}{H_{2}SO_{4}}\xrightarrow{110^{o}C}\underset{\text{Ethyl hydrogensulphate}}{CH_{3}CH_{2}HSO_{4}}+H_2O$$
The products $$(A)$$ and $$(B)$$ are:
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Explanation
Acidic cleavage of ether takes place and products formed are $$PhCH_2I$$ and 4-hydroxyphenol. The latter product does not add more $$I$$ even in the presence of excess $$HI$$ as the reaction is not feasible.
Hence correct option is (B).
Which of the following is the best synthesis of the ether $$(A)$$ shown below:
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$$Ph - O - PH \underset{\text{(Dinitration)}}{\xrightarrow[+H_2SO_4]{HNO_3}}$$
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Explanation
The best reaction for the synthesis of such ether is Williamson's ether synthesis. $$PhONa$$ is more suitable $$PhO^-$$ attacks the C-atom where $$F$$ is attached leading to the desired product.
Hence option (B) is correct.
Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields.
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o-Cresol
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m-Cresol
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2, 4-Dihydroxytoluene
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Benzyl alcohol
Explanation
$$\text{Halogenation in the presence of light follows free radical pathway and thus will react with the alkyl group to give haloalkyl. }$$
Thus formed haloalkyl in the presence of alkaline medium will undergo substitution reaction and will form alcohol by replacing halogen atom in the haloalkyl group.
The reaction with toluene can be represented as:
$$\underset{toluene}{C_6H_5CH_3}\xrightarrow{Cl_2,hν}C_6H_5CH_2Cl\xrightarrow{aq.NaOH}\underset{benzyl alcohol}{C_6H_5CH_2OH}$$
Thus, the monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields benzyl alcohol.
Which of the following reactions will yield phenol?
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