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CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 14 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Alcohols, Phenols And Ethers
Quiz 14
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.
Picric acid is $$ (at 25^o C) $$
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A white solid
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A colourless liquid
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A gas
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A bright yellow solid
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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$$ CH_3CH(OH)CH_3 $$
Which of the following statement is correct?
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Phenol is less acidic than ethyl alcohol
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Phenol is more acidic than ethyl alcohol
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Phenol is more acidic than carboxylic acid
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Phenol is more acidic than carbonic acid
Explanation
In phenol, hydroxyl group directly attach to $$sp^2$$ hybridised carbon which withdraw the electron pair of oxygen to wards itself. Due to this, electron density on oxygen atom decreases and icreases the polarity of $$O-H$$ bond, which leads to the formation of phenoxide ion.
In phenoxide ion, negative charge is delocalised, which makes phenoxide ion stable. But in alkoxide ion, the negtive charge is localised on oxygen only.
Thus, phenol is more acidic than alcohol.
Hence, option $$B$$ is correct.
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
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.
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
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.
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 $$A$$.
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.
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.
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.
$$\underset{\text{Propylene}}{CH_2 = CH - CH_3 + H_2O} \xrightarrow[]{H_2SO_4} \underset{\text{Isopropyl alcohol}}{CH_3 - {\overset{OH}{\overset{|}{C}}H - CH_3}}$$
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.
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.
$$\underset{\text{2 Methyl-2-hydroxypropane}}{CH_3 - \overset{CH_3}{\overset{|}{\underset{OH}{\underset{|}{C}}}}} \!\!\!\!\!\!\!\!\!\! - CH_3 \xrightarrow[\text{Dehydration}]{H_2SO_4} \underset{\text{Ethane}}{CH_3 - \overset{CH_3}{\overset{|}{C}} = CH_2 + H_2O}$$
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
$$\underset{Ethanol}{CH_{3}-CH_{2}OH}+\underset{conc.}{H_{2}SO_{4}}\xrightarrow{110^{o}C}\underset{\text{Ethyl hydrogensulphate}}{CH_{3}CH_{2}HSO_{4}}+H_2O$$
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 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.
Given :
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Explanation
The order of acidic strength
$$(-COOH)> $$ phenol with $$EWG(-{NO}_{2}gp.)>$$ phenol$$>(C\equiv C)$$ (sp character)
Two moles o base $$\overset { \ominus }{ N } {H}_{2})$$ would abstract 2 mol of most acidic $${ H }^{ \oplus }$$ out of the four gtoups ie., from $$(COOH)$$ and phenol with $$EWG(-{NO}_{2}gp.)$$.
Hence, after the abstraction of $$EWG(-{NO}_{2}gp.)$$ from $$(COOH)$$ and phenolic $$(OH)$$ with $$({NO}_{2})$$ group at m-position, the structure of the product would be $$(A).$$
Which of the following statements is/are correct?
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Phenol is more easily oxidised than benzene.
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$$Benzaldehyde$$ and $$PhCl$$ can be distinguished by $$NaHSO_{3}$$
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$$p-Cresol$$ and benzoic acid can be distinguished by $$NaOH$$
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Phenol and benzoic acid can be distinguished by adding fry ice aqueous $$NaOH$$ solution.
Explanation
$$a.$$ The $$\bar{e}-donating$$ effect of $$(OH)$$ makes the ring of phenol very $$\bar{e}-rich$$, enabling it to readily donate $$\bar{e}'s$$ to oxidising agents.
$$b.$$ Benzaldehyde reacts with $$NaHSO_3$$ but $$PhCl$$ does not.
$$c.$$ Both react with NaOH to form sodium salts.
$$d.$$ Dry ice $$(CO_2)$$ reacts with aq. $$NaOH$$ to form $$NaHCO_3$$, which reacts with acid to form $$PhCOONa+CO_2+H_2O$$.
Hence, Options "A", "B" & "D" are correct answers.
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason is incorrect
Explanation
When phenol is treated with chlorine in presence of iron, chlorination takes place at ortho and/or para positions. In aqueous medium, tri-substituted product is obtained and in organic medium, mono-substituted product is obtained.
Hence, both Assertion and Reason are incorrect.
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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Practice Class 12 Medical Chemistry Quiz Questions and Answers
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