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CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 10 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Alcohols,Phenols And Ethers
Quiz 10
$$A\:(C_7H_8O)+Na/\Delta\longrightarrow$$ No gaseous product
$$A+Br_2/FeBr_3\longrightarrow\:C_7H_7OB_r$$ (Two isomers)
$$A$$ can be:
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Explanation
When compound A is heated with sodium no gaseous product is obtained.
When phenol is heated with sodium, hydrogen gas is evolved.
Thus presence of phenolic OH group is ruled out. Thus options A, B and C are ruled out.
Option D (anisole) is the correct answer.
Anisole on bromination with bromine in presence of ferric bromide gives two isomers ortho and para bromo derivatives of anisole.
$$C(CH_3)_2=C(CH_3)_2\xrightarrow[H_2O]{Cl_2} A\xrightarrow[]{OH^-} B$$
The compound $$B$$ is:
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$$(CH_3)_2\underset{OH}{\underset{|}{C}-} \underset{Cl}{\underset{|}{C}} (CH_3)_2$$
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$$(CH_3)_2\underset{OH}{\underset{|}{C}-} \!\!\underset{\,\,OH}{\underset{|}{C}} \!\!(CH_3)_2$$
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none of the above
Explanation
The given reaction is base catalyzed cyclization of vicinal chlorohydrins.
Alkene reacts with chlorine in presence of water to form vicinal chlorohydrin.
This is followed by elimination of $$HCl$$ in presence of base to form epoxide.
The compound $$A$$ and $$B$$ are :
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None of the above
Explanation
Alkene is oxidized with cold alkaline potassium permanganate solution to form diol $$A$$.
Of the two hydroxyl groups, the secondary hydroxyl group is oxidized with chromium trioxide in presence of acetic acid to form hydroxy ketone $$B$$. Under these, conditions, tertiary hydroxyl group is not oxidized as it requires drastic conditions.
Identify the structure of A:
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$$CH_3 - CH_2 - CH_2 - \underset{\!\!\!OH}{\underset{\!\!\!\!\!\!|}{CH}} - CH_3$$
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$$(CH_3)_2CHCHOHCH_3$$
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$$(CH_3)_3C-CH_2-OH$$
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$$CH_3 - CH_2 - \underset{\!\!\!OH}{\underset{\!\!\!\!\!|}{CH}} - CH_2 - CH_3$$
Explanation
The structure of A in the following reaction is option B.
Which of the following reaction is most exothermic?
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$$\overset{\bullet}{C}H_3 + CH_3CH_2CH_3 \rightarrow CH_4 + CH_3 \overset{\bullet}{C}HCH_3$$
Explanation
The reactions involving the neutralization of an acid with a base are most exothermic reactions.
Thus, the neutralization of phenol with sodium hydroxide is highly exothermic.
Which of the following will be the correct product (P) for the given reaction?
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Explanation
Dehydration of alcohol involves the formation of carbocation which further undergoes rearrangement as the tertiary carbocation is more stable than secondary carbocation, which then accepts the electron from $$\pi$$ bond of cyclohexene and forms bicyclic compound.
Option C is correct.
The addition of mercuric acetate in the presence of water is called oxymercuration. The adduct obtained gives alcohol on reduction with $$NaBH_4$$ in alkaline medium. This is known as demercuration. Oxymercuration-demercuration allows Markownikoff's addition of H, OH without rearrangement. The net result is the addition of $$H_2O$$.
Based on the above information, find product A in the given reaction.
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Give the expected major product of the above reaction.
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Which of the following reagents should be used to prepare tert-butyl ethyl ether?
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$$tert$$-butyl bromide and sodium ethoxide
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$$tert$$-butyl alcohol and ethyl bromide
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$$tert$$-butyl alcohol and ethanol
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potassium $$tert$$-butoxide and ethyl bromide
Explanation
$$tert$$-butyl ethyl ether can be prepared by the reaction of potassium $$tert$$-butoxide with ethyl bromide.
The process is Williamson synthesis.
This reaction follows $$S_{N}{2}$$ mechanism.
Option D is correct.
Select schemes A, B, C out of the given below.
I. Acid catalysed hydration
II. Hydroboration oxidation
III. Oxymecruation-demercuration
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I in all cases
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I, II, III
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II, III, I
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III, I, II
What is $$A$$ ?
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$$ TsO-CH_2 - C \equiv CH$$
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$$ TsO_2-CH_2 - C \equiv CH$$
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$$ TsO-CH_2 - C - CH_3$$
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none of the above
Explanation
Catalytic p-toluenesulfonic acid $$(TsOH)$$ reacts as acid and gives elimination products with the formation of $$TsO-CH_2-C\equiv CH$$.
Which of the following is the correct method for synthesizing methyl-t-butyl ether?
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$$\displaystyle \left ( CH_{3} \right )_{3}CBr+NaOMe\rightarrow $$
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$$\displaystyle \left ( CH_{3} \right )_{3}CONa+MeBr\rightarrow $$
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$$\displaystyle \left ( CH_{3} \right )_{3}CBr+HOMe\rightarrow $$
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Both A and B
Explanation
In the synthesis of n-alkyl ter-alkyl ether by Williamson's method,
$$\displaystyle 1^{\circ}RX\:and\:3^{\circ} RONa$$ are used.
If $$\displaystyle 3^{\circ}RX$$ is used, it will undergo dehydrohalogenation to form alkene.
Hence, the correct method for synthesising methyl-t-butyl ether is:
$$\displaystyle \left ( CH_{3} \right )_{3}CONa+CH_3Br\rightarrow \left ( CH_{3} \right )_{3}COMe+NaBr$$
Which of the following are possible products in the above reaction?
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Identify $$B$$ and $$C$$ respectively in following conversion.
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None of the above
Explanation
Osmium tetraoxide is an oxidising agent oxidize alkene into diols and diols in acidic medium accept proton to form good leaving group water and to sablize carbocation, ring expansion occur.
In given reaction A is :
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Which method is useful for the synthesis of ether?
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Primary alcohols can be prepared from alkenes by:
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mercuration and demercuration of alkenes
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direct hydration of alkenes
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hydroboration of alkenes
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all of the above
Explanation
Primary alcohols can be prepared from alkene by hydroboration oxidation.
The hydroboration of propene followed by oxidation gives propanol.
The reagent used for hydroboration is diborane.
The reagent used for oxidation is alkaline hydrogen peroxide solution.
$$6CH_3CH=CH_2+B_2H_6 \rightarrow 2(CH_3CH_2CH_2)_3B$$
$$(CH_3CH_2CH_2)_3B+3H_2O_2 \xrightarrow {OH^-} 3CH_3CH_2CH_2OH+B(OH)_3$$
Reactions I and II are limited mainly to naphthalene compounds. These reactions are called, respectively:
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bucherer reaction and its reversal
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bischler-Napieralski reaction and its reversal
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birnbaum-Simonini reaction and its reversal
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borodine-Hunsdiecker reaction and its reversal
Explanation
The Bucherer reaction is the reversible conversion of a naphthylamine to a naphthol in the presence of an aqueous sulfite or bisulfite.
The reaction is given as shown above.
Which of the following statements is/are correct?
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The compounds (B) is (I).
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The compounds (B) is (II)
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The compounds (C) is (I)
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The compounds (C) is (II)
Explanation
Resonance structure of 2-naphthoxide ion is an ambident nucleophile and a better nucleophile than given structure of A because negative charge is on more electronegative oxygen atom. In the presence of polar non-protic solvent (DMF, dimethyl formamide), O-alkylation is favoured to give product II (C). In the presence of protic solvent, hydrogen bonding with the O atom occurs, thus reducing the nucleophilicity of oxygen. Therefore, C-alkylation is favoured to give product I (B).
The compound (A) is:
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Explanation
Red $$P$$ and $$HI$$ is a traditional reducing agent to convert alcohols into alkanes. $$H_3PO_3$$ form in this reduction reaction. The product $$A$$ of reaction is cyclobutandicarboxylic acid.
Which statement is correct for the conversion of $$ROH$$ to $$RX$$ by reagents $$(A)$$ $$ ( SOCl_{2},\ PBr_{3}$$ $$ PCl_{3}$$ $$ Pl_{3},\:TsCl )$$ compared to using $$HX$$?
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No rearrangement occurs with predictable stereochemistry by reagents $$(A)$$
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Reasgents $$(A)$$ are not useful for $$\displaystyle 3^{\circ}$$ alcohols, while $$HX$$ reacts easily with $$\displaystyle 3^{\circ}$$ alcohols involving no rearrangement.
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Reagents $$(A)$$ must be used in anhydrous conditions because all react vigorously with $$\displaystyle H_{2}O$$ and they produce harmful gases $$\displaystyle \left ( SO_{2},HCl,HBr,\:and\:HI \right )$$
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All of the above
Explanation
All the three statements (A), (B) and (C) are correct. Hence, (D) is the correct option.
(A) No rearrangement occurs with predictable stereochemistry by reagents $$(A)$$. However if $$HX$$ is used, the rearrangement is possible.
(B) Reagents $$(A)$$ are not useful for tertiary alcohols, while $$HX$$ reacts easily with tertiary alcohols involving no rearrangement.
(C) Reagents $$(A)$$ must be used in anhydrous conditions because all react vigorously with $$H_2O$$ and they produce harmful gases ($$SO_2$$, $$HCl$$, $$HBr$$ and $$HI$$).
For example, phosphorous trichloride reacts with water to form orthophosphorous acid and $$HCl$$.
$$PCl_3 + 3H_2O \rightarrow P(OH)_3 + 3HCl$$
Which is the best method for the conversion of (A) $$pentan-3-ol\ to\ 3-bromopentane$$ (B)?
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$$\displaystyle \left ( A \right )\xrightarrow {HBr}\left ( B \right )$$
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None of these
Which of the following reaction is feasible for the preparation of 1-propoxy-2 methyl propane?
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Williamson's synthesis of $$\displaystyle \left ( Me_{2}CHCH_{2}ONa+MeCH_{2}CH_{2}Cl \right )$$
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Williamson's synthesis of $$\displaystyle \left ( Me_{2}CHCH_{2}Cl+MeCH_{2}CH_{2}ONa \right )$$
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Alkoxymercuration-demercuration of $$\displaystyle \left ( Me_{2}C= CH_{2}+MeCH_{2}CH_{2}OH \right )$$
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Alkoxy mercuration demercuration of propene with $$\displaystyle Me_{2}CHCH_{2}OH$$
Explanation
The reactions in options A and B are feasible for the preparation of 1-propoxy-2 methyl propane.
Alkoxy mercuration-demercuration of $$\displaystyle \displaystyle \left ( Me_{2}C= CH_{2}+MeCH_{2}CH_{2}OH \right )$$ or Alkoxy mercuration-demercuration of propene with $$\displaystyle \displaystyle Me_{2}CHCH_{2}OH$$ will give the regioisomer of 1-propoxy-2 methyl propane as alkoxy group will be attached to more substituted $$C$$ atom of $$C=C$$ double bond.
Options A and B are correct.
Give the decreasing order of reactivity of the following compounds with $$HBr$$.
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III > IV > II > I
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III > II > IV > I
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III> II > I > IV
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II > III > IV > I
Explanation
The decreasing order of reactivity of the following compounds with HBr is $$ \displaystyle III> II > I > IV$$.
The $$ \displaystyle -OH$$ group of alcohol is protonated with $$ \displaystyle HBr$$ followed by loss of water molecule to form carbocation.
$$ \displaystyle Ar-CH_2-OH + H^+ \rightarrow Ar-CH_2-OH_2^+$$
$$ \displaystyle Ar-CH_2-OH_2^+ \rightarrow Ar-CH_2^+ + H_2O$$
Electron donating substituents (such as methyl and methoxy group in para position) on benzene ring stabilizes the carbocation and decreases the reactivity.
Electron withdrawing substituent (such as methoxy group in meta position) on benzene ring destabilizes the carbocation and decreases the reactivity.
Consider the following reaction sequence.
$$\displaystyle { CH }_{ 2 }={ CH }_{ 2 }\xrightarrow [ dil.{ KMnO }_{ 4 } ]{ cold } A\xrightarrow [ { PCl }_{ 5 } ]{ excess } B$$
The products $$(A)$$ and $$(B)$$ are, respectively:
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$$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }Cl$$
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$$\displaystyle { CH }_{ 3 }{ CHO }$$ and $$\displaystyle { CH }_{ 3 }{ CHCl }_{ 2 }$$
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$$\displaystyle { CH }_{ 2 }OH { CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 2 }Cl{ CH }_{ 2 }Cl$$
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$$\displaystyle { CH }_{ 2 }OH{ CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 2 }OH{ CH }_{ 2 }Cl$$
Explanation
The products (A) and (B) are $$\displaystyle { CH }_{ 2 }OH { CH }_{ 2 }OH$$ and $$\displaystyle { CH }_{ 2 }Cl{ CH }_{ 2 }Cl$$ respectively.
$$\displaystyle { CH }_{ 2 }={ CH }_{ 2 }\xrightarrow [ dil.{ KMnO }_{ 4 } ]{ cold } \underset {A}{{ CH }_{ 2 }OH { CH }_{ 2 }OH} \xrightarrow [ { PCl }_{ 5 } ]{ excess } \underset {B}{{ CH }_{ 2 }Cl{ CH }_{ 2 }Cl}$$
When ethylene is treated with cold dilute potassium permanganate solution, two $$-OH$$ groups are added to $$C$$ atoms joined by double bond to form a vicinal diol.
Thus, it is an example of dihydroxylation.
In the next step, on treatment with phosphorous pentachloride, the $$-OH$$ groups are replaced with $$Cl$$ atoms to form vicinal dichloride.
Option C is correct.
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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 are 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 product (B) is :
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Explanation
The cleavage of (C- OH) bond is feasible rather than cleavage of (Ar-OH) bond because of resonance stabilistion in ArOH.
$$PhCH=C{ H }_{ 2 }\overset { ArC{ O }_{ 3 }H }{ \underset { C{ H }_{ 2 }{ Cl }_{ 2 } }{ \longrightarrow } } A\overset { 1.LiAl{ H }_{ 4 }\cdot { Et }_{ 2 }O }{ \underset { 2.{ H }_{ 2 }O }{ \longrightarrow } } B$$
The end product (B) is
:
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$$PhCH(OH)C{ H }_{ 3 }$$
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$$PhC{ H }_{ 2 }C{ H }_{ 2 }COAr$$
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$$PhC{ H }_{ 2 }C{ H }_{ 2 }OH$$
Explanation
In the first step, the O atom is added across C=C bond to form epoxide.
In the second step, the epoxide ring is reduced to form secondary alcohol.
$$\displaystyle PhCH=C{ H }_{ 2 }\overset { ArC{ O }_{ 3 }H }{ \underset { C{ H }_{ 2 }{ Cl }_{ 2 } }{ \longrightarrow } } \underset {\text {A, 2-phenyloxirane}}{C_8H_8O}
\overset { 1.LiAl{ H }_{ 4 }\cdot { Et }_{ 2 }O }{ \underset { 2.{ H }_{ 2 }O }{ \longrightarrow } } \underset {\text {B}}{PhCH(OH)C{ H }_{ 3 }} $$
Option B is correct.
Which of the following will result in the formation of an ether?
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$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C\overset { - }{ O } \overset { + }{ Na } +C{ H }_{ 3 }C{ H }_{ 2 }Br\longrightarrow $$
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$${ \left( C{ H }_{ 3 } \right) }_{ 3 }CBr+{ C }_{ 2 }{ H }_{ 5 }\overset { - }{ O } \overset { + }{ N } a\longrightarrow $$
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$${ C }_{ 6 }{ H }_{ 5 }ONa+C{ H }_{ 3 }Br\longrightarrow $$
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$${ C }_{ 6 }{ H }_{ 5 }Br+C{ H }_{ 3 }\overset { - }{ O } \overset { + }{ N } a\longrightarrow $$
Explanation
The following will result in the formation of an ether
$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C\overset { - }{ O } \overset { + }{ Na } +C{ H }_{ 3 }C{ H }_{ 2 }Br\longrightarrow $$ and $${ C }_{ 6 }{ H }_{ 5 }ONa+C{ H }_{ 3 }Br\longrightarrow $$
The following will not result in the formation of an ether $${ \left( C{ H }_{ 3 } \right) }_{ 3 }CBr+{ C }_{ 2 }{ H }_{ 5 }\overset { - }{ O } \overset { + }{ N } a\longrightarrow $$ The tertiary alkyl bromide will undergo elimination to form an alkene.
$${ C }_{ 6 }{ H }_{ 5 }Br+C{ H }_{ 3 }\overset { - }{ O } \overset { + }{Na} \rightarrow$$
The nucleophilic aromatic substitution requires drastic conditions.
Propene is allowed to react with m-chloroperoxybenzoic acid. The product (A) is then reduced with $$\displaystyle { LiAlH }_{ 4 }$$ in dry ether to give (B).
$$\displaystyle { CH }_{ 3 }CH={ CH }_{ 2 }\xrightarrow { m-CPBA }\ A\xrightarrow [ 2.{ H }_{ 3 }{ O }^{ + } ]{ 1.{ LiAlH }_{ 4 } } B$$
The structure of the product (B) is:
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$$\displaystyle { CH }_{ 3 }{ CH(OH)CH }_{ 2 }OH$$
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$$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }OH$$
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$$\displaystyle { CH }_{ 3 }{ CH }(OH)CH_{ 3 }$$
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Explanation
The product (A) is 2-methyloxirane and the product (B) is propan-2-ol. Alkenes reacts with peracids to form oxiranes. The oxirane on reduction forms alcohols.
Option C is correct.
For the reaction given, what is product $$(A)$$?
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Explanation
Product (A) is benzene. During reaction, 3 molecules of water are lost. The product is obtained by rearrangement of carbon skeleton.
Which statement is correct for the conversion of ROH to RX by reagents $$\displaystyle \left ( SOCl_{2},PBr_{3},PCl_{3},PI_{3} \right )$$ and TsCl compared to using HX?
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No rearrangement occurs with predictable stereo-chemistry by reagents
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Reagents are not useful for $$\displaystyle 3^{\circ}$$ alcohols, while HX reacts easily with $$\displaystyle 3^{\circ}$$ alcohols involving no rearrangement
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Reagents must be used in anhydrous conditions because all react vigrously with $$\displaystyle H_{2}O$$ and they produce harmful gases ($$\displaystyle SO_{2}$$, HCl, HBr, and HI)
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All
Explanation
All the statements given in options A, B and C are correct for the conversion. Hence option D is correct.
The product (B) is:
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all
Consider the reaction:
$$\displaystyle C_{3}H_{7}-OH+Et{O^{\bigoplus }}BF_{4}^{\circleddash }\rightarrow C_{3}H_{7}-O-Et+EtOEt$$.
Which of the following statements is wrong?
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The nucleophile in the reaction is $$\displaystyle C_{3}H_{7}O^-$$
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The nucleophile in the reaction is $$\displaystyle BF_{4}^{{\circleddash }}$$
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The leaving group is $$\displaystyle Et_{2}O$$
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$$\displaystyle SN^{2}$$ reaction occurs
A compound (X) reacts with thionyl chloride to give a compound (Y). (Y) reacts with $$\displaystyle Mg$$ to form a Grignard reagent, which is treated with acetone and the product is hydrolysed to give 2-methyl-2- butanol. What are structural formulae of (X) and (Y) ?
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X = Acetic acid Y = Ethyl Chloride
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X = Acetaldehyde Y = Ethyl Chloride
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X = Ethyl Alcohol Y = Ethyl Chloride
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X = Propyl alcohol Y = Propyl Chloride
Explanation
The reaction can be written as, shown in the above image.
Hence, $$X$$ is Ethyl alcohol.
and $$Y$$ is Ethyl chloride.
Select the incorrect synthesis.
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Explanation
The synthesis of option D is incorrect as it is nucleophilic aromatic substitution reaction which requires drastic conditions. Under these conditions, the sodium alkoxide undergoes elimination reaction to form an alkene.
In which of the following solvents, KI has highest solubility? The dielectric constant $$(\epsilon)$$ of each liquid is given in parentheses.
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$$ C_{6}H_{6}(\epsilon =0)$$
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$$(CH_{3})_{2}CO(\epsilon =21)$$
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$$CH_{3}OH(\epsilon =32)$$
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$$CCl_{4}(\epsilon =0)$$
Explanation
KI is an ionic compound. Hence, it will be most soluble in apolar solvent, which has a high value of dielectric constant. Hence, $$CH_3OH$$ with the highest value of dielectric constant among the given options will provide high solubility.
For the reaction the major product is:
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Explanation
A ring expansion creates the six membered carbocation intermediate.
Complete the blanks by identifying $$X$$ and $$Y$$.
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Ethanal, ethene
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Ethanol, ethene
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Ethanal, ethyne
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Ethanol, ethyne
Explanation
Since, $$Y$$ undergo hydrogenation and give alkane with two carbon.
Hence, $$Y$$ must contain double bond and it is Ethene.
Since, $$KMnO_4$$ is an oxidizing agent, it oxidizes alcohol to aldehyde and further oxidises into acetic acid.
Option B is correct.
The boiling point of phenols are higher than the hydrocarbons of comparable masses due to:
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more polarising power
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presence of hydrogen bonding
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resonance stabilisation
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acidic character
Explanation
The boiling point of compound is effected by forces of attraction existing between molecules. Hydrogen bonding is a stronger force of attraction existing between molecules of phenols as compared to vander waals forces of attraction existing between alkane molecules. (they are weaker forces)
$$\therefore$$ Boiling point of phenol is more than hydrocarbon due to presence of hydrogen bonding.
m-Methoxyphenol is a weaker acid than phenol.
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True
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False
Correct statement(s) in case of n-butanol and t-butanol is (are):
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both are having equal solubility in water.
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t-butanol is more soluble in water than n-butanol.
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boiling point of t-butanol is lower than n-butanol.
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boiling point of n-butanol is lower than t-butanol.
Explanation
More branching means less boiling point and high solubility.
Which of the following has lowest boiling point?
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$$p$$-nitrophenol
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$$m$$-nitrophenol
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$$o$$-nitrophenol
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phenol
Explanation
o-nitrophenol forms intramolecular Hydrogen bond and no association occurs whereas molecules of p-nitrophenol get associated through intermolecular Hydrogen bonding and boils relatively at higher temperature (above the boiling point of water). Hence, only ortho-nitrophenol is steam volatile.
Identify the correct sequence of given steps for the conversion of calcium carbide to methyl alcohol.
(a) Reaction with aqueous $$KOH$$
(b) Hydrolysis
(c) Reaction with soda lime
(d) Reaction with $$Ha{SO}_{4}/{H}_{2}{SO}_{4}$$
(e) Reaction with $$Hg{SO}_{4}$$
(f) Reaction with $$P{Cl}_{5}$$
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$$b\ d\ e\ c\ f\ a$$
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$$b\ d\ c\ a\ f\ e$$
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$$b\ f\ d\ e\ a\ c$$
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$$d\ b\ c\ e\ f \ a$$
Explanation
Steps for the conversion of calcium carbide to methyl alcohol:
(b) Hydrolysis
(d) Reaction with $$Ha{SO}_{4}/{H}_{2}{SO}_{4}$$
(e) Reaction with $$Hg{SO}_{4}$$
(c) Reaction with soda lime
(f) Reaction with $$P{Cl}_{5}$$
(a) Reaction with aqueous $$KOH$$
$$CaC_2 + 2 H_2O\rightarrow C_2H_2 + Ca(OH)_2 \Rightarrow C_2H_2 + H_2SO_4/ HgSO_4 \rightarrow CH_3CHO
\Rightarrow CH_3CHO + H_2SO_4\rightarrow CH_3COOH \Rightarrow CH_3COOH + NaOH + CaO \rightarrow CH_3COONa
\Rightarrow CH_3COONa + PCl_5 \rightarrow CH_3Cl\Rightarrow CH_3Cl + KOH \rightarrow CH_3OH$$
Ethanol is produced by/from :
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ethene
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fermentation of sugars
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both A and B
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none of the above
Explanation
Formation of Ethanol:
$$\mathrm{\underset{Ethene}{H_2C=CH_2} \xrightarrow[H_2O]{H_2SO_4} \underset{Ethanol}{H_3C-CH_2OH}}$$
$$\mathrm{\underset{Glucose(Sugar)}{C_6H_12O_6} \xrightarrow{Zymase \ enzyme} \underset{Ethanol}{C_2H_5OH}}$$
So, Ethanol can be formed from Ethene and Fermentation of sugar.
Hence, Option "C" is the correct answer.
$$R-OH\xrightarrow{PBr_3}A\xrightarrow[]{Mg/Dry \ ether}B\xrightarrow[]{HCHO}C\xrightarrow[]{H^+/H_2O}D$$
What are $$A,B,C$$ & $$D$$ respectively?
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$$R-Br,R-Br-Mg,R-CH_2Mg-Br,R-CH_2CHO$$
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$$R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2CHO$$
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$$R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2OH$$
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$$R-Br,R-Mg-Br,R-CH_2-Br,R-CH_2OH$$
Explanation
$$R-OH\overset{PBr_3}{\rightarrow}R-Br\xrightarrow[]{Mg/Dry
ether}R-Mg-Br\xrightarrow[]{HCHO}R-CH_2-O-Mg-Br\xrightarrow[]{H^+/H_2O}RCH_2OH$$
So the correct order is $$R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2OH$$
Hence option C is correct.
Which one of the following is produced when acetone is saturated with $$HCl$$ gas?
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Acetone alcohol
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Phorone
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Mesityl oxide
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Benzene
$$CH_3CH_2-OH\xrightarrow{A}CH_3-CH_2-Cl\xrightarrow{B}CH_2=CH_2.$$
A and B in this sequence of reactions are
:
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KOH(aq) and $$PCl_5$$
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$$PCl_5$$ and KOH(aq)
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$$Cl_2$$ and KOH(alc)
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$$PCl_5$$ and KOH(alc)
Explanation
A and B in this sequence of reactions are $$\displaystyle PCl_5$$ and KOH(alc).
$$\displaystyle CH_3CH_2OH\xrightarrow{PCl_5(A)}CH_3CH_2Cl\xrightarrow{alc.KOH(B)}CH_2=CH_2$$
Ethanol reacts with $$PCl_5$$ to give ethyl chloride. Ethyl chloride on heating with alcoholic potash undergoes dehydrohalogenation to give ethene.
Hence the correct option is D.
$$R-OH\xrightarrow {P+I_2}A\xrightarrow[alc. \ KCN]{Heating}B\xrightarrow[Na/alcohol]{LiAlH_4}C\xrightarrow[NaNO_2+dil\ HCl]{HNO_2}D$$
What are $$A,B,C$$ & $$D$$ respectively?
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$$R-I$$, $$R-CN$$, $$R-NH_2$$, $$R-CH_2OH$$
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$$R-I$$, $$R-CN$$, $$R-CH_2-NH_2$$, $$R-CH_2OH$$
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$$R-I$$, $$R-NC$$, $$R-CH_2-NH_2$$, $$R-CH_2OH$$
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$$R-I$$, $$R-CN$$, $$R-CH_2-NO_2$$, $$R-CH_2OH$$
Explanation
$$R-OH\xrightarrow{P+I_2}R-I\xrightarrow[alc KCN]{Heating}R-CN\xrightarrow[Na/alchohol]{LiAlH_4}R-CH_2-NH_2\xrightarrow[NaNO_2+dil\ HCl]{HNO_2}R-CH_2OH$$
Hence option B is correct.
$$R-OH\xrightarrow {PBr_3}A\xrightarrow[]{Mg/Dry \ ether}B\xrightarrow[CH_2-CH_2-O]{Grignard\ reaction}C\xrightarrow[]{H^+/H_2O}D$$
What are $$A,B,C$$ and $$D$$ are respectively?
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$$R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2CH_2-OH$$
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$$R-Br,R-Mg-Br,R-CH_2-O-CH_2-Mg-Br,R-CH_2CH_2-OH$$
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$$R-Br,R-Mg-Br,R-CH_2-CH_2-O-Mg-Br,R-CH_2CH_2-OH$$
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$$R-Br,R-Mg-Br,R-CH_2-CH_2-O-Mg-Br,R-CH_2OH$$
Explanation
$$R-OH\overset{PBr_3}{\rightarrow}R-Br\xrightarrow[]{Mg/Dry\ ether}R-Mg-Br\xrightarrow[CH_2-CH_2-O]{Grignard\ reaction}R-CH_2-CH_2-O-Mg-Br\\ \quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \xrightarrow[]{H^+/H_2O}RCH_2-CH_2-OH$$
$$PBr_3$$ converts primary or secondary alcohol to alkyl halide. Mg and dryether forms Grignard reagent R-Mg-X , which on reaction with epoxide like ethylene oxide ($$CH_2CH_2O$$) forms $$R-CH_2CH_2-O-Mg-Br$$ and then in presence of acid gives alcohol.
So the order is $$R-Br, R-Mg-Br, \ R-CH_2-CH_2-O-Mg-Br,\ R-CH_2CH_2-OH$$.
Hence option C is correct,
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