Question 12 Marks
Carry out the following conversions:
Propanone to 2-Methylpropan-2-ol.
Propanone to 2-Methylpropan-2-ol.
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Question 22 Marks
How will you bring the following conversions?
Phenol to benzyl alcohol.
Phenol to benzyl alcohol.
Question 32 Marks
How will you bring the following conversions?
Phenol to m-bromophenol.
Phenol to m-bromophenol.
Question 42 Marks
Give Reasons:
Relative ease of dehydration of alcohols is 3° > 2° > 1°.
Relative ease of dehydration of alcohols is 3° > 2° > 1°.
Answer
View full question & answer→The dehydration of alcohols occurs through the formation of carbocation intermediate. As the stability of carbocations decreases in the order, 3° > 2° > 1°, therefore, the dehydration of alcohols follows the order, 3° > 2° > 1°.
Question 52 Marks
Arrange the following compounds in increasing order of acidity and give a suitable explanation. Phenol, $o-$nitrophenol, $o-$cresol.
Answer
View full question & answer→Ncreasing order of acidity. $o-$Cresol
In cresol,
, the electron donating group $(-CH3)$ gives electrons and intensify the charge on phenoxide ion and therefore makes it unstable. Therefore, $o-$cresol is less acidic than phenol. In $o-$nitrophenol, the electron withdrawing $(-NO_2)$ group withdraws electrons and disperses the $-$ve charge and stabilizes the phenoxide ion. Therefore, $o-$nitrophenol is more acidic than phenol.
In cresol,
, the electron donating group $(-CH3)$ gives electrons and intensify the charge on phenoxide ion and therefore makes it unstable. Therefore, $o-$cresol is less acidic than phenol. In $o-$nitrophenol, the electron withdrawing $(-NO_2)$ group withdraws electrons and disperses the $-$ve charge and stabilizes the phenoxide ion. Therefore, $o-$nitrophenol is more acidic than phenol.Question 62 Marks
Ethers can be prepared by Williamson synthesis in which an alkyl halide is reacted with sodium alkoxide. Di-tert-butyl ether can’t be prepared by this method. Explain.
Answer
In tert-butyl halides, the elimination reaction is favoured over substitution. So, alkene is the only reaction product formed and not the ether.
View full question & answer→In tert-butyl halides, the elimination reaction is favoured over substitution. So, alkene is the only reaction product formed and not the ether.
Question 72 Marks
How will you bring the following conversions?
Phenol to aspirin.
Phenol to aspirin.
Question 82 Marks
Preparation of alcohols from alkenes involves the electrophilic attack on alkene carbon atom. Explain its mechanism.
Answer
View full question & answer→The mechanism of acid catalysed addition of water $($hydration$)$ to alkenes involves the following three steps$:$
Step $1:$ Electrophilic attack by hydronium ion $\ce{(H_3O^+)}$ on alkene gives an intermediate, carbocation.
Step $2:$ Nucleophilic attack by water on carbocation to yield protonated alcohol.
Step $3:$ Deprotonatio $($loss of proton$)$ to from an alcohol.
Step $1:$ Electrophilic attack by hydronium ion $\ce{(H_3O^+)}$ on alkene gives an intermediate, carbocation.
Step $2:$ Nucleophilic attack by water on carbocation to yield protonated alcohol.
Step $3:$ Deprotonatio $($loss of proton$)$ to from an alcohol.
Question 92 Marks
A compound ‘A’ is optically active. On mild oxidation, it gives a compound ‘B’ but on vigorous oxidation gives another compound ‘C’. C along with D is also formed from B by reaction with iodine and alkali. Deduce the structures of A, B, C, D.
Question 102 Marks
Give the major products that are formed by heating each of the following ethers with HI.
Question 112 Marks
Explain the following behaviours:
- $-OH$ group attached to a carbon of benzene ring activates it towards electrophilic substitution.
- Reactivity of all the three classes of alcohols with conc. $HCl$ and $ZnCl_2$ is different.
- Anisole reacts with $HI$ to give phenol and methyl iodide and not iodobenzene and methyl alcohol.
Answer
View full question & answer→- Due to $+R$ effect of the $OH$ group, the electron density in the benzene ring increases thereby facilitating the attack by an electrophile. Further, the electron density is relatively higher at the $o-$ and $p-$positions, therefore, electrophilic substitution occurs mainly at $o-$ and $p-$positions.
- The reaction of alcohols with conc. $HCl $ and $ZnCl_2$ takes place through intermediate formation of carbonium ions. Greater the stability of carbonium ion greater is the reactivity of alcohol. Due to $+ve$ I effect of alkyl groups the stability of carbonium ion follows the order $1^\circ < 2^\circ < 3^\circ .$ As a result of this reactivity of alcohols towards conc. $HCl$ and $ZnCl_2$ follows the same order i.e., $1^\circ < 2^\circ < 3^\circ .$
- Protonation of anisole gives oxonium ion $\text{C}_6\text{H}_5-\stackrel{{. \ .}}{\hbox{O}}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}$ The bond between $O-CH_3$ is weaker than the bond between $O-C_6H_5$ because the carbon of phenyl group is $sp_2$ hybridised and there is partial double bond character. Therefore, the attack by $I^- $ ion breaks $O-CH_3$ bond to give methyl iodide and phenol.
Question 122 Marks
Give reasons for the following:
p-nitrophenol is more acidic than o-nitrophenol.
p-nitrophenol is more acidic than o-nitrophenol.
Answer
Intramolecular H-bonding in o-nitrophenol makes loss of proton difficult. Therefore, p-nitrophenol is more acidic than o-nitrophenol.
View full question & answer→Intramolecular H-bonding in o-nitrophenol makes loss of proton difficult. Therefore, p-nitrophenol is more acidic than o-nitrophenol.
Question 132 Marks
Write complete reaction for the bromination of phenol in aqueous and non aqueous medium.
Answer
When phenol is treated with bromine water. 2,4,6-tribromophenol is formed as white precipitate:
View full question & answer→When phenol is treated with bromine water. 2,4,6-tribromophenol is formed as white precipitate:
Question 142 Marks
Write the mechanism of the reaction of $HI$ with methoxybenzene.
Answer
View full question & answer→In case of alkyl aryl ethers, the products are always phenol and an alkyl halide because due to resonance $C_6H_5–O$ bond has partial double bond character.
$\text{C}_6\text{H}_5-\text{O}-\text{CH}_3+\text{HI}\rightarrow\text{C}_6\text{H}_5\text{OH}+\text{CH}_3\text{I}$
$\text{Methoxybenzene} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Phenol}$
Mechanism:
Step$-1:$ Ether molecule first gets protonated to give methylphenyl oxonium ion.
In this ion, the bond between $0 - CH2$ is weaker than the bond between $O-CH,$ which has partial double bond character. This partial double bond character is due to the resonance between the lone pair of electrons on the $O-$atom and the sp’ hybridised carbon atom of the phenyl group.
Step$-2:$ Protonated ether undergoes $S_N2$ attack by lion to give methyl iodide and phenol.
$\text{C}_6\text{H}_5-\text{O}-\text{CH}_3+\text{HI}\rightarrow\text{C}_6\text{H}_5\text{OH}+\text{CH}_3\text{I}$
$\text{Methoxybenzene} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Phenol}$
Mechanism:
Step$-1:$ Ether molecule first gets protonated to give methylphenyl oxonium ion.
In this ion, the bond between $0 - CH2$ is weaker than the bond between $O-CH,$ which has partial double bond character. This partial double bond character is due to the resonance between the lone pair of electrons on the $O-$atom and the sp’ hybridised carbon atom of the phenyl group.
Step$-2:$ Protonated ether undergoes $S_N2$ attack by lion to give methyl iodide and phenol.
Question 152 Marks
Account for the following:
- Phenol does not react with $\ce{NaHCO_3}$ whereas carboxylic acids react.
- Phenol is more easily nitrated than benzene.
Answer
Such acid$-$base reaction takes place only if the acid formrd is weaker than the reacting acid. In other words, phenol is not a strong enough acid to react with $\ce{NaHCO_3}.$
View full question & answer→- $\text{RCOOH}+\text{NaHCO}_3\xrightarrow{ \ \ \ \ \ \ \ \ \ }\text{RCOONa}+\text{H}_2\text{CO}_3$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Weaker acid}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{than RCOOH}$
Such acid$-$base reaction takes place only if the acid formrd is weaker than the reacting acid. In other words, phenol is not a strong enough acid to react with $\ce{NaHCO_3}.$
- Nitration involves attack of electrophile nitronium ion $\stackrel{{+ \ \ \ }}{\hbox{ NO}_2}$ on benzene ring. Due to $+R$ effect of $-OH$ group electron density on benzene increases. Therefore, phenol is more easily nitrated as compared to benzene.
Question 162 Marks
Explain why low molecular mass alcohols are soluble in water.
Answer
The lower members of alcohols are highly soluble in water but the solubility decreases with increase in molecular weight. The solubility of lower alcohols in water is due to the formation of hydrogen bonds between alcohols and water molecules.
View full question & answer→The lower members of alcohols are highly soluble in water but the solubility decreases with increase in molecular weight. The solubility of lower alcohols in water is due to the formation of hydrogen bonds between alcohols and water molecules.
Question 172 Marks
Phenol is more easily nitrated than benzene.
Answer
View full question & answer→Nitration involves attack of electrophile nitronium ion $\ce{(NO_2)}$ on benzene ring. Due to $+R$ effect of $-\ce{OH}$ group electron density on benzene increases. Therefore, phenol is more easily nitrated as compared to benzene,
Question 182 Marks
Give the equations of reactions for the preparation of phenol from cumene.
Answer
View full question & answer→Cumene is oxidised into cumene hydro peroxide by air at high pressure which on decomposition by aqueous acid gives phenol and acetone.
The cumene required in the above reaction is prepared from benzene and propene by Friedel-crafts reaction.
The cumene required in the above reaction is prepared from benzene and propene by Friedel-crafts reaction.
Question 192 Marks
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer
View full question & answer→Acid catalysed dehydration of primary alcohols to ethers occurs by $SN_2$ reaction involving nucleophilic attack by the alcohol molecule on the protonated alcohol molecule.
Under these conditions $, 2^\circ$ and $3^\circ$ alcohols, however, give alkenes rather than ethers.
The reason being that due to steric hindrance, nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not occur.
Instead protonated $2^\circ $ and $3^\circ$ alcohols lose a molecule of water to form stable $2^\circ$ and $3^\circ$ carbocation.
These carbocations prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecules to form ethers.
Similarly $, 3^\circ$ alcohols give alkenes rather than ethers.
Under these conditions $, 2^\circ$ and $3^\circ$ alcohols, however, give alkenes rather than ethers.
The reason being that due to steric hindrance, nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not occur.
Instead protonated $2^\circ $ and $3^\circ$ alcohols lose a molecule of water to form stable $2^\circ$ and $3^\circ$ carbocation.
These carbocations prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecules to form ethers.
Similarly $, 3^\circ$ alcohols give alkenes rather than ethers.
Question 202 Marks
Give structures of the products you would expect when each of the following alcohol reacts with HCl –ZnCl2
- Butan-1-ol
- 2-Methylbutan-2-ol
Answer
Primary alcohols do not react appreciably with Lucas’ reagent (HCl -ZnCl2) at room temperature.
View full question & answer→- $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}\xrightarrow{\text{HCl}-\text{ZnCl}_2}\text{No reaction}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Butan-1-ol}$
Primary alcohols do not react appreciably with Lucas’ reagent (HCl -ZnCl2) at room temperature.
- $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_3\xrightarrow{\text{HCl}-\text{ZnCl}_2}\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_3+\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\2-\text{Methylbutan}-2-\text{ol}(3^\circ)\ \ \ \ \ \ \ \ \ \ \ 2-\text{Chloro}\ 2-\text{Methylbutan}(\text{White turbidity})$
Tertiary alcohols react immediately with Lucas’ reagent.
Question 212 Marks
Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer
In aryl alkyl ethers, the +R-effect of the alkoxy (OR) group increases the electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.
Since the electron density increases more at the two ortho and one para position as compared to meta position therefore, electrophilic substitution reactions mainly occur at o-and -positions.
View full question & answer→In aryl alkyl ethers, the +R-effect of the alkoxy (OR) group increases the electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.
Since the electron density increases more at the two ortho and one para position as compared to meta position therefore, electrophilic substitution reactions mainly occur at o-and -positions.
Question 222 Marks
Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer
The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:
Step 1:
Protonation of ethanol to form ethyl oxonium ion:
Step 2:
Formation of carbocation (rate determining step):
Step 3:
Elimination of a proton to form ethene:
The acid consumed in step 1 is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.
View full question & answer→The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:
Step 1:
Protonation of ethanol to form ethyl oxonium ion:
Step 2:
Formation of carbocation (rate determining step):
Step 3:
Elimination of a proton to form ethene:
The acid consumed in step 1 is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.
Question 232 Marks
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:2-Methoxy-2-methylpropane.
Answer
View full question & answer→$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \text{CH}_3-\text{C}-\text{O}^-\text{Na}^++\text{CH}_3-\text{Br}\xrightarrow{\ \Delta\ }\text{CH}_3-\text{C}-\text{OCH}_3+\text{NaBr}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Bromoethane} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\text{Sodium-2-methyl-2-propoxide}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{2-methyl-2-methoxypropane}$
Question 242 Marks
How is $1-$propoxypropane synthesised from propan$-1-ol$? Write mechanism of this reaction.
Answer
View full question & answer→- Williamson's synthesis,
- $3\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}+\text{PBr}_3\xrightarrow{\ \ \ \ \ }3\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}+\text{H}_3\text{PO}_3\\\ \ \ \text{Propan-1-ol}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{1-Bromopropane}$
- $2\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}+2\text{Na}\xrightarrow{\ \ \ \ \ \ \ }2\text{CH}_3\text{CH}_2\text{CH}_2\text{O}^-\text{Na}^++\text{H}_2\\ \ \ \ \ \ \ \ \text{Propan-1-ol}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium propoxide}$
- By dehydration of $l-$propanol with conc. $\ce{H_2SO_4}$ at $413 K.$
Question 252 Marks
Write the mechanism of the reaction of HI with methoxymethane.
Answer
When equimolar amounts of HI and methoxy methane are reacted, a mixture of methyl alcohol and methyl iodide is formed by the following mechanism:
If however, excess of HI is used, methyl alcohol formed in step (b) is also converted into methyl iodide by followmg mechanism:
View full question & answer→When equimolar amounts of HI and methoxy methane are reacted, a mixture of methyl alcohol and methyl iodide is formed by the following mechanism:
If however, excess of HI is used, methyl alcohol formed in step (b) is also converted into methyl iodide by followmg mechanism:
Question 262 Marks
Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Answer
The -OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.
As a result, the benzene ring is activated towards electrophilic substitution.
View full question & answer→The -OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.
As a result, the benzene ring is activated towards electrophilic substitution.
Question 272 Marks
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer
View full question & answer→In Williamsons’s synthesis, the alkyl halide should be primary. Thus, the alkyl halide should be derived from ethanol and the alkoxide ion from 3-methylpentan-2-ol. The synthesis is as follows$\text{CH}_3\text{CH}_2\text{OH}+\text{HBr}\xrightarrow{\ \ \ \Delta\ \ \ }\text{CH}_3\text{CH}_2\text{Br}+\text{H}_2\text{O}\\\text{CH}_3\text{CH}_2\text{CH}-\text{CH}-\text{OH}+\text{Na}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{CH}_2-\text{CH}-\text{CH}-\text{O}^-\text{Na}^++\text{H}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{CH}_3$
$\text{CH}_3\text{CH}_2\text{CH}-\text{CH}-\text{O}^-\text{Na}^++\text{CH}_3\text{CH}_2\text{Br}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{CH}_3\\\xrightarrow{\ \ \ \ \text{S}_{\text{N}^2}\ \ \ \ \ \ }\text{CH}_3\text{CH}_2-\text{CH}-\text{CH}-\text{OCH}_2\text{CH}_3+\text{NaBr}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{3-Ethyl-3-methyl}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Pentane}$
$\text{CH}_3\text{CH}_2\text{CH}-\text{CH}-\text{O}^-\text{Na}^++\text{CH}_3\text{CH}_2\text{Br}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{CH}_3\\\xrightarrow{\ \ \ \ \text{S}_{\text{N}^2}\ \ \ \ \ \ }\text{CH}_3\text{CH}_2-\text{CH}-\text{CH}-\text{OCH}_2\text{CH}_3+\text{NaBr}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{3-Ethyl-3-methyl}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Pentane}$
Question 282 Marks
Give structures of the products you would expect when each of the following alcohol reacts with $\ce{SOCl_2}$
$(i)$ Butan$-1-ol$
$(ii) \ 2-$Methylbutan$-2-ol$
$(i)$ Butan$-1-ol$
$(ii) \ 2-$Methylbutan$-2-ol$
Answer
View full question & answer→- $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}+\text{SOCl}_2\xrightarrow[]{}\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}+\text{SO}_2+\text{HCl}\\ \ \ \ \ \ \ \ \text{Butan-1-ol}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{1-chlorobutane}$
- $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_3+\text{SOCl}_2\xrightarrow[]{}\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_3+\text{SO}_2+\text{HCl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \text{2-Methylbutan-2-ol}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{2-Chloro-2-Methylbutane}$
Question 292 Marks
Write the mechanism of hydration of ethene to yield ethanol.
Answer
View full question & answer→The mechanism of hydration of ethene to form ethanol involves three steps.
Step $1:$
Protonation of ethene to form carbocation by electrophilic attack of $H_3O+:$
$\text{H}_2\text{O}+\text{H}^+\xrightarrow{\ \ \ \ \ \ \ }\text{H}_3\text{O}^+$
Step $2:$
Nucleophilic attack of water on carbocation:
Step $3:$
Deprotonation to form ethanol:
Step $1:$
Protonation of ethene to form carbocation by electrophilic attack of $H_3O+:$
$\text{H}_2\text{O}+\text{H}^+\xrightarrow{\ \ \ \ \ \ \ }\text{H}_3\text{O}^+$
Step $2:$
Nucleophilic attack of water on carbocation:
Step $3:$
Deprotonation to form ethanol:
Question 302 Marks
How can phenol be converted to aspirin?
Answer
View full question & answer→Phenol is converted into salicylic acid.
The reaction is usually carried out by allowing sodium phenoxide to absorb carbon dioxide and then heating the product to $400K$ and $4-7$ atm pressure. First unstable intermediate is formed which undergoes a proton shift to form sodium salicylate. The subsequent acidification of sodium salicylate gives,
Then aspirin is obtained by acetylating salicylic acid with acetic anhydride and cons. $H_2SO_4.$
The reaction is usually carried out by allowing sodium phenoxide to absorb carbon dioxide and then heating the product to $400K$ and $4-7$ atm pressure. First unstable intermediate is formed which undergoes a proton shift to form sodium salicylate. The subsequent acidification of sodium salicylate gives,
Then aspirin is obtained by acetylating salicylic acid with acetic anhydride and cons. $H_2SO_4.$
Question 312 Marks
Alcohols react both as nucleophiles as well as electrophiles. Write one reaction of each type and describe its mechanism.
Answer
Alcohols as nucleophiles: The bond between O-H is broken when alcohols react as nucleophiles.
Alcohols as electrophiles: The bond C-O is broken when alcohols react as electrophiles. Protonated alcohols react in this manner.
View full question & answer→Alcohols as nucleophiles: The bond between O-H is broken when alcohols react as nucleophiles.
Alcohols as electrophiles: The bond C-O is broken when alcohols react as electrophiles. Protonated alcohols react in this manner.
Question 322 Marks
Match the starting materials given in Column $I$ with the products formed by these $($Column $II)$ in the reaction with $II.$
|
Column $I$
|
|
Column $II$
|
|
| $(i)$ | $CH_3-o-CH_3$ | $(a)$ |
 |
| $(ii)$ |
 |
$(b)$ |
$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{CH}_3-\text{C}-\text{I}+\text{CH}_3\text{OH} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
|
| $(iii)$ |
$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{H}_3\text{C}-\text{C}-\text{O}-\text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
|
$(c)$ |
 |
| $(iv)$ |
 |
$(d)$ | $CH_3-OH + CH_3-I$ |
|
|
|
$(e)$ |
 |
|
|
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$(f)$ |
 |
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|
|
$(g)$ |
$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{CH}_3-\text{C}-\text{OH}+\text{CH}_3\text{I} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
|
Answer
$\ce{CH_3-O-CH_3}$ is a symmetrical ether so the products are $CH_3I$ and $CH_2OH.$
View full question & answer→| Column $I$ | Column $I$ | ||
| $(i)$ | $CH3-O-CH3$ | $(d)$ | $CH_3-OH + CH_3-I$ |
| $(ii)$ |  |
$(e)$ |  |
| $(iii)$ | $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{H}_3\text{C}-\text{C}-\text{O}-\text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ | $(b)$ | $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{CH}_3-\text{C}-\text{I}+\text{CH}_3\text{OH} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ |
| $(iv)$ |  |
$(a)$ |  |
- In $\ce{(CH_3)_2CH-O-CH_3}$ unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows $S_N^2$ mechanism. Thus, the halide ion attacks the smaller alkyl group and the products are
- In this case, one of the alkyl group is tertiary and the other is primary. It follows $S_N^1$ mechanism and halide ion attacks the tertiary alkyl group and the products are $(CH_3)_3 C-I$ and $CH_3OH.$
- Here, the unsymmetrical ether is alkyl aryl ether. In this ether $0-CH_3$ bond is weaker than $0-C_6H_5$ bond which has partial double bond character due to resonance. So, the halide ion attacks on alkyl group and the products are $C_6H_5-OH$ and $CH_3I.$
Question 332 Marks
Give structures of the products you would expect when each of the following alcohol reacts with HBr
- Butan-1-ol
- 2-Methylbutan-2-ol
Answer
View full question & answer→- $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}+\text{HBr}\xrightarrow[-\text{H}_{2}\text{O}]{}\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\\ \ \ \ \ \ \ \ \text{Butan-1-ol} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{1-Bromobutane}$
- $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_3+\text{HBr}\xrightarrow[]{}\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_3+\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\text{2-Methylbutan-2-ol}(3^\circ)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{2-Bromo-2-Methylbutane}$
Question 342 Marks
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\xrightarrow{\ \ \ \text{HBr} \ \ }\text{CH}_3-\text{C}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Answer
The given reaction involve the following mechanism,
Step 1: Protonation to 3-methylbutan-2-ol.
$\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\xrightarrow{\ \ \text{H}^+\ \ }\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ _{_+^{\text{OH}_2}} \\ \ \ \text{3-Methylbutan-2-ol}$
Step 2: Formation of secondary carbocation by the elimination of a water molecule.
Step 3: Rearrangement by the hydride-ion shift.
Step 4: Nucleophilic attack.
View full question & answer→The given reaction involve the following mechanism,
Step 1: Protonation to 3-methylbutan-2-ol.
$\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\xrightarrow{\ \ \text{H}^+\ \ }\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ _{_+^{\text{OH}_2}} \\ \ \ \text{3-Methylbutan-2-ol}$
Step 2: Formation of secondary carbocation by the elimination of a water molecule.
Step 3: Rearrangement by the hydride-ion shift.
Step 4: Nucleophilic attack.
Question 352 Marks
Write the equations involved in the following reactions:
- Reimer - Tiemann reaction.
- Kolbe’s reaction.
Answer
View full question & answer→- Reimer-Tiemann reaction
- Kolbe’s reaction
Question 362 Marks
Which of the following is an appropriate set of reactants for the preparation of $1-$methoxy$-4-$nitrobenzene and why?
Answer
View full question & answer→In structure $(a),$ group sodium methoxide $\ce{(CH_3ONa)}$ is a strong nucleophile also it is a strong base.
Hence, an elimination reaction predominates over a substitution reaction.
Hence, an elimination reaction predominates over a substitution reaction.
Question 372 Marks
Match the structures of the compounds given in Column I with the name of the compounds given in Column II.
| Column I | Column II | ||
| (i) |  |
(a) | Hydroquinone |
| (ii) |  |
(b) | Phenetole |
| (iii) |  |
(c) | Catechol |
| (iv) |  |
(d) | o-Cresol |
| (v) |  |
(e) | Quinone |
| (vi) |  |
(f) | Resorcinol |
| (g) | Anisole. |
Answer
Explanation:
View full question & answer→| Column I | Column I | ||
| (i) |  |
(d) | o-Cresol |
| (ii) |  |
(c) | Catechol |
| (iii) |  |
(f) | Resorcinol |
| (iv) |  |
(a) | Hydroquinone |
| (v) |  |
(g) | Anisole |
| (vi) |  |
(b) | Pheneto |
- Cresols are organic compounds which are methyl phenols. There are three forms of cresol: o-cresol, p-cresol and w-cresol.
- Catechol is also known as pyrocatechol. Its IUPAC name is 1, 2-dihydrobenzene. It is used in the production of pesticides, perfumes and pharmaceuticals.
- Its IUPAC name is 1, 3-dihydroxybenzene. Resorcinol is used to treat acne, seborrheic dermatitis and other skin disorder.
- Hydroquinone is also known as quinol. Its IUPAC name is 1, 4-dihydroxybenzene. Itisawhite granularsolid.lt isagoodreducing agent.
- Anisole or methoxy benzene, is a colourless liquid with a smell reminiscent of anise seed.Phenetole is an organic compound. It is also known as ethylphenyl ether. It is volatile in nature and its vapour are explosive in nature.
Question 382 Marks
Give a chemical test to distinguish between the following pairs of compounds.
Answer
View full question & answer→- Phenol gives violet colouration with $\ce{FeCI_3}$ Solution While cyclohexanol does not.
- ISpropyl alcohal when warmed with $\ce{NaOI (I_2/ NaOH)}$ gives yellow precipitate of iodoform, in contrast benzyl alcohal does not respond to iodoform test.
Question 392 Marks
Write steps to carry out the conversion of phenol to aspirin.
Question 402 Marks
Out of 2-chloroethanol and ethanol which is more acidic and why?
Answer
View full question & answer→The acidic character of alcohols is due to the polar nature of O-H bond. 2-Chloroethanol, is more acidic due to -I effect of chlorine atom. It increases the polarity of O- H bond and increases the acidic strength.
Question 412 Marks
A compound $‘A\ ’$ having molecular formula $\ce{C_4H_{10}O}$ is found to be soluble in concentrated sulphuric acid. It does not react with sodium metal or potassium permanganate. On heating with excess of $HI,$ it gives a single alkyl halide. Deduce the structure of compound $A$ and explain all the reactions.
Answer
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}$
$\ \ \ \ \ \ \ \ \ \ \text{Diethyl ether} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Soluble oxonium salt}$
$\text{CH}_3\text{ CH}_2\text{ OCH}_2\text{ CH}_3\xrightarrow{ \ \ \ \ \ \ \text{m} \ \ \ \ \ \ \ }2\text{CH}_3\text{ CH}_2-\text{I}$
$$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta \ \ \ \ \ \ \ \ \ \text{Ethyl iodide}$
$\ \ \ \ \ \ \ \ \ \ \ \ \text{Diethyl ether}$
View full question & answer→- As compound $A$ does not react with sodium metal or potassium permanganate, it cannot be an alcohol.
- As compound $A$ dissolves in conc. $H_2SO_4$, it may be an ether.
- As compound $A$ on heating with excess of $HI$ gives a single alkyl halide, therefore, compound $A$ must be a symmetrical ether.
- The only symmetrical ether having molecular formula $C_4H_{10}O$ is diethyl ether. Thus compound $'A\ '$ is diethylether, $\ce{CH_3-CH_2-O-CH_2-CH_3.}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}$
$\ \ \ \ \ \ \ \ \ \ \text{Diethyl ether} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Soluble oxonium salt}$
$\text{CH}_3\text{ CH}_2\text{ OCH}_2\text{ CH}_3\xrightarrow{ \ \ \ \ \ \ \text{m} \ \ \ \ \ \ \ }2\text{CH}_3\text{ CH}_2-\text{I}$
$$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta \ \ \ \ \ \ \ \ \ \text{Ethyl iodide}$
$\ \ \ \ \ \ \ \ \ \ \ \ \text{Diethyl ether}$
Question 422 Marks
Carry out the following conversions:
Phenol to benzoquinone.
Phenol to benzoquinone.
Question 432 Marks
Give the major products that are formed by heating the following ethers with HI.
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{O}-\text{C}-\text{CH}_2-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{O}-\text{C}-\text{CH}_2-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Answer
View full question & answer→$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\\text{CH}_3 \ \text{CH}_2 \ \text{CH}_2\text{OH}+\text{CH}_3\text{CH}_2-\text{C}-\text{I}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Question 442 Marks
o-nitrophenol is more acidic than $o-$methoxyphenol.
Answer
View full question & answer→$-R$ and $-I$ effect of $-NO_2$ group decreases the electron density in $O-H$ bond and make loss of proton easy in $O-$nitrophenol whereas $+R$ effect of $-\ce{OCH_3}$ group increases the electron density in $O-H$ bond and makes release of proton difficult in $o-$methoxyphenol.
That is, why $o-$nitrophenol is stronger acid than $o-$methoxyphenol.
That is, why $o-$nitrophenol is stronger acid than $o-$methoxyphenol.
Question 452 Marks
Explain why alcohols and ethers of comparable molecular mass have different boiling points?
Answer
View full question & answer→Ethers have low polarity and as a result do not show any association by intermolecular hydrogen bonding.
Therefore, ethers have low boiling points and lower than that of isomeric alcohols and almost same as those of alkanes of comparable molecular masses.
The large difference in boiling points of alcohols and ethers is due to the presence of hydrogen bonding in alcohols.
Therefore, ethers have low boiling points and lower than that of isomeric alcohols and almost same as those of alkanes of comparable molecular masses.
| $\ce{CH_3CH_2OCH_2CH_3}$ | $\ce{CH_3CH_2CH_2CH_2OH}$ | $\ce{CH_3CH_2CH_2CH_2CH_3}$ |
| Diethyl ether | $n-$Butyl alcohol | $n-$Pentane |
| $307.6 K$ | $390 K$ | $309.1 K$ |
Question 462 Marks
How are the following conversions carried out?
Propanol to 1-propoxypropane
Propanol to 1-propoxypropane
Answer
View full question & answer→$\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{OH}\xrightarrow{ \ \ \ \ \ \text{H}_2\text{SO}_4 \ \ \ \ }\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3+\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propenol} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 140^{\circ}\text{C} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1-\text{Propoxypropane}$
Question 472 Marks
Why is the reactivity of all the three classes of alcohols with conc. $\ce{HCl}$ and $\ce{ZnCl_2} ($Lucas reagent$)$ different?
Answer
View full question & answer→Reaction with hydrogen halides: Alcohols react with hydrogen halides to form alkyl halides.
$\ce{ROH + HX R-X + H_2O}$
The difference in reactivity of three classes of alcohols with $\ce{HCl}$ distinguishes them from one another $($Lucas test$).$
Alcohols are soluble in Lucas reagent $($conc. $\ce{HCl}$ and $\ce{ZnCl_2)}$ while their halides are immiscible and produce turbidity in solution.
In case of tertiary alcohols, turbidity is produced immediately as they form the halides easily.
Primary alcohols do not produce turbidity at room temperature.
The reaction is followed by the formation of carbocation since tertiary carbocation is most stable.
Thus, the order of reactivity will be $3^\circ > 2^\circ > 1^\circ .$
$\ce{ROH + HX R-X + H_2O}$
The difference in reactivity of three classes of alcohols with $\ce{HCl}$ distinguishes them from one another $($Lucas test$).$
Alcohols are soluble in Lucas reagent $($conc. $\ce{HCl}$ and $\ce{ZnCl_2)}$ while their halides are immiscible and produce turbidity in solution.
In case of tertiary alcohols, turbidity is produced immediately as they form the halides easily.
Primary alcohols do not produce turbidity at room temperature.
The reaction is followed by the formation of carbocation since tertiary carbocation is most stable.
Thus, the order of reactivity will be $3^\circ > 2^\circ > 1^\circ .$
Question 482 Marks
Write the structures of the isomers of alcohols with molecular formula $\ce{C_4H_{10}O}$. Which of these exhibits optical activity?
Answer
View full question & answer→Isomers of alcohols $\ce{(C_4H_{10}O)}$ are:
- $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \\ \ \ \text{Butan}-1-\text{ol}(1^\circ)$
- $\ \ \ \ \text{CH}_3-\text{CH}-\text{CH}_2\text{OH} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ 2\text{Methylpropan}-1-\text{ol}(1^\circ)$
- $\ \ \ \ \text{CH}_3-\text{CH}-\text{CH}_2\text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH} \\ \ \ \ \ \ \text{Butan}-2-\text{ol}(2^\circ) \\ \text{Butan}-2\text{ol os optically active}.$
- $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \text{CH}_3-\text{C}-\text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH} \\ 2-\text{methylpropan}-2-\text{ol}(3^\circ)$
Question 492 Marks
Why is the $\ce{C-O-H}$ bond angle in alcohols slightly less than the tetrahedral angle whereas the $\ce{C-O-C}$ bond angle in ether is slightly greater?
Answer
View full question & answer→The bond angle in alcohol is slightly less than the tetrahedral angle $(109^{\circ}-28\ ’).$ It is due to the repulsion between the unshared electron pairs of oxygen.
In ethers, the four electron pairs, i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement.
The bond angle is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky $(-R)$ groups.
In ethers, the four electron pairs, i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement.
The bond angle is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky $(-R)$ groups.
Question 502 Marks
Explain a process in which a biocatalyst is used in industrial preparation of a compound known to you.
Answer
View full question & answer→Biocatalysts are complex organic compounds which act as catalysts in reaction taking place in living organism. These biocatalysts (enzymes) are used in the manufacture of ethanol.
Ethanol from sugar solution $($molasses$):$
Molasses is a non$-$crystalline form of sugar obtained as the mother liquor after crystallisation of sugar from sugar solution. This contains about $50\%$ sugar. It is diluted to about $10\%$ solution and yeast is added and kept for about $2-3$ days. Yeast supplies the enzymes invertase and zymase. The enzyme invertase hydrolyses sucrose to glucose and fructose. The enzyme zymase $($found in yeast$)$ converts glucose and fructose to ethanol.
$\text{C}_{12}\text{H}_{22}\text{O}_\text{11}+\text{H}_2\text{O}\xrightarrow{\text{Invertase}}\text{C}_6\text{H}_{12}\text{O}_6+\text{C}_6\text{H}_{12}\text{O}_6 \\ \ \ \text{Sucrose} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Glucose} \ \ \ \ \ \ \text{Fructose}$
$\ \ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6 \ \ \ \ \ \ \ \xrightarrow{\text{Zymase}} \ \ \ \ \ \ \ 2\text{C}_2\text{H}_5\text{OH} \ \ \ \ \ \ + \ \ \ \ \ \ 2\text{CO}_2\\ \text{Glucose or fructose } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethanol}$
In wine making, grapes are the source of sugars and yeast. As grapes ripen, the quantity of sugar increases and the yeast grows on the skin of the grapes. When the grapes are crushed, sugar and enzyme come in contact and fermentation starts. Fermentation takes place under anaerobic conditions $($i.e., in the absence of air$).$ During fermentation $CO_{2 }$ is released.
The action of enzyme is inhibited when the concentration of alcohol exceeds $14\%.$ If air enters the fermentation mixture, the $O_2$ of the air oxidised ethanol to ethanoic acid which spoils the taste of alcoholic drinks and makes it sour.
Ethanol from sugar solution $($molasses$):$
Molasses is a non$-$crystalline form of sugar obtained as the mother liquor after crystallisation of sugar from sugar solution. This contains about $50\%$ sugar. It is diluted to about $10\%$ solution and yeast is added and kept for about $2-3$ days. Yeast supplies the enzymes invertase and zymase. The enzyme invertase hydrolyses sucrose to glucose and fructose. The enzyme zymase $($found in yeast$)$ converts glucose and fructose to ethanol.
$\text{C}_{12}\text{H}_{22}\text{O}_\text{11}+\text{H}_2\text{O}\xrightarrow{\text{Invertase}}\text{C}_6\text{H}_{12}\text{O}_6+\text{C}_6\text{H}_{12}\text{O}_6 \\ \ \ \text{Sucrose} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Glucose} \ \ \ \ \ \ \text{Fructose}$
$\ \ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6 \ \ \ \ \ \ \ \xrightarrow{\text{Zymase}} \ \ \ \ \ \ \ 2\text{C}_2\text{H}_5\text{OH} \ \ \ \ \ \ + \ \ \ \ \ \ 2\text{CO}_2\\ \text{Glucose or fructose } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethanol}$
In wine making, grapes are the source of sugars and yeast. As grapes ripen, the quantity of sugar increases and the yeast grows on the skin of the grapes. When the grapes are crushed, sugar and enzyme come in contact and fermentation starts. Fermentation takes place under anaerobic conditions $($i.e., in the absence of air$).$ During fermentation $CO_{2 }$ is released.
The action of enzyme is inhibited when the concentration of alcohol exceeds $14\%.$ If air enters the fermentation mixture, the $O_2$ of the air oxidised ethanol to ethanoic acid which spoils the taste of alcoholic drinks and makes it sour.