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Chemistry 3 Marks Question

Alcohols, Phenols, and Ethers · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 12 questions.

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Question 13 Marks
  1. Draw the structures of all isomeric alcohols of molecular formula $C_5H_{12}O$ and give their IUPAC names.
  2. Classify the isomers of alcohols in question $11.3\ (i)$ as primary, secondary and tertiary alcohols.
Answer
  1. Eight isomeric alcohols are possible:
  1. $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}\\ \ \ \ \text{Pentan-1-ol}(1^\circ)$
  2. $\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \text{2-Methylbutan-1-ol}(1^\circ)$
  3. $\text{CH}_3-\text{CH}-\text{CH}_2-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \text{3-Methylbutan-1-ol}(1^\circ)$
  4. $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \text{2,2-Dimethylpropan-1-ol}(1^\circ)$
  5. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}-\text{CH}_3\\ \text{Pentan-2-ol}(2^\circ)$
  6. $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\\ \text{3-Methylbutan-2-ol}(2^\circ)$
  7. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2-\text{CH}_3\\ \text{Pentan-3-ol}(2^\circ)$
  8. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \text{2-Methylbutan-2-ol}(3^\circ)$
  1. Primary alcohol: Pentan-1-ol; methyl butan-1-ol; 3-methyl-1-ol; 2,2Dimethyl propan-t-ol.
Secondary alcohol: Pentan-2-ol; 3-methylbutan-2-ol; pentan-3-ol.
Tertiary alcohol: 2-methylbutan-2-ol.
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Question 23 Marks
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer

The acidic nature of phenol can be represented by the following two reactions:
  1. Phenol reacts with sodium to give sodium phenoxide, liberating $H_2$.
  1. Phenol reacts with sodium hydroxide to give sodium phenoxide and water as byproducts.

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas ethoxide ion does not.
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Question 33 Marks
Match the items of column I with items of column II.
 
Column I
 
Column II
(i)
Methanol
(a)
Conversion of phenol to o-hydroxysalicylic acid
(ii)
Kolbe’s reaction
(b)
Ethyl alcohol
(iii)
Williamson’s synthesis
(c)
 Conversion of phenol to salicylaldehyde
(iv)
 Conversion of $2°$ alcohol to ketone
(d)
 Wood spirit
(v)
 Reimer-Tiemann reaction
(e)
 Heated copper at $573K$
(vi)
 Fermentation
(f)
 Reaction of alkyl halide with sodium alkoxide
Answer
 
Column I
 
Column II
(i)
Methanol
(d)
Wood spirit
(ii)
Kolbe's reaction
(a)
Conversion of phenol to o-hydroxysalicylic acid
(iii)
Williamson's synthesis
(f)
 Reaction of alkyl halide with sodium alkoxide
(iv)
 Conversion of 2° alcohol to ketone
(e)
 Heated copper at $573K$
(v)
 Reimer-Tiemann reaction
(c)
 Conversion of phenol to salicylaldehyde
(vi)
 Fermentation
(b)
 Ethyl alcohol.
Explanation:
  1. Methanol is also known as 'wood spirit' as it was produced by the destructive distillation of wood.
  2. In Kolbe’s reaction, 2-hydroxy benzoic acid (salicylic acid) is prepared by the reaction of phenol with $CO_2$ gas.
  1. Williamson,s synthesis is am important method for the preparation of ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.
R-X + R-ONa → ROR + NaX
  1. When a 2 alcohol is allowed to pass over heated copper at 573 K, dehydrogenation takes place and an ketone is formed.
$\text{R}-\text{CH}-\text{R}'\xrightarrow[573\text{ K}]{\text{Cu}}\text{R}-\text{C}-\text{R} \\ \ \ \ \ \ \ \ \ \ {|} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {||} \\ \ \ \ \ \ \ \ \ \text{OH} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ O}$
  1. On trating phenol with chloroform in the presence of NaOH, an aldehydic group is introduced at ortho position of benzene ring
  1. Ethanol is prepered by the fermentation of sugars.
$\text{C}_{12}\text{H}_22\text{O}_{11}+\text{H}_2\text{O}\xrightarrow{\text{Invertase}}\text{C}_6\text{H}_{12}\text{O}_6+\text{C}_6\text{H}_{12}\text{O}_6$
$\text{C}_6\text{H}_{12}\text{O}_6\xrightarrow{\text{Zymase}}2\text{C}_2\text{H}_5\text{OH}+2\text{CO}_2$
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Question 43 Marks
Arrange water, ethanol and phenol in increasing order of acidity and give reason for your answer.
Answer

Phenol > Water > Ethanol

An alkoxide ion is a better proton acceptor than hydroxide ion, which suggests that alkoxides are stronger bases (sodium ethoxide is a stronger base than sodium hydroxide). The reaction of phenol with aqueous sodium hydroxide indicates that phenols are stronger acids than alcohol and water.
The ionisation of an alcohol and a phenol takes place as follows:

In alkoxide ion, the negative charge is localised on oxygen while in phenoxide ion, the charge is delocalised. The delocalisation of negative charge (structures I-V) makes.
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Question 53 Marks
Identify the product of the following reaction:$ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\\text{CH}_3-\text{C}-\text{CH}_2-\text{Br}\xrightarrow{\stackrel{{}}{\hbox{Na}}\stackrel{{}}{\hbox{OH}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta\\ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Answer

Neopentyl bromide ionises to form first a 1° carbocation which rearranges to form the more stable 3° carbocation. This is attacked by weak nucleophile ethanol followed by loss of proton to yield ethyl tert-pentyl ether.
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Question 63 Marks
Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer

The electron withdrawing groups are more effective in increasing the acidic strength at the para-position relative to the ortho position because of greater dispersal of charge on oxygen atom.
Resonance structure of phenoxide ion:

Resonance structure of paranitrophenol:

Resonance structures of ortho nitro phenol:

Thus, presence of nitro group at ortho and para position incerease the acidic character.
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Question 73 Marks
Which is a stronger acid-phenol or cresol? Explain.
Answer
All the cresols are weaker acids than phenols. Methyl group has +I effect (positive inductive effect) as well as hyperconjugation effect but the hyperconjugation effect predominates over the +I effect. Since both these effects increase the electron density in the O-H bond and hence all the cresols are weaker acids than phenols
As hyperconjugation effect can operate only through ortho and para positions and not through meta positions, therefore, meta-cresol is stronger acid than ortho and para-cresols. However, due to stronger +I effect at ortho position than at para position (+I effect decreases with distance), ortho-cresol is a weaker acid than para-cresol. Thus, the order of acidic strength in increasing order is:
ortho-cresol < para-cresol < meta-cresol < phenol
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Question 83 Marks
Match the items of column I with items of column II.
 
Column I
 
Column II
(i)
Antifreeze used in car engine
(a)
Neutral ferric chloride
(ii)
Solvent used in perfumes
(b)
Glycerol
(iii)
Starting material for picric acid
(c)
Methanol
(iv)
Wood spirit
(d)
Phenol
(v)
Reagent used for detection of phenolic group
(e)
Ethleneglycol
(vi)
By product of soap industry used in cosmetics
(f)
Ethanol.
Answer
 
Column I
 
Column II
(i)
Antifreeze used in car engine
(e)
Ethleneglycol
(ii)
Solvent used in perfumes
(f)
Ethanol
(iii)
Starting material for picric acid
(d)
Phenol
(iv)
Wood spirit
(c)
Methanol
(v)
Reagent used for detection of phenolic group
(a)
 Neutral ferric chloride
(vi)
By product of soap industry used in cosmetics
(b)
Glycerol.
Explanation:
  1. Ethleneglycol (IUPAC name of ethylene glycol is ethane-1,2-diol a small percentage of it is used in antifreeze formulation).
  2. Ethanol (it is less irritating to skin so it is used in perfumes).
  3. Phenol (by the reaction of phenol with conc. $HNO_3$ phenol can be converted into picric acid).
  4. Methanol (methanol is known as wood spirit as it was obtained by destructive distillation of wood).
  5. Neutral ferric chloride (Neutral ferric chloride gives violet colour when treated with phenol).
  6. Glycerol (it is the by product in soap industry and is used in cosmetics).
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Question 93 Marks
How are the following conversions carried out?
Propanol to propan-2-ol
Answer
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{OH}\xrightarrow{ \ \ \ \ \ \text{H}_2\text{SO}_4 \ \ \ \ }\text{CH}_3-\text{CH}=\text{CH}_2\xrightarrow{\text{H}_2\text{O}/\text{H}^{+}}\text{CH}_3-\text{CH}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \text{Propanol} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 170^{\circ}\text{C} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propan-2-ol}$
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Question 103 Marks
An organic compound ' $A$ ' having molecular formula $\mathrm{C}_3 \mathrm{H}_6$ on treatment with aqueous $\mathrm{H}_2 \mathrm{SO} 4$ gives ' $B$ ' which on treatment with $\mathrm{HCl} / \mathrm{ZnCl}_2$ gives ' C '. The compound C on treatment with ethanolic KOH gives back the compound ' A '. Identify the compounds A, B, C.
Answer
$\text{A}=\text{CH}_3-\text{CH}=\text{CH}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propene}$
$\text{B}=\text{CH}_3-\text{CH}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propan-2-ol}$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\\text{C}=\text{CH}_3-\text{CH}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ 2-\text{chloropropane}$
$\text{CH}_3-\text{CH}=\text{CH}_2\xrightarrow{ \ \ \ \ \ \ \text{aq}.\text{H}_2\text{SO}_4 \ \ \ \ \ }\text{CH}_3-\text{CH}-\text{CH}_3\xrightarrow{\text{HCl}/\text{ZnCl}_2}\\ \ \ \ \ \ \ \ \ \ \text{Propene} \ \ \ \ \ \ \ \ \ (\text{Hydration}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propan-2-ol}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{B}$
$\text{CH}_3-\text{CH}-\text{CH}_3\xrightarrow{\text{Ethanolic KOH}}\text{CH}_3-\text{CH}=\text{CH}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{|} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propene}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\2-\text{Chlorophane}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}$
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Question 113 Marks
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer
Williamson's synthesis is a versatile method for the synthesis of both symmetrical and unsymmetrical ethers. However, for the synthesis of unsymmetrical ethers, a proper choice of reactants is necessary. Since Williamson's synthesis occurs by $SN_2$ mechanism and primary alkyl halides are most reactive in $Sn_2$ reaction, therefore, best yields of unsymmetrical ethers are obtained when the alkyl halides are primary and the alkoxide may be primary, secondary or tertiary. For example, tert-butylethyl ether is prepared by treating ethyl bromide with sodium tert-butoxide.$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta+ \ \ \ \ \ \delta- \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \text{CH}_3-\text{C}-\text{O}^-\text{Na}^++\text{CH}_3\text{CH}_2-\text{Br}\xrightarrow[]{\ \ \Delta\ }\text{CH}_3-\text{C}-\text{OCH}_2\text{CH}_3+\text{Na}^+\text{Br}^-\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ (3^\circ\text{ Alkoxide})$
The above ether cannot be prepared by treating sodium ethoxide with tert-butyl chloride or bromide since under these condition an alkene, i.e., isobutylene is the main product.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\$\text{CH}_3)_3-\text{C}-\text{Br}+\text{C}_2\text{H}_5\text{O}^-\text{Na}^+\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{CH}_3-\text{C}=\text{CH}_2+\text{NaBr}+\text{C}_2\text{H}_5\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{2-Methylpropene}$
Aryl and vinyl halides cannot be used as substrates because they are less reactive in nucleophilic substitution.
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Question 123 Marks
Answer the following question:
An organic compound A having molecular formula $\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}$ gives a characteristic colour with aqueous $\mathrm{FeCl}_3$ solution. A on treatment with $\mathrm{CO}_2$ and NaOH at 400 K under pressure gives B which on acidification gives a compound C . The compound $C$ reacts with acetyl chloride to give $D$ which is a popular pain killer. Deduce the structure of $A, B, C$ and $D$.
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