Careers360 Logo
NCERT Exemplar Class 12 Chemistry Solutions chapter 11 Alcohols, Phenols and Ethers

NCERT Exemplar Class 12 Chemistry Solutions chapter 11 Alcohols, Phenols and Ethers

Edited By Sumit Saini | Updated on Sep 16, 2022 05:55 PM IST | #CBSE Class 12th

In NCERT exemplar Class 12 Chemistry solutions chapter 11, students will learn the uses of alcohol in medicines in various forms as antiseptic disinfectant, antidote, etc. Alcohols are also used in Mouthwash, cosmetics, syrups and so on. Ethers are used in perfumes, resins, dyes, insecticides, etc. Phenols are used as surgical antiseptics and disinfectants in household cleaners. They are also used in explosives. Every topic in this NCERT exemplar Class 12 Chemistry chapter 11 solutions is fairly and accurately explained.

These functional groups are essential parts of organic chemistry. Phenols are similar to alcohols but form stronger hydrogen bonds.
Also, check - NCERT Solutions for Class 12, other subjects

NCERT Exemplar Class 12 Chemistry Solutions Chapter 11: MCQ (Type 1)

Question:1

Monochlorination of toluene in sunlight followed by hydrolysis with aq. \text {NaOH} yields.
(i) o-Cresol
(ii) m-Cresol
(iii) 2, 4-Dihydroxytoluene
(iv) Benzyl alcohol
Answer:

The answer is the option (iv). The monochlorination of toluene in sunlight results in halogenation that follows a free radical pathway and the product will react with the alkyl group forming haloalkyl.
And in the alkaline medium, the substitution reaction will take place resulting in the deduction of a hydrogen atom from the haloalkyl group. The end product formed will be alcohol; therefore, the monochlorination of toluene in light in the presence of aq. NaOH forms benzyl alcohol. So, option (iv) is the correct answer.

Question:2

How many alcohols with molecular formula \text {C}_{4}\text {H}_{10}\text {O} are chiral?
(i) 1
(ii) 2
(iii) 3
(iv) 4
Answer:

The answer is the option (i). To identify the chirality of any compound in chemistry, the carbon present should be surrounded by four different groups and should not have any plane of symmetry.
The possible alcohols with the molecular formula \text {C}_{4}\text {H}_{10}\text {O} are –
- \text {1-Butanol}
- \text {2-Butanol}
- \text { 2-methyl-1-propanol}
- \text { 2-methyl-1-propanol}
If all the alcohol's chemical structure is drawn of the possible compounds, then only one of them contains carbon with four different groups that are \text {2-Butanol} , so the right answer is the option (i) 1.

Question:3

What is the correct order of reactivity of alcohols in the following reaction?
\text {R-OH+HCl} \overset{ZnCl_2}{\rightarrow} \text {R-Cl}+\text {H}_{2}\text {O}

(i) 1^{o}>2^{o}>3^{o}
(ii) 1^{o}<2^{o}>3^{o}
(iii) 3^{o}>2^{o}>1^{o}
(iv) 3^{o}>1^{o}>2^{o}

Answer:

The answer is option (iii). When dry anhydrous \text {ZnCl}_{2} is mixed with concentrated \text {HCl}, it forms Lucas reagent and the reaction involved is knows as the Lucas test. Formation of an intermediate carbocation takes place during the reaction, this reaction being a slow step of the mechanism, this carbocation dictates the rates of reaction.
According to the \text {S}_N{1} mechanism, the alcohol reactivity order is 3^{o}>2^{o}. But tertiary alcohol reacts very fast with Lucas reagent, so the reactivity order of alcohols in this particular reaction is –
\text {R-OH+HCl} \overset{ZnCl_2}{\rightarrow} \text {R-Cl}+\text {H}_{2}\text {O}
3^{o}>2^{o}>1^{o}
So the correct option is (iii) 3^{o}>2^{o}>1^{o}

Question:4

\text {CH}_{3}\text {CH}_{2}\text {OH} can be converted into \text {CH}_{3}\text {CHO} by ______________.
(i) catalytic hydrogenation
(ii) treatment with \text {LiAIH}_{4}
(iii) treatment with pyridinium chlorochromate
(iv) treatment with \text {KMnO}_{4}
Answer:

The answer is the option (iii). The compound \text {CH}_{3}\text {CH}_{2}\text {OH} oxidises and forms \text {CH}_{3}\text {CHO} as the end product. In the oxidation reaction process, the addition of pyridinium chlorochromate, a strong oxidising agent, breaks the cycle by forming aldehydes and preventing complete oxidation of \text {CH}_{3}\text {CH}_{2}\text {OH}. As the reaction is halted at the aldehyde stage, the end product formed is \text {CH}_{3}\text {CHO}.
So \text {CH}_{3}\text {CH}_{2}\text {OH} can be converted into \text {CH}_{3}\text {CHO} by option (iii) treatment with pyridinium chlorochromate

Question:5

The process of converting alkyl halides into alcohols involves_____________.
(i) addition reaction
(ii) substitution reaction
(iii) dehydrohalogenation reaction
(iv) rearrangement reaction
Answer:

The answer is the option (ii). To covert alkyl halides into alcohols, the reaction involved is –
\text {R-X}\rightarrow \text {R-OH}
In order to get the required result, the replacement of \text {-X} into \text {-OH} is necessary. And as replacing one of the groups is required, the type of reaction involved is known as substitution reaction. Thus, the correct option is (ii) substitution reaction.

Question:6

Which of the following compounds is aromatic alcohol?

(i) A, B, C, D
(ii) A, D
(iii) B, C
(iv) A
Answer:

The answer is the option (iii). The characteristics of an aromatic compound are that they have an \text {-OH} group indirectly attached to the benzene ring. So, Option (iii) B, C is the right answer.

Question:7

Give IUPAC name of the compound given below.

(i) \text {2-Chloro-5-hydroxyhexane}
(ii)\text { 2-Hydroxy-5-chlorohexane}
(iii) \text { 5-Chlorohexan-2-ol}
(iv) \text { 2-Chlorohexan-5-ol}
Answer:

The answer is the option (iii). To determine the IUPAC name of any chemical compound, the steps of nomenclature included are –
- Identifying the functional group and their correct order of priority. -OH having a higher priority, the suffix used will be '-ol'.
- Recognise the parental chain, which is the longest carbon chain containing the functional group. In this care, the
the chain consists of 6 carbons hence 'hexyl'.
- Numbering of the chain by giving the smallest number to the functional group and other side chains; thus, '2-ol'.
- Organise the side chain atoms/molecules and give, so they are chloro and 5-chloro in the mentioned compound.
- The resulting IUPAC name is 5-Chlorohexan-2-ol.
The conclusion of the correct option is (iii) 5-Chlorohexan-2-ol

Question:8

IUPAC name of m-cresol is ___________.
(i) 3-methyl phenol
(ii) 3-chlorophenol
(iii) 3-methoxy phenol
(iv) benzene-1,3-diol
Answer:

The answer is the option (i). The structure of m-cresol consists of a benzene ring with \text {-OH} functional and \text {-CH}_{3} substituent attached. m-cresol is the common name assigned to the compound, and according to the rules of nomenclature, the IUPAC name is of m-cresol is 3-methyl phenol. So, Option (i) 3-methyl phenol is the right answer.

Question:9

IUPAC name of the compound is ______________.

(i) \text {1-methoxy-1-methylethane}
(ii) \text {2-methoxy-2-methylethane}
(iii) \text {2-methoxypropane}
(iv) \text {isopropylmethyl ether}
Answer:

The answer is the option (iii). To name the compounds containing carbon as their substituent group, steeps of nomenclature involved are –
1. Identify the longest carbon chain, also known as the parent chain. Calculate the number of carbons present and assign its root name.
2. Numbering of the parent chain by assigning the smallest number to the first substituent.
3. Recognise the position and name of all substituent carbon groups.
4. In the presence of identical groups, use the group prefix such as di, tri, tetra, etc.
5. Alphabetically arrange the substituent position and names in front of the root name. While making the order, prefix like di, tri, tetra should not be accounted for except iso and cyclo.
By following these rules, the IUPAC name of the given compound is \text {2-methoxypropane}. So the correct option is (iii) \text {2-methoxypropane}

Question:10

Which of the following species can act as the strongest base?

Answer:

The answer is the option (ii). The strength of a base is determined by the corresponding conjugate acid. If the acid is weak, the base is strong.
\text {OH+H}\rightarrow \text {H}_{2}\text {O}\left ( \text {Water} \right )
\text {OH+H}\rightarrow \text {ROH}\left ( \text {alcohol} \right )
\text {OC}_{6}\text {H}_{5}+\text {H}\rightarrow \text {C}_{6}\text {H}_{5}\text {OH}\left ( \text {phenol} \right )
\text {OC}_{6}\text {H}_{5}\left ( \text {m-NO}_{2} \right )+\text {H}\rightarrow \text {HOC}_{6}\text {H}_{5}\left ( \text {m-NO}_{2} \right )\left ( \text {m-nitrophenol} \right )
In the mentioned options, the phenol having a benzene ring has higher acidity than aliphatic alcohols. Since the acidity of water is more than alcohol, it makes \text {ROH} the least acidic product. Since, according to the above-mentioned point, the weaker acid makes a strong base. \text {RO} will be the strongest base in all four options. This makes option (ii) as is the right answer.

Question:11

Which of the following compounds will react with sodium hydroxide solution in water?
(i) C_{6}H_{5}OH
(ii) C_{6}H_{5}CH_{2}OH
(iii) \left ( CH_{3} \right )_{3}COH
(iv) C_{2}H_{5}OH
Answer:

The answer is the option(i). Sodium hydroxide is a strong base and to be able to make the reaction feasible; the other acidic reactant should be strong. Option (ii), (iii), and (iv) are alcohols and are weak in nature due to less affinity towards water thus they are able to dissociate into ions only to a certain extent. Whereas option (i) phenol is a strong acid due to its higher affinity towards water. This makes option (i) as the right answer

Question:12

Phenol is less acidic than ______________.
(i) ethanol
(ii) o-nitrophenol
(iii) o-methyl phenol
(iv) o-methoxy phenol
Answer:

The answer is the option (ii). The acidity of phenol is determined by its electron-withdrawing or donating ability. The (NO_{2}) present in phenol is an electron-withdrawing group, so it increases the acidity whereas groups like (-CH_{3}, -OCH_{3}) being electron-donating groups, they decrease the acidity of phenol is less acidic compared to o-nitrophenol from the given options. Thus, the correct option is (ii) o-nitrophenol

Question:13

Which of the following is most acidic?
(i) Benzyl alcohol
(ii) Cyclohexanol
(iii) Phenol
(iv) m-Chlorophenol
Answer:

The answer is the option (iv). The addition of electron-withdrawing groups in phenol increases its acidity, whereas the addition of electron-donating groups decreases the acidity. And as a known fact, that the (+R) effect duet to donating is dominant over the (+I) effect. But this is only feasible in the case of substituents at ortho and para positions.
Keeping the above-mentioned point in mind, the m-Chlorophenol compound is the most acidic of all four options. So, option (iv) is the answer.

Question:14

Mark the correct order of decreasing acid strength of the following compounds.

\\(i) e > d > b > a > c\\ (ii) b > d > a > c > e\\ (iii) d > e > c > b > a\\ (iv) e > d > c > b > a
Answer:

The answer is the option (ii). According to the (+R) effect in the cyclic compound, the presence of an electron-withdrawing group at para position to -OH makes the acidity level maximum of the compound whereas having an electron-donating group at the para position to -OH makes the acidity level minimum. So, according to this concept, the correct decreasing order of acidity of the given compounds is Option (ii) b > d > a > c > e

Question:15

Mark the correct increasing order of reactivity of the following compounds with HBr/HCl.

(i) a < b < c
(ii) b < a < c
(iii) b < c < a
(iv) c < b < a
Answer:

The answer is the option (iii). The type of reaction taking place in this process is the nucleophilic substitution reaction. The mechanism followed is S_{N}1, so the first slow step of the mechanism involves the formation of carbocation that determines the rate of reaction and then the addition of a nucleophile. The carbocation formed also controls the reactivity of this product formed. But in case of the presence of a highly electron-withdrawing group (-NO_{2}), the carbocation will be destabilised, and the reaction will not be favoured.
Whereas the high electron-withdrawing nature of the (-Cl) group due to lone pair presence, its tendency of electron-withdrawing is slightly low than (-NO_{2}), it will also manage to destabilise the carbocation and do not favour the reaction.
In the case of the first compound, C_{6}H_{5}CH_{2}OH, the absence of any destabilising factors makes the reaction occurrence favourable. Thus, the concluded increasing order of reactivity will be Option (iii) b < c < a

Question:16

Arrange the following compounds in increasing order of boiling point.
Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol

\\(i) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol\\ (ii) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol\\ (iii) Pentan-1-ol, butan-2-ol, butan-1-ol, propan-1-ol\\ (iv) Pentan-1-ol, butan-1-ol, butan-2-ol, propan-1-ol
Answer:

The answer is the option (i). To find out the order of boiling point in alcohols, the important factor is molecular mass. With an increasing molecular mass of the compound, the boiling point also increases. So, the compound with a maximum boiling point will be pentan-1-ol, and the compound with the least boiling point is Propan-1-ol.
The correct order would be Option (i) \text {Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol}

NCERT Exemplar Class 12 Chemistry Solutions Chapter 11: MCQ (Type 2)

Question:17

Which of the following are used to convert RCHO into RCH_{2}OH ?
(i) {H_{2}}/{Pd}
(ii) LiAlH_{4}
(iii) NaBH_{4}
(iv) Reaction with RMgX followed by hydrolysis
Answer:

The answer is the option (i), (ii), and (iii). The above-mentioned reactant is an aldehyde that is converted into alcohol. This type of reaction is called a reduction reaction. The reduction reaction is carried out in the presence of hydrogen metal catalysts to increase the rate of the reaction. The metal catalyst used is {H_{2}}/{Pd}. If it is replaced by NaBH_{4} or LiAlH_{4}, the reaction will still take place to get the exact result. This makes option (i), (ii), and (iii) as the correct answers.

Question:18

Which of the following reactions will yield phenol?

Answer:

The answer is the option (A), (B), (C). To carry out these reactions, the conditions required should be easy to maintain. The reaction occurring in (D) is a nucleophilic substitution of chlorobenzene which is not feasible as it requires drastic temperature and pressure conditions.
The reaction given in (A) is the Dow process, (B) is aniline diazotisation and (C) option contains a reaction from benzene sulphonic acid. All these reactions are feasible due to manageable reaction conditions. The correct options are Option (A), (B), and (C)

Question:19

Which of the following reagents can be used to oxidise primary alcohols to aldehydes?
(i) CrO_{3} in anhydrous medium.
(ii) KMnO_{4} in acidic medium.
(iii) Pyridinium chlorochromate.
(iv) Heat in the presence of Cu at 573K

Answer:

The answer is the option (i), (iii), and (iv). The complete oxidation of primary alcohols results in the formation of carboxylic acid unless a mild oxidising agent is added to terminate the reaction at the aldehyde formation step. The most common mild oxidising agent used is PCC (Pyridinium chlorochromate). Some other compounds used for this purpose are CrO_{3} in an anhydrous medium and Heat in the presence of Cu at 573K. So, the correct answers are an option (i), (iii), and (iv)

Question:20

Phenol can be distinguished from ethanol by the reactions with _________.
(i) {Br_{2}}/{water}
(ii) Na
(iii) Neutral FeCl_{3}
(iv) All the above
Answer:

The answer is the option (i) and (iii). Phenol can easily be distinguished from ethanol by adding bromine water as it results in a white ppt of tri-bromo phenol or Neutral FeCl_{3} can also be used for this purpose. But ethanol being aliphatic alcohol will not react with either of the compounds. So, the right answer is option (i) and (iii)

Question:21

Which of the following are benzylic alcohols?

Answer:

The answer is the option (ii) and (iii). The most prominent way to identify benzylic alcohol is to check if an -OH group is attached to a sp^{3} hybridised carbon atom on an aromatic ring. As in the case of option (ii) and (iii), the sp^{3} hybridised carbon is attached to a benzene ring these are benzylic alcohols but in option (i) and (iv), the sp^{3} hybridised carbon is not attached to a benzene ring. So, option (ii) and (iii) are the answers.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 11: Short Answer Type

Question:22

What is the structure and IUPAC name of glycerol?
Answer:

Glycerol comes under the category of trihydric alcohol and the IUPAC name according to the nomenclature rules is \text {Propane-1,2,3-triol}

Question:23

Write the IUPAC name of the following compounds.

Answer:

The systematic structure of naming of organic compounds in chemistry is given International Union of Pure and Applied Chemistry (IUPAC).
The IUPAC name of the mentioned compounds are
(A) \text {3-Ethyl-5-methyl hexane-2,4-diol}.
(B) \text {1-Methoxy-3-nitrocyclohexane.}

Question:24

Write the IUPAC name of the compound given below.

Answer:

After following the laws of nomenclature designed by IUPAC, the name of the compound is 3-Methylpent-2-ene-1,2-diol.

Question:25

Name the factors responsible for the solubility of alcohols in water.
Answer:

The high affinity of alcohols with water comes due to the following listed reasons –
I) Hydrogen bonds – with a greater number of H-bonds, the solubility of the compound increases.
II) Size of the alkyl or aryl groups- the size of alkyl or aryl groups attached determines the solubility as larger size contributes to a smaller number of H-bonds that decrease the solubility factor.
III) The molecular mass of the Alcohols – the larger molecular mass makes the compound less soluble due to difficulty in dissociation.

Question:26

What is denatured alcohol?
Answer:

To prevent alcohol stealing during the manufacturing process, it is mixed with additional components like copper sulfate and pyridine to make it unfit for drinking purposes. These substances change the colour and make the odour foul.

Question:27

Suggest a reagent for the following conversion.

Answer:

The above-given reaction is the formation of ketone from secondary alcohol through the process of oxidation. The most common oxidising agents used in this reaction are Pyridinium chlorochromate (PCC), chromic anhydride (CrO_{3}), etc.

Question:28

Out of 2-chloroethanol and ethanol, which is more acidic and why?
Answer:

The acidity of the compound depends on the type of group attached to the compound, due to Cl being an electron-withdrawing group, the acidity of 2-chloroethanol is more. This leads to a decrease in electron density in the -O-H bond as a (-I) ( negative inductive effect) effect is created, This makes it very easy for 2-chloroethanol to release a proton

Question:29

Suggest a reagent for conversion of ethanol to ethanal.
Answer:

The reaction involved is the conversion of primary alcohol to an aldehyde by oxidation. The most feasible reagent for the above reaction is Pyridinium chlorochromate (PCC)

Question:30

Suggest a reagent for conversion of ethanol to ethanoic acid.
Answer:

To covert primary alcohol to a carboxylic acid through an oxidation reaction, the most efficient reagent is Acidified \text {KMnO}_{4}.

Question:31

Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.
Answer:

The bonds present in o-nitrophenol are intramolecular H-bonds, but in the case of p-nitrophenol, the bonds formed are intermolecular H-bonds. When these compounds are heated, the boiling temperature of p-nitrophenol gets elevated due to strong intermolecular H-bonds, but the same does not happen with o-nitrophenol. This makes the o-nitrophenol more volatile.

Question:32

Out of o-nitrophenol \; and\; o-cresol, which is more acidic?
Answer:

The presence of an electron-withdrawing group at para and ortho position increases the acidity of the cyclic compounds. As NO_{2} is present in o-nitrophenol, it makes it more acidic, but in o-cresol, the electron releasing group decreases its acidity.

Question:33

When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of the compound formed.
Answer:

Treating phenol with bromine water makes a white precipitate as a by-product. The compound formed is named as 2,4,6-tribromophenol, and the structure of this compound is given below -

Question:34

Arrange the following compounds in increasing order of acidity and give a suitable explanation.
Phenol, o-nitrophenol, o-cresol

Answer:

The acidity factor of these compounds depends on the type of groups attached. If the group is electron-withdrawing, they increase the acidity of the compound due to (-I) and (-R) effect, which is exactly the case in o-nitrophenol, making it most acidic. Also, due to the (+I) effect depicted by o-cresol, it is weaker than phenol.
So, the correct Increasing order of the acidity of the given compounds: o-cresol < Phenol < o-nitrophenol.

Question:35

Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides.Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.
Answer:

All the alkyl groups exhibit the (+I) effect and with an increasing number of alkyl groups from Primary, secondary to tertiary alcohols, the -OH bond strength of increases due to elevation in the electron density. There is a low reactivity of sodium with tertiary alcohol due to steric hindrance and difference in nature. So, the correct decreasing order sodium metal reactivity towards alcohol is Primary alcohols > secondary alcohols > tertiary alcohols.

Question:36

What happens when benzene diazonium chloride is heated with water?
Answer:

During the treatment of benzene diazonium chloride in water, multiple by-products Nitrogen gas and Hydrochloric acid are formed. The major product formed is phenol.

Question:37

Arrange the following compounds in decreasing order of acidity.
H_{2}O, ROH, HC \equiv CH

Answer:

To determine the acidity of these given compounds, they are treated with water. So, when RONa is treated with water, it gets replaced by a stronger acid, and a similar happens with sodium ethynide on treatment with water and alcohol. The result is acetylene; this indicates that the strongest acids are water and alcohol. So the decreasing order of acidity of these compounds is- H_{2}O > ROH > HC \equiv CH

Question:38

Name the enzymes and write the reactions involved in the preparation of ethanol from sucrose by fermentation.
Answer:

The enzymes used in the fermentation process of converting ethanol from sucrose are invertase and zymase.
The enzyme invertase processes sucrose into glucose and fructose and later in the presence of zymase, glucose and fructose are converted into ethanol. The reaction involved in the process is given below:

Question:39

How can propane-2-one be converted into tert- butyl alcohol?
Answer:

When ketones are treated with dry ether (Grignard's Reagent), they get converted into tertiary alcohols after hydrolysis. Thus to convert Propane-2-one to tert- butyl alcohol in the presence of Grignard's Reagent, you need to add CH_{3}MgBr and also perform hydrolysis to get the desired result.

Question:40

Write the structures of the isomers of alcohols with molecular formula C_{4}H_{10}O. Which of these exhibits optical activity?
Answer:

To become an optically active, compounds are required to be chiral. To check the chirality of any compound, they must have a carbon atom attached with four different groups and have no plane of symmetry. From the above-mentioned compounds, only Butan-2-ol is optically active as the rest are achiral compounds.

Question:41

Explain why is OH group in phenols is more strongly held as compared to OH group in alcohols.
Answer:

Even though both alcohol and phenol groups contain -OH group, the bond strength is much higher in phenols as compared to alcohols due to the direct attachment of -OH group to an sp^{2} hybridised carbon in phenol. The bond size is relatively smaller, and because of resonance & charge distribution, it acquires a partial double bond character in phenolic compounds.

Question:42

Explain why nucleophilic substitution reactions are not very common in phenols.
Answer:

For the feasibility of a nucleophilic substitution reaction, the compound present should be able to uptake the nucleophile involved, but in the case of phenols, the resonance effect on -OH bond gives its electron to the benzene ring making the electron density very high. Due to this high electron density, nucleophiles are repelled and are not able to substitute any atom from the phenol compound.

Question:43

Preparation of alcohols from alkenes involves the electrophilic attack on an alkene carbon atom. Explain its mechanism.

Answer:

To prepare alcohols from alkene in the presence of sulphuric acid through the process of hydration of alkenes. The reactions take place under the influence of Markownikoff's rule, and there are major 3 steps involved in the mechanism.
Step (1) Protonation of alkene and formation of a carbocation

Step (2) Nucleophilic attack of water

Step (3) Deprotonation occurs, and alcohol is formed. H_{3}O^{+} is released.

Question:44

Explain why is O=C=O non-polar while R-O-R is polar.
Answer:

The dipole movement of both compounds decides whether they are polar or non-polar. As O=C=O is a linear molecule, it is non-polar in nature because of equal distribution of its diploe movement in both directions. This makes the net dipole movement zero. But in the case of ethers, the dipole movement might be in opposite directions, but the net movement is not zero, which makes it a non-polar molecule.

Question:45

Why is the reactivity of all the three classes of alcohols with cone. \text {HCl} and \text {ZnCl}_{2} (Lucas reagent) different?
Answer:

The reactivity factor of all alcohols depends on the carbocation formed. Since the reaction mechanism followed by these compounds is S_{N}1, the higher stability of intermediate carbocation formed in the first slow step of reaction makes the compound more reactive. The tertiary carbocation formed are most stable, making these compound most reactive. But in case of secondary alcohols, they do not show any turbidity at room temperature, but on heating, they exhibit turbidity, so the carbocation formed is slightly more stable than the primary alcohols.

Question:46

Write steps to carry out the conversion of phenol to aspirin.
Answer:

The most common method of preparing aspirin is reacting salicylic acid with acetic anhydride. Reacting phenol with CO_{2} and NaOH yields salicylic acid. The name of the reaction is Kolbe's reaction. Later in the reaction, the acetylation of -OH group takes place by replacing hydrogen with an acetyl group.

Question:47

Nitration is an example of aromatic electrophilic substitution, and its rate depends upon the group already present in the benzene ring. Out of benzene and phenol, which one is more easily nitrated and why?
Answer:

The hydroxyl group present in phenol makes more nitrate. And because of the resonance of -OH bond, it makes the benzene ring rich in electron density which further makes it feasible for electrophilic substitution reaction. This indicates that the nitration, also known as electrophilic substitution, takes place in compounds with high electron density.

Question:48

In Kolbe's reaction, instead of phenol, phenoxide ion is treated with carbon dioxide. Why?
Answer:

The phenoxide ion used in the reaction is more reactive as compared to phenol due to the ability to donate its lone pair. This increases the compounds' affinity towards the electrophilic substitution reaction. These abilities make the compound a stronger nucleophile compared to phenol, and it is able to react easily with CO_{2}

Question:49

The dipole moment of phenol is smaller than that of methanol. Why?
Answer:

The dipole movement is dependent on the polarity of the compound. They are directly proportional, and in the case of phenol, the carbon atom is sp^{2} hybridised, creating an electron-withdrawing effect on the benzene ring. But in methanol, the hybridisation of the carbon is sp^{3}, which results in an electron releasing effect. Due to this scenario, the bond polarity of the C-O bond is more in phenol as compared to methanol. This directly results in smaller dipole movement in phenol.

Question:50

Ethers can be prepared by Williamson synthesis in which an alkyl halide is reacted with sodium alkoxide. Di-tert-butyl ether cannot be prepared by this method. Explain.
Answer:

Alkaloids being strong base as well as nucleophiles can react with tertiary alkyl halides through elimination reaction. During the reaction of tert-butyl-bromide with tert-butoxide rather than the expected substitution reaction, it undergoes the elimination reaction. Due to this mishappening, the product formed is Iso butylene instead of Di-tert-butyl ether.

Question:51

Why is the C-O-H bond angle in alcohols slightly less than the tetrahedral angle, whereas the C-O-C bond angle in ether is slightly greater?
Answer:

Bond angle of C-O-H being slightly less in alcohol compared to the tetrahedral bond angle. This phenomenon occurs because-
The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle because of the repulsion between the two lone pairs of electrons on oxygen atoms as these pairs pushes the C-O bonds closer to each other.
In case of ethers, the oxygen atom is attached to bulky alkyl groups, hence repulsion between these R groups pushes the two C-O bonds away from each other and hence the bond angle is slightly greater than the tetrahedral angle.

Question:52

Explain why low molecular mass alcohols are soluble in water.
Answer:

The occurrence of intermolecular hydrogen bonds due to the H-bonds in alcohol compounds plays a major role. With the increasing number of alkyl groups, the molecular mass increases, due to which the polar nature of these compounds gets suppressed. Hence, the solubility factor of alcohol is indirectly proportional to the molecular mass.

Question:53

Explain why p-nitrophenol is more acidic than phenol.
Answer:

The presence of the electron-withdrawing group in cyclic compounds is responsible for providing the compound its acidic strength, and in para-phenol, a strong electron-withdrawing group (NO_{2}) is present at par position. There is no group present in phenol. This presence increases the acidic strength as well as stabilises the phenoxide ion.

Question:54

Explain why alcohols and ethers of comparable molecular mass have different boiling points?
Answer:

Due to the presence of intermolecular H-bonds in alcohol, the energy required to break H-bonds is more than the energy required to break simple dipole-dipole bonds present in ethers. Thus, the boiling point of alcohol is higher than the boiling point of ethers.

Question:55

The carbon-oxygen bond in phenol is slightly stronger than that in methanol. Why?
Answer:

The higher bond strength in the compound is due to the resonance effect in the molecule that mimics double bond characters in C-O bonding. Therefore, the size of this bond is decreased in phenol. Also, the oxygen atom in phenol is directly attached to the sp^{3} hybridised carbon atom by single bond. Hence the carbon-oxygen bond in phenol is slightly stronger than that in methanol.

Question:56

Arrange water, ethanol and phenol in increasing order of acidity and give a reason for your answer.
Answer:

When all the above molecules are treated with water, the phenol molecule easily forms phenoxide ion and also gets stabilised duet to resonance. These features increase its acidity, but in the case of ethanol, the ion formed is ethoxide ion which is very unstable; thus, the acidity of the compound is very less. But water being a good proton donor is slightly more acidic than ethanol. The increasing order of acidity is: Ethanol < water < phenol

NCERT Exemplar Class 12 Chemistry Solutions Chapter 11: Matching Type

Question:57

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

Apply to Aakash iACST Scholarship Test 2024

Applications for Admissions are open.

Answer:

The correct match sequence of structure with their names us –
(i \rightarrow d), (ii \rightarrow c), (iii \rightarrow f), (iv \rightarrow a), (v \rightarrow g), (vi \rightarrow b)
(i) Cresols are methyl phenols that exist in three types namely: w-cresol, o-cresol, and p-cresol.
(ii) The common name of Catechol is pyrocatechol. It is the main constituent in many pesticides and perfumes. The IUPAC name is 1, 2-dihydrobenzene.
(iii). The compound Resorcinol is very commonly used to aid acne, skin disorders. ITSS IUPAC name is 1, 3-dihydroxybenzene.
(iv). The IUPAC name of Hydroquinone is 1, 4-dihydroxybenzene, and it is commonly known as quinol. The compound is a good reducing agent and looks like white granular solid.
(v) It is a colourless liquid also known as methoxy benzene, and the odour is similar to anise seed
(vi) Phenetole is commonly called ethyl phenyl ether; it is an organic compound. With a volatile nature, its vapours have explosive nature.

Question:58

Match the starting materials given in Column I with the products formed by these (Column II) in the reaction with HI.

Column I

Column II

(i)

\text {CH}_{3}-\text {O}-\text {CH}_{3}

(a)

(ii)

(b)

(iii)

(c)

(iv)

(d)

\text {CH}_{3}-\text {OH+CH}_{3}-\text {I}



(e)



(f)



(g)

Answer:

When the given compounds are reacted with HI, their correct product formation is –(i \rightarrow d), (ii \rightarrow e), (iii \rightarrow b), (iv \rightarrow a)
(i) Due to the symmetrical structure of \text {CH}_{3}-\text {O}-\text {CH}_{3}, the end products of the reaction are \text {CH}_{3}\text {I} and \text {CH}_{2}\text {OH}.
(ii) The compound being asymmetrical follows the SN^{2} mechanism; therefore it forms the following product

(iii) The alkyl group attached to the compound are primary and tertiary; thus, it follows the SN^{1} mechanism. In the process, the halide group attacks the tertiary and forms \left ( \text {CH}_{3} \right )_{3}\text {C-I} and \text {CH}_{3}\text {OH}
(iv) The structure mentioned is alkyl aryl ether which is an asymmetrical ether in nature. And having the partial bond structure, the products formed are \text {C}_{6}\text {H}_{5}-\text {OH} and \text {CH}_{3}\text {I}.

Question:59

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)

Ethylene glycol

(vi)

By product of soap industry used in cosmetics

(f)

Ethanol

Answer:

The correct matching pattern of the given compound and their properties is mentioned below – (i \rightarrowe), (ii \rightarrow f), (iii \rightarrow d), (iv \rightarrow c), (v \rightarrow a), (vi \rightarrow b)
(i) Mainly used in the manufacturing of polyester fibres industry as a raw material its IUPAC name is ethane-1, 2-diol.
(ii) The key properties of ethanol include being a good solvent for fatty substances. Being safe for skin use, it is a good constitute of perfumes.
(iii)When phenol reacts with cone. HNO_{3} it gets converted into picric acid (2, 4, 6-trinitro-phenol)
(iv) The widely known 'wood spirit' or Methanol, CH_{3}OH is made by destructive distillation of wood.
(v) Neutral ferric chloride is a well know reagent used for phenolic group detection, and it imparts purple/red colour when mixed with phenols.
(vi). The standard soap preparation is done by reacting NaOH with fatty acids.

Question:60

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

(d)

Ethyl alcohol

(iii)

Williamson's synthesis

(c)

Conversion of phenol to salicyaldehyde

(iv)

Conversion of 2o alcohol to ketone

(d)

wood spirit

(v)

Reimer-Tiemann reaction

(e)

Heated copper at 573 K

(vi)

Fermentation

(f)

Reaction of alkyl halide with sodium alkoxide

Answer:

The right match options for column 1 are -
(i \rightarrow d), (ii \rightarrow a), (iii \rightarrow f), (iv \rightarrow e), (v \rightarrow c), (vi \rightarrow b)
(i) The destructive distillation method is used to produce methanol commonly known as 'wood spirit.'
(ii) When phenol reacts with CO_{2} gas, the reaction results in 2-hydroxy benzoic acid (salicylic acid). The name of the process is Kolbe's reaction.

(iii). Another important method of ether preparation of ether is Williamson's Synthesis. In this preparation method, an alkyl halide is made to react with sodium alkoxide.
R-X+R-ONa\rightarrow ROR+NaX
(iv) In this method, secondary alcohol passes over highly heated copper up to 573K; this results in dehydrogenation and formation of a ketone.

(v) An aldehyde group is introduced at the ortho position of the benzene ring when phenol is treated with chloroform in the presence of NaOH.

(vi) Preparation of ethanol is done via fermentation of sugars.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 11: Assertion and Reason Type

Question:61

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Addition reaction of water to but-1-ene in acidic medium yields-butan-1-ol.
Reason (R): Addition of water in acidic medium proceeds through the formation of a primary carbocation.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(ii) When water is added to but-l-ene in an acidic medium it forms butan-2-ol. The addition of water in the process crosses secondary carbocation formation.

Question:62

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): p-Nitrophenol is more acidic than phenol.
Reason (R): Nitro group helps in the stabilisation of the phenoxide ion by dispersal of negative charge due to resonance.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(i) p-Nitrophenol has a higher acidic nature compared to phenol; this is due to the stabilising of phenoxide ion by the nitro group by negative charge dispersal.

Question:63

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): IUPAC name of the compound

Reason (R): In IUPAC nomenclature, ether is regarded as hydrocarbon derivative in which a hydrogen atom is replaced by -OR or -OAr group [where R = alkyl group and Ar = aryl group]

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:


(iv) The IUPAC name of the given compound according to the nomenclature rules is 2-propoxypropane.

Question:64

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Bond angle in ethers is slightly less than the tetrahedral angle.
Reason (R): There is a repulsion between the two bulky (-R) groups.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(iv) Because there is repulsion in between two bulky alkyl groups, tetrahedral angle in ethers is slightly less than bond angle.

Question:65

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Boiling points of alcohols and ethers are high.
Reason (R): They can form intermolecular hydrogen bonding.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(ii) Ethers have lower boiling points than alcohols. In hydrogen bonding, ethers cannot form intermolecular whereas alcohols can.

Question:66

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Like bromination of benzene, bromination of phenol is also carried out in the presence of Lewis acid.
Reason (R): Lewis acid polarises the bromine molecule.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(iv) Lewis acid is not useful for the bromination of phenol as its absence won't do anything.

Question:67

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): o-Nitrophenol is less soluble in water than the m- and p-isomers.
Reason (R): m-Nitrophenol and p-Nitrophenol exists as associated molecules.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(v) Intramolecular H-bonding is present in o-nitrophenol.

Thus, m-nitrophenol and p-nitrophenol form H-bonds with H2O but o-nitrophenol does not. Also, m-nitrophenol and p-nitrophenol exist as associated molecules because of the presence of intermolecular H-bonding.

Question:68

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Ethanol is a weaker acid than phenol.
Reason (R): Sodium ethoxide may be prepared by the reaction of ethanol with aqueous NaOH

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(iii) Because ethoxide ion cannot be stabilised by resonance, as compared to phenoxide ion; that's why ethanol is a weaker acid than phenol. The reaction between ethanol with sodium is done to obtain Sodium ethoxide.

Question:69

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Phenol forms 2,4, 6-tribromophenol on treatment with Br_{2} in carbon disulphide at 273 K.
Reason (R): Bromine polarises in carbon disulphide.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(ii) When bromine is treated with phenol in water, it forms 2, 4, 6-tribromophenol. Even when the Lewis acid is absent, the polarisation of bromine can still take place in phenols.

Question:70

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Phenols give o-nitrophenol and p-nitrophenol on nitration with cone. HNO_{3} and H_{2}SO_{4} mixture.
Reason (R):-OH group in phenol is o-,p-directing.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.
(ii) Assertion and reason both are wrong statements.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Assertion is wrong statement but reason is correct statement.
(v) Both assertion and reason are correct statements but reason is not correct explanation of assertion..
Answer:

(iv) When nitration of Phenols is done with dil. HNO_{3}, they give o, p-nitrophenol and 2,4, 6-trinitrophenol is formed with cone. HNO_{3}.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 11: Long Answer Type

Question:71

Write the mechanism of the reaction of HI with methoxybenzene.
Answer:

Phenol and an alkyl halide are always the products of alkyl aryl ethers because C_{6}H_{5}-O bond has a partial double bond character when resonance takes place.
Mechanism :
Step 1 – The protonation of ether molecules forms a methylphenyl oxonium ion

The above-mentioned ion has a weaker O-CH_{3} bond as compared to the O-C_{3}H_{5} as it has a partial double bond character. This double bond character is exhibited due to the resonance between the lone pair of electrons present on O-atom and the carbon atom of the phenyl group, which is sp^{2} hybridised.
Step 2 – The protonated ether gives methyl iodide and phenol after interacting with I ion in SN^{2} mechanism.

Question:72

(a) Name the starting material used in the industrial preparation of phenol.
(b) Write complete reaction for the bromination of phenol in aqueous and non-aqueous medium.
(c) Explain why Lewis acid is not required in bromination of phenol ?

Answer:

Solution: (a) The main initiating material used in the industrial preparation of phenol is Cumene (Isopropyl benzene)

(c) In the normal case scenario, Lewis acid is necessary to carry out the halogenation reaction, and FeBr_{3} is responsible for the polarisation of the halogen molecule. But when phenol is taken for halogenation reaction, due to its -OH group no activation agent is required as it is a high activation group present on the benzene ring.
The mechanism involved is that phenoxide ion is given by phenol after ionisation in aqueous solution. And the presence of negative charge on the oxygen atom dictates it to give its electron to the benzene ring. This makes the ring trisubstituted product due to its high activation. But in non-polar solvents, phenol does not ionise to a greater extent. As a consequence, electrons are donated to a small extent by the -OH group. The ring is activated slightly only contributes to mono-substitution.

Question:73

How can phenol be converted to aspirin?
Answer:

The chemical name of aspirin is salicylic acid. The mechanism of converting phenol into aspirin involves carbon dioxide being absorbed by sodium phenoxide and maintaining the temp at 400 K and 4-7 atm pressure. The first intermediate generated form of sodium salicylate by protonation. The acidification of sodium salicylate results in –

The acetylation of salicylic acid and acetic anhydride in the presence of conc. Sulphuric acid results in the formation of aspirin.

Question:74

Explain a process in which a biocatalyst is used in the industrial preparation of a compound known to you.
Answer:

Biocatalysts, as the name suggests, are the organic compounds that play the role of catalysts by reducing its activation energy and speeding up the process. These are a series of enzymes found in living organisms that catalyse their biochemical reactions. The most famous use of biocatalyst is in ethanol production from sugar solution (molasses):
When sugar solution is crystallised, the left-out solution or the mother liquor is termed as molasses which is the non-crystallised part. Its sugar concentration is 50%. It is kept for about 2-3 days after diluting it with 10% solution and yeast. Yeast has a role in providing enzymes invertase and zymase. The invertase gets broken down into sucrose to glucose and fructose after hydrolyses. Whereas the zymase gets helps in converting fructose and glucose to ethanol.

The winemaking process is very different; the source of sugar and yeast are grapes, not molasses. With time the sugar content increases due to the ripening of the grapes. After crushing of grapes, fermentation take its toll and occurs under anaerobic conditions and releases CO_{2}.
When the concentration of alcohol reaches up to 14%, the fermentation stops. If in case there is the presence of air in the fermented mixture, ethanol gets oxidised to ethanoic acid, which degrades the taste and quality of the alcohol

Apart from being informative, NCERT exemplar solutions for class 12 Chemistry chapter 11 is also fun and interesting for all those chemistry lovers who love to find different reactions and compounds. NCERT exemplar class 12 Chemistry solutions chapter 11 pdf download can be easily accessed for preparation.

Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

  • Alcohols, Phenols and Ethers
  • Classification
  • Nomenclature
  • Structures of Functional Groups
  • Alcohols and Phenols
  • Some Commercially Important Alcohols
  • Ethers
Aakash iACST Scholarship Test 2024

Get up to 90% scholarship on NEET, JEE & Foundation courses

ALLEN Digital Scholarship Admission Test (ADSAT)

Register FREE for ALLEN Digital Scholarship Admission Test (ADSAT)

NCERT Exemplar Class 12 Chemistry Solutions

Important Topics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

· These are a few important aspects of Alcohols, Phenols and Ethers.

Alcohols:

  • Grignard synthesis

  • Sp3 Hybridization

  • Tetrahedral position

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Phenols:

  • Crystalline solids

  • Carbolic acids

JEE Main high scoring chapters and topics

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Ethers:

  • Polar but insoluble in water

  • Low boiling point

  • V shape.

. These topics are explained in the NCERT exemplar class 12 Chemistry solutions chapter 11 ·

NCERT Exemplar Class 12 Solutions

Check NCERT Solutions for Class 12 Chemistry

Must read NCERT Solution subject wise

Read more NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. Are these solutions available in PDF format?

These NCERT exemplar Class 12 Chemistry solutions chapter 11 are available in webpage format. But one can download them as pdf using online free tools.

2. What are the aspects of Alcohol?

Below are the aspects of Alcohol:

  • Grignard synthesis

  • Sp3 Hybridization

  • Tetrahedral position 

3. What can one learn from this NCERT exemplar solutions for Class 12 Chemistry solutions Chapter 11?

 Students will learn about the significance and importance of these functional groups. It will also act as an easy-to-understand guide to Alcohols, phenols and Ethers.

 

Articles

Explore Top Universities Across Globe

University of Essex, Colchester
 Wivenhoe Park Colchester CO4 3SQ
University College London, London
 Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
 Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
 Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
 University Park, Nottingham NG7 2RD
Lancaster University, Lancaster
 Bailrigg, Lancaster LA1 4YW

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello,


Yes, you can give improvement exams for 2-3 subjects in CBSE (Central Board of Secondary Education). CBSE allows students who have appeared for their Class 12 board exams to improve their scores by re-appearing for exams in up to five subjects the following year.


Hope this helps,


Thank you

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


Best CBSE schools in Delhi: Click Here


In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top