Class 12 Chemistry Sample Paper With Solutions Set B

Please refer to Class 12 Chemistry Sample Paper With Solutions Set B below. These Class 12 Chemistry Sample Papers will help you to get more understanding of the type of questions expected in the upcoming exams. All sample guess papers for Chemistry Class 12 have been designed as per the latest examination pattern issued by CBSE. Please practice all CBSE Sample Papers for Chemistry in Standard 12.

Sample Paper Class 12 Chemistry With Solutions Set B

Topic-1
Haloalkanes and their Properties

Very Short Answer-Objective Type Questions

A. Multiple choice Questions:

Question. Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

Question. A primary alkyl halide would prefer to undergo_____________.
(a) S_N1 reaction
(b) S_N2 reaction
(c) a–Elimination
(d) Racemisation

Question. Which reagent will you use for the following reaction?
CH₃CH₂CH₂CH₃ → CH₃CH₂CH₂CH₂Cl+ CH₃CH₂CHClCH₃
(a) Cl₂/UV light
(b) NaCl + H₂SO₄
(c) Cl₂ gas in dark
(d) Cl₂ gas in the presence of iron in dark

Question. The order of reactivity of following alcohols with halogen acids is :

(a) (i) > (ii) > (iii)
(b) (iii) > (ii) > (i)
(c) (ii) > (i) > (iii)
(d) (i) > (iii) > (ii)

Question. Arrange the following compounds in increasing order of their boiling points.

(a) (ii) < (i) < (iii)
(b) (i) < (ii) < (iii)
(c) (iii) < (i) < (ii)
(d) (iii) < (ii) < (i)

B. Match the following :

Question. Match the species given in Column I with those mentioned in Column II.

Answer. (a) → (iii)
(b) → (v)
(c) → (i)
(d) → (ii)
(e) → (iv)

C. Answer the following:

Question. Amongst the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields a single monochloride.
Answer. Neopentane or 2, 2-Dimethylpropane.

Question. Write the structure of an isomer of compound C₄H₉Br which is most reactive towards S_N1 reaction.
Answer.

Question. Write the IUPAC name of the following :

Answer.

Question. Identify the chiral molecule in the following pair :

Answer.

Question. Write the structure of 1-Bromo-4-chlorobut-2-ene.
Answer. BrCH₂CH = CHCH₂Cl 

Question.

which is more reactive towards S_N1 reaction and why ?

Answer.

is more reactive as being secondary halide it forms more stable carbocation intermediate than primary halides.

Question. Which would undergo S_N2 reaction faster in the following pair and why ?

Answer. CH₃ — CH₂ — Br would undergo S_N2 reaction faster it has less steric hindrance than tert – butyl bromide. 
CH₃ — CH₂ — Br
Because it is a primary halide / (1˚) halide

Question. Out of

which is an example of allylic halide? 
Answer.

Question. Write the structure of 2,4-dinitrochlorobenzene.
Answer.

Question. Predict the major product formed when sodium ethoxide reacts with tert.Butyl chloride.
Answer. 2-Methylprop-1-ene / isobutene / structure

Question. Among the isomers of pentane (C₅H₁₂), write the one which on photochemical chlorination yields a single monochloride.
Answer. Neopentane / C(CH₃)₄

Question. Write the structure of 3-Bromo-2-methylprop-1-ene. 
Answer. BrCH₂(CH₃)C = CH₂ 

Question.

which is an example of benzylic halide?
Answer.

Question. Which would undergo S_N1 reactions faster in the following pair

Answer.

Question. In the following pair of halogen compounds,which will react faster by S_N1 mechanism ?

Answer. The 3° alkyl halides are most reactive, because the intermediate carbonation formed in their case is most stable in S_N1 mechanism. The most stable intermediate is formed at faster rate.

Question.How can methyl bromide be preferentially converted to methyl isocyanide ?
Answer. CH₃Br + alc. AgCN   →   CH₃NC + AgBr
                                               Methyl isocyanide 

Question. What happens when bromine attacks :
CH₂ = CH—CH₂—C ≡ CH ?
Answer.

The reaction of bromine with pent-1-en-4-yne
gives 1, 1, 2, 2, 4, 5 hexabromopentane as the main product. 

Question. Identify the compound that on hydrogenation produces an optically active compound from the following compounds:

Answer. B

Question. Out of

which is an example of vinylic halide?
Answer.

Short Answer Type Questions 

Question. Answer the following questions :
(i) What is meant by chirality of a compound ? Give an example.
(ii) Which one of the following compound is more easily hydrolyzed by KOH and why ?
CH₃CHClCH₂CH₃ Or CH₃CH₂CH₂Cl
Answer. (i) A carbon atom surrounded by 4 different groups is termed as chiral carbon and the property of being chiral is chirality.
Eg. 2-Butanol (or any other example). 
(ii) CH₃CHClCH₂CH₃ will be easily hydrolysed
(towards S_N1) as the cation formed in this case will be secondary, which is more stable.

Question. Which compound in each of the following pairs will react faster in S_N2 reaction with –OH ?
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl
OR
Write the equations for the preparation of 1-bromobutane from :
(i) 1-butanol
(ii) but-1-ene 
Answer. (i) CH₃I as I– ion is better leaving group than Br– ion. 
(ii) CH₃Cl as 1° alkyl halides are more reactive than tert-alkyl halides in S_N2 reaction with OH–. 
OR

Question. (i) Which alkyl halide from the following pair is chiral and undergoes faster S_N2 reaction ?

(ii) Out of SN1 and S_N2, which reaction occurs with
(a) Inversion of configuration
(b) Racemisation
Answer. (i) (b) is chiral 
(a) undergoes faster S_N2 
(ii) (a) S_N2 
(b) S_N1 

Question. Which one of the following compounds is more reactive towards S_N2 reaction and why?
CH₃CH(Cl)CH₂CH₃ or CH₃CH₂CH₂Cl
Answer. CH₃CH₂CH₂Cl , due to primary halide which has less steric hindrance.

Question. Which alkyl halide from the following pair is (i) Chiral and (ii) undergoes SN1 reaction faster?
(a) (CH₃)₃CBr
(b) CH₃CH₂CHBrCH₃
Answer. (i) (b) is chiral. 
(ii) (a) 

Question. (i) Allyl cholride can be distinguished from Vinyl chloride by NaOH and silver nitrate test. Comment.
(ii) Alkyl halide reacts with Lithium aluminium hydride to give alkane. Name the attacking reagent
which will bring out this change.
Answer. (i) Vinyl chloride does not respond to NaOH and silver nitrate test because of partial double bond character due to resonance. 
(ii) Hydride ion / H– 

Question. Account for the following :
(i) The C — Cl bond length in chlorobenzene is shorter than that in CH₃ — Cl.
(ii) Chloroform is stored in closed dark brown bottles.
Answer. (i) In chlorobenzene each carbon atom is sp² hybridised due to resonance there is a partial double bond character, so bond length is short.
(ii) In the presence of air and sunlight, chloroform get oxidised to phosgene (Carbonyl chloride – COCl₂) a poisonous compound. Thus, to avoid the formation of phosgene, chloroform is stored in dark brown bottles filled up to brim.

Question. Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same. 
Answer. The two reasons are as follows :
(i) Resonance effect : The electron pair of chlorine atom is involved in conjugation with the π-electrons of the benzene ring. As a result, electrons of C—Cl bond get delocalized and a partial double bond character develops in the bond and hence, it becomes difficult for the nucleophile to cleave the C— Cl bond. 
(ii) Increased electron density : A repulsion is suffered by the nucleophile due to increased electron density on the benzene ring which prohibits the nucleophile to make a close access for the attack on the molecule.

Question. Write a test to detect the presence of double bond in a molecule.
Answer. Presence of a double bond in a molecule can be detected by bromine water test and Baeyer’s test.
The molecules with a double bond decolourise bromine water and aqueous KMnO₄.

Long Answer Type Questions-I

Question. Give reasons :
(i) C– Cl bond length in chlorobenzene is shorter than C– Cl bond length in CH₃–Cl.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii) S_N1 reactions are accompanied by racemization in optically active alkyl halides.
Answer. (i) In chlorobenzene each carbon atom is sp² hybridised due to resonance there is a partial double bond character, so bond length is short.
(ii) In chlorobenzene, carbon to which chlorine is attached to sp² hybridised and is more electronegative than the corresponding carbon
in cyclohexyl chloride which is sp³ hybridised.
So the dipole moment is lower in chlorobenzene. In chlorobenzene, –I and +R effect oppose each other while in the other only –I effect is the only contributing factor resulting in lower dipole moment of cyclohexyl chloride.
(iii) In S_N1 reaction, carbocation intermediate formed is a planar molecule which will lead to form dand l- products. Hence, racemization occurs. 

Question. Following compounds are given to you :
2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane
(i) Write the compound which is most reactive towards S_N2 reaction.
(ii) Write the compound which is optically active.
(iii) Write the compound which is most reactive towards β-elimination reaction.
Answer. (i) 1-Bromopentane 
(ii) 2-Bromopentane 
(iii) 2-Bromo-2-methylbutane

Question. Explain the following :
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) Alkyl halides, though polar, are immiscible with water.
(iii) Grignard’s reagents should be prepared under anhydrous conditions.
Answer.  In chlorobenzene, carbon to which chlorine is attached to sp² hybridised and is more electronegative than the corresponding carbon
in cyclohexyl chloride which is sp³ hybridised.
So the dipole moment is lower in chlorobenzene. In chlorobenzene, –I and +R effect oppose each other while in the other only –I effect is the only contributing factor resulting in lower dipole moment of cyclohexyl chloride.
(ii) Alkyl halides though polar are immiscible with water because less energy is released when new attractions are set up between the alkyl halide and the water molecules as they are not as strong as the original hydrogen bonds in the water as a result the solubility of alkyl halides in water is low. 
(iii) Grignard reagents should be prepared under anhydrous conditions because Grignard reagents are highly reactive. In the presence of moisture they react to give alkanes.
RMgX + H₂O      →      R—H + Mg(OH)X
Grignard reagent         Alkane

Question. What happens when :
(i) CH₃–Cl is treated with aqueous KOH ?
(ii) CH₃–Cl is treated with KCN ?
(iii) CH₃–Br is treated with Mg in the presence of dry ether ?
Answer.

Question. Give the IUPAC name of the product formed when :
(i) 2-Methyl-1-bromopropane is treated with sodium is the presence of dry ether.
(ii) 1-Methyl cyclohexene is treated with HI.
(iii) Chloroethane is treated with silver nitrite. 
Answer. (i) 2,5-Dimethyl hexane.
(ii) 1-Methyl-1-iodocyclohexane
(iii) Nitroethane

Long Answer Type Questions-II

Question.1. Some alkylhalides undergo substitution whereas some undergo elimination reaction on treatment with bases. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
Answer. Primary alkyl halides prefer to undergo substitution reactions by S_N2 mechanism. A transition state is formed in which carbon is bonded to nucleophile and finally halogen atom is pushed out.

In tert-n-butyl halides prefer to undergo elimination reactions due to the formation of stable carbocations. 3^o alkyl halides follow S_N1 mechanism by forming tertiary carbocations. Now, if the reagent used is a weak base then substitution occurs while if it is a strong base then instead of substitution elimination occurs.

Here the reagent used i.e., aq. KOH It is a weak base, so substitution takes place.

Here the reagent used i.e., aq. KOH is a strong base,so elimination competes over substitution and alkene is formed.
Secondary alkyl halide can undergo substitution or elimination depending on type of solvent and temperature conditions.

Topic-2
Haloarenes and Polyhalogen Compounds

A. Multiple choice Questions:

Question. Identify the compound Y in the following reaction

Question.Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is
(a) Electrophilic elimination reaction
(b) Electrophilic substitution reaction
(c) Free radical addition reaction
(d) Nucleophilic substitution reaction

Question. Arrange the following compounds in the increasing order of their densities

(a) (i) < (ii) < (iii) < (iv)
(b) (i) < (iii) < (iv) < (ii)
(c) (iv) < (iii) < (ii) < (i)
(d) (ii) < (iv) < (iii) < (i)

Question. What is ‘A’ in the following reaction?

Question. Match the species given in Column I with those mentioned in Column II.

Answer. (a) → (ii)
(b) → (i)
(c) → (iv)
(d) → (iii)

C. Answer the following :

Question. Write the IUPAC name of the given compound :

Answer. 2-Phenylethanol. 

Question. Write the IUPAC name of the given compound :

Answer. 2, 4,6-Tribromoaniline / 2,4,6-Tribromobenzenamine.

Question. Write IUPAC name of the following compound ?

Answer. 1, 4-dichloro-3-methylbenzene. 

Question. Draw the structure of major monohalo product in the following reaction :

Answer.

Question. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? 
Answer. Benzyl chloride; 
Due to resonance, stable benzyl carbocation is formed.

Short Answer Type Questions

Question. Draw the structure of major monohalo product in each of the following reactions :

Answer.

Question. (i) Draw the structures of major monohalo products in each of the following reactions :

(ii) Which halogen compound in each of the following pairs will react faster in S_N2 reaction :
(a) CH₃Br or CH₃I
(b) (CH₃)₃ C—Cl or CH₃—Cl 
Answer.

(ii) (a) CH₃I will react faster in S_N2 reaction because CH₃I has a better leaving group than CH₃Br.
(b) CH₃Cl will react faster in S_N2 reaction because 1° alkyl halide undergoes faster reaction than 3° alkyl halide.

Question. How will you carry out the following conversion :
(i) 2-Bromopropane to 1-bromopropane
(ii) Benzene to p-chloronitrobenzene
Answer.

Question. Although chlorine is an electron withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why it is so ?
Answer. Although chlorine is an electron withdrawing group, yet it is ortho-para directing in nature in electrophilic aromatic substitution because when chlorine is present in benzene ring, it releases electron by resonance whereas it acts as withdrawing group only through inductive effect. By inductive effect, chlorine atom destabilizes the intermediate carbocation formation but by resonance, chlorine atom stabilises the intermediate carbocation and effect is more at o-and p-positions. Resonance effect opposes inductive effect.
Inductive is stronger than resonance effect due to which reactivity is controlled by inductive effect and orientation by resonance

Question. Haloalkanes undergo nucleophilic substitution whereas haloarenes undergo electrophilic substitution. Explain.
Answer. In haloalkanes, the halogen atom is attached to carbon atom. As the halogen atom is more electronegative than carbon, the bond between carbon and halogen is polar in character.
C^δ+—X^δ–
Due to presence of partial positive charge on carbon atom, the nucleophiles can attack on electron deficient carbon thereby resulting in the displacement of weaker nucleophile, thus typical reactions of alkyl halides are nucleophilic substitution reactions :

Whereas in haloarenes, as the halogen atom attached to benzene ring releases electron towards benzene ring (+m effect) thereby the ortho and para positions become site of attack for electrophiles more than nucleophiles.

Question. Give chemical tests to distinguish between the following pairs of compounds :
(i) Benzyl chloride and chlorobenzene
(ii) Chloroform and carbon tetrachloride
Answer. (i) On adding sodium hydroxide and silver nitrate to both the compounds, benzyl chloride forms white precipitate but chlorobenzene does not form white precipitate.
C₆H₅CH₂Cl + NaOH → C₆H₅CH₂OH + NaCl
NaCl + AgNO₃ → AgCl(s) + NaNO₃
                        (white ppt.)

(ii) On heating chloroform and carbon tetrachloride with aniline and ethanolic potassium hydroxide separately chloroform forms pungent smelling isocyanide but carbon tetrachloride does not form this compound.

Question. Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark.
Answer. Direct chlorination or bromination of benzene or other aromatic hydrocarbon gives chloroarenes and bromoarenes. These reactions are carried out in the presence of Lewis acids such as ferric or aluminium halides (FeCl₃, AlCl₃) in the dark, at ordinary temperatures (310-320 K). The reaction is electrophilic aromatic substitution. The halogen itself is not a strong electrophile for the attack of benzene ring. The Lewis acid promotes the electrophilic aromatic substitution by polarizing the halogen molecule. Actually, it generates a better electrophile for the attack of benzene ring.

Question. Diphenyls are potential threat to the environment.
How are these produced from arylhalides?
Answer. Diphenyls such as p,p-dichlorodiphenyl trichloroethane (DDT) are posing serious environment problems because of its chemical stability and its fat solubility. Its residue accumulates in environment and its long-term effects could be  highly dangerous.
Diphenyls can be prepared from aryl halides by the following two methods :

Long Answer Type Questions-I

Question. Give the IUPAC names of the following compounds :

Answer. (i) 2-bromobutane 
(ii) 1, 3-dibromobenzene 
(iii) 3-chloropropene

Question. Name the following according to IUPAC system:

Answer. (i) Butan – 2 – ol 
(ii) 2 – bromotoluene 
(iii) 1-chloro-2, 2-dimethylpropane

Question. How do you convert : (i) Chlorobenzene to biphenyl, (ii) Propene to 1-iodopropane, (iii) 2-bromobutane to but-2-ene.
OR
Write the major product (s) in the following :

Answer. (i) Chlorobenzene to biphenyl : When two chlorobenzene combine with sodium metal in the presence of dry ether it forms biphenyl.

Question. How can the following conversion be carried out :
(i) Aniline to bromobenzene
(ii) Chlorobenzene to 2-chloroacetophenone
(iii) Chloroethane to butane
OR
What happens when
(i) Chlorobenzene is treated with Cl₂/FeCl₃,
(ii) Ethyl chloride is treated with AgNO₂,
(iii) 2-bromopentane is treated with alcoholic KOH ?
Write the chemical equations in support of your answer. 
Answer. (i)

(ii) CH₃CH₂Cl + AgNO₂ → CH₃CH₂NO₂ + AgCl 
(iii) CH₃CH₂CH₂CH(Br)CH₃ + KOH (alc.) → CH₃CH₂CH=CH CH₃

Question. Draw the structures of the major monohalo product for each of the following reactions :

Answer. 

Question. (i) Account for the following :
(a) Electrophilic substitution reactions in haloarenes occur slowly.
(b) Haloalkanes, though polar, are insoluble in water.
(ii) Arrange the following compounds in increasing order of reactivity towards SN2 displacement : 2–Bromo–2–Methylbutane, 1-Bromopentane,2–Bromopentane
Answer. (i) Due to-I effect of X, the ring set partially deactivated.
(b) They fail to form Hydrogen bonds with water / more energy is required to break hydrogen 
(ii) 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

Question. (i) Write the structural formula of A, B, C and D in the following sequence of reaction :

(ii) Illustrate Sandmeyer’s reaction with the help of a suitable example.

Answer. (i) A: CH₃–CH=CH₂I
B: CH₃–CH₂–CH₂Br
C: CH₃–CH₂–CH₂I
D: CH₃–CH₂–CH₂MgI

Question. (a) Identify the chiral molecule in the following pair :

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
(c) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1- methylcyclohexane with alcoholic KOH.
Answer.

Question. Which one of the following compounds will undergo faster hydrolysis reaction by S_N1 mechanism? Justify your answer.

OR
A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane.Write the structures of the isomers possible. Give the IUPAC name of the isomer which can exhibit enantiomerism.
Answer. C₆H₅CH₂Cl will undergo SNl reaction faster.
The carbocation formed by C₆H₅CH₂Cl gets stabilized through reasonance.
Greater the stability of carbocation, greater will be its ease of formation from the respective halide.

Question. Give reasons :
(i) n-Butyl bromide has higher boiling point than t-butyl bromide.
(ii) Racemic mixture is optically active.
(iii) The presence of nitro group (–NO₂)at O/P positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer. (i) Larger surface area, higher van der Waals’ forces,higher the boiling point. 
(ii) Rotation due to one enantiomer is cancelled by another enantiomer. 
(iii) –NO₂ acts as Electron withdrawing group or –I effect.

Question. Write the product(s) formed when
(i) 2-Bromopropane undergoes dehydrohalogenation reaction.
(ii) Chlorobenzene undergoes nitration reaction.
(iii) Methylbromide is treated with KCN.
Answer. (i) Propene 
(ii) 4-nitrochlorobenzene and 2-nitrochlorobenzene / structures 
(iii) Methylcyanide / Ethanenitrile / structure

Long Answer Type Question-II 

Question. Some halogen containing compounds are useful in daily life. Some compounds of this class are responsible for exposure of flora and fauna to more and more of UV light which causes destruction to a great extent. Name the class of these halocompounds. In your opinion, what should be done to minimise harmful effects of these compounds.
Answer. There are a number of halogen containing compounds such as chloroform, iodoform, carbon tetrachloride, DDT, benzene hexachloride, freons, etc, which are useful in our daily life. But these compounds have adverse effects on health of humans, animals and other living beings. Compounds like freons containing chloroflorocarbons cause the depletion of ozone layer. Consequently, UV rays reach the earth which harm flora and fauna.
(i) Tetrachloromethane: In contact with air, carbon tetrachloride rises to the atmosphere and depletes the ozone layer. Depletion of the ozone layer increases human exposure to UV rays leading to increased skin cancer, eye diseases and disorders, and possible disruption of the immune system.
These UV rays cause damages to plants, and reduction of plankton populations in the ocean’s euphotic zone.
(ii) Freons: Freon-113 remains in the air long enough to reach the upper atmosphere. It provides chlorine atoms which damage the ozone layer. Because of this depletion, UV rays enter in our atmosphere and become responsible for the damage to great extent.
(iii) p-p’-dichlorodiphenyltrichloroethane (DDT): It is non-biodegradable so gets deposited in fatty tissues.If ingestion continues for a long time, DDT builds up within the animal and affect the reproductive system.
To minimize the harmful effect of these compounds the use of equipments like, air conditioner, refrigerator, aerosol that produce such compounds should be discouraged. Government has tightened laws on use of DDT. Such rules should be strictly followed.