Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 14

A diabasic organic acid (A) on heating gives (B). The compound (B) on soda lime treatment gives the lowest alkane which can be prepared by Wurtz synthesis. The compound (A) is

Explanation

The lowest alkane which can be prepared by wurrtz's synthesis is ethane

$\mathrm{(H_3C-CH_3)}$ .

Decarboxylation of compound in option:

A) gives

$\mathrm{CH_4}$ as product.

B) gives

$\mathrm{C_2H_4}$ as product.

C) gives gltaric anhydride as product.

D) gives

$\mathrm{H_3C-CH_3}$ as product.

Hence, Option "D" is the correct answer.

The major product of the above reaction is:

Which of the following has highest melting point?

0%
$BeCl_2$
0%
$MgCl_2$
0%
$CaCl_2$
0%
$BaCl_2$

The complex / complex ion, which shows optical activity is :

0%
$[Cr (H_2O)_4Cl_2]^+$
0%
$[Co(H_2O)_2(NH_3)_2Cl_2]^+$
0%
$[Co(NH_3)_6]^{3+}$
0%
$[Co(CN)_5(NC)]$

Which order is/are correct regarding melting point?

Decreasing order of melting point of compound

$I - IV$ follows:

0%
$IV > III > II > I$
0%
$I > II > III > IV$
0%
$II > III > IV > I$
0%
$II > IV > III > I$

The nitration of a compound is due to the

0%
$NO_2$
0%
$NO_3$
0%
$NO$
0%
$NO_2^+$

Explanation

The process of nitration takes place as below

$HONO_2+ 2H_2SO_4 \rightleftarrows H_3O^+ +2HSO_4^- +\underset{(nitronium\ ion)}{ NO_2^+}$

The electrophile responsible for nitration is

$NO_2^+$ ion.

Meso-tartaric acid is

0%
optically inactive
0%
optically active because of molecular symmetry
0%
optically inactive due to external compensation
0%
optically active because of asymmetric carbon atom

Which is optically active

0%
$CH_2Cl_2$
0%
$CHCl_3$
0%
Meso form of tartaric acid
0%
Glyceraldehyde

The IUPAC name of:

$\begin{matrix} { ClH }_{ 2 }C-C=C-{ CH }_{ 2 }Cl \\ |\qquad | \\ Br\qquad Br \end{matrix}$

0%
$2,3$ dibromo $-1,4-$ dichlorobut $-2-$ ene
0%
$1,4-$ dichloro $-2,3-$ bromobutane $-2$
0%
Dichlorodibromobutane
0%
Dichlorodibromobutane

The compound

$C_8H_9Cl (A)$ on treatment with

$KCN$ followed by hydrolysis gives

$C_9H_{10} O_2 (B) .$ Ammonium salt of

$B$ on dry distillation yields

$C$ , which reacts with an alkaline solution of bromine to gives

$C_8H_{11}N. (D).$ Another compound

$E (C_8H_{10}O)$ is obtained by the action of nitrous acid on

$D$ or by the action of aqueous potash on A. E on oxidation gives

$F (C_8H_6O_4)$ which gives the inner anhydride G on heating. The compound D is reaction with

$CHCl_3+NaOH$ gives a compound H.

The compound

$A$ on reaction with

$AgCN$ gives :

0%
Compound H
0%
Compound B
0%
Compound G
0%
Compound D

Explanation

The compound D is primary amine and primary amine reacts with chloroform and NaOH gives isocyanide and when compound A reacts with

$AgCN$ it also gives same product.

Option A is correct.

1595106_1748856_ans_121b8655df4d4f10aa165de619b50325.png

In the reaction

$CH_3 -Br + 2Na + Br - CH_3 \rightarrow$ , the product called

0%
Wurtz reaction
0%
Aldol condensation
0%
Perth's reaction
0%
Levit reaction

Explanation

this is the

$Wurtz reaction.$

Typically, you would react two alkyl halides to form an alkane that consists of their respective R groups. So, as an example, you would have:

$CH_3Br+CH_3Cl$

$\xrightarrow[]{Na(s)}$

$CH_3CH_3$

(with loss of NaCl and then NaBr )

The

$IUPAC$ name of

$n-$ butyl chloride is:

0%
$1-$ chlorobutane
0%
$n-$ chlorobutane
0%
tre-butylchloride
0%
$2-$ methylbutane

Dehydrohalogenation of an alkyl halide is

0%
An addition reaction
0%
A substitution reaction
0%
An elimination reaction
0%
An oxidation reaction

Explanation

Alkyl halide gives alkene on Dehydrohalogenation, reaction takes place in presence of alc.

$KOH.$ This reaction is called an elimination reaction.

$CH_3CH_2Br + Alc. KOH \rightarrow CH_2 = CH_2 + KBr + H_2O$

There are two types of elimination reactions.

(a)

$E_1 \rightarrow$ Unimolecular elimination

(b)

$E_2 \rightarrow$ Bimolecular elimination

What would be the product formed when 1-Bromo-3-chlorocyclobutane reacts with two equivalents of metallic sodium in ether?

The final product formed when ethyl bromide is treated with an excess of alcoholic KOH is

0%
Ethylene
0%
Ethane
0%
Ethyne
0%
Vinyl bromide

Explanation

$CH_{3}-CH_{2}-Br + KOH\rightarrow \underset{\text{Ethylene}}{CH_{2}=CH_{2}} + KBr + H_{2}O$

This reaction is known as dehydrohalogenation.

A mixture of ethyl iodide and n-propyi iodide is subjected to Wurtz reaction. The hydrocarbon that will not be formed is

0%
n-butane
0%
n-propane
0%
n-pentane
0%
n-hexane

Explanation

Ethyl iodide and n-propyl iodide are allowed to undergo Wurtz reaction and they form butane and hexane as self-addition products and pentane as cross addition product. ∴ So,

$n- propane$ will not be formed.

The reaction ,

$CH_3Br + Na \rightarrow Product ,$ is called

0%
Perkin reaction
0%
Levit reaction
0%
Wurtz reaction
0%
Aldol condensation

Explanation

It is a common method to prepare alkanes. Methane cannot be prepared by Wurtz reaction.

$CH_3Br + 2Na + BrCH_3 \rightarrow \underset{ethane}{C_2H_6} + 2NaBr$

Option C is correct.

Which one of the following compounds cannot be prepared by Wurtz reaction?

0%
$CH_4$
0%
$C_2H_6$
0%
$C_3H_8$
0%
$C_4H_{10}$

Explanation

It is not possible to prepare

$CH_4$ by Wurtz reaction. By Wurtz reaction we get minimum 2 C product.

Example:

$2CH_3-X+2Na \xrightarrow{Dry\ ether} H_3C-CH_3+2 NaX$

The reaction of an aromatic halogen compound with an alkyl halides in presence of sodium and ether is called

0%
Wurtz reaction
0%
Sandmeyer's reaction
0%
Wurtz-fittig reaction
0%
Kolbe reaction

Explanation

It is Wurtz fittig reaction

$C_6H_5Br + CH_3Br \xrightarrow[Ether]{Na} C_6H_5CH_3 + 2NaBr$

Which of the following methods is not suitable for the preparation of

$RX$ ?

0%
Halogenation of alkane
0%
$ROH$ and $PX_3$
0%
$ROH$ and $HX$
0%
Alkene and $HX$

Explanation

$R-H+X_2 \underset{High\ temp. 400^oC}{\underrightarrow {\ U.V.light}}\ R-X+HX$

It is a drastic method as it requires high temperature or catalyst

$CuCl_2, FeCl_3, FeBr_3$ etc.

$ROH+PX_3 \to 3RX+H_3PO_3$

$ROH+HX\to R-X+H_2O$

(2), (3), and (4) are feasible.

Option A is correct.

1820515_1768709_ans_03a3439e61dd4d0aaca246b6ce3ddc6c.png

Which of the following would be hydrolysed most readily?

0%
$EtBr$
0%
$MeBr$
0%
0%

Explanation

Hydrolysis occurs through

$S_N1$ mechanism

Allyl halide undergoes substitution (

$S_N1$ ) very easily.

Option D is correct.

$C_3H_7Cl \xrightarrow {KOH(alc.)} (A) \xrightarrow {770\ K}{Cl_2 (g)} (X)$
(X) can be:

0%
Vinyl chloride
0%
Allyl chloride
0%
Ethyl chloride
0%
Ethyl iodide

Explanation

$CH_3-CH_2-CH_2-Cl\xrightarrow {alc.KOH}(A) \Rightarrow CH_3-CH=CH_2$

$(X) \Rightarrow$ Allylic chlorination

$Cl-CH_2-CH=CH_2$

Allyl chloride

Option B is correct.

Which method is feasible for the preparation of compound (X) only ?

The decreasing order of the acidic character of the following is :

0%
(I) > (II) > (III) > (IV)
0%
(I) > (III) > (II) > (IV)
0%
(IV) > (III) > (II) > (I)
0%
(IV) > (II) > (III) > (I)

The chemical reaction

${ { (CH }_{ 3 }) }_{ 3 }CBr\xrightarrow [ { -H }_{ 2 }O,-KBr ]{ alc.KOH } { { (CH }_{ 3 }) }_{ 2 }C={ CH }_{ 2 }$
is an example of:

0%
Nucleophilic subsitution
0%
Electrophilic subsitution
0%
Free radical subsitution
0%
$\beta$ -Elimination

Explanation

$E1$ and

$E2$ are both called

$\beta$ -elimination because

$\beta-H$ (most acidic) is removed by base.

But

$E2$ is also called anti-or trans-elimination.

Option D is correct.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Which statement is correct about the above reaction?

0%
Both products (B) and (C) are optically active.
0%
Both products (B) and (C) are optically inactive.
0%
Product (B) is optically inactive but product (C) is optically active.
0%
Product (B) is formed by anti-addition but product (C) is formed by syn-addition of $H_2$ .

Explanation

Both reagents convert alkyne to alkene

P-2catalyst adds by $syn-addition$ , while Birch reduction adds by $anti-addition$ .

hence option (C) is correct.

1680689_1770669_ans_fc49c852ba2d4d47a5c52622803e9995.png

The compound (A) is:

0%
0%
0%
0%
None of these

Arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution.

0%
$(a)< (b)< (c)$
0%
$(a)< (c)< (b)$
0%
$(c)< (b)< (a)$
0%
$(b)< (c)< (a)$

Number of optically active isomer is / are :

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Choose the correct option from the following:

0%
In the electrophilic substitution of toluene with $Cl_2 /$ Iron (III) chloride, Iron (III) chloride acts as a Lewis acid.
0%
In the reaction of toluene with $Cl_2/FeCl_3$ , ortho and para isomers are easily separated.
0%
Similar reaction with iodine is reversible in nature.
0%
All of these

Explanation

In the electrophilic substitution of toluene with

$Cl_2 /$ Iron (III) chloride, Iron (III) chloride acts as a Lewis acid.

In the reaction of toluene with

$Cl_2/FeCl_3$ , ortho and para isomers are easily separated. These isomers are easily separated due to a large difference in their melting points.

Reaction:

A similar reaction with iodine is reversible in nature.

Hence, Option "A" is the correct answer.

In the question arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution.

0%
(I) < (II) < (III)
0%
(I) < (III) < (II)
0%
(III) < (II) < (I)
0%
(II) < (III) < (I)

Explanation

Presence of electron donating group (-CH

$_3$ ) increases the reactivity towards nucleophilic substitution and the effect of o-position is greater than at m-position because of hyperconjugation and inductive effects.

So, the order becomes (II) < (III) < (I)

Hence, Option "D" is the correct answer.

Ethyl alcohol is obtained when ethyl chloride is boiled with

0%
alcoholic KOH
0%
aqueous KOH
0%
water
0%
aqueous $KMnO_4$

Explanation

Ethyl alcohol is obtained when ethyl chloride is boiled with aqueous KOH.

Reaction:

$C_2H_5Cl + KOH_{(aq)} \rightarrow C_2H_5OH + KCl$

Hence, Option "B" is the correct answer.

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 14 - MCQExams.com

The ethanolysis of primary alkyl halide generally proceeds
via $S_{N} 2$ mechanism. In case it proceeds via $S_{N} 1$ mechanism, the intermediate
carbocation is unstable, since it has no adjacent carbon atom for resonance.
Hence, both assertion and reason are incorrect.

0:0:3

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 14 - MCQExams.com

The ethanolysis of primary alkyl halide generally proceedsvia SN​2 mechanism. In case it proceeds via SN​1 mechanism, the intermediatecarbocation is unstable, since it has no adjacent carbon atom for resonance.Hence, both assertion and reason are incorrect.

0:0:3

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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The ethanolysis of primary alkyl halide generally proceeds
via $S_{N} 2$ mechanism. In case it proceeds via $S_{N} 1$ mechanism, the intermediate
carbocation is unstable, since it has no adjacent carbon atom for resonance.
Hence, both assertion and reason are incorrect.