MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 13

When alkyl halide is heated with an aqueous solution of silver nitrite,it gives:

0%
Nitroalkane only
0%
Alkyl nitrite only
0%
Alkyl amine
0%
None of these

Explanation

On treating ethanolic solution of nitroalkane with silver nitrite

$Ag-O-N=O$ nitroalkane is formed because since the bond between

$Ag-O$ is covalent, the lone pair on nitrogen acts as an attacking site for nucleophilic substitution.

$R-X + AgNO_2 \rightarrow R-NO_2 +AgX$

Hence the correct option is A.

By Wurtz reaction which of the following will give you n - hexane?

0%
$1$ -chloro propane
0%
$2$ -chloro propane
0%
Both of these
0%
None of these

In chlorobenzene,

$-Cl$ group attached to benzene acts as activating ortho-para directing due to:

0%
(-I) effect = (+) R effect
0%
(-I) effect > (+ R) effect
0%
(-I) effect < (+ R) effect
0%
(-I) effect < (- R) effect

Product

$X$ is :

The total number of optically compounds formed in the following reaction is:-

0%
Four
0%
Two
0%
Six
0%
Zero

The major product of the following reaction is:

$C_6H_5CH_2-\underset{\,\,\,Br}{\underset{|}{\overset{\,\,CH_3}{\overset{|}{C}}}}-CH_2-CH_3 \xrightarrow[C_2H_5OH]{C_2H_5ONa}$

0%
$C_6H_5CH_2-\underset{\,\,\,\,\,\,\,OC_2H_5}{\underset{|}{\overset{\,\,\,\,\,CH_3}{\overset{|}{C}}}}-CH_2CH_3$
0%
$C_6H_5CH=\underset{\,\,\,CH_3}{\underset{|}{C}}-CH_2CH_3$
0%
$C_6H_5CH_2-\underset{\,\,\,CH_3}{\underset{|}{C}}=CHCH_3$
0%
$C_6H_5CH_2-\underset{\,\,\,\,\,\,\,\,\,\,CH_2CH_3}{\underset{|}{C}}=CH_2$

IUPAC name of $(CH_3)_3CCl$ is:

0%
$1-butyl chloride$
0%
$3-chloro butane$
0%
$2-chloro-2-methylpropane$
0%
$2-butyl chloride$

The general formula of alkyl halides is:

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${ C }_{ n }{ H }_{ 2n }X$
0%
${ C }_{ n }{ H }_{ 2n+1 }{ X }$
0%
${ C }_{ n }{ H }_{ 2n }{ X }_{ 2 }$
0%
${ C }_{ n }{ H }_{ 2n-1 }{ X }$

Rank the alkyl halides in order of increasing reactivity in

$E_2$ reaction.

0%
0%
0%
0%
none

Write the product :

${C_6}{H_5}-C{H_2}-CH=CH_2+HBr\rightarrow$

0%
${C_6}{H_5} - CH(Br) - CH = C{H_2}$
0%
${C_6}{H_5} - C{H_{}} - C{H_2}-CH_2(Br)$
0%
$C_6H_5-CH_2-CH_1(Br)-CH_3$
0%
none of the above

Which of the following do not form Grignard reagent?

0%
$CH_{3}F$
0%
$CH_{3}Cl$
0%
$CH_{3}Br$
0%
$CH_{3}I$

When

$Me_{3}C - O - CH_{3}$ react with excess of

$HI$ gives

$CH_{3}I$ and

$Me_{3}CI$ . It follows the type of mechanism is:

0%
Only $S_{N}1$
0%
Only $S_{N}2$
0%
$S_{N}2$ and $S_{N}1$
0%
$S_{N}1$ and $S_{N}2$

Acetaldehyde undergoes aldol condensation giving

$\beta$ - hydroxyl carbonyl compound,which on dehydration in basic medium giving

$\alpha,\beta$ unsaturated carbonyl compound,this dehydration in basic medium is an example for

0%
$E_1$ elimination
0%
$E_2$ elimination
0%
$E_1$ cb elimination
0%
$E_1$ or $E_2$

$C_6H_5CCL_5\xrightarrow[Fe]{Cl_2}X$
In the above reaction X is

0%
0%
0%
0%
None of these

In which of the following compound the electrophile attack on o- and p- position :

Identify the false statement

0%
Chloro benzene with chlorine in presence o $AlCl_3$ gives o and p- di chloro benzene
0%
Benzene with acetyl chloride in presence o $AlCl_3$ gives Acetophenone
0%
Benzene on onolysis gives three moles of glycol
0%
Benzene on chlorination in presence of sunlight forms an addition product which is used as insecticide

Which of the following is not optically active?

0%
$[Co(en)_{3}]^{3+}$
0%
$[Cr(ox)_{3}]^{3-}$
0%
$cis[CoCl_{2}(en)_{2}]^{+}$
0%
$trans-[CoCl_{2}(en)_{2}]^{+}$

Among the following in which compound

$Br$ atom has the highest locant number during

$IUPAC$ nomenclature:

0%
Only $(a)$
0%
Only $(b)$
0%
Only $(c)$
0%
Both $(a)$ and $(c)$

The major product of the following reaction is:

$CH_{3}CH_{2}Cl\overset{NaCN}{\rightarrow}X\overset{Ni/H_{2}}{\rightarrow}Y\xrightarrow[anhydride]{Acetic}Z$

Z in the above reaction sequence is

0%
$CH_{3}CH_{2}CH_{2}NHCOCH_{3}$
0%
$CH_{3}CH_{2}CH_{2}NH_{2}$
0%
$CH_{3}CH_{2}CH_{2}CONHCH_{3}$
0%
$CH_{3}CH_{2}CH_{2}CONHCOCH_{3}$

What is the IUPAC name of the following compound?

0%
2-Bromo 3- methylpent-3-ene
0%
4-Bromo-3-methylpent-2-ene
0%
3-Bromo-1,2,dimethylbut-1-ene
0%
3-Bromo-3-methyl-1,2-dimethylprop-1-ene

Which one of the following is an optically active?

The major product of the reaction given below is:

The major product obtained on the monobromination with

$Br_{2}/FeBr_{3}$ :

Explanation

In the given reactant, the two groups which are present on the benzene rings $(-OCH_3 \ and \ -CH_3 )$ are electron-donating and ortho- para- directing groups.

The ortho- para- directing nature of the groups are in the following order:-

$NH_2 > -NR_2> -OH > -OCH_3 > -NHAc> -CH_3 > -X$

Thus, they both will direct $Br$ group on ortho- and para- position. But due to the steric hindrance, the compound at the para position will be a major product.

Hence, option B is correct.

1715354_1348039_ans_12b26c543fa048828056fc1e62016e1d.PNG

Which is suitable reagent for Wurtz reaction?

0%
Na/ether
0%
Na/alcohol
0%
Zn/ether
0%
Zn/alcohol

Which of the following is most reactive toward $Ag^{\oplus} NO_3^{\ominus}$ or rate of $S_{N_1}$ reaction?

Which of the following combinations is correctly matched?

What is the

$IUPAC$ name of the given compound?

0%
$1-$ chloro cyclohex $-1-$ ene
0%
$2-$ chloro cyclohex $-2-$ ene
0%
$3-$ chloro cyclohex $-1-$ ene
0%
$4-$ chloro cyclohex $-3-$ ene

$+HBr \longrightarrow$ Product :

Explanation

cyclopropane when reacts with

$HBr$ ,it opens the ring and is

$H$ and

$Br$ are added at both ends of opened ring.

1218831_1369387_ans_88b12954be6b4c92add283e3f8b6beb6.png

Which of the following compounds will give yellow precipitate on shaking with aqueous solution of

$NaOH$ followed by the addition of

$Ag{NO}_{3}$ solution?

Find the major product of the following reaction

Among the following the compounds which can undergo an

$S_N1$ reaction in an aqueous solution, are:

0%
I and IV only
0%
II and IV only
0%
II and III only
0%
II, III and IV only

Explanation

The compound which can form stable carbocation will undergo

$S_N1$ reaction more easily.

Due to the presence of 3 methyl groups on carbon attached to halogen option

$II$ undergoes

$SN^{I}$ and also due to the presence of

$OMe$ electron donation group in option

$III$ undergoes

$SN^{I}$ .

It is because

$II$ and

$III$ carbocation is stabilized by hyperconjugation and resonance.

In (I) structure the carbocation is formed at unsaturated carbon atom is highly unstable and in (Iv) the carbocation formed at the bridge which is highly unstable.

Hence option C is correct.

1433075_1688418_ans_0c08c7bca9574b22b8b7479f69fff1f0.png

$CH_3Br + Nu^- \longrightarrow CH_3 -Nu + Br^-$
The decreasing order of the rate of the above reaction with
nucleophile

$(Nu^- )$ Ato D is

$[Nu^- : (A)PhO^- , (B) AcO^- , (C) HO^- ,(D) CH_3O^- ]$

0%
D>C>A>B
0%
D>C>B>A
0%
A >B >C>D
0%
B >D>C>A

$S_N1$ mechanism for the hydrolysos of an alkyl halide involves the formation of intermediate:

0%
Free radical
0%
Carbocation
0%
Carbanion
0%
none of these

Explanation

$S_{N}1$ is a 2 step mechanism. The first step involves the breaking of the

$C-X$ bond to form a

$carbocation\ intermediate,$ while the second step involves the attack of nucleophile on carbocation.

Option B is correct.

1824487_1716836_ans_0a5c31d9834449108a0527fdeae02029.jpg

1-Bromopropane and 2-bromopropane on treatment with sodium in presence of ether gives:

0%
n-hexane
0%
2,3-dimethylbutane
0%
2-methylpentane
0%
a mixture of all these different alkanes '

Explanation

According to the

$\rightarrow$

$2RX + 2Na \xrightarrow[]{ether} R-R+ 2NaX$

$CH_3-CH_2-CH_2-Br+CH_3-CH(Br)-CH_3 \xrightarrow[Na]{ether} \underset {\displaystyle {2-methylbutane}}{CH_3 -CH_2-CH_2-\underset{\underset{\,\,\,\, \displaystyle {\,\,\,\,CH_3}}|}{CH}-CH_3}$

$+\underset{\displaystyle {hexane}}{CH_3-CH_2-CH_2-CH_2-CH_2-CH_3} + \underset{\displaystyle {2,3-dimethylbutane}}{CH_3-\underset{\underset{\,\,\,\, \displaystyle {\,\,\,\,CH_3}}|}{CH}- \underset{\underset{\,\,\,\, \displaystyle {\,\,\,\,CH_3}}|}{CH}-CH_3}$

Limitation:-

1. Wurtz reaction does not give odd number of carbon chain product

$Hence \;the \;correct \;option \;is\; D.$

Secondary butyl chloride on boiling with alcoholic potash gives:

0%
only 1- butene
0%
only 2- butene
0%
isobutylene
0%
a mixture of 1-butene and 2- butene

Explanation

$CH_3CH(Cl)-CH_2CH_3 + KOH(alc)\rightarrow CH_3-CH=CH-CH_3 + CH_3-CH_2CH=CH_2$

According to Saytzeff rule the major product is

$CH_2-CH=CH-CH_3$ which is more stable compare to 1-butene

$Hence \;the \;correct \;option \;is\; D.$

Which of the following alkyl halide is hydrolysed by

$S_N^1$ mechanism?

0%
$(CH_3)_3CX$
0%
$CH_3CH_2X$
0%
$CH_3CH_2CH_2X$
0%
$(CH_3)_2CHX$

Which one of the following compounds is stable?

0%
$CH_3CH(OH)_2$
0%
$(CH_3)_2C(OH)_2$
0%
$CCl_3CH(OH)_2$
0%
None of these

Explanation

Here we noticed that two

$OH$ group attached with one carbon atom at all three compounds which is called the geminal diol compound.

$CH_3-CH(OH)_2$

$(CH_3)_2C(OH)_2$

$CCl_3CH(OH)_2$

Dehydration reaction

$1. CH_3-CH(OH)_2\xrightarrow[]{H^{+}} CH_3CH(OH_2^{+})-O-H \xrightarrow[-H_2O]{H^{+}}CH_3-CHO$

$2.(CH_3)_2C(OH)_2 \xrightarrow[]{H^{+}} (CH_3)2C(OH_2)-O-H\xrightarrow[-H_2O]{H^{+}} CH_3-CH_2CHO$

$3. CCL_3CH(OH)_2\xrightarrow[H^{+}] CCl_3CH(OH_2)-O-H\xrightarrow[-H_2O]{H^{+}} CH_3-C-CHO$

$\because$

$+I$ group = geminal diol compound stability decreases

$\because$

$-I$ group = geminal diol compound stability increases

$Hence \;the\; correct \;option \;is\; C.$

Which of the following is optically active?

0%
Butane
0%
4-Methyl heptane
0%
3-Methyl heptane
0%
2-Methyl heptane

Tert. butyl chloride preferably undergoes hydrolysis by:

0%
$S_{N^1}$ mechanism
0%
$S_{N^2}$ mechanism
0%
Any of (A) and (B)
0%
None of these

Explanation

In general, primary alkyl halides react via an

$S_N^2$ mechanism whereas

$\text{tertiary alkyl halides react via }S_N^1\text{ mechanism}$ because the formation of stable

$3^o$ carbocation takes place therefore Tert. butyl chloride preferably undergoes hydrolysis by an

$S_{N}^1$ mechanism.
Option A is correct.

Which of the following is liquid at room temperature?

0%
$CH_3I$
0%
$CH_3Br$
0%
$C_2H_5CI$
0%
$CH_3F$

Which of the following alkyl halides is hydrolysed by

$S_N 1$ mechanism?

0%
$CH_3 Cl$
0%
$CH_3 CH_2 Cl$
0%
$CH_3 CH_2 CH_2 Cl$
0%
$(CH_3)_3 CCl$

Explanation

Tertiary Alkyl halide are more prone to

$\mathrm{S_N1}$ mechanism.

Among the given options, only

$\mathrm{(CH_3)_3Cl}$ is a tertiary halide.

Hence, Option "D" is the correct answer.

The catalyst used in the preparation of an alkyl chloride by the action of dry HCl on alcohol is:

0%
Anhydrous $AlCl_3$
0%
$FeCl_3$
0%
Anhydrous $ZnCl_2$
0%
$Cu$

Which of the following will be least reactive in nucleophilic substitution?

0%
$CH_3CH_2CH_2Cl$
0%
$(CH_3)_3 C-Cl$
0%
$CH_3-CHClCH_3$
0%
$H_2C =CHCH_2Cl$
0%
$H_2C=CHCl$

An ethereal solution of alkyl halide is heated with sodium metal. The reaction is known as:

0%
Frankland's reaction
0%
Sandmeyer's reaction
0%
Wurtz-Fittig reaction
0%
Wurtz reaction

The hydrolysis of

$(CH_3)_3 C- Cl$ aqueous NaOH follows

$S_N1$ mechanism.

0%
True
0%
False

Explanation

$(CH_3)_3 C- Cl$ , a tertiary halide undergoes hydrolysis through

$S_N1$ mechanism.

Hence, the given statement is True

$(X) +KCN \longrightarrow CH_3CN \quad \xrightarrow {2H_2/Ni} CH_3CH_2NH_2$ what is (X) ?

0%
$CH_3CH_2Cl$
0%
$CH_3Cl$
0%
$CH_3CH_2CH_2Cl$
0%
$(CH_3)_2CHCl$

Which of the following is the least reactive towards
nucleophile?

0%
$CH_3CH_2Cl$
0%
$CH_3Cl$
0%
$H_2C=CH=CH_2Cl$
0%
$C_6H_5Cl$
0%
$CH_3CH(Cl)CH_3$

Explanation

Due to delocalization of electrons in benzene ring , it is difficult to attach nucleophile .In

$C_6H_5Cl$ , there is no -M group .So, it does not participate in nucleophilic substitution reaction.Therefore,

$C_6H_5Cl$ is the least reactive towards nucleophile.

Hence, Option D is correct.

Identify the correct order of reactivity in electrophilic substitution reaction of the following compounds.

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

Explanation

Order of reactivity of given compounds in electrophilic substitution reaction is same as the order of electron density in the given compounds.

Order of electron density of the given compounds.

$\mathrm{II (due \ to \ +I \ effect \ of \ C_2H_5) > I > III(due \ to \ -R \ effect \ of \ Cl) > IV (due \ to \ -R \ effect \ of \ NO_2 }$

Hence, Option "C" is the correct answer.

Here P is / are:

0%
0%
0%
0%
All of these

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 13 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 13 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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