Haloalkanes and Haloarenes Test

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Haloalkanes and Haloarenes Test - 80

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following compounds is stable?

A
$$ CH_3CH(OH)_2 $$

B
$$ (CH_3)_2C(OH)_2 $$

C
$$ CCl_3CH(OH)_2 $$

D
None of these

Solution

Here we noticed that two $$OH$$ group attached with one carbon atom at all three compounds which is called the geminal diol compound.
1. $$CH_3-CH(OH)_2$$
2. $$(CH_3)_2C(OH)_2$$
3. $$CCl_3CH(OH)_2$$

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
$$Hence \;the\; correct \;option \;is\; C.$$
Question 2
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$$ (X) +KCN \longrightarrow CH_3CN \quad \xrightarrow {2H_2/Ni} CH_3CH_2NH_2 $$ what is (X) ?

A
$$ CH_3CH_2Cl $$

B
$$ CH_3Cl $$

C
$$ CH_3CH_2CH_2Cl $$

D
$$ (CH_3)_2CHCl $$

Solution

It is a nucleophilic substitution reaction (SN2) in which cyanide ion will displace chloride ion.

Thus methyl chloride will react with potassium cyanide to form methyl cyanide.

$C H_{3} - C l + K C N \to C H_{3} - C N + K C l$
$Thus, X =CH3Cl$
$Since, Option B is correct.$
Question 3
1 / -0

Which of the following is liquid at room temperature?

A
$$ CH_3I $$

B
$$ CH_3Br $$

C
$$ C_2H_5CI $$

D
$$ CH_3F $$

Solution
Question 4
1 / -0

Which of the following is the least reactive towards
nucleophile?

A
$$ CH_3CH_2Cl $$

B
$$ CH_3Cl $$

C
$$ H_2C=CH=CH_2Cl $$

D
$$ C_6H_5Cl $$

E
$$ CH_3CH(Cl)CH_3 $$

Solution

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.
Question 5
1 / -0

Which of the following is the most reactive towards nucleophilic substitution reaction?

A
$$R-Cl$$

B
$$ (RCO)_2O $$

C
$$RCOCl$$

D
$$ RCONH_2 $$
Solution
- $$\text{Rate of nucleophilic substitution reaction is directly proportional to positive charge density }$$
among all in $$R-\overset{\overset{O}\parallel}{C}-Cl$$ chlorine because of having highest electronegativity it can easily pull out electron cloud towards

it and form stable carbocation $$R-\overset{+}{C}=O$$ and proceed towards nucleophilic substitution reaction easily.

Thus, Option C is correct.
Question 6
1 / -0

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

A
I > II > III > IV

B
IV > III > II > I

C
II > I > III > IV

D
II > III > I > IV

Solution

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.
Question 7
1 / -0

Identify the optically active compounds from the following.

A
$$[Co(en)_3]^{3+}$$

B
trans-$$[Co(en)_2Cl_2]^+$$

C
cis-$$[Co(en)_2Cl_2]^+$$

D
$$[Cr(NH_3)_5Cl]$$

Solution

Optical Isomerism
As Illustrated diagrametically above,
Optical isomers are mirror that cannot be superimposed on one another such molecule or ions are also termed as chiral. the the complex molecule $$ [Co(en)_3] $$ as in option
(i) Satisfies the condition and is , therefore an optically active compound.
Further in a coordination entity of the type cis $$ - [Co(en)_2Cl_2]^{2+} $$ only the cis-isomer can show optical activity, this condition is satisfied at option (iii) .
Question 8
1 / -0

The product W is:

A

B

C

D

Solution
Question 9
1 / -0

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

A

B

C

D

Solution

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.
Question 10
1 / -0

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 :

A
Compound H

B
Compound B

C
Compound G

D
Compound D

Solution

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.