Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 49

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Question 1
1 / -0

For the reaction, the major product is formed by:

A
An E$$1$$ reaction

B
An E$$2$$ reaction

C
$$S_N1$$ reaction

D
$$S_N2$$ reaction

Solution

Since, $$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}$$ is a moderate base, so there is no possibility
of $$E_{1}$$ reaction and in $$E_{1 C B}$$, there takes place the carbanion
formation. Hex, there is no anion stabilising group, so $$E_{1} C B$$ is not possible.
E. mechanism is possible because of presence of $$\mathrm{H}$$ -atom anti
to the leaving group $$(B r)$$.
So, option(B) is correct.
Question 2
1 / -0

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is:

A
$$CH_3 - CH_3$$

B
$$CH_2 = CH_2$$

C
$$CH \equiv CH$$

D
$$CH_4$$

Solution

$$R-H+{ Br }_{ 2 }\quad \underrightarrow { Substitution } \quad R-Br+HBr$$
$$2R-Br+2Na\quad \underrightarrow { wurtz\quad reaction } \quad R-R+2{ Na }^{ + }{ Br }^{ - }$$
In substitution reaction alkanes produces alkyl bromides and in Wurtz reaction two alkyl halides are reacted with sodium $$(Na)$$ metal to form higher alkane.
Since after Wurtz reaction, the gaseous hydrocarbon has less than $$4$$ carbon atoms. So, $$'R'$$ should contain only $$1$$ carbon atom. So $$A$$ is $${ CH }_{ 4 }$$.
Question 3
1 / -0

Which of the following solvents is most suitable for $$S_N1$$ reaction?

A
Polar protic

B
Polar aprotic

C
Non-polar

D
All of these
Question 4
1 / -0

$$CH_3CHI_2 \xrightarrow{KCN}\, \xrightarrow[\Delta]{H_2O}$$ ?;
Here the end product would be :

A
2 - cyanopropionic acid

B
ethane - 1, 1 - dicarboxylic acid

C
2 - methyl ethanoic acid

D
propionic acid
Solution
- $$CN^{-}$$ will substitute the iodine,
- Hydrolysis of cyanide results into $$COOH$$ group,
- Heat will eliminate one $$COOH$$ group from the structure.
Question 5
1 / -0

What are possible product obtained in the following reaction?
$$(CH_3)_3-C-Br\overset{80\% EtOH}{\underset{20\% H_2O}{\rightarrow}}$$.

A
$$(CH_3)_2=CH_2$$ via E$$1$$

B
$$(CH_3)_3-C-OH$$ via $$S_N1$$

C
$$(CH_3)_3C-O-Et$$ via $$S_N1$$

D
$$(CH_3)_2=CH_2$$ via $$E2$$

Solution

In Tertiary Alkyl Halides E2 reactions are favored over SN2 reactions. The greater the alkyl substitution, the faster the reaction, since in the Transiton stage, a double bond is formed partially. A greater substituted alkene is lower in energy. Hence the activating energy is reduced, making the reaction faster.
Question 6
1 / -0

$$ CH_{ 3 }CH_{ 2 }Cl\quad \xrightarrow [ ]{ NaCN } \times \xrightarrow [ ]{ Ni/H_{ 2 } } Y $$
Y in the above reacting sequence is:

A
$$ CH_3CH_2CH_2NHCOCH_3 $$

B
$$CH_3CH_2CH_2NH_2$$

C
$$CH_3CH_2CH_2CONHCH_3$$

D
$$CH_3CH_2CH_2CONHCOCH_3$$

Solution

$$Ni/H_2$$ helps in reduction -CN reduces to $$-CN_2NH_2$$
Question 7
1 / -0

On reaction of chlorobenzene with acetyl chloride in presence of anhydrous $${ AICl }_{ 3 }$$, the major product formed is:

A

B

C

D

Solution
Question 8
1 / -0

Which of the following is most reactive towards the $${ S }_{ N }1$$ reaction?

A

B

C

D
A and B reacts at similar rate
Question 9
1 / -0

The addition of HBr is easiest with:

A
$$CH_2 = CH- Cl$$

B
$$Cl - CH = CH - Cl$$

C
$$CH_3 - CH = CH_2$$

D
$$(CH_3)_2C = CH_2$$

Solution

Addition of $$H-X$$ (Where $$x$$ is halogen) on alkenes is
electrophilic addition reaction
The rate of reaction depends on the stability of intermediate
carbocation.
Thus, that carbocation is more stable for cempound (2)
So, correct option is (D)
Question 10
1 / -0

The correct statement for $$\alpha $$ elimination is?

A
It forms cyclic compounds

B
It forms carbine or substituted carbine

C
Two atoms are removed from $$\alpha $$ and $$\beta$$ positions

D
In $$CHCl_{3};\alpha $$ elimination is not possible

Solution

α elimination (eliminations in which both the proton and the leaving group are located on the same atom) follow a mechanism akin to an E1cB β-elimination. A strong base removes an acidic proton adjacent to an electron withdrawing group to give a carbanion. Loss of a leaving group from the carbanion creates a carbene. One of the best known elimination reactions occurs when chloroform is treated with base, forming a dichlorocarbene.