Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Identify the product $$X$$.

A

B

C

D

Solution

It has mechanism like tautomerism and dihydroxy product forms as product.
Question 2
1 / -0

Which will give white ppt. with $$AgNO_{3}$$?

A

B

C

D
Both A and C

Solution

As both A and C provide chloride ion, due to formation of stable carbocation (allylic and benzylic), they both give white ppt. of $$AgCl$$ with silver nitrate.
Question 3
1 / -0

Predict the product formed in the reaction.

A

B

C

D

Solution

First a 5 membered ring carbocation forms as intermediate which converts into 6-membered ring and product forms.
Question 4
1 / -0

$$X\xrightarrow{Na + dry\hspace{1mm}ether}Z$$
What can be used as $$X$$ if $$Z$$ is$$\displaystyle \:CH_{3}-CH_{3}$$?

A
$$CH_{3}-Cl$$

B
$$HCHO$$

C
$$CH_3CHO$$

D

Solution

Chloromethane $$ { CH }_{ 3 }Cl $$ $$(X)$$ undergoes Wurtz reaction in the presence of metal $$Na$$ and solvent dry ether which couples to give ethane $$ { CH }_{ 3 }-{ CH }_{ 3 } $$ $$(Z)$$.
Question 5
1 / -0

Select the incorrect statement about the product mixture in the following reaction.

A
It is optically active.

B
It is racemic mixture.

C
It is a resolvable mixture.

D
It is a mixture of erythro compounds.

Solution

In this reaction, a single optically active compound is formed as product.
Question 6
1 / -0
Consider the following groups:
1. $$-OAc$$
2. $$-OMe$$
3. $$-O-SO_{2}-Me$$
4. $$-O-SO_{2}-CF_{3}$$
The order of leaving group nature is:
A
$$I> II> III> IV$$

B
$$IV> III> I> II$$

C
$$III> II> I> IV$$

D
$$II> III> IV>I$$

Solution

If the negative charge on the leaving group is stabilized by resonance, it is a better leaving group. In the methoxy group, no such leaving group is present.

Hence it is the least preferred leaving group (among the given groups)

In $$-O-SO_{2}-CF_{3}$$ apart from resonance due to $$O-SO_{2}$$ group, the strong electron-withdrawing effect of the trifluoromethyl group is also present. Hence, it is the most preferred leaving group.

The order of leaving group nature is $$IV>III>I>II$$
or $$-O-SO_{2}-CF_{3}>-O-SO_{3}-Me>-OAc>-OMe$$
Question 7
1 / -0

In the following pair of compounds, which compound is more nucleophilic in a polar-protic solvent?

$${ CH }_{ 3 }Cl$$ or $${ CH }_{ 3 }OH$$

A
$${ CH }_{ 3 }Cl$$

B
$${ CH }_{ 3 }OH$$

C
$${ CH }_{ 3 }Cl = { CH }_{ 3 }OH$$

D
None of these

Solution

Hydrogen bonding takes place in the polar-protic solvent when solute compound includes one of FON (Flourine, Oxygen and Nitrogen) atom.

So, in case of $$CH_3OH$$ hydrogen bonding will take place between O and H atom, which decreases its nucleophilicity. Polar protic-solvent consists of $$H^+$$ ions which can form H-bond with methanol and shield the solute molecule from donating electron thus decreases its nucleophilicity. However, there is no H-bonding present as such in case of methyl chloride $$(CH_3OH)$$.

Therefore, $$CH_3Cl$$ is more nucleophilic than $$CH_3OH$$.
Hence, the correct option is A.
Question 8
1 / -0

Product is :

A

B

C

D

Solution

This reaction is called as dow's process formation of phenol takes place.

Therefore, option A is correct.
Question 9
1 / -0

For the given pair of $${{ S }_{ N }}1$$ reactions, the ______ reaction is faster.

A
first

B
second

C
both progress at the same rate

D
cannot be determined

Solution

The given pair of $$ { S }_{ N }1 $$ reactions, the first reaction in $$ { S }_{ N }1 $$ reaction is faster because, after heterolysis of $$C-Br$$ bond, the primary carbocation formed undergoes rearrangement via hydride shift which leads to the formation of a more stable tertiary carbocation.
Question 10
1 / -0

Which of the following relation is correct for recativity?

$${ CH }_{ 3 }COO^{ - }$$ or $${ CH }_{ 3 }{ CH }_{ 2 }S^{ - }$$ with $${ CH }_{ 3 }I$$ in ethanol.

A
$${ CH }_{ 3 }{ CH }_{ 2 }S^{ - } > { CH }_{ 3 }COO^{ - }$$

B
$${ CH }_{ 3 }{ CH }_{ 2 }S^{ - } < { CH }_{ 3 }COO^{ - }$$

C
$${ CH }_{ 3 }{ CH }_{ 2 }S^{ - } = { CH }_{ 3 }COO^{ - }$$

D
None of these

Solution

Higher electronegativity means lower nucleophilicity because the role of a nucleophile is to share e-. If the atom is more electronegative it is less willing to share its e- and wants to hold onto them. In the polar protic solvent like ethanol $$CH_3COO^-$$ participating in Hydrogen bonding, the nucleophile is considerably less reactive than $$CH_3CH_2S^-$$ as it interacts less to hydrogen ion. Thus the reaction will be faster for $$CH_3CH_2S^-$$ than $$CH_3COO^-$$.