Haloalkanes and Haloarenes Questions — NEET Chemistry

Q1

The following reaction method is not suitable for the preparation of the corresponding haloarene products, due to high reactivity of halogen, when X is :

A F

B I

C Cl

D Br

Correct Answer

Option A

Solution

$\bullet$ Aryl chlorides and bromides can easily be prepared by electrophilic substitution of arenes (toluene) with Cl

_2

and Br

_2

respectively in the presence of Lewis acid catalyst (Fe in dark). $\bullet$ Reaction with I

_2

is reversible and requires the presence of oxidising agent.

$\bullet$ Corresponding fluoroarene is not prepared by this method due to high reactivity of fluorine.

Hence, ‘X’ is F.

Q2

Identify the major product C formed in the following reaction sequence :

\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{I} \xrightarrow{\mathrm{NaCN}} \mathrm{A}

\mathrm{\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Partial\,hydrolysis}^{O{H^ - }}} B\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{B{r_2}}^{NaOH}} \mathop C\limits_{(major)}}

A propylamine

B butylamine

C butanamide

D

\alpha

-bromobutanoic acid

Correct Answer

Option A

Solution

$\bullet$ Step-l is

\mathrm{S}_{\mathrm{N}}

reaction with

\stackrel{\ominus}{\mathrm{C}} \mathrm{N}

nucleophile. $\bullet$ Step-II will give amide. $\bullet$ Step-III is Hoffmann bromamide degradation reaction.

Q3

The major product formed in dehydrohalogenation reaction of 2-Bromo pentane is Pent-2-ene. This product formation is based on?

A Huckel's Rule

B Saytzeff's Rule

C Hund's Rule

D Hofmann Rule

Correct Answer

Option B

Solution

Major product formed in dehydrohalogenation reaction of 2-bromopentane is pent-2-ene because according to Saytzeff's Rule, in dehydrohalogenation reactions, the preferred product is that alkene which has greater number of alkyl group(s) attached to the doubly bonded carbon atoms.

Q4

Which of the following alkane cannot be made in good yield by Wurtz reaction?

A 2, 3-Dimethylbutane

B n-Heptane

C n-Butane

D n-Hexane

Correct Answer

Option B

Solution

Wurtz reaction is used to prepare symmetrical alkanes like R 1 $-$ X + 2Na + X $-$ R 1

\overset{{Dry\;ether}}\longrightarrow

R 1 $-$ R 1 + R 1 $-$ R 2 + R 2 $-$ R 2 + 2NaX Using Wurtz reaction, n-Heptane can not be produced in good production because it is unsymmetrical alkane.

Q5

Elimination reaction of 2-Bromo-pentane to form pent-2-ene is : (A)

\beta

-Elimination reaction (B) Follows Zaitsev rule (C) Dehydrohalogenation reaction (D) Dehydration reaction

A (A), (C), (D)

B (B), (C), (D)

C (A), (B), (D)

D (A), (B), (C)

Correct Answer

Option D

Solution

Since $\beta$ -hydrogen is abstracted it is $\beta$ -elimination.

Since more substituted alkene is formed (stable and major product), it follows Zaitsev's rule or Saytzev's rule.

Since H and Br are removed, it is dehydrohalogenation.

Q6

Consider the reaction, CH 3 CH 2 CH 2 Br + NaCN

\to

CH 3 CH 2 CH 2 CN + NaBr This reaction will be the fastest in

A ethanol

B methanol

C N, N' -dimethylformamide (DMF)

D Water

Correct Answer

Option C

Solution

The reaction, CH 3 CH 2 CH 2 Br + NaCN $\to$ CH 3 CH 2 CH 2 CN + NaBr follows S N 2 mechanism which is favoured by polar aprotic solvent i.e., N, N'–dimethylformamide (DMF).

Q7

In an S N 1 reaction on chiral centers, there is

A inversion more than retention leading to partial racemisation

B 100% retention

C 100% inversion

D 100% racemisation

Correct Answer

Option A

Solution

In case of optically active alkyl halides, S N 1 reaction is accompanied by racemisation.

The carbocation formed in the slow step being sp 2 hybridised is planar and attack of nucleophile may take place from either side resulting in a mixture of products, one having the same configuration and other having inverted configuration.

The isomer corresponding to inversion is present in slight excess because S N 1 also depends upon the degree of shielding of the front side of the reacting carbon.

Q8

Which of the following compounds will undergo racemisation when solution of KOH hydrolyses?

A (i) and (ii)

B (ii) and (iv)

C (iv)

D (i) and (iv)

Correct Answer

Option C

Solution

Out of the given four compounds only (iv) compound is chiral and hence only this compound will undergo racemisation.

Q9

In the replacement reaction The reaction will be most favourable if M happens to be

A Na

B K

C Rb

D Li

Correct Answer

Option C

Solution

Tertiary alkyl halides shows S N 1 mechanism to the greater extent.

In the given reaction negative ion will attack on carbocation.

Thus, greater the tendency of ionization (greater ionic character in M-F bond) more favourable will be reaction.

In the given options Rb-F is most ionic and hence it will be most favourable for S N 1 mechanism.

Q10

Consider the reactions. The mechanisms of reactions (i) and (ii) are respectively

A S N 1 and S N 2

B S N 1 and S N 1

C S N 2 and S N 2

D S N 2 and S N 1

Correct Answer

Option C

Solution

These reactions are purely S N 2 reactions as in reaction (i) and (ii) there is no rearrangement takes place (rearrangement occurs in S N 1 mechanism).

Simple substitution of nucleophile takes place.