Chemistry MCQs for NEET — Practice Questions with Answers

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If an ether with a tertiary alkyl group on one side and a methyl group on the other ($CH_3OC(CH_3)_3$) reacts with HI, which bond will preferentially cleave?

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Explanation

While the NCERT example shows specific cleavage at the less hindered side for $S_N2$, for tertiary alkyl groups, the reaction with HI often proceeds via an $S_N1$ mechanism. The tertiary carbocation formed is highly stable, leading to the formation of the tertiary alkyl iodide. The context implies the general principle of bond cleavage with HI.

Considering the reaction of methyl phenyl ether (anisole) with HI, why is iodobenzene not a product?

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Explanation

The text states: 'Phenols do not react further to give halides because the $sp^2$ hybridised carbon of phenol cannot undergo nucleophilic substitution reaction needed for conversion to the halide.' This directly explains why iodobenzene is not formed from the phenyl group.

In the electrophilic substitution reaction of anisole with bromine in ethanoic acid, what is the major product obtained and in what yield?

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Explanation

The NCERT text explicitly states: 'Para isomer is obtained in 90% yield' for the bromination of anisole.

Which of the following statements is true regarding the reactivity of the aromatic ring in ethers towards electrophilic substitution compared to ethers?

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Explanation

The text states, 'The alkoxy group (-OR) ... activates the aromatic ring towards electrophilic substitution in the same way as in phenol.'

Consider the reaction: $CH_3CH_2OCH_2CH_3 + HI \xrightarrow{Heat}$. The products formed are:

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Explanation

When symmetrical ethers react with HI, the C-O bond breaks, forming an alcohol and an alkyl iodide. If excess HI is present and heated, the alcohol can further react to form alkyl iodide. However, with limited HI, ethanol and iodoethane are the primary products. The provided example 7.7 (i) shows $CH_3CH_2 – O – CH_2CH_3$ reacting with HI to give $CH_3CH_2OH$ and $CH_3CH_2I$.

Which of the following represents the general formula for diazonium salts?

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Explanation

According to the NCERT text, 'The diazonium salts have the general formula $ ext{R} ext{N}_2^+ ext{X}^-$ where R stands for an aryl group and $ ext{X}^-$ ion may be $ ext{Cl}^-$, $ ext{Br}^-$, $ ext{HSO}_4^-$, $ ext{BF}_4^-$, etc.'

Which of the following is NOT a common anion found in diazonium salts?

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Explanation

The NCERT text states that the $ ext{X}^-$ ion in diazonium salts 'may be $ ext{Cl}^-$, $ ext{Br}^-$, $ ext{HSO}_4^-$, $ ext{BF}_4^-$, etc.' $ ext{OH}^-$ is not listed as a common anion in the general formula, although it can replace the diazo group in a reaction.

What is the common name for $ ext{C}_6 ext{H}_5 ext{N}_2^+ ext{Cl}^-$?

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Explanation

The NCERT text explicitly states, 'For example, $ ext{C}_6 ext{H}_5 ext{N}_2^+ ext{Cl}^-$ is named as benzenediazonium chloride'.

Primary aliphatic amines form alkyldiazonium salts that are described as:

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Explanation

The NCERT text mentions, 'Primary aliphatic amines form highly unstable alkyldiazonium salts'.

Arenediazonium salts are stable for a short time in solution at what temperature range?

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Explanation

The NCERT text states, 'Primary aromatic amines form arenediazonium salts which are stable for a short time in solution at low temperatures (273-278 K)'.

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