Hydrocarbons's Question 44 | Chemistry XI - Part 2 | NCERT Exemplar Questions

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MCQ (Single Correct Answer)

Assertion (A) The compound tetraene has the following structural formula.

It is cyclic and has conjugated $8 \pi$-electron system but it is not an aromatic compound. Reason (R) $(4 n+2) \pi$ electrons rule does not hold good and ring is not planar.

Both A and R are correct and R is the correct explanation of A

Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

Both A and R are not correct

A is not correct but R is correct

MCQ (Single Correct Answer)

Assertion (A) Toluene on Friedal Crafts methylation gives $o$ - and $p$ xylene. Reason (R) $\mathrm{CH}_3$-group bonded to benzene ring increases electron density at $o$ - and $p$ - position.

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

Both A and R are correct but R is not the correct explanation of A

Both A and R are not correct

A is not correct but R is correct

MCQ (Single Correct Answer)

Assertion (A) Nitration of benzene with nitric acid requires the use of concentrated sulphuric acid.

Reason (R) The mixture of concentrated sulphuric acid and concentrated nitric acid produces the electrophile, $\mathrm{NO}_2^{+}$.

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

Both A and R are correct but R is not the correct explanation of A

Both $A$ and $R$ are not correct

A is not correct but R is correct

MCQ (Single Correct Answer)

Assertion (A) Among isomeric pentanes, 2, 2-dimethylpentane has highest boiling point.

Reason (R) Branching does not affect the boiling point.

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

Both A and R are not correct

$A$ is not correct but $R$ is correct

Subjective

An alkyl halide $\mathrm{C}_5 \mathrm{H}_{11}$ (A) reacts with ethanolic KOH to give an alkene ' $B$ ', which reacts with $\mathrm{Br}_2$ to give a compound ' $C$ ', which on dehydrobromination gives an alkyne ' $D$ '. On treatment with sodium metal in liquid ammonia, one mole of ' $D$ ' gives one mole of the sodium salt of ' $D$ ' and half a mole of hydrogen gas. Complete hydrogenation of ' $D$ ' yields a straight chain alkane. Identify $A, B, C$ and $D$. Give the reactions involved.

Explanation

The reaction scheme involved in the problem is

Hydrogenation of alkyne $(D)$ gives straight chain alkane hence all the compounds $(A),(B)$, $(C)$ and (D) must be straight chain compounds. Alkyne (D) form sodium salt which proves that it is terminal alkyne. Involved reactions are as follows

It is important point that alkyl halide $(A)$ can not be 2-bromopentane because dehydrobromination of $(A)$ would have given 2-pentene as the major product in accordance with Markownikoff's rule.

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