Question 38

In the following reaction 'X' is
CH$$_3$$(CH$$_2$$)$$_4$$CH$$_3$$ $$\xrightarrow{\text{Anhy. AlCl}_3 / \text{HCl}, \Delta}$$ X (major product)

Solution

The reactant is $$n$$-hexane ($$\text{CH}_3(\text{CH}_2)_4\text{CH}_3$$), a straight-chain alkane.

The reagent used is anhydrous $$\text{AlCl}_3 / \text{HCl}$$ with heating ($$\Delta$$).

$$\text{HCl}$$ and $$\text{AlCl}_3$$ react to form a strong Lewis acid complex ($$\text{H}^+\text{AlCl}_4^-$$), which helps abstract a hydride ion from $$n$$-hexane to form a secondary carbocation.

The carbocation undergoes rearrangements (hydride and methyl shifts) to form more stable or branched intermediate carbocations.

Finally, a hydride transfer yields the stable branched alkane isomers.

For $$n$$-hexane, the rearrangement yields a mixture of branched isomers, primarily:

2-methylpentane (Major product)
3-methylpentane

$$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \xrightarrow{\text{Anhy. AlCl}_3/\text{HCl}, \Delta} \text{CH}_3-\underset{\substack{| \\ \text{CH}_3}}{\text{CH}}-\text{CH}_2-\text{CH}_2-\text{CH}_3$$

In the following reaction 'X' is
CH$$_3$$(CH$$_2$$)$$_4$$CH$$_3$$ $$\xrightarrow{\text{Anhy. AlCl}_3 / \text{HCl}, \Delta}$$ X (major product)

Solution

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In the following reaction 'X' isCH$$_3$$(CH$$_2$$)$$_4$$CH$$_3$$ $$\xrightarrow{\text{Anhy. AlCl}_3 / \text{HCl}, \Delta}$$ X (major product)

Solution

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Video Solution

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In the following reaction 'X' is
CH$$_3$$(CH$$_2$$)$$_4$$CH$$_3$$ $$\xrightarrow{\text{Anhy. AlCl}_3 / \text{HCl}, \Delta}$$ X (major product)