Problem 11 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The compound that will react most readily with \(\mathrm{NaOH}\) to form methanol is (a) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{I}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+} \mathrm{I}^{-}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)

Step-by-Step Solution

Answer

(c) \((\mathrm{CH}_3)_3 \mathrm{S}^+ \mathrm{I}^−\) is the compound that reacts most readily with \(\mathrm{NaOH}\) to form methanol.

Step 1: Understand the Reaction

The question asks which compound will react most readily with \(\mathrm{NaOH}\) to form methanol. Methanol is \(\mathrm{CH}_3\mathrm{OH}\). So, we need a compound that can donate a \(\mathrm{CH}_3\) group when reacted with \(\mathrm{NaOH}\).

Step 2: Analyze Each Option

Let's consider how each option might react with \(\mathrm{NaOH}\):(a) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{I}^{-}\) - Generally does not release a \(\mathrm{CH}_3\) group easily.(b) \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) - An ether that does not break to form methanol with \(\mathrm{NaOH}\).(c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+} \mathrm{I}^{-}\) - Can release \(\mathrm{CH}_3\)I when reacted with \(\mathrm{NaOH}\), which can form methanol.(d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) - Generally requires a different mechanism like \(\text{SN1}\) to react.

Step 3: Evaluate Reactivity

Option \(c\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+} \mathrm{I}^{-}\) is a good candidate because when reacted with \(\mathrm{NaOH}\), it can potentially form \(\mathrm{CH}_3\)I. The \(\mathrm{CH}_3\)I can further react with \(\mathrm{NaOH}\) to form methanol \(\mathrm{CH}_3\mathrm{OH}\).

Step 4: Conclusion

The compound \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+} \mathrm{I}^{-}\) is most likely to release a \(\mathrm{CH}\) group to form methanol when reacted with \(\mathrm{NaOH}\). This makes it the best option compared to the others.

Key Concepts

Reactivity with NaOH

To understand which compound is most reactive with sodium hydroxide (\(\mathrm{NaOH}\)) for forming methanol, we must explore the general behavior of substances when reacting with bases like \(\mathrm{NaOH}\). Sodium hydroxide is a strong base capable of breaking bonds in certain organic compounds.
In this context, we must look for a compound that can be broken down by \(\mathrm{NaOH}\) to release a \(\mathrm{CH}_3\) (methyl) group, which can then become methanol, \(\mathrm{CH}_3\mathrm{OH}\). Methanol is composed of a methyl group attached to a hydroxyl group.

• Substances with weak bonds or unstable groups that can dissociate in the presence of a strong base are more reactive.
• An understanding of the formation of leaving groups (such as halogens or positively charged ions) is essential as these are more likely to interact with \(\mathrm{NaOH}\).
• Option \(c)\), \(\left(\mathrm{CH}_3\right)_3 \mathrm{~S}^{+} \mathrm{I}^{-}\), becomes particularly reactive because it can release \(\mathrm{CH}_3\) groups (as \(\mathrm{CH}_3\mathrm{I}\)) which is a suitable precursor to methanol formation.

Methanol Formation

Methanol is a simple alcohol with the formula \(\mathrm{CH}_3\mathrm{OH}\). It is formed when a methyl group combines with a hydroxide ion from \(\mathrm{NaOH}\). The generation of methanol from a reaction depends heavily on the availability of a free methyl group.
Upon reacting \(\mathrm{NaOH}\) with compounds containing methyl groups, the following can occur:

• A positive ion, such as \(\left(\mathrm{CH}_3\right)_3 \mathrm{~S}^{+} \), loses its methyl group with ease in the presence of \(\mathrm{NaOH}\).
• This methyl group then combines with the hydroxide ion from \(\mathrm{NaOH}\) to form methanol.

Therefore, compounds easily releasing their methyl groups are excellent candidates for producing methanol when \(\mathrm{NaOH}\) is involved.

Reaction Mechanisms in Organic Chemistry

Understanding reaction mechanisms in organic chemistry, such as those involving \(\mathrm{NaOH}\), is crucial. These mechanisms explain how reactions proceed at a molecular level, providing insight into which compounds can become products like methanol.
Typically, a mechanism involves a series of steps:

• The first step often involves the breaking of a bond within the compound, facilitated by \(\mathrm{NaOH}\).
• Compounds formed with easily breakable bonds in a salty or electron-deficient state, like \(\left(\mathrm{CH}_3\right)_3 \mathrm{~S}^{+} \mathrm{I}^{-}\), are more susceptible.
• A leaving group departs (e.g., \(\mathrm{I}^-\)) resulting in the formation of intermediates.
• The intermediate methyl groups then attach to the hydroxide ions, creating methanol.

A thorough understanding of these steps helps in recognizing which compounds will effectively yield methanol when reacting with \(\mathrm{NaOH}\).