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Which of the following represent the correct order of nucleophillic addition for (I) \(\mathrm{HCHO}\), (II) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), (III) \(\mathrm{CH}_{3} \mathrm{CHO}\), (IV) \(\mathrm{CH}_{3} \mathrm{COC}_{2} \mathrm{H}_{5}\) (a) \(\mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV}\) (b) \(\mathrm{I}>\mathrm{III}>\mathrm{II}>\mathrm{IV}\) (c) \(\mathrm{IV}>\mathrm{II}>\mathrm{III}>\mathrm{I}\) (d) \(\mathrm{I}>\mathrm{IV}>\mathrm{III}>\mathrm{II}\)

Short Answer

Expert verified
The correct order is (b) I > III > II > IV.

Step by step solution

01

Identify the Functionality

The given compounds are aldehydes and ketones. The reactivity for nucleophilic addition reactions generally depends on the steric and electronic effects. Aldehydes are typically more reactive than ketones due to less steric hindrance and increased positive charge on the carbonyl carbon.
02

Determine Reactivity Order of Aldehydes

Among aldehydes, formaldehyde (HCHO) is the most reactive because it has no alkyl groups attached, leading to minimal steric hindrance and a strong partial positive charge on the carbonyl carbon. Acetaldehyde (CH_{3}CHO) follows, as it has one methyl group which offers more steric hindrance compared to formaldehyde.
03

Consider Reactivity of Ketones

Ketones such as acetone (CH_{3}COCH_{3}) have more steric hindrance due to the presence of two alkyl groups compared to aldehydes, making them less reactive towards nucleophilic addition. Ethyl methyl ketone (CH_{3}COC_{2}H_{5}) is even less reactive than acetone due to its larger size.
04

Arrange Based on Reactivity

Following the analysis: formaldehyde is the most reactive, followed by acetaldehyde, then acetone, and finally ethyl methyl ketone. This corresponds to the order I (HCHO), III (CH_{3}CHO), II (CH_{3}COCH_{3}), IV (CH_{3}COC_{2}H_{5}).
05

Confirm the Correct Option

Comparing the order obtained with the options provided, option (b) I > III > II > IV matches with our arrangement based on reactivity. Therefore, option (b) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reactivity Order
In the realm of nucleophilic addition reactions, the reactivity order is crucial. This refers to how quickly different compounds react when a nucleophile is introduced.
Generally, aldehydes react more rapidly than ketones. This is due to certain structural and electronic properties of these compounds.
Aldehydes contain a carbonyl group attached to an alkyl group and a hydrogen atom. Ketones, on the other hand, have two alkyl groups attached to the carbonyl group. The presence of these groups influences the reactivity of compounds because they exert what is known as steric effects and electronic effects.
In the given exercise, formaldehyde (0HCHO1) is mentioned as the most reactive, with acetaldehyde (0CH_{3}CHO1) following. Next in line is acetone (0CH_{3}COCH_{3}1), and finally, ethyl methyl ketone (0CH_{3}COC_{2}H_{5}1). This reactivity sequence is essential for predicting how these compounds will respond in a nucleophilic attack.
Aldehydes and Ketones
Aldehydes and ketones are important functional groups in organic chemistry. They both contain a carbonyl group (0C=O1), but their structures differ slightly, which significantly affects their chemical behavior.
An aldehyde has at least one hydrogen atom attached to the carbonyl carbon, making the carbonyl group more accessible, thus enhancing its reactivity to nucleophiles. This is why in terms of nucleophilic addition reactivity, aldehydes come out on top compared to ketones.
Ketones have two alkyl groups attached to the carbonyl carbon, which makes them less reactive. The alkyl groups create more steric hindrance and reduce the partial positive charge on the carbonyl carbon.
This structural difference is the key reason why formaldehyde, with no obstructive alkyl groups, is more reactive than a common ketone like acetone. Understanding these nuances lets us predict and explain reaction mechanisms and reactivity trends in organic synthesis.
Steric Effects
Steric effects play a significant role in the reactivity of aldehydes and ketones in nucleophilic addition reactions. These effects refer to the physical obstacles that molecules face, influencing how they react with one another.
In the context of aldehydes and ketones, steric effects arise from the size and number of groups attached to the carbonyl carbon.
For example, formaldehyde has no alkyl groups adjacent to the carbonyl carbon, which means there is little to no steric hindrance, allowing nucleophiles easy access to the carbonyl carbon.
On the contrary, as you look at acetone with its two methyl groups, or ethyl methyl ketone with a methyl and an ethyl group, the steric hindrance increases. This makes it more challenging for nucleophiles to approach and attack the carbonyl carbon.
The greater the steric hindrance, the less reactive the compound tends to be to nucleophilic addition. This fundamental concept explains the reactivity order discussed and aids in designing reactions and predicting outcomes in organic chemistry.

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Most popular questions from this chapter

Which of the following order/statement (s) is/are correct? (a) In \(S_{N} 1\) reactions, \(\mathrm{AgNO}_{3}\) increases the rates of solvolysis (b) Soft bases have enhanced nucleophilicities towards \(\mathrm{S}_{\mathrm{N}} 2\) reactions. (c) \(\mathrm{PhSO}_{3}^{-}>\mathrm{Cl}_{3} \mathrm{CCOO}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{PhO}^{-}\)(leav- ing group ability) (d) HCOOH may lead a tertiary halide towards \(\mathrm{S}_{\mathrm{N}}^{2} .\)

In which of the following molecules lone pair of nitrogen atom is not participating in resonance? (a) C=C(N)c1ccccc1 (b) c1ccncc1 (c) Cc1cccc(C)c1N(C)C (d) NCc1ccccc1

Most stable carbocation is (a) \(\mathrm{CH}_{2}=\mathrm{C} \mathrm{H}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}\) (c) [C]C1C=CCCC1 (d) CC1C=CCCC1

The order of effectiveness of o- and p- directing groups is (a) \(\mathrm{OH}>\mathrm{NR}_{2}>\mathrm{NH} \cdot \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{CH}_{3}\) (b) \(\mathrm{OH}>\mathrm{Cl}>\mathrm{CH}_{3}>\mathrm{NR}_{2}>\mathrm{NH} . \mathrm{COCH}_{3}\) (c) \(\mathrm{OH}>\mathrm{NH} . \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{NR}_{2}>\mathrm{CH}_{3}\) (d) \(\mathrm{OH}>\mathrm{CH}_{3}>\mathrm{NH} . \mathrm{COCH}_{3}>\mathrm{Cl}>\mathrm{NR}_{2}\)

Consider the following carbanions: (1) \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}} \mathrm{H}_{2}\) (2) \(\mathrm{CH}_{3}=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}\) (3) \(\mathrm{CH} \equiv \stackrel{\ominus}{\mathrm{C}}\) Correct order of stability of these carbanions in decreasing order is (a) \(1>2>3\) (b) \(2>1>3\) (c) \(3>2>1\) (d) \(3>1>2\)

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