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Tertiary alkyl halides are practically inert to substitution by \(\mathrm{SN}^{2}\) mechanism because of (a) insolubility (b) instability (c) inductive effect (d) steric hindrance

Short Answer

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(d) steric hindrance

Step by step solution

01

Understand SN2 Mechanism

The \(\text{S}_N2\) mechanism involves a nucleophile attacking the electrophilic carbon at the opposite side of the leaving group, leading to a bimolecular substitution. This requires a backside attack and is significantly hindered by steric factors.
02

Analyze Tertiary Alkyl Halides

Tertiary alkyl halides are characterized by a central carbon atom bonded to three other carbon groups and a halogen. This structure creates significant steric hindrance, making it difficult for a nucleophile to approach the carbon center from the backside.
03

Identify the Effect of Steric Hindrance

Due to the bulky groups surrounding the central carbon in tertiary alkyl halides, there is simply not enough space for the nucleophile to effectively carry out a backside attack. This steric hindrance is the primary reason these halides are inert to the \( \text{S}_N2 \) mechanism.

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Key Concepts

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

Tertiary Alkyl Halides
Tertiary alkyl halides are a type of organic molecule where a central carbon atom is bonded to three other carbon groups and a halogen. The designation 'tertiary' arises from this carbon's connection to three alkyl groups. These compounds are quite different from primary or secondary alkyl halides due to their structural characteristics. This difference in structure plays a significant role in their chemical reactivity, particularly in substitution reactions such as the \( \text{S}_N2 \) mechanism.

In an \( \text{S}_N2 \) reaction, the nucleophile attempts to attack the electrophilic carbon. However, in tertiary alkyl halides, the carbon is closely surrounded by bulky alkyl groups, which makes the approach of a nucleophile difficult. This dense packing around the electrophilic center impacts the reaction pathway significantly and limits the reaction's feasibility.
Steric Hindrance
Steric hindrance is a term used to describe the physical blockage that hinders the approach of reactants in chemical reactions. It is a crucial factor affecting the reactivity of molecules, especially in \( \text{S}_N2 \) mechanisms. When large groups are present around a reactive site, they physically block the path of approaching reactants. This means that even if the reactants are chemically compatible, the spatial arrangement prevents them from coming together effectively.

In the case of tertiary alkyl halides, steric hindrance is pronounced. The bulky alkyl groups surrounding the central carbon create a barrier that prevents nucleophiles from completing a backside attack necessary for the \( \text{S}_N2 \) mechanism. As a consequence, tertiary alkyl halides tend to be inert to \( \text{S}_N2 \) reactions, opting instead for \( \text{S}_N1 \) mechanisms where the steric factor is less of an issue.
Nucleophile
A nucleophile is a chemical species that donates an electron pair to form a new chemical bond. Nucleophiles are rich in electrons and tend to be attracted to positively charged or electron-deficient centers in molecules, such as the carbon attached to a halogen in alkyl halides.

In the \( \text{S}_N2 \) mechanism, the nucleophile plays a critical role as it approaches the electrophilic carbon from the opposite side of the leaving halide group. This process, called a backside attack, is crucial for the substitution to occur. However, when it comes to tertiary alkyl halides, the nucleophile's path is blocked by bulky carbon groups, making it almost impossible for the attack to take place.
  • This blockage explains why tertiary alkyl halides do not easily undergo \( \text{S}_N2 \) reactions.
  • The nucleophile cannot access the necessary site to form a new bond due to steric hindrance.
Understanding the characteristics and limitations of nucleophiles helps in predicting the outcome of substitution reactions involving different types of alkyl halides.

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Most popular questions from this chapter

The increasing order of stability of the following free radicals is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}\) (b) \(\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (c) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\)

Which of the following heterocyclic compound would not have aromatic character ? (a) c1cncnc1 (b) c1cn[nH]c1 (c) c1c[nH]cn1 (d) c1cnccn1

Arrange the following compounds in the order of increasing tendency to undergo electrophillic substitution: (1) O=[N+]([O-])c1ccccc1 (2) c1ccccc1(3) Oc1ccccc1 (4) Cc1ccccc1 (5) (a) \(5<1<2<4<3\) (b) \(3>2>1>4>5\) (c) \(1>2>5>3>4\) (d) \(5<1<4<2<3\)

Four structures (1) - (4) of different alcohols are given below: (1) CCC(O)c1ccccc1 (2) CC(O)Cc1ccccc1 (3) CCC(O)CC (4) CC(C)CCO The order of facility, from fastest to slowest, of acid catalysed dehydration will be (a) \(2>1>3>4\) (b) \(1>2>3>4\) (c) \(4>3>2>1\) (d) \(2>3>4>1\)

Given below are the four compound which are stabilized either by inductive, mesomeric or hyperconjugation effects.. (a) CC(=O)C1=C(C)CCCC1 (b) BrC1=CCCCC1 (c) CC1=CCCCC1C (d) CC=CCl

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