The hydrolysis of 2-bromo-3-methyl butane by ${{\text{S}}_{\text{N}}}\text{1}$ mechanism gives mainly:A. 3-Methyl-2-butanolB. 2-Methyl-2-butanolC. 2,2-Dimethyl-1-propanolD. 2-Methyl-1-butanol (2024)

Hint: Hydrolysis means the reaction of compounds with water. ${{\text{S}}_{\text{N}}}\text{1}$ mechanism mainly follows two to three steps, formation of carbocation, followed by shifting of positive charge and then deprotonation of nucleophile. It is a nucleophilic substitution reaction.

Complete step by step answer:
Let us find the final product using ${{\text{S}}_{\text{N}}}\text{1}$ mechanism:
Step (1)- The carbon-bromine bond is polar because of the difference in electronegativity of the elements. Bromine is more polar than carbon. So, the bond breaks to form the carbocation and the leaving group will be bromide ion.

Step (2)- The positive charge on the carbon is shifted as there can be formation of more stable carbocation. The stability of a carbocation seen by hyperconjugation effect. The more the number of hyper conjugating structures the more will be the stability of carbocation. This carbocation has 4 $\alpha -$hydrogens. Hyper conjugating structures are counted by counting the number of $\alpha -$hydrogens. So, there will be hydride shift or shifting of hydrogen atom.

This is more stable as it has 6 $\alpha -$hydrogens.
Step (3)- Attack of nucleophile that is water on the carbocation, so, oxonium ion is formed. There will be positive signs on electronegative oxygen atoms.

The hydrolysis of 2-bromo-3-methyl butane by ${{\text{S}}_{\text{N}}}\text{1}$ mechanism gives mainly:A. 3-Methyl-2-butanolB. 2-Methyl-2-butanolC. 2,2-Dimethyl-1-propanolD. 2-Methyl-1-butanol (3)

Step (4)- The positive charge on the oxygen atom is removed as ${{\text{H}}^{+}}$ ion gets removed from the attached water molecule. The final product formed will be 2-methyl-2-butanol.

The hydrolysis of 2-bromo-3-methyl butane by ${{\text{S}}_{\text{N}}}\text{1}$ mechanism gives mainly:A. 3-Methyl-2-butanolB. 2-Methyl-2-butanolC. 2,2-Dimethyl-1-propanolD. 2-Methyl-1-butanol (4)

The correct answer is option ‘b’, 2-methyl-2-butanol.
So, the correct answer is “Option B”.

Note: The role of solvent is very important for ${{\text{S}}_{\text{N}}}\text{1}$ reaction. The preferred solvents for ${{\text{S}}_{\text{N}}}\text{1}$ reaction are both polar and protic solvents. The polar nature of the solvent helps in stabilizing the ionic intermediates formed whereas the protic nature of the solvent helps to solvate the leaving group.

The hydrolysis of 2-bromo-3-methyl butane by ${{\text{S}}_{\text{N}}}\text{1}$ mechanism gives mainly:A. 3-Methyl-2-butanolB. 2-Methyl-2-butanolC. 2,2-Dimethyl-1-propanolD. 2-Methyl-1-butanol (2024)

FAQs

What is the hydrolysis of 2 bromo 3 methyl butane? ›

Hydrolysis of 2-bromo-3-methylbutane ( 2 ∘ ) gives only 2-methyl-2-butanol ( 3 ∘ ).

What is the hydrolysis of 2 bromo 2 methyl propane? ›

Hydrolysis of 2-bromo-2-methylpropane (tert-butyl bromide) yields 2-methylpropan-2-ol. (CH3) 3CBr + 2H2O rightarrow (CH3) COH + H3O+ + Br- Give the Sn1 mechanism. Draw structures - including electrons and charges - and add curved arrows.

What is the major product of 2-bromo-3-Methylbutane on hydrolysis with aqueous Koh? ›

2-Methyl-2-butanol.

What is the elimination of 2-bromo-3-Methylbutane? ›

2-bromo-3-methylbutane is a type of halogenoalkane, and when it reacts with ethanolic potassium hydroxide, an elimination reaction occurs. Ethanolic potassium hydroxide is a strong base and favors elimination over substitution. Because we have secondary bromo group attached, we get E2 elimination.

What is obtained by hydrolysis of optically active to bromo butane? ›

The hydrolysis of optically active 2−bromobutane with aqueous NaOH will result in the formation of both (+) and (−) butan−2−ol as the reaction will proceed via SN1 mechanism.

What happens to methanol formed during hydrolysis of methyl 2 hydroxybenzoate? ›

The methanol produced remains in the filtrate because it is completely miscible (capable of mixing) with water. The reaction of concentrated hydrochloric acid with the excess sodium hydroxide and the sodium salt of the salicylic acid is exothermic.

What is the mechanism of the SN1 reaction? ›

SN1 reaction mechanism follows a step-by-step process wherein first, the carbocation is formed from the removal of the leaving group. Then the carbocation is attacked by the nucleophile. Finally, the deprotonation of the protonated nucleophile takes place to give the required product.

What are the products of the reaction between 2-bromo-2-methylpropane and water? ›

Question: 2-bromo-2-methylpropane undergoes reaction with water to form 2-methyl-2-propanol whereas 1-bromo-2, 2-dimethylpropane undergoes the same reaction to produce 2-methyl-2- butanol.

What is the isomerism of 2-methyl propane? ›

Methyl propane is an isomer of Butane. Butane has the chemical formula C4H10 and it has two structure isomers also called as unbranched butane, or normal butane, and isobutane, or i-butane. As per the IUPAC nomenclature practice, the isomers of methyl propane called as the butane and 2-methyl propane.

What will be the major product formed when 2 methyl butane undergoes bromination in presence of light? ›

In presence of sun light, free radical mechanism takes place and most stable 30 free radical forms in intermediate so 2-bromo-2-methyl butane forms as major product.

What is the major product in the dehydrohalogenation of 2 bromo 3 3 dimethylbutane? ›

The major product in the dehydrohalogenation of 2-Bromo-3,3-dimethyl-butane is: 3,3-dimethyl but-1-ene.

What is the major product of 2 bromo 2-methylbutane? ›

The major product obtained in the photo-catalyzed bromination of 2-methyl butane is: 2-bromo-2-methylbutane.

What is the elimination of 2 bromo butane? ›

In the elimination of 2-bromobutane, for example, we find that trans-2-butene is produced in a 6:1 ratio with its cis-isomer. The Zaitsev Rule is a good predictor for simple elimination reactions of alkyl chlorides, bromides and iodides as long as relatively small strong bases are used.

What is the elimination reaction of 3 bromo 3 Methylpentane? ›

Answer and Explanation:

In such a reaction, 3-Bromo-3-methyl pentane reacts with sodium methoxide and methanol and produces the major elimination product 3-Methyl-2-pentene. Sodium methoxide acts as a base whereas methanol is used as a solvent. Thus, the major elimination product is 3-Methyl-2-pentene.

What is the equation for 2-bromo-3-methylbutane? ›

2-Bromo-3-methylbutane | C5H11Br | ChemSpider.

What is the hydrolysis of methyl propane? ›

2-Chloro-2-methylpropane slowly hydrolyzes in water, releasing hydrochloric acid. Therefore, the addition of 2-chloro-2-methylpropane to an aqueous solution of sodium hydroxide containing an acid–base indicator causes a color change when the solution becomes acidic.

What is the hydrolysis reaction of sodium borohydride? ›

Kinetics of the NaBH4 hydrolysis reaction increases with γ-Al2O3 nanoparticles and the calculated activation energy is 29 kJ/moles. This study also reports that a combined dual-solid-fuel system is highly efficient in terms of hydrogen storage capacities compared with a single hydride based system.

What is the structural formula of 2 bromo 3 methyl butane? ›

2-Bromo-3-methylbutane | C5H11Br | CID 529978 - PubChem.

What is hydrolysis of primary Haloalkanes? ›

Hydrolysis of halogenoalkanes is a chemical reaction where a halogenoalkane (also known as a haloalkane or alkyl halide) reacts with water, resulting in the formation of an alcohol and a halide ion.

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