Reaction of tert−butyl pentyl ether [CH3CH2CH2CH2CH2OC(CH3)3] with HBr forms 1−bromopentane (CH3CH2CH2CH2CH2Br) and compound H.
H has a molecular ion in its mass spectrum at 56 and gives peaks in its IR spectrum at 3150−3000, 3000−2850, and 1650 cm−1. Draw a structure for H and arrange the correct stepwise mechanism that accounts for its formation.
1 answer:
Answer:The compound H is is 2-Methyl prop-1-ene and its structure can be found in attachment. The m/z value of 2-Methyl prop-1-ene is 56 which clearly matches with the mas spectrum data. The structure can also be ascertained using the provided IR data. The IR data has absorption stretching frequency in the region of 1650cm⁻¹ which is due to the C=C double bond. The stretching frequency at 3150-3000cm⁻¹ is due to the unsaturated C-H bond. The stretching frequency at 3000-2850cm⁻¹ is due to the saturated C-H bonds. Kindly refer attachment for the mechanism.
Explanation:
The reaction of tert-butyl ether with HBr leads to the formation of 1-bromopentane and tertbutyl alcohol.
The tert butyl alcohol formed undergoes E1 elimnation reaction to give 2-Methyl prop-1-ene.
The mass spectra and IR data available for the compound H completely matches with that of 2-Methyl prop-1-ene hence we can ascertain that the compound H is 2-Methyl prop-1-ene.
The m/Z value of 2-Methyl prop-1-ene is 56 which is in compete accordance with the provided data for compound H .
The IR data also completely matches with the structure of 2-Methyl prop-1-ene as the following Infrared absorption peaks are provided which matches with that of 2-Methyl prop-1-ene :
The absorption stretching frequency in the region of 1650cm⁻¹ corresponds to the C=C double bond which is clearly evident in 2-Methyl prop-1-ene.
The stretching frequency at 3150-3000cm⁻¹ corresponds to unsaturated C-H bond.
The stretching frequency at 3000-2850cm⁻¹ is due to the saturated C-H bonds.
The mechanism of the reaction involves the following steps:
1. The oxygen atom in tert-butyl pentyl ether is protonated by treating it with Hydrogen bromide and Br⁻ is lost from hydrogen bromide.
2. Now since the oxygen atom is protonated it turns into a good leaving group and can leave as tertiary butyl alcohol. The eliminated Br⁻ now attacks in a SN2 manner from the back side at the primary carbon center which leads to the formation of 1-Bromopentane and tertiary-butyl alcohol
3.The tert-butyl alcohol formed further reacts with HBr present to give elimiantion product 2-Methyl prop-1-ene through E1 elimination mechanism. The OH is protonated and further it gets eliminated as H₂O leading to formation of a tertiary carbocation. The tertiary carbocation formed gives an elimination product of 2-Methyl prop-1-ene .
Kindly refer the attachment for complete reaction mechanism.
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Answer : The rate of the reaction if the concentration of is doubled is, 0.006 M/s
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
The balanced equations will be:
In this reaction, and
are the reactants.
The rate law expression for the reaction is:
As we are given that:
= concentration of
= 0.100 M
= concentration of
= 0.100 M
Rate = 0.0030 M/s
Now put all the given values in the above expression, we get:
Now we have to calculate the rate of the reaction if the concentration of is doubled.
Thus, the rate of the reaction if the concentration of is doubled is, 0.006 M/s