Compound a with molecular formula c4h9br is treated with aqueous koh solution. the rate of reaction depends on concentration of a only. when another optically active isomer b is treated with aqueous koh the rate or reaction is found to depend on b and oh–. write down formulas of a and b. out of these two compounds which would be converted to inverse configuration and racemisation?
Answers
Compound B will undergo inversion.
Compound A will be 2 - Bromo - 2 - methyl propane. As it's a tertiary molecule , it will follow Sn1 nucleophilic substitution reaction .
Compound B will be 2 - bromobutane and an optically active compound and it' ll follow Sn2 nucleophilic substitution reaction and this molecule will undergo inversion.
The reaction is given in the following image.
Answer:
As the rate of reaction depends upon the concentration of compound 'A' (C4H9Br)(C4H9Br) only therefore, the reaction is proceeded by SN1 mechanism and the given compound will be tertiary alkyl halide i.e., 2-bromo-2-methylpropane and the structure is as follow
CH3
|
H3C − C − Br (A)
|
CH3
Optically active isomer of (A) is 2-bromobutane (B) and its structural formula is
CH3−CH2−CHCH3
|
Br
(ii) As compound (B) Is opically active therefore, compound (B) must be 2-bromabutane Since, the rate of reaction of compound (B) depends both upon the Concentration compound (B) and KOH, hence, the reaction follow SN2 mechanism, In SN2 reaction nucleophile attack from, the back side, therefore, the product of hydrolysis will have opposite configuration
*The picture of here is above
Explanation:
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