An optically active compound having molecular formula C7H15Br reacts with aqueous KOH to give a racemic mixture of products. Write the mechanism involved for this reaction.
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C7H15-Br + KOH . the product is C7H15-OH + KBr.
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● Answer with explaination -
C5H15Br on reaction with aq.KOH forms racemic mixture. SN1 reactions involve racemization, so this must be SN1 reaction.
Further, 3° alkyl halides are more susceptible to SN1 reaction , means C5H15Br is tertiary alkyl halide. Also the compound is optically active so all functional groups must be different.
Optically active tertiary alkyl halide with molecular formula C5H15Br is most 3-bromo-3-methylhexane.
[Mechanism of SN1 reaction is given in image attachment.]
Hope this helps you...
● Answer with explaination -
C5H15Br on reaction with aq.KOH forms racemic mixture. SN1 reactions involve racemization, so this must be SN1 reaction.
Further, 3° alkyl halides are more susceptible to SN1 reaction , means C5H15Br is tertiary alkyl halide. Also the compound is optically active so all functional groups must be different.
Optically active tertiary alkyl halide with molecular formula C5H15Br is most 3-bromo-3-methylhexane.
[Mechanism of SN1 reaction is given in image attachment.]
Hope this helps you...
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