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Why does 1-Bromo-2-methylbutane not form a racemic mixture on nucleophilic substitution by OH- ion even though its chiral?
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Posted by
Marty Lemish
Why does 1-Bromo-2-methylbutane not form a racemic mixture on nucleophilic substitution by OH- ion even though its chiral?
Well, look at the substrate … [math]H_{3}stackrel{4}C-CH_{2}stackrel{ ext{*}}CH(CH_{3})stackrel{1}CH_{2}Br[/math]
[math]C2[/math] as written is CHIRAL … i.e. it is substituted by hydrogen, a bromo-methyl group, a methyl group, and by an ethyl group … nucleophilic substitution at the bromo-methyl [math]C1[/math] does NOT affect the chirality of the backbone…
i.e. the product… [math]H_{3}stackrel{4}C-CH_{2}stackrel{ ext{*}}CH(CH_{3})stackrel{1}CH_{2}OH[/math]
Well, look at the substrate … [math]H_{3}stackrel{4}C-CH_{2}stackrel{ext{*}}CH(CH_{3})stackrel{1}CH_{2}Br[/math]
[math]C2[/math] as written is CHIRAL … i.e. it is substituted by hydrogen, a bromo-methyl group, a methyl group, and by an ethyl group … nucleophilic substitution at the bromo-methyl [math]C1[/math] does NOT affect the chirality of the backbone…
i.e. the product… [math]H_{3}stackrel{4}C-CH_{2}stackrel{ext{*}}CH(CH_{3})stackrel{1}CH_{2}OH[/math]
carbcation formed is an intermediate and is planer so nucleophile can attack from either direction.
but the direction from which leaving group departure the substrate is not that much supporting nucleophile to attack on it .means there is recimization but inversion product will be present in heigher amount than retension product.
carbcation formed is an intermediate and is planer so nucleophile can attack from either direction.
but the direction from which leaving group departure the substrate is not that much supporting nucleophile to attack on it .means there is recimization but inversion product will be present in heigher amount than retension product.
Well, look at the substrate … [math]H_{3}stackrel{4}C-CH_{2}stackrel{ ext{*}}CH(CH_{3})stackrel{1}CH_{2}Br[/math]
[math]C2[/math] as written is CHIRAL … i.e. it is substituted by hydrogen, a bromo-methyl group, a methyl group, and by an ethyl group … nucleophilic substitution at the bromo-methyl [math]C1[/math] does NOT affect the chirality of the backbone…
i.e. the product… [math]H_{3}stackrel{4}C-CH_{2}stackrel{ ext{*}}CH(CH_{3})stackrel{1}CH_{2}OH[/math]
…RETAINS the chirality of the substrate …
Well, look at the substrate … [math]H_{3}stackrel{4}C-CH_{2}stackrel{ext{*}}CH(CH_{3})stackrel{1}CH_{2}Br[/math]
[math]C2[/math] as written is CHIRAL … i.e. it is substituted by hydrogen, a bromo-methyl group, a methyl group, and by an ethyl group … nucleophilic substitution at the bromo-methyl [math]C1[/math] does NOT affect the chirality of the backbone…
i.e. the product… [math]H_{3}stackrel{4}C-CH_{2}stackrel{ext{*}}CH(CH_{3})stackrel{1}CH_{2}OH[/math]
…RETAINS the chirality of the substrate …
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SN1 reaction is a two steped process
step 1 formation of carbcation
step 2 attack of nucleophile on carbcation
carbcation formed is an intermediate and is planer so nucleophile can attack from either direction.
but the direction from which leaving group departure the substrate is not that much supporting nucleophile to attack on it .means there is recimization but inversion product will be present in heigher amount than retension product.
SN1 reaction is a two steped process
step 1 formation of carbcation
step 2 attack of nucleophile on carbcation
carbcation formed is an intermediate and is planer so nucleophile can attack from either direction.
but the direction from which leaving group departure the substrate is not that much supporting nucleophile to attack on it .means there is recimization but inversion product will be present in heigher amount than retension product.
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