xrightarrow[Delta]{KOH text{ (alc.) }} major | vedclass

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Question

(A) The reaction is a dehydrohalogenation reaction using alcoholic $KOH$ (an elimination reaction,$E2$).The substrate is $1	ext{-phenyl-2-bromo-3-met

Vedclass · Accepted Answer

$1-$phenyl$-3-$methylbut$-1-$ene

Vedclass · Answer

(A) The reaction is a dehydrohalogenation reaction using alcoholic $KOH$ (an elimination reaction,$E2$).The substrate is $1	ext{-phenyl-2-bromo-3-methylbutane}$.Elimination of $HBr$ occurs from the $\beta$-carbon atoms adjacent to the carbon bearing the $Br$ atom.There are two possible $\beta$-hydrogens:$1$. From $C-1$ (benzylic position),which leads to a conjugated alkene $(1	ext{-phenyl-3-methylbut-1-ene})$.$2$. From $C-3$,which leads to a less substituted alkene $(1	ext{-phenyl-3-methylbut-2-ene})$.According to Saytzeff's rule,the more substituted and conjugated alkene is the major product.Therefore,the major product $P$ is $1	ext{-phenyl-3-methylbut-1-ene}$.

$\xrightarrow[\Delta]{KOH \text{ (alc.) }}$ major product $P$
Product $P$ is

Similar Questions

Dehydrohalogenation of $2$-bromobutane results in the formation of:

Which among the following reactions does not occur?

What is $E$?
Cyclohexanone $\xrightarrow{Br_2 + CH_3COOH} A$ $\xrightarrow{Alc. KOH} B$ (Product with six-membered ring) $\xrightarrow[(2) H_2O]{(1) CH_3MgBr} C$ $\xrightarrow{H^{\oplus} + \Delta} D$ $\xrightarrow{HBr (1 \text{ mole})} E$

$CH_3-CH_2-CH(Br)-CH_3 \xrightarrow{Alc. KOH} X$ (major)
$CH_3-CH_2-CH(NMe_3^+) - CH_3 \xrightarrow{EtONa, \Delta} Y$ (major)
Product $(X)$ and $(Y)$ respectively are:

Which of the following reaction$(s)$ produce the Saytzeff product as a major product?

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$\xrightarrow[\Delta]{KOH \text{ (alc.) }}$ major product $P$Product $P$ is