Which one of the following compounds will giv | vedclass

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(A) In the dehydrohalogenation of $2$-bromobutane $(CH_3-CH_2-CH(Br)-CH_3)$ with alcoholic $KOH$,the $Br^{-}$ ion is a good leaving group. According t

Vedclass · Accepted Answer

$CH_3-CH_2-CH(Br)-CH_3 \xrightarrow[alc. KOH]{\Delta}$

Vedclass · Answer

(A) In the dehydrohalogenation of $2$-bromobutane $(CH_3-CH_2-CH(Br)-CH_3)$ with alcoholic $KOH$,the $Br^{-}$ ion is a good leaving group. According to $Saytzeff$'s rule,the elimination of a $\beta$-hydrogen occurs from the more substituted carbon to form the more stable alkene,$2$-butene $(CH_3-CH=CH-CH_3)$,as the major product. In contrast,alkyl fluorides (option $B$),quaternary ammonium salts (option $C$),and esters (option $D$) typically follow $Hofmann$'s rule,yielding the less substituted alkene ($1$-butene) as the major product due to the poor leaving group ability or steric/electronic factors in their respective transition states.

Which one of the following compounds will give a $Saytzeff$ alkene as the major product in an elimination reaction?

Similar Questions

Assertion $(A)$: $S_N1$ hydrolysis of optically active $2-$bromooctane results in the formation of $(\pm)-$octan$-2-$ol.
Reason $(R)$: The reaction proceeds through a planar carbocation which can be attacked by the nucleophile from either side.
The correct answer is

The dehydrobromination of $2$-bromobutane results in the formation of which of the following?

The reactivity of the halogen atom is minimum in:

The decreasing order of the rate of $S_{N^1}$ reaction for the following compounds is:
$I$. $Ph-C(Br)(CH_3)-Ph$
$II$. $Ph-CH(Br)-Ph$
$III$. $Ph-CH(Br)-CH_3$
$IV$. $Ph-CH_2Br$

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