Consider the following reaction :
\( {\left( {C{H_3}} \right)_3}CCH\left( {OH} \right)C{H_3}\buildrel {conc.\,\,\,{H_2}S{O_4}} \over \longrightarrow \)
\( {\left( {C{H_3}} \right)_2}CHCH\left( {Br} \right)C{H_3}\buildrel {alc.\,\,KOH} \over \longrightarrow \)
\( {\left( {C{H_3}} \right)_2}{\rm{ }}CHCH\left( {Br} \right)C{H_3}\buildrel {{{(C{H_3})}_3}{O^\Theta }{K^ \oplus }} \over \longrightarrow \)
\( {\left( {C{H_3}} \right)_2}\mathop C\limits_{\mathop |\limits_{OH} } - C{H_2} - CHO\buildrel \Delta \over \longrightarrow \)
Which of these reaction(s) will not produce Saytzeff product ?
\( {\left( {C{H_3}} \right)_3}CCH\left( {OH} \right)C{H_3}\buildrel {conc.\,\,\,{H_2}S{O_4}} \over \longrightarrow \)
\( {\left( {C{H_3}} \right)_2}CHCH\left( {Br} \right)C{H_3}\buildrel {alc.\,\,KOH} \over \longrightarrow \)
\( {\left( {C{H_3}} \right)_2}{\rm{ }}CHCH\left( {Br} \right)C{H_3}\buildrel {{{(C{H_3})}_3}{O^\Theta }{K^ \oplus }} \over \longrightarrow \)
\( {\left( {C{H_3}} \right)_2}\mathop C\limits_{\mathop |\limits_{OH} } - C{H_2} - CHO\buildrel \Delta \over \longrightarrow \)
Which of these reaction(s) will not produce Saytzeff product ?
Show Hint
- (d) only
- (b) and (d)
- (a), (c) and (d)
- (c) only
The Correct Option is D
Solution and Explanation
(b) (CH3)2CHCH(Br)CH3 with alc. KOH: small base, gives the more substituted 2-methylbut-2-ene: Saytzeff product.
(c) The same bromide with bulky (CH3)3CO-K+: removes the less hindered H and gives 3-methylbut-1-ene: Hofmann product, not Saytzeff.
(d) (CH3)2C(OH)CH2CHO on heating: dehydration gives the more substituted conjugated alkene (CH3)2C=CH-CHO: Saytzeff product.
Only (c) does not give the Saytzeff product.
Answer: Option 4 (c) only
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