The dipole of CCl₄, CHCl₃ and CH₄ are in the order :

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 ?

Given that the standard potentials (E^o) of Cu²⁺/Cu and Cu⁺/Cu are 0.34 V and 0.522 V respectively, the E^o of Cu²⁺/Cu⁺ :

Consider the following reaction :


The product 'X' is used :