One equivalent of H+ adds to buta-1,3-diene, CH2=CH-CH=CH2. The preferred cation formed in the first step is:

CH3-CH^+-CH=CH2 (but-3-en-2-yl cation), delocalised to CH3-CH=CH-CH2+

CH2=CH-CH2-CH2+ (primary, not conjugated)

CH2=CH-CH^+-CH3 treated as a localised secondary cation with no resonance

JEE Advanced levelSingle correct optionDifficulty 3 of 3g3040

One equivalent of $H X^{+}$ adds to buta-1,3-diene, $CH X_{2} = CH - CH = CH X_{2}$ . The preferred cation formed in the first step is:

A $CH X_{3} - CH X^{+} - CH = CH X_{2}$ (but-3-en-2-yl cation), delocalised to $CH X_{3} - CH = CH - CH X_{2} X^{+}$
B $CH X_{2} = CH - CH X_{2} - CH X_{2} X^{+}$ (primary, not conjugated)
C $CH X_{2} = CH - CH X^{+} - CH X_{3}$ treated as a localised secondary cation with no resonance
D $CH X_{3} - CH X_{2} - CH = CH X^{+}$ (vinylic)

Adding

H X^{+}

to a terminal carbon leaves the charge on C2, adjacent to the remaining C=C, so it is allylic and resonance stabilised:

CH X_{3} - CH X^{+} - CH = CH X_{2}

and

CH X_{3} - CH = CH - CH X_{2} X^{+}

. Adding

H X^{+}

to C2 would give a primary cation without conjugation.