Arrange the following groups in decreasing CIP priority: -C#CH, -CH=CH2, -CH(CH3)2, -CH2CH3.

-C#CH > -CH=CH2 > -CH(CH3)2 > -CH2CH3

-CH(CH3)2 > -C#CH > -CH=CH2 > -CH2CH3

-CH=CH2 > -C#CH > -CH(CH3)2 > -CH2CH3

-C#CH > -CH(CH3)2 > -CH=CH2 > -CH2CH3

JEE Main levelSingle correct optionDifficulty 2 of 3st024

Arrange the following groups in decreasing CIP priority: $- C \equiv CH$ , $- CH = CH X_{2}$ , $- CH (CH X_{3}) X_{2}$ , $- CH X_{2} CH X_{3}$ .

A $- C \equiv CH > - CH = CH X_{2} > - CH (CH X_{3}) X_{2} > - CH X_{2} CH X_{3}$
B $- CH (CH X_{3}) X_{2} > - C \equiv CH > - CH = CH X_{2} > - CH X_{2} CH X_{3}$
C $- CH = CH X_{2} > - C \equiv CH > - CH (CH X_{3}) X_{2} > - CH X_{2} CH X_{3}$
D $- C \equiv CH > - CH (CH X_{3}) X_{2} > - CH = CH X_{2} > - CH X_{2} CH X_{3}$

Compare the substituent sets on the first carbon, duplicating atoms for multiple bonds: ethynyl

(C, C, C)

; vinyl

(C, C, H)

; isopropyl

(C, C, H)

; ethyl

(C, H, H)

. Ethynyl is first and ethyl is last. Vinyl and isopropyl tie, so go to the next sphere: the vinyl branch carbon carries

(C, H, H)

while an isopropyl methyl carries

(H, H, H)

, so vinyl wins.