CH_3 −CH_∣_(CH_3 ) −CH=C^∣^(C^(∣CH_3 ) H_2 ) −CH_∣_(CH_∥_(CH_2 ) ) −CH_∣_Δ −CH_∣_(Br) −CH


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Question Number 164386 by mathls last updated on 16/Jan/22

${C H}_{3} - C \underset{\underset{{C H}_{3}}{∣}}{H} - C H = \overset{\overset{{\overset{∣ {C H}_{3}}{C H}}_{2}}{∣}}{C} - C \underset{\underset{C \underset{\underset{{C H}_{2}}{∥}}{H}}{∣}}{H} - C \underset{\underset{Δ}{∣}}{H} - C \underset{\underset{B r}{∣}}{H} - {C H}_{3}$ $w h a t i s t h e i u p a c n a m e ?$
Commented by mathls last updated on 16/Jan/22

$p l e a s e h e l p m e$
Commented by mathls last updated on 16/Jan/22

$p l e a s e s o l v e m r W!$
Commented by mathls last updated on 16/Jan/22

$w h o c a n h e l p m e!$
Commented by mr W last updated on 16/Jan/22

$i {c a n}^{'} t ?$
Commented by mathls last updated on 16/Jan/22

$y e s y o u c a n b e c a u s e y o u a r e t h e s t a r o f$ $m a t h e d i t o r a p p .$
Commented by nikif99 last updated on 16/Jan/22

$w h a t m e a n s Δ i n t h e 6 t h a t o m o f C ?$
Commented by mr W last updated on 16/Jan/22

$y o u m e a n t w i t h ‘ ‘ w h o c a n h e l p m e!^{″}$ $‘ ‘ w h o c a n h e l p m e ?^{″} .$ $s o i m e a n t w i t h ‘ ‘ i {c a n}^{'} t ?^{″} ‘ ‘ i {c a n}^{'} t!^{″} .$ $t h a t m e a n s i {d i d n}^{'} t a s k i f i c a n, b u t$ $s a i d t h a t i r e a l l y {c a n}^{'} t .$
Commented by mathls last updated on 16/Jan/22

$i t m e a n s c y c l o p r o p y l e$
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